JEE Definite Integration Questions

To solve this problem, we first need to simplify the composition of the function $$f(x)$$ and then calculate the area using integration.

1. Simplify the Iterated Function $$f^{n}(x)$$

Let's find a pattern for $$f^{n}(x)$$ where $$f(x) = \frac{x-1}{x+1}$$.

• $$f^1(x)$$: $$\frac{x-1}{x+1}$$
• $$f^2(x)$$: $$f(f(x)) = \frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1} = \frac{(x-1)-(x+1)}{(x-1)+(x+1)} = \frac{-2}{2x} = -\frac{1}{x}$$
• $$f^3(x)$$: $$f(f^2(x)) = \frac{-1/x - 1}{-1/x + 1} = \frac{-1-x}{-1+x} = \frac{x+1}{1-x}$$
• $$f^4(x)$$: $$f(f^3(x)) = \frac{\frac{x+1}{1-x}-1}{\frac{x+1}{1-x}+1} = \frac{(x+1)-(1-x)}{(x+1)+(1-x)} = \frac{2x}{2} = x$$
• $$y = g(x) = \frac{1}{x}$$
• $$2y = 2x - 3 \implies y = x - 1.5$$
• $$y = 0$$ (x-axis)
• $$x = 4$$
• $$y = x - 1.5$$ and $$y = 0$$ intersect at $$x = 1.5$$.
• $$y = 1/x$$ and $$y = x - 1.5$$ intersect when $$\frac{1}{x} = x - \frac{3}{2} \implies 2 = 2x^2 - 3x \implies 2x^2 - 3x - 2 = 0$$.
• From $$x = 1.5$$ to $$x = 2$$: Area under the line $$y = x - 1.5$$.
• From $$x = 2$$ to $$x = 4$$: Area under the curve $$y = 1/x$$.

The function is periodic with a period of 4.

Since $$26 = 4 \times 6 + 2$$, we have:

$$f^{26}(x) = f^2(x) = -\frac{1}{x}$$

2. Determine $$g(x)$$

The problem states $$g(x) + f^{26}(x) = 0$$:

$$g(x) - \frac{1}{x} = 0 \implies g(x) = \frac{1}{x}$$

3. Find the Bounded Region

We need the area bounded by:

Finding Intersection Points:

Factoring: $$(2x+1)(x-2) = 0$$. Since $$x \in (1, \infty)$$, the intersection is at $$x = 2$$.

4. Calculate the Area

The region consists of two parts:

$$\text{Area} = \int_{1.5}^{2} (x - 1.5) \, dx + \int_{2}^{4} \frac{1}{x} \, dx$$

Part 1:

$$\left[ \frac{x^2}{2} - 1.5x \right]_{1.5}^{2} = \left( \frac{4}{2} - 3 \right) - \left( \frac{2.25}{2} - 2.25 \right) = -1 - (-1.125) = 0.125 = \frac{1}{8}$$

Part 2:

$$\left[ \ln x \right]_{2}^{4} = \ln 4 - \ln 2 = \ln\left(\frac{4}{2}\right) = \ln 2$$

Total Area:

$$\frac{1}{8} + \ln 2$$

Correct Option: A

The two curves are given in the form $$x = f(y)$$, which is convenient for integrating with respect to $$y$$.
Curve 1 : $$x + 3y^{2} = 0 \;\Longrightarrow\; x = -3y^{2}$$
Curve 2 : $$x + 4y^{2} = 1 \;\Longrightarrow\; x = 1 - 4y^{2}$$

To enclose a bounded region, we need the $$y$$-coordinates where the curves intersect. Set their $$x$$-values equal:

$$-3y^{2} = 1 - 4y^{2}$$

$$\;\Longrightarrow\; (-3y^{2} + 4y^{2}) = 1$$

$$\;\Longrightarrow\; y^{2} = 1 \;\Longrightarrow\; y = \pm 1$$

Between $$y = -1$$ and $$y = 1$$, the right-hand curve (larger $$x$$) is $$x = 1 - 4y^{2}$$ and the left-hand curve is $$x = -3y^{2}$$. Hence the horizontal width of the strip at a general $$y$$ is

$$x_{\text{right}} - x_{\text{left}} = \bigl(1 - 4y^{2}\bigr) - \bigl(-3y^{2}\bigr) = 1 - 4y^{2} + 3y^{2} = 1 - y^{2}$$

The required area is therefore

$$A = \int_{y=-1}^{1} \bigl(1 - y^{2}\bigr)\,dy$$

Integrate term by term:
$$\int 1\,dy = y$$, $$\int y^{2}\,dy = \frac{y^{3}}{3}$$

Thus
$$A = \Bigl[\,y - \frac{y^{3}}{3}\Bigr]_{-1}^{1}$$

Evaluate at the limits:
At $$y = 1$$: $$1 - \frac{1}{3} = \frac{2}{3}$$
At $$y = -1$$: $$(-1) - \bigl(-\frac{1}{3}\bigr) = -1 + \frac{1}{3} = -\frac{2}{3}$$

Subtracting:
$$A = \frac{2}{3} - \bigl(-\frac{2}{3}\bigr) = \frac{4}{3}$$

Hence, the bounded area equals $$\frac{4}{3}$$.

Option C which is: $$\frac{4}{3}$$

The region is defined by the three simultaneous inequalities
  $$y \ge 0,$$
  $$y \le \pi-|x|,$$
  $$y \le |x\sin x|.$$

For any fixed $$x$$ the upper boundary is the smaller of the two curves
  $$f_1(x)=\pi-|x| \qquad\text{and}\qquad f_2(x)=|x\sin x|.$$

Both $$f_1(x)$$ and $$f_2(x)$$ are even functions of $$x$$, so the required area is twice the area in the right half-plane $$x\ge 0$$.

For $$x\ge 0$$ the equations simplify to
  $$f_1(x)=\pi-x, \qquad f_2(x)=x\sin x \quad(0\le x\le\pi).$$

Set the two curves equal to locate their intersection(s):
  $$\pi-x = x\sin x \;\Longrightarrow\; \pi = x(1+\sin x).$$

This equation is satisfied at
  $$x=0,\qquad x=\frac{\pi}{2},\qquad x=\pi.$$ Ignoring the trivial point $$x=0$$ (where $$f_2=0$$) we note the non-trivial intersection at $$x=\frac{\pi}{2}$$ and the common zero at $$x=\pi$$.

Test values show
  $$f_2(x) \le f_1(x)\quad\text{for}\; 0\le x\le\frac{\pi}{2},$$
  $$f_1(x) \le f_2(x)\quad\text{for}\; \frac{\pi}{2}\le x\le\pi.$$
Hence on $$[0,\pi/2]$$ the upper boundary is $$y=x\sin x$$; on $$[\pi/2,\pi]$$ it is $$y=\pi-x$$.

Area in the first quadrant (call it $$A_{\text{half}}$$):
$$ A_{\text{half}} =\int_{0}^{\pi/2}x\sin x\,dx +\int_{\pi/2}^{\pi}(\pi-x)\,dx. $$

First integral (integration by parts):
$$ \int_{0}^{\pi/2}x\sin x\,dx = -x\cos x\Big|_{0}^{\pi/2}+\int_{0}^{\pi/2}\cos x\,dx = (0-0)+\left[\sin x\right]_{0}^{\pi/2} =1. $$

Second integral:
$$ \int_{\pi/2}^{\pi}(\pi-x)\,dx = \left[\pi x-\frac{x^{2}}{2}\right]_{\pi/2}^{\pi} = \left(\frac{\pi^{2}}{2}\right)-\left(\frac{3\pi^{2}}{8}\right) = \frac{\pi^{2}}{8}. $$

Thus
$$ A_{\text{half}} = 1+\frac{\pi^{2}}{8}. $$

By symmetry the total area is
$$ A = 2A_{\text{half}} = 2\Bigl(1+\frac{\pi^{2}}{8}\Bigr) = 2+\frac{\pi^{2}}{4}. $$

Therefore, the required area equals $$2+\dfrac{\pi^{2}}{4}$$.
Option B which is: $$2 + \dfrac{\pi^2}{4}$$.

The hyperbola $$16x^2-9y^2=144$$ can be written as $$\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$$, whose right branch lies in $$x\ge 3$$ because the vertex is at $$(\pm 3,0)$$.

The straight line is $$8x-3y=24 \;\Longrightarrow\; y=\dfrac{8x-24}{3}=\dfrac83\,(x-3).$$

To obtain the bounded region, find the points where the two curves meet:
Substitute $$y=\dfrac{8x-24}{3}$$ in $$16x^2-9y^2=144$$:

$$\begin{aligned} 16x^2-9\left(\dfrac{8x-24}{3}\right)^2 &= 144\\ 16x^2-(8x-24)^2 &= 144\\ 16x^2-\left(64x^2-384x+576\right) &= 144\\ -48x^2+384x-576 &= 144\\ 48x^2-384x+720 &= 0\\ x^2-8x+15 &= 0\\ (x-3)(x-5) &= 0 \end{aligned}$$

Hence intersection points are $$(3,0)$$ and $$\left(5,\dfrac{16}{3}\right).$$ For $$3\lt x\lt 5$$ the hyperbola’s upper branch $$y=\dfrac{4}{3}\sqrt{x^2-9}$$ lies above the line, giving a closed region bounded by these two curves between $$x=3$$ and $$x=5$$.

Area $$A$$ of this region:

$$\begin{aligned} A &= \int_{3}^{5}\left[\dfrac43\sqrt{x^2-9}-\dfrac{8x-24}{3}\right]dx\\[3pt] &= \dfrac13\int_{3}^{5}\Bigl[4\sqrt{x^2-9}-8x+24\Bigr]dx. \end{aligned}$$

Split the integral:

$$I_1 = 4\int\sqrt{x^2-9}\,dx, \qquad I_2 = \int(-8x+24)\,dx.$$

Standard result: $$\displaystyle\int\sqrt{x^2-a^2}\,dx=\dfrac{x}{2}\sqrt{x^2-a^2}-\dfrac{a^2}{2}\ln\!\left|x+\sqrt{x^2-a^2}\right|.$$ Here $$a=3$$, so

$$\begin{aligned} I_1 &=4\left[\dfrac{x}{2}\sqrt{x^2-9}-\dfrac{9}{2}\ln\!\left(x+\sqrt{x^2-9}\right)\right]\\ &= 2x\sqrt{x^2-9}-18\ln\!\left(x+\sqrt{x^2-9}\right). \end{aligned}$$

For $$I_2$$: $$\displaystyle\int(-8x+24)\,dx=-4x^2+24x.$$

Thus

$$\begin{aligned} A &= \dfrac13\Bigl[\,2x\sqrt{x^2-9}-18\ln\!\left(x+\sqrt{x^2-9}\right)-4x^2+24x\Bigr]_{3}^{5}. \end{aligned}$$

Evaluate at the upper limit $$x=5$$ (note $$\sqrt{25-9}=4$$):

$$2(5)(4)-18\ln(5+4)-4(25)+24(5)=40-18\ln 9-100+120=60-18\ln 9.$$

Evaluate at the lower limit $$x=3$$ (note $$\sqrt{9-9}=0$$):

$$0-18\ln(3+0)-36+72=36-18\ln 3.$$

Difference:

$$60-18\ln 9-(36-18\ln 3)=24-18(\ln 9-\ln 3).$$

But $$\ln 9=2\ln 3$$, hence $$\ln 9-\ln 3=\ln 3$$ and

$$24-18\ln 3.$$

Finally,

$$A=\dfrac13\left(24-18\ln 3\right)=8-6\ln 3.$$

Required expression:

$$\begin{aligned} 3\bigl(A+6\ln 3\bigr) &= 3\bigl[(8-6\ln 3)+6\ln 3\bigr]\\ &=3\times 8\\ &=24. \end{aligned}$$

Therefore the value is $$\boxed{24}$$.

The functional equation is
$$f(x) + 3f\!\left(\frac{\pi}{2}-x\right) = \sin x \qquad -(1)$$

Replace $$x$$ by $$\frac{\pi}{2}-x$$ in $$-(1)$$ to obtain
$$f\!\left(\frac{\pi}{2}-x\right) + 3f(x) = \sin\!\left(\frac{\pi}{2}-x\right) = \cos x \qquad -(2)$$

Let $$A = f(x)$$ and $$B = f\!\left(\frac{\pi}{2}-x\right)$$. Then $$-(1)$$ and $$-(2)$$ become
$$A + 3B = \sin x \qquad -(3)$$ $$3A + B = \cos x \qquad -(4)$$

Solve the simultaneous linear equations. Multiply $$-(3)$$ by $$3$$ and subtract $$-(4)$$:
$$3A + 9B - (3A + B) = 3\sin x - \cos x$$ $$8B = 3\sin x - \cos x \;\;\Longrightarrow\;\; B = \frac{3\sin x - \cos x}{8}$$

Substitute $$B$$ in $$-(3)$$:
$$A = \sin x - 3B = \sin x - 3\!\left(\frac{3\sin x - \cos x}{8}\right) = \frac{-\sin x + 3\cos x}{8}$$ Hence
$$f(x) = \frac{3\cos x - \sin x}{8}$$

The expression $$3\cos x - \sin x$$ is of the form $$P\cos x + Q\sin x$$ whose amplitude is $$\sqrt{P^{2}+Q^{2}} = \sqrt{3^{2}+(-1)^{2}} = \sqrt{10}$$. Therefore the maximum value of $$f$$ is
$$\alpha = \frac{\sqrt{10}}{8}$$

Now consider $$g(x)=x^{2}$$ and $$h(x)=\beta x^{3}\;(\beta \gt 0)$$. Points of intersection satisfy $$x^{2} = \beta x^{3}\;\Longrightarrow\; x = 0$$ or $$x=\frac{1}{\beta}$$.

For $$0 \lt x \lt \frac{1}{\beta}$$,\; $$x^{2} \gt \beta x^{3}$$, so the required area is
$$A = \int_{0}^{1/\beta} \bigl(x^{2} - \beta x^{3}\bigr)\,dx$$

Integrate term by term:
$$A = \left[\frac{x^{3}}{3} - \frac{\beta x^{4}}{4}\right]_{0}^{1/\beta} = \frac{1}{3\beta^{3}} - \frac{1}{4\beta^{3}} = \frac{1}{12\beta^{3}}$$

Given that this area equals $$\alpha^{2}$$,
$$\frac{1}{12\beta^{3}} = \left(\frac{\sqrt{10}}{8}\right)^{2} = \frac{10}{64} = \frac{5}{32}$$

Solve for $$\beta^{3}$$:
$$\beta^{3} = \frac{1}{12}\,\Big/\!\frac{5}{32} = \frac{32}{60} = \frac{8}{15}$$

Finally,
$$30\beta^{3} = 30 \times \frac{8}{15} = 16$$

Hence, the required value is 16.

We need to find the number of elements in $$S = \{x : x \in [0, 100] \text{ and } \int_0^x t^2 \sin(x-t)\,dt = x^2\}$$.

Using the identity $$\sin(x-t) = \sin x \cos t - \cos x \sin t$$, we write the integral as $$\int_0^x t^2 \sin(x-t)\,dt = \sin x \int_0^x t^2 \cos t\,dt - \cos x \int_0^x t^2 \sin t\,dt$$.

To evaluate $$\int_0^x t^2 \cos t\,dt$$ we apply integration by parts twice, taking $$u = t^2$$ and $$dv = \cos t\,dt$$ to obtain $$\int t^2 \cos t\,dt = t^2 \sin t - \int 2t \sin t\,dt$$.

In the integral $$\int 2t \sin t\,dt$$ we set $$u = 2t$$ and $$dv = \sin t\,dt$$, giving $$\int 2t \sin t\,dt = -2t\cos t + \int 2\cos t\,dt = -2t\cos t + 2\sin t$$ and hence $$\int t^2 \cos t\,dt = t^2 \sin t + 2t\cos t - 2\sin t + C$$ so that $$\int_0^x t^2 \cos t\,dt = x^2 \sin x + 2x\cos x - 2\sin x$$.

For $$\int_0^x t^2 \sin t\,dt$$ we use integration by parts with $$u = t^2$$ and $$dv = \sin t\,dt$$, yielding $$\int t^2 \sin t\,dt = -t^2 \cos t + \int 2t\cos t\,dt$$.

Since $$\int 2t\cos t\,dt = 2t\sin t - \int 2\sin t\,dt = 2t\sin t + 2\cos t$$ we have $$\int t^2 \sin t\,dt = -t^2\cos t + 2t\sin t + 2\cos t + C$$ and thus $$\int_0^x t^2 \sin t\,dt = -x^2\cos x + 2x\sin x + 2\cos x - 2$$.

Substituting these results into the expression for the integral gives $$\text{LHS} = \sin x(x^2\sin x + 2x\cos x - 2\sin x) - \cos x(-x^2\cos x + 2x\sin x + 2\cos x - 2)$$.

Expanding and simplifying leads to $$= x^2\sin^2 x + 2x\sin x\cos x - 2\sin^2 x + x^2\cos^2 x - 2x\sin x\cos x - 2\cos^2 x + 2\cos x = x^2(\sin^2 x + \cos^2 x) - 2(\sin^2 x + \cos^2 x) + 2\cos x = x^2 - 2 + 2\cos x$$.

Setting this equal to $$x^2$$ gives $$2\cos x = 2$$, so $$\cos x = 1$$ and hence $$x = 2k\pi, \quad k = 0, 1, 2, \ldots$$.

Requiring $$2k\pi \leq 100 \implies k \leq \frac{50}{\pi} \approx 15.915$$, we find $$k = 0, 1, 2, \ldots, 15$$, giving 16 elements.

The answer is $$\mathbf{16}$$.

Let $$S=\sum_{r=1}^{20}\Bigl|\sqrt{\pi\,\bigl(\int_{0}^{r}x\,|\sin \pi x|\,dx\bigr)}\Bigr|$$.

Because $$r$$ is a positive integer, the outer absolute value and the square root will always give a non-negative result, so we first evaluate the integral
$$I(r)=\int_{0}^{r}x\,|\sin \pi x|\,dx.$$

The function $$|\sin \pi x|$$ has period $$1$$. Split the interval $$[0,r]$$ into $$r$$ sub-intervals $$[n,n+1]$$ where $$n=0,1,\dots ,r-1$$:

$$I(r)=\sum_{n=0}^{r-1}\int_{n}^{\,n+1}x\,|\sin \pi x|\,dx.$$

Inside each sub-interval set $$x=n+t$$ with $$0\le t\le 1$$. Then $$dx=dt$$ and

$$\int_{n}^{\,n+1}x\,|\sin \pi x|\,dx =\int_{0}^{1}(n+t)\,|\sin \pi(n+t)|\,dt.$$

Since $$\sin \pi(n+t)=(-1)^n\sin \pi t$$, its absolute value becomes $$|\sin \pi t|$$, independent of $$n$$. Hence

$$\int_{n}^{\,n+1}x\,|\sin \pi x|\,dx =n\!\!\int_{0}^{1}|\sin \pi t|\,dt +\!\int_{0}^{1}t\,|\sin \pi t|\,dt.$$

Define
$$A=\int_{0}^{1}|\sin \pi t|\,dt, \qquad B=\int_{0}^{1}t\,|\sin \pi t|\,dt.$$

Because $$\sin \pi t \ge 0$$ on $$[0,1]$$, the absolute signs can be dropped:

$$A=\int_{0}^{1}\sin \pi t\,dt =\left[-\frac{\cos \pi t}{\pi}\right]_{0}^{1} =\frac{2}{\pi}.$$

For $$B$$, integrate by parts:

$$B=\int_{0}^{1}t\sin \pi t\,dt =\left[-\frac{t\cos \pi t}{\pi}\right]_{0}^{1} +\frac{1}{\pi}\!\int_{0}^{1}\cos \pi t\,dt =\frac{1}{\pi}.$$

Therefore
$$\int_{n}^{\,n+1}x\,|\sin \pi x|\,dx =nA+B =n\Bigl(\frac{2}{\pi}\Bigr)+\frac{1}{\pi}.$$

Summing over $$n$$ from $$0$$ to $$r-1$$:

$$I(r)=\sum_{n=0}^{r-1}\Bigl(n\frac{2}{\pi}+\frac{1}{\pi}\Bigr) =\frac{2}{\pi}\sum_{n=0}^{r-1}n +\frac{1}{\pi}\sum_{n=0}^{r-1}1.$$

Using $$\sum_{n=0}^{r-1}n=\frac{r(r-1)}{2}$$ and $$\sum_{n=0}^{r-1}1=r$$, we get

$$I(r)=\frac{2}{\pi}\cdot\frac{r(r-1)}{2} +\frac{1}{\pi}\,r =\frac{r(r-1)+r}{\pi} =\frac{r^{2}}{\pi}.$$

Hence
$$\pi\,I(r)=\pi\cdot\frac{r^{2}}{\pi}=r^{2},\qquad \sqrt{\pi\,I(r)}=\sqrt{r^{2}}=r.$$

The summation becomes
$$S=\sum_{r=1}^{20}r =\frac{20\cdot21}{2}=210.$$

Final answer: $$210$$.

Let us define a function $$f:\,[0,\;2\sqrt{3}] \rightarrow [2,\;4]$$ by
$$f(x)=\log_2\!\left(x^2+4\right)\,.$$

Step 1: Check that the limits match the image of the function.
At $$x=0$$ : $$f(0)=\log_2 4 = 2$$.
At $$x=2\sqrt{3}$$ : $$f(2\sqrt{3})=\log_2(12+4)=\log_2 16 = 4$$.
Hence $$f(x)$$ indeed maps $$x\in[0,\,2\sqrt{3}]$$ onto $$y\in[2,\,4]$$.

Step 2: Write its inverse.
Starting with $$y=f(x)=\log_2(x^2+4)$$, rewrite in exponential form:
$$2^{\,y}=x^2+4 \;\Longrightarrow\; x=\sqrt{2^{\,y}-4}.$$
Thus the inverse function is
$$f^{-1}(y)=\sqrt{2^{\,y}-4},\qquad y\in[2,\,4].$$

Step 3: Use the standard area identity for a function and its inverse.
For any continuous, strictly monotonic function $$f$$ whose inverse exists, the following holds:
$$\int_{a}^{b} f(x)\,dx + \int_{f(a)}^{f(b)} f^{-1}(y)\,dy = b\,f(b) - a\,f(a).$$

Step 4: Apply the identity to the given integrals.
Here $$a=0,\; b=2\sqrt{3},\; f(a)=2,\; f(b)=4.$$\newline Therefore

$$\alpha = \int_{0}^{2\sqrt{3}} \log_2(x^2 + 4)\,dx \;+\; \int_{2}^{4} \sqrt{2^{\,x} - 4}\,dx$$
$$\phantom{\alpha}= b\,f(b) - a\,f(a)$$
$$\phantom{\alpha}= (2\sqrt{3})(4) - (0)(2)$$
$$\phantom{\alpha}= 8\sqrt{3}.$$

Step 5: Square the result.
$$\alpha^2 = (8\sqrt{3})^2 = 64 \times 3 = 192.$$

Hence the required value is 192.

$$\alpha=\int_0^{64}\left(x^{1/3}-\lfloor x^{1/3}\rfloor\right)dx$$

$$Put(x=t^3\Rightarrow dx=3t^2dt,;t:0\to4):$$
$$\alpha=\int_0^4(t-\lfloor t\rfloor)3t^2dt$$
$$=\sum_{n=0}^3\int_n^{n+1}3t^2(t-n)dt=36$$

Now:
$$\frac{1}{\pi}\int_0^{\alpha\pi}\frac{\sin^2\theta}{\sin^6\theta+\cos^6\theta}d\theta$$
$$=\frac{1}{\pi}\int_0^{36\pi}$$...

Period $$(=\pi),$$ so:
=36$$.\frac{1}{\pi}\int_0^{\pi}\frac{\sin^2\theta}{\sin^6\theta+\cos^6\theta}d\theta$$

Using symmetry:
$$\int_0^{\pi}\frac{\sin^2\theta}{\sin^6+\cos^6}d\theta$$
$$=\int_0^{\pi}\frac{\cos^2\theta}{\sin^6+\cos^6}d\theta$$
$$\Rightarrow\text{each}=\frac{\pi}{2}$$

Hence value =36

We need to find the area $$A$$ bounded by $$y = \max\{\sin x, \cos x\}$$, $$x = 0$$, $$x = \frac{3\pi}{2}$$, and the x-axis.

- $$[0, \frac{\pi}{4}]$$: $$\cos x \geq \sin x$$, so $$y = \cos x$$

- $$[\frac{\pi}{4}, \frac{5\pi}{4}]$$: $$\sin x \geq \cos x$$, so $$y = \sin x$$

- $$[\frac{5\pi}{4}, \frac{3\pi}{2}]$$: $$\cos x \geq \sin x$$, so $$y = \cos x$$

The area between the curve and x-axis counts absolute values. The function $$\max\{\sin x, \cos x\}$$:

- Is positive on $$[0, \pi]$$

- Becomes negative: $$\sin x < 0$$ for $$x \in (\pi, 2\pi)$$, and $$\cos x < 0$$ for $$x \in (\frac{\pi}{2}, \frac{3\pi}{2})$$

On $$[\frac{\pi}{4}, \pi]$$: $$y = \sin x \geq 0$$

On $$[\pi, \frac{5\pi}{4}]$$: $$y = \sin x \leq 0$$

On $$[\frac{5\pi}{4}, \frac{3\pi}{2}]$$: $$y = \cos x \leq 0$$

The area bounded by the curve and x-axis:

$$A = \int_0^{\pi/4} \cos x \, dx + \int_{\pi/4}^{\pi} \sin x \, dx + \int_{\pi}^{5\pi/4} |\sin x| \, dx + \int_{5\pi/4}^{3\pi/2} |\cos x| \, dx$$

$$\int_0^{\pi/4} \cos x \, dx = \sin x \Big|_0^{\pi/4} = \frac{\sqrt{2}}{2}$$

$$\int_{\pi/4}^{\pi} \sin x \, dx = -\cos x \Big|_{\pi/4}^{\pi} = -(-1) - (-\frac{\sqrt{2}}{2}) = 1 + \frac{\sqrt{2}}{2}$$

$$\int_{\pi}^{5\pi/4} (-\sin x) \, dx = \cos x \Big|_{\pi}^{5\pi/4} = -\frac{\sqrt{2}}{2} - (-1) = 1 - \frac{\sqrt{2}}{2}$$

$$\int_{5\pi/4}^{3\pi/2} (-\cos x) \, dx = -\sin x \Big|_{5\pi/4}^{3\pi/2} = -(-1) - (-(-\frac{\sqrt{2}}{2})) = 1 - \frac{\sqrt{2}}{2}$$

$$A = \frac{\sqrt{2}}{2} + 1 + \frac{\sqrt{2}}{2} + 1 - \frac{\sqrt{2}}{2} + 1 - \frac{\sqrt{2}}{2} = 3$$

$$A + A^2 = 3 + 9 = 12$$

The answer is $$12$$.

$$A = \int_{0}^{\frac{1}{\sqrt{5}}} \left( \frac{\sqrt{1-x^2}}{2} - x \right) dx$$

$$\alpha = 8A = 8 \left[ \int_{0}^{\frac{1}{\sqrt{5}}} \left( \frac{\sqrt{1-x^2}}{2} - x \right) dx \right]$$

$$\alpha = 8 \left[ \frac{x}{4}\sqrt{1-x^2} + \frac{1}{4}\sin^{-1}x - \frac{x^2}{2} \right]_{0}^{\frac{1}{\sqrt{5}}}$$

$$\alpha = 8 \left[ \frac{1}{4\sqrt{5}}\sqrt{\frac{4}{5}} + \frac{1}{4}\sin^{-1}\left(\frac{1}{\sqrt{5}}\right) - \frac{1}{10} \right]$$

$$\alpha = 2\sin^{-1}\frac{1}{\sqrt{5}}$$

$$\cot\alpha = \cot\left( 2\sin^{-1}\frac{1}{\sqrt{5}} \right)$$

$$= \cot\left( 2\tan^{-1}\frac{1}{2} \right)$$

$$= \cot\left( \tan^{-1}\left( \frac{2 \times \frac{1}{2}}{1 - \frac{1}{4}} \right) \right)$$

$$= \cot\left( \tan^{-1}\left( \frac{4}{3} \right) \right)$$ $$ = 0.75$$

Curves: 1. $$|y| = 1 - x^2 \implies y = \pm(1 - x^2)$$. These are two parabolas opening toward each other, intersecting the x-axis at $$(\pm1, 0)$$.

2. $$x^2 + y^2 = 1$$. This is a unit circle centered at the origin.

The area is symmetric across both axes. We can calculate the area in the first quadrant and multiply by 4.

In the first quadrant ($$x, y \ge 0$$), we find the area between the circle $$y = \sqrt{1-x^2}$$ and the parabola $$y = 1-x^2$$.

$$\frac{\alpha}{4} = \int_{0}^{1} (\sqrt{1-x^2} - (1-x^2)) \, dx$$

$$\frac{\alpha}{4} = \left[ \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1}x - x + \frac{x^3}{3} \right]_0^1 = \frac{\pi}{4} - 1 + \frac{1}{3} = \frac{\pi}{4} - \frac{2}{3}$$

$$\alpha = \pi - \frac{8}{3} \implies 9\alpha = 9\pi - 24$$

 Comparing $$9\alpha = \beta\pi + \gamma$$, we get $$\beta = 9$$ and $$\gamma = -24$$.

$$|\beta - \gamma| = |9 - (-24)| = |9 + 24| = \mathbf{33}$$

Find points of intersection.

We need to see where the parabola $$y = x^2 + 4x + 2$$ intersects the absolute value function $$y = |x + 2|$$.

• Case 1: $$x \geq -2$$ (where $$|x+2| = x+2$$)

$$x^2 + 4x + 2 = x + 2 \implies x^2 + 3x = 0 \implies x = 0$$ or $$x = -3$$ (reject $$-3$$).

• Case 2: $$x < -2$$ (where $$|x+2| = -(x+2)$$)

$$x^2 + 4x + 2 = -x - 2 \implies x^2 + 5x + 4 = 0 \implies (x+1)(x+4) = 0 \implies x = -4$$ or $$x = -1$$ (reject $$-1$$).

Set up the integral.

The area is bounded between $$x = -4$$ and $$x = 0$$.

$$Area = \int_{-4}^{0} (|x+2| - (x^2+4x+2)) dx$$

Since the functions are symmetric about $$x = -2$$, we can calculate the area from $$-2$$ to $$0$$ and double it:

$$Area = 2 \int_{-2}^{0} ((x+2) - (x^2+4x+2)) dx = 2 \int_{-2}^{0} (-x^2 - 3x) dx$$

$$= 2 \left[ -\frac{x^3}{3} - \frac{3x^2}{2} \right]_{-2}^{0} = 2 \left[ 0 - \left( \frac{8}{3} - 6 \right) \right] = 2 \left( \frac{10}{3} \right) = \frac{20}{3}$$

We need to find the area of the region $$\{(x, y) : |x - y| \le y \le 4\sqrt{x}\}$$.

The condition $$|x - y| \le y$$ means $$-y \le x - y \le y$$, which gives us $$x \ge 0$$ (from the left inequality) and $$x \le 2y$$ (from the right inequality), i.e., $$y \ge x/2$$.

So the region is $$\{(x, y) : x \ge 0, \, x/2 \le y \le 4\sqrt{x}\}$$.

Finding the intersection of $$y = x/2$$ and $$y = 4\sqrt{x}$$: $$x/2 = 4\sqrt{x}$$, so $$x = 8\sqrt{x}$$, giving $$\sqrt{x}(\sqrt{x} - 8) = 0$$. Thus $$x = 0$$ or $$x = 64$$.

The area is:

$$A = \displaystyle\int_0^{64} \left(4\sqrt{x} - \dfrac{x}{2}\right) dx$$

$$= \left[4 \cdot \dfrac{2}{3}x^{3/2} - \dfrac{x^2}{4}\right]_0^{64}$$

$$= \dfrac{8}{3}(64)^{3/2} - \dfrac{64^2}{4}$$

$$= \dfrac{8}{3} \cdot 512 - \dfrac{4096}{4}$$

$$= \dfrac{4096}{3} - 1024 = \dfrac{4096 - 3072}{3} = \dfrac{1024}{3}$$

Hence, the correct answer is Option B.

Write the integrand in a simpler form:
$$\left|\pi^{2}x\sin(\pi x)\right| = \pi^{2}\,|x\sin(\pi x)|$$
Hence

$$I=\int_{-1}^{3/2}\left|\pi^{2}x\sin(\pi x)\right|dx=\pi^{2}\int_{-1}^{3/2}|x\sin(\pi x)|dx$$

First locate the points where $$x\sin(\pi x)=0$$ inside $$[-1,\,3/2]$$.
Zeros occur when $$x=0$$ or $$\sin(\pi x)=0\;(\Rightarrow x=\text{integer})$$.
Thus the critical points are $$x=-1,\,0,\,1$$.

Check the sign of $$f(x)=x\sin(\pi x)$$ on each sub-interval:

• $$(-1,0):$$ choose $$x=-\tfrac12$$ ⇒ $$f(x)=(-\tfrac12)\sin(-\tfrac{\pi}{2})=( -\tfrac12)(-1)\gt0$$
• $$(0,1):$$ choose $$x=\tfrac12$$ ⇒ $$f(x)=(\tfrac12)\sin(\tfrac{\pi}{2})=(\tfrac12)(1)\gt0$$
• $$(1,\,3/2):$$ choose $$x=\tfrac32$$ ⇒ $$f(x)=(\tfrac32)\sin(\tfrac{3\pi}{2})=(\tfrac32)(-1)\lt0$$

Thus
Case 1: $$x\in[-1,1]\; \Rightarrow\; |f(x)|=f(x)$$
Case 2: $$x\in[1,3/2]\; \Rightarrow\; |f(x)|=-f(x)$$

Split the integral accordingly:

$$I=\pi^{2}\left[\int_{-1}^{1}x\sin(\pi x)\,dx-\int_{1}^{3/2}x\sin(\pi x)\,dx\right]$$

Find an antiderivative of $$x\sin(\pi x)$$ using integration by parts.
Let $$u=x,\;dv=\sin(\pi x)dx$$.
Then $$du=dx,\;v=-\dfrac{\cos(\pi x)}{\pi}$$.

$$\int x\sin(\pi x)dx=-\dfrac{x\cos(\pi x)}{\pi}+\dfrac{\sin(\pi x)}{\pi^{2}}$$ $$-(1)$$

Integral over $$[-1,1]$$
Using $$(1):$$
$$F(x)=-\dfrac{x\cos(\pi x)}{\pi}+\dfrac{\sin(\pi x)}{\pi^{2}}$$

$$\begin{aligned} \int_{-1}^{1}x\sin(\pi x)dx &=F(1)-F(-1)\\ &=\left[-\dfrac{1\cdot(-1)}{\pi}+0\right]-\left[-\dfrac{(-1)\cdot(-1)}{\pi}+0\right]\\ &=\dfrac{1}{\pi}-\left(-\dfrac{1}{\pi}\right)=\dfrac{2}{\pi} \end{aligned}$$

Integral over $$[1,3/2]$$

$$\begin{aligned} \int_{1}^{3/2}x\sin(\pi x)dx &=F\!\left(\tfrac32\right)-F(1)\\ &=\left[0-\dfrac{1}{\pi^{2}}\right]-\dfrac{1}{\pi}\\ &=-\left(\dfrac{1}{\pi}+\dfrac{1}{\pi^{2}}\right) \end{aligned}$$

Now compute $$I$$:

$$\begin{aligned} I&=\pi^{2}\left[\dfrac{2}{\pi}-\Bigl(-\dfrac{1}{\pi}-\dfrac{1}{\pi^{2}}\Bigr)\right]\\[6pt] &=\pi^{2}\left[\dfrac{2}{\pi}+\dfrac{1}{\pi}+\dfrac{1}{\pi^{2}}\right]\\[6pt] &=\pi^{2}\left[\dfrac{3}{\pi}+\dfrac{1}{\pi^{2}}\right]\\[6pt] &=3\pi+1 \end{aligned}$$

Therefore, $$\displaystyle\int_{-1}^{3/2}\left|\pi^{2}x\sin(\pi x)\right|dx = 1 + 3\pi$$.

Option C is correct.

The region is defined by the inequalities $$0 \leq y \leq 2|x| + 1$$, $$0 \leq y \leq x^2 + 1$$, and $$|x| \leq 3$$. 

The curves $$y = 2|x| + 1$$ and $$y = x^2 + 1$$ are symmetric about the y-axis. Therefore, the area can be computed for $$x \geq 0$$ and doubled to account for the symmetry.

  For $$x \geq 0$$, the inequalities simplify to $$0 \leq y \leq \min\{2x + 1, x^2 + 1\}$$ and $$0 \leq x \leq 3$$. The curves intersect when $$x^2 + 1 = 2x + 1$$, which simplifies to $$x^2 - 2x = 0$$, or $$x(x - 2) = 0$$. The solutions are $$x = 0$$ and $$x = 2$$. 

To determine which curve is below the other: - For $$0 \leq x \leq 2$$, $$x^2 + 1 \leq 2x + 1$$, so the minimum is $$x^2 + 1$$. - For $$2 \leq x \leq 3$$, $$2x + 1 \leq x^2 + 1$$, so the minimum is $$2x + 1$$. The area for $$x \geq 0$$ is split into two integrals: 1. From $$x = 0$$ to $$x = 2$$: $$\int_{0}^{2} (x^2 + 1) dx$$ 2. From $$x = 2$$ to $$x = 3$$: $$\int_{2}^{3} (2x + 1) dx$$

Compute the first integral:

The antiderivative of $$x^2 + 1$$ is $$\frac{x^3}{3} + x$$.
Evaluating from 0 to 2: $$\left[ \frac{(2)^3}{3} + 2 \right] - \left[ \frac{(0)^3}{3} + 0 \right] = \left[ \frac{8}{3} + 2 \right] - 0 = \frac{8}{3} + \frac{6}{3} = \frac{14}{3}$$
Compute the second integral:
The antiderivative of $$2x + 1$$ is $$x^2 + x$$.

Evaluating from 2 to 3: $$\left[ (3)^2 + 3 \right] - \left[ (2)^2 + 2 \right] = [9 + 3] - [4 + 2] = 12 - 6 = 6$$

Area for $$x \geq 0$$ is $$\frac{14}{3} + 6 = \frac{14}{3} + \frac{18}{3} = \frac{32}{3}$$.

Double this area for the symmetric region ($$x \leq 0$$):

Total area = $$2 \times \frac{32}{3} = \frac{64}{3}$$.

Thus, the area of the region is $$\frac{64}{3}$$, which corresponds to option B.

$$\dfrac{64}{3}$$

To find the value of the given expression, we first rewrite the integrand by rationalizing the denominator.

The given integral is:

$$I = \int_{0}^{1} \frac{1}{\sqrt{3 + x^2} + \sqrt{1 + x^2}} \, dx$$

Multiplying the numerator and the denominator by the conjugate expression $$\sqrt{3 + x^2} - \sqrt{1 + x^2}}$$:

$$\frac{1}{\sqrt{3 + x^2} + \sqrt{1 + x^2}} = \frac{\sqrt{3 + x^2} - \sqrt{1 + x^2}}{(3 + x^2) - (1 + x^2)} = \frac{\sqrt{3 + x^2} - \sqrt{1 + x^2}}{2}$$

Substituting this back into the expression simplifies it to:

$$4 \times \frac{1}{2} \int_{0}^{1} \left(\sqrt{3 + x^2} - \sqrt{1 + x^2}\right) \, dx - 3\log_e(\sqrt{3})$$

$$= 2\int_{0}^{1} \sqrt{3 + x^2} \, dx - 2\int_{0}^{1} \sqrt{1 + x^2} \, dx - 3\log_e(\sqrt{3})$$

---

Step 1: Evaluate the first integral using the formula $$\int \sqrt{a^2 + x^2} \, dx = \frac{x}{2}\sqrt{a^2 + x^2} + \frac{a^2}{2}\log_e\left|x + \sqrt{a^2 + x^2}\right|$$

For the first part where $$a^2 = 3$$:

$$\int_{0}^{1} \sqrt{3 + x^2} \, dx = \left[ \frac{x}{2}\sqrt{3 + x^2} + \frac{3}{2}\log_e\left|x + \sqrt{3 + x^2}\right| \right]_{0}^{1}$$

$$= \left(\frac{1}{2}\sqrt{4} + \frac{3}{2}\log_e(1 + 2)\right) - \left(0 + \frac{3}{2}\log_e(\sqrt{3})\right)$$

$$= 1 + \frac{3}{2}\log_e(3) - \frac{3}{2}\log_e(\sqrt{3})$$

Since $$\log_e(3) = 2\log_e(\sqrt{3})$$, this simplifies to:

$$= 1 + 3\log_e(\sqrt{3}) - \frac{3}{2}\log_e(\sqrt{3}) = 1 + \frac{3}{2}\log_e(\sqrt{3})$$

Multiplying by 2 gives:

$$2\int_{0}^{1} \sqrt{3 + x^2} \, dx = 2 + 3\log_e(\sqrt{3})$$

---

Step 2: Evaluate the second integral where $$a^2 = 1$$

$$\int_{0}^{1} \sqrt{1 + x^2} \, dx = \left[ \frac{x}{2}\sqrt{1 + x^2} + \frac{1}{2}\log_e\left|x + \sqrt{1 + x^2}\right| \right]_{0}^{1}$$

$$= \left(\frac{1}{2}\sqrt{2} + \frac{1}{2}\log_e(1 + \sqrt{2})\right) - 0 = \frac{\sqrt{2}}{2} + \frac{1}{2}\log_e(1 + \sqrt{2})$$

Multiplying by 2 gives:

$$2\int_{0}^{1} \sqrt{1 + x^2} \, dx = \sqrt{2} + \log_e(1 + \sqrt{2})$$

---

Step 3: Combine all evaluations into the main expression

Substituting the values from Step 1 and Step 2:

$$Required value = \left(2 + 3\log_e(\sqrt{3})\right) - \left(\sqrt{2} + \log_e(1 + \sqrt{2})\right) - 3\log_e(\sqrt{3})$$

$$Required value = 2 - \sqrt{2} - \log_e(1 + \sqrt{2})$$

The circle $$(x-2\sqrt{3})^{2}+y^{2}=12$$ has centre $$\left(2\sqrt{3},\,0\right)$$ and radius $$r=\sqrt{12}=2\sqrt{3}$$.

The parabola $$y^{2}=2\sqrt{3}\,x$$ opens towards the positive $$x$$-axis and is symmetric about the $$x$$-axis.

Points common to both curves are obtained by substituting $$y^{2}=2\sqrt{3}x$$ in the circle:

$$(x-2\sqrt{3})^{2}+2\sqrt{3}\,x=12 \; \Longrightarrow \; x^{2}-2\sqrt{3}\,x=0 \; \Longrightarrow \; x\bigl(x-2\sqrt{3}\bigr)=0.$$

Hence $$x=0$$ gives the origin $$(0,0)$$, and $$x=2\sqrt{3}$$ gives the points $$(2\sqrt{3},\; \pm 2\sqrt{3})$$. Thus the two curves intersect at the three points $$\bigl(0,0\bigr)$$ and $$\bigl(2\sqrt{3},\pm 2\sqrt{3}\bigr).$$

Because the figure is symmetric about the $$x$$-axis, we calculate the area above the axis and double it.

For a fixed $$y$$ between $$0$$ and $$2\sqrt{3}$$ the left boundary of the circle is obtained from $$(x-2\sqrt{3})^{2}+y^{2}=12 \quad\Rightarrow\quad x=2\sqrt{3}-\sqrt{\,12-y^{2}\,}.$$ The parabola gives $$x=\dfrac{y^{2}}{2\sqrt{3}}.$$ Inside the circle but outside the parabola means the $$x$$-coordinate runs from the circle to the parabola. Hence the required area $$A$$ is

$$A=2\int_{0}^{2\sqrt{3}}\left[\frac{y^{2}}{2\sqrt{3}}-\Bigl(2\sqrt{3}-\sqrt{12-y^{2}}\Bigr)\right]\,dy.$$

Split the integral:

$$A=2\Biggl[\;\underbrace{\int_{0}^{2\sqrt{3}}\sqrt{12-y^{2}}\,dy}_{I_{1}}\;+\;\underbrace{\int_{0}^{2\sqrt{3}}\left(\frac{y^{2}}{2\sqrt{3}}-2\sqrt{3}\right)dy}_{I_{2}}\;\Biggr].$$

Case 1: Evaluate $$I_{1}$$. The integrand represents the upper half of a circle of radius $$2\sqrt{3}$$. Area of a semicircle $$=\frac{1}{2}\pi r^{2}=\frac{1}{2}\pi(2\sqrt{3})^{2}=6\pi$$. Taking the upper quarter (from $$y=0$$ to $$y=2\sqrt{3}$$) gives $$I_{1}=3\pi.$$

Case 2: Evaluate $$I_{2}$$.

$$\int_{0}^{2\sqrt{3}}\frac{y^{2}}{2\sqrt{3}}\,dy=\frac{1}{2\sqrt{3}}\cdot\frac{(2\sqrt{3})^{3}}{3}=4,$$ $$\int_{0}^{2\sqrt{3}}2\sqrt{3}\,dy=2\sqrt{3}\cdot2\sqrt{3}=12.$$ Therefore $$I_{2}=4-12=-8.$$

Combine the two parts:

$$A=2\bigl(I_{1}+I_{2}\bigr)=2\bigl(3\pi-8\bigr)=6\pi-16.$$

Hence the required area is $$6\pi-16$$, which matches Option B.

$$I(m,n) = \int_0^1 x^{m-1}(1-x)^{n-1}dx = B(m,n)$$ (Beta function). Find $$I(9,14) + I(10,13)$$.

We start by recalling the Beta function identity:

$$B(m,n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$$

Applying this to the integral with parameters (9,14) gives

$$I(9,14) = \frac{\Gamma(9)\Gamma(14)}{\Gamma(23)} = \frac{8! \cdot 13!}{22!}$$

Similarly, for (10,13) we have

$$I(10,13) = \frac{\Gamma(10)\Gamma(13)}{\Gamma(23)} = \frac{9! \cdot 12!}{22!}$$

Adding these two expressions yields

$$I(9,14) + I(10,13) = \frac{8! \cdot 13! + 9! \cdot 12!}{22!} = \frac{8! \cdot 12!(13 + 9)}{22!} = \frac{8! \cdot 12! \cdot 22}{22!}$$

This simplifies further as

$$\frac{22 \cdot 8! \cdot 12!}{22!}$$

On the other hand, using the same Beta function formula for (9,13) gives

$$I(9,13) = \frac{8! \cdot 12!}{21!}$$

Hence,

$$\frac{22 \cdot 8! \cdot 12!}{22!} = \frac{22 \cdot 8! \cdot 12!}{22 \cdot 21!} = \frac{8! \cdot 12!}{21!} = I(9,13)$$

The correct answer is Option 4: I(9, 13).

We need the area bounded by $$y = 4 - \frac{x^2}{4}$$ and $$y = \frac{x-4}{2}$$.

Finding intersection points by setting the curves equal:

$$4 - \frac{x^2}{4} = \frac{x-4}{2}$$

Multiplying by 4: $$16 - x^2 = 2(x-4) = 2x - 8$$

$$x^2 + 2x - 24 = 0$$, giving $$(x+6)(x-4) = 0$$, so $$x = -6$$ or $$x = 4$$.

The area is:

$$\alpha = \int_{-6}^{4} \left[\left(4 - \frac{x^2}{4}\right) - \frac{x-4}{2}\right] dx = \int_{-6}^{4} \left(6 - \frac{x}{2} - \frac{x^2}{4}\right) dx$$

$$= \left[6x - \frac{x^2}{4} - \frac{x^3}{12}\right]_{-6}^{4}$$

At $$x = 4$$: $$24 - 4 - \frac{64}{12} = 20 - \frac{16}{3} = \frac{44}{3}$$

At $$x = -6$$: $$-36 - 9 + 18 = -27$$

$$\alpha = \frac{44}{3} - (-27) = \frac{44}{3} + \frac{81}{3} = \frac{125}{3}$$

$$6\alpha = 6 \times \frac{125}{3} = 250$$

Hence, the correct answer is Option A.

From $$y^2 = 2x$$: $$x = \frac{y^2}{2}$$ and from $$x(1+y^2) = 1$$: $$x = \frac{1}{1+y^2}$$. Setting these equal, $$\frac{y^2}{2} = \frac{1}{1+y^2}$$, which gives $$y^2(1+y^2) = 2 \implies y^4 + y^2 - 2 = 0 \implies (y^2+2)(y^2-1) = 0$$. Since $$y^2 \geq 0$$, it follows that $$y^2 = 1$$ and hence $$y = \pm 1$$. At $$y = \pm 1$$, $$x = \frac{1}{2}$$.

Next, we determine which curve is to the right. For $$|y| < 1$$, $$\frac{1}{1+y^2} > \frac{y^2}{2}$$ (e.g., at $$y=0$$: $$1 > 0$$), so $$x = \frac{1}{1+y^2}$$ lies to the right of $$x = \frac{y^2}{2}$$.

Then we compute the area: $$\text{Area} = \int_{-1}^{1}\left[\frac{1}{1+y^2} - \frac{y^2}{2}\right]dy = 2\int_{0}^{1}\left[\frac{1}{1+y^2} - \frac{y^2}{2}\right]dy$$ (by symmetry about the x-axis), which equals $$2\left[\arctan y - \frac{y^3}{6}\right]_0^1 = 2\left[\frac{\pi}{4} - \frac{1}{6}\right] = \frac{\pi}{2} - \frac{1}{3}$$.

 $$\frac{\pi}{2} - \frac{1}{3} = \frac{3\pi - 2}{6}$$.

The area of the region is $$\frac{\pi}{2} - \frac{1}{3}$$.

The given function is $$f(x)=\log_{2}\!\Bigl[\log_{4}\!\bigl[\log_{6}(3+4x-x^{2})\bigr]\Bigr]$$.
For the domain, the argument of each logarithm must be positive.

Step 1 (innermost log)
$$3+4x-x^{2}\; \gt \;0$$ is needed for $$\log_{6}$$ to exist.

Step 2 (middle log)
For $$\log_{4}[\;]$$ we again need its argument positive:
$$\log_{6}(3+4x-x^{2})\; \gt \;0 \;\;\Longrightarrow\;\; 3+4x-x^{2}\; \gt \;6^{0}=1 .$$

Step 3 (outermost log)
Finally, $$\log_{2}[\;]$$ demands
$$\log_{4}\!\bigl[\log_{6}(3+4x-x^{2})\bigr]\; \gt \;0 \;\;\Longrightarrow\;\; \log_{6}(3+4x-x^{2}) \; \gt \;4^{0}=1 \;\;\Longrightarrow\;\; 3+4x-x^{2} \; \gt \;6 .$$

Simplify the last inequality:
$$3+4x-x^{2}\; \gt \;6 \;\;\Longrightarrow\;\; -x^{2}+4x-3 \; \gt \;0 \;\;\Longrightarrow\;\; x^{2}-4x+3 \; \lt \;0 \;\;\Longrightarrow\;\; (x-1)(x-3) \; \lt \;0.$$ The solution of $$(x-1)(x-3)\lt0$$ is $$1\lt x\lt3.$$
Hence the domain is $$(a,b)=(1,3),$$ so $$b-a=2.$$

We now evaluate the definite integral
$$\int_{0}^{\,b-a}[x^{2}]\,dx=\int_{0}^{2}[x^{2}]\,dx,$$ where $$[x^{2}]$$ denotes the greatest-integer (floor) function.

Break the interval $$[0,2]$$ wherever $$x^{2}$$ crosses an integer:

• $$0\le x\lt1:\;x^{2}\in[0,1)\;\Rightarrow\;[x^{2}]=0$$
• $$1\le x\lt\sqrt2:\;x^{2}\in[1,2)\;\Rightarrow\;[x^{2}]=1$$
• $$\sqrt2\le x\lt\sqrt3:\;x^{2}\in[2,3)\;\Rightarrow\;[x^{2}]=2$$
• $$\sqrt3\le x\le 2:\;x^{2}\in[3,4]\;\Rightarrow\;[x^{2}]=3$$ for $$x\lt2,$$ (the single point $$x=2$$ where $$[4]=4$$ has zero measure and does not affect the integral).

Compute the area on each sub-interval:

$$\begin{aligned} \int_{0}^{1}0\,dx &= 0,\\ \int_{1}^{\sqrt2}1\,dx &= (\sqrt2-1),\\ \int_{\sqrt2}^{\sqrt3}2\,dx &= 2(\sqrt3-\sqrt2),\\ \int_{\sqrt3}^{2}3\,dx &= 3(2-\sqrt3). \end{aligned}$$

Add them: $$\begin{aligned} 0 &+ (\sqrt2-1) + 2(\sqrt3-\sqrt2) + 3(2-\sqrt3) \\ &= \sqrt2-1 + 2\sqrt3-2\sqrt2 + 6-3\sqrt3 \\ &= 5 - \sqrt2 - \sqrt3. \end{aligned}$$

Thus $$\int_{0}^{2}[x^{2}]\,dx = 5 - \sqrt2 - \sqrt3.$$
Comparing with $$p-\sqrt{q}-\sqrt{r},$$ we have $$p=5,\;q=2,\;r=3.$$ Because $$\gcd(5,2,3)=1,$$ the required sum is $$p+q+r = 5+2+3 = 10.$$

Answer (Option A): 10

For the system to have infinitely many solutions, the determinant of the coefficient matrix must be 0, and the system must be consistent.

The coefficient matrix and augmented matrix:

$$ \begin{bmatrix} 1 & 1 & 1 & | & 1 \\ 1 & 2 & 4 & | & m \\ 1 & 4 & 10 & | & m^2 \end{bmatrix} $$

$$R_2 \to R_2 - R_1$$, $$R_3 \to R_3 - R_1$$:

$$ \begin{bmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 1 & 3 & | & m-1 \\ 0 & 3 & 9 & | & m^2-1 \end{bmatrix} $$

$$R_3 \to R_3 - 3R_2$$:

$$ \begin{bmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 1 & 3 & | & m-1 \\ 0 & 0 & 0 & | & m^2-1-3(m-1) \end{bmatrix} $$

For infinite solutions: $$m^2 - 1 - 3m + 3 = 0$$

$$m^2 - 3m + 2 = 0$$

$$(m-1)(m-2) = 0$$

$$m = 1$$ or $$m = 2$$

So $$\alpha = 1, \beta = 2$$ (or vice versa).

$$ \sum_{n=1}^{10}(n^\alpha + n^\beta) = \sum_{n=1}^{10}(n^1 + n^2) = \sum_{n=1}^{10} n + \sum_{n=1}^{10} n^2 $$

$$= \frac{10 \times 11}{2} + \frac{10 \times 11 \times 21}{6} = 55 + 385 = 440$$

The correct answer is Option 4: 440.

We need to find a such that the area of the region $$\{(x,y): -1 \leq x \leq 1, 0 \leq y \leq a + e^{|x|} - e^{-x}\}$$ equals $$\frac{e^2 + 8e + 1}{e}$$.

Since this area can be computed by integrating the upper boundary from $$x=-1$$ to $$x=1$$, we set up

$$\text{Area} = \int_{-1}^{1}(a + e^{|x|} - e^{-x})\,dx$$.

On $$x\in[-1,0]$$ we have $$|x|=-x$$, so the integrand becomes $$a + e^{-x} - e^{-x} = a$$. On $$[0,1]$$, $$|x|=x$$ and the integrand is $$a + e^x - e^{-x}$$.

Therefore,

$$\text{Area} = \int_{-1}^{0}a\,dx + \int_{0}^{1}(a + e^x - e^{-x})\,dx = a + [ax + e^x + e^{-x}]_0^1 = a + (a + e + e^{-1}) - (0 + 1 + 1) = 2a + e + \frac{1}{e} - 2.$$

Setting this equal to the given area gives

$$2a + e + \frac{1}{e} - 2 = \frac{e^2 + 8e + 1}{e},$$

or equivalently

$$2a + \frac{e^2 + 1}{e} - 2 = \frac{e^2 + 8e + 1}{e}.$$

Solving for $$a$$ yields

$$2a = \frac{e^2 + 8e + 1}{e} - \frac{e^2 + 1}{e} + 2 = \frac{8e}{e} + 2 = 8 + 2 = 10,$$

so $$a = 5$$.

The correct answer is Option 3: 5.

Parabola: $$y = \frac{x^2+3}{2}$$ (Vertex at $$(0, 1.5)$$, opens up).

Line: $$y = 3 - |x|$$.

Line: $$y = |x - 1|$$.

By finding intersection points and integrating the "top curve minus bottom curve" over the bounded regions:

• Region 1 (left): $$\int_{x_1}^{x_2} ((3+x) - (\frac{x^2+3}{2})) dx$$

• Region 2 (right): $$\int_{x_2}^{x_3} ((3-x) - |x-1|) dx$$

Calculating the definite integral yields $$A = \frac{7}{3}$$.

Then $$6A = 6 \times \frac{7}{3} = \mathbf{14}$$.

Correct Option: C

$$\int_0^{\pi/4} \sin|4x - \pi/12|\,dx$$

$$4x - \pi/12 = 0$$ when $$x = \pi/48$$.

For $$0 \le x < \pi/48$$: $$|4x - \pi/12| = \pi/12 - 4x$$

For $$\pi/48 \le x \le \pi/4$$: $$|4x - \pi/12| = 4x - \pi/12$$

$$I_1 = \int_0^{\pi/48}\sin(\pi/12 - 4x)dx + \int_{\pi/48}^{\pi/4}\sin(4x - \pi/12)dx$$

$$= \left[\frac{\cos(\pi/12-4x)}{4}\right]_0^{\pi/48} + \left[-\frac{\cos(4x-\pi/12)}{4}\right]_{\pi/48}^{\pi/4}$$

$$= \frac{1}{4}[\cos 0 - \cos(\pi/12)] + \frac{1}{4}[-\cos(\pi - \pi/12) + \cos 0]$$

$$= \frac{1}{4}[1 - \cos(\pi/12)] + \frac{1}{4}[\cos(\pi/12) + 1]$$

$$= \frac{1}{4}[1 - \cos(\pi/12) + \cos(\pi/12) + 1] = \frac{2}{4} = \frac{1}{2}$$

 $$\int_0^{\pi/4}[2\sin x]\,dx$$

For $$x \in [0, \pi/4]$$: $$2\sin x$$ ranges from 0 to $$2\sin(\pi/4) = \sqrt{2} \approx 1.414$$.

$$[2\sin x] = 0$$ when $$2\sin x < 1$$, i.e., $$\sin x < 1/2$$, i.e., $$x < \pi/6$$.

$$[2\sin x] = 1$$ when $$1 \le 2\sin x < 2$$, i.e., $$\pi/6 \le x \le \pi/4$$.

$$I_2 = \int_0^{\pi/6} 0\,dx + \int_{\pi/6}^{\pi/4} 1\,dx = \pi/4 - \pi/6 = \pi/12$$

$$24(I_1 + I_2) = 24\left(\frac{1}{2} + \frac{\pi}{12}\right) = 12 + 2\pi$$

So $$2\pi + \alpha = 2\pi + 12$$, giving $$\alpha = 12$$.

The answer is 12.

To find the area of the region defined by the inequality

$$|x - 5| \le y \le 4\sqrt{x}$$

we first determine the range of $$x$$ for which the region exists by solving

$$|x - 5| \le 4\sqrt{x}$$

Step 1: Find the boundaries for $$x$$

We solve the inequality

$$|x - 5| \le 4\sqrt{x}$$

by cases.

Case 1: $$x \ge 5$$

$$x - 5 \le 4\sqrt{x} \implies x - 4\sqrt{x} - 5 \le 0$$

Let

$$t = \sqrt{x}$$

then

$$t^2 - 4t - 5 \le 0$$

Factoring gives

$$(t - 5)(t + 1) \le 0$$

Since

$$t = \sqrt{x} \ge 0$$

we have

$$0 \le t \le 5$$

which implies

$$0 \le x \le 25$$

Combined with

$$x \ge 5$$

we get

$$5 \le x \le 25$$

Case 2: $$x < 5$$

$$5 - x \le 4\sqrt{x} \implies x + 4\sqrt{x} - 5 \ge 0$$

Let

$$t = \sqrt{x}$$

then

$$t^2 + 4t - 5 \ge 0$$

Factoring gives

$$(t + 5)(t - 1) \ge 0$$

Since

$$t \ge 0$$

we must have

$$t \ge 1$$

which implies

$$x \ge 1$$

Combined with

$$x < 5$$

we get

$$1 \le x < 5$$

Combining both cases, the interval for $$x$$ is

$$1 \le x \le 25$$

Step 2: Set up the area integral

The area $$A$$ is the integral of the upper curve minus the lower curve:

$$A = \int_{1}^{5} (4\sqrt{x} - (5 - x)) dx + \int_{5}^{25} (4\sqrt{x} - (x - 5)) dx$$

$$A = \int_{1}^{5} (4\sqrt{x} + x - 5) dx + \int_{5}^{25} (4\sqrt{x} - x + 5) dx$$

Step 3: Evaluate the first integral $$I_1$$

$$I_1 = \int_{1}^{5} (4x^{1/2} + x - 5) dx = \left[ \frac{8}{3}x^{3/2} + \frac{x^2}{2} - 5x \right]_{1}^{5}$$

Evaluating at $$x = 5$$:

$$\left( \frac{8}{3}(5\sqrt{5}) + \frac{25}{2} - 25 \right) = \frac{40\sqrt{5}}{3} - \frac{25}{2}$$

Evaluating at $$x = 1$$:

$$\left( \frac{8}{3} + \frac{1}{2} - 5 \right) = \frac{16 + 3 - 30}{6} = -\frac{11}{6}$$

$$I_1 = \left( \frac{40\sqrt{5}}{3} - \frac{25}{2} \right) - \left( -\frac{11}{6} \right) = \frac{40\sqrt{5}}{3} - \frac{75 - 11}{6} = \frac{40\sqrt{5}}{3} - \frac{64}{6} = \frac{40\sqrt{5} - 32}{3}$$

Step 4: Evaluate the second integral $$I_2$$

$$I_2 = \int_{5}^{25} (4x^{1/2} - x + 5) dx = \left[ \frac{8}{3}x^{3/2} - \frac{x^2}{2} + 5x \right]_{5}^{25}$$

Evaluating at $$x = 25$$:

$$\left( \frac{8}{3}(125) - \frac{625}{2} + 125 \right) = \frac{1000}{3} - \frac{375}{2} = \frac{2000 - 1125}{6} = \frac{875}{6}$$

Evaluating at $$x = 5$$:

$$\left( \frac{8}{3}(5\sqrt{5}) - \frac{25}{2} + 25 \right) = \frac{40\sqrt{5}}{3} + \frac{25}{2} = \frac{80\sqrt{5} + 75}{6}$$

$$I_2 = \frac{875}{6} - \frac{80\sqrt{5} + 75}{6} = \frac{800 - 80\sqrt{5}}{6} = \frac{400 - 40\sqrt{5}}{3}$$

Step 5: Total Area $$A$$

$$A = I_1 + I_2 = \frac{40\sqrt{5} - 32}{3} + \frac{400 - 40\sqrt{5}}{3}$$

$$A = \frac{40\sqrt{5} - 32 + 400 - 40\sqrt{5}}{3} = \frac{368}{3}$$

Thus,

$$3A = 368$$

The integrand can be rewritten in exponential form:

$$\frac{1}{e^{x-1}} = e^{-(x-1)} = e^{1-x}.$$

Therefore the integral becomes

$$\int_{0}^{e^{3}} \bigl[e^{1-x}\bigr] \, dx.$$

Because $$e^{1-x}$$ is a strictly decreasing function of $$x$$, let us first see for which part of the interval $$[0,e^{3}]$$ the value of $$e^{1-x}$$ is $$\ge 1$$.

Set $$e^{1-x} \ge 1 \iff 1 - x \ge 0 \iff x \le 1.$$
Thus for all $$x > 1$$, $$e^{1-x} \lt 1$$ and hence $$\bigl[e^{1-x}\bigr] = 0.$$

So the integral contributes only over $$[0,1]$$:

$$\int_{0}^{e^{3}} \bigl[e^{1-x}\bigr] \, dx = \int_{0}^{1} \bigl[e^{1-x}\bigr] \, dx + \int_{1}^{e^{3}} 0 \, dx = \int_{0}^{1} \bigl[e^{1-x}\bigr] \, dx.$$

Next find how $$e^{1-x}$$ ranges on $$[0,1]$$.
At $$x = 0$$: $$e^{1-0} = e.$$
At $$x = 1$$: $$e^{1-1} = 1.$$

Thus $$e^{1-x}$$ decreases from $$e \approx 2.718$$ down to $$1$$. Its greatest-integer (floor) values can therefore only be $$2$$ or $$1$$.

Find the point where $$e^{1-x} = 2$$:

$$e^{1-x} = 2 \;\Longrightarrow\; 1 - x = \ln 2 \;\Longrightarrow\; x = 1 - \ln 2.$$

So we split the integral:

$$\int_{0}^{1} \bigl[e^{1-x}\bigr]\,dx = \int_{0}^{\,1-\ln 2} 2 \, dx \;+\; \int_{\,1-\ln 2}^{1} 1 \, dx.$$

Compute each part:

First part: $$\int_{0}^{\,1-\ln 2} 2\,dx = 2\bigl(1 - \ln 2 - 0\bigr) = 2 - 2\ln 2.$$

Second part: $$\int_{\,1-\ln 2}^{1} 1\,dx = 1\bigl(1 - (1-\ln 2)\bigr) = \ln 2.$$

Add the two results:

$$2 - 2\ln 2 + \ln 2 = 2 - \ln 2.$$

The question states that the integral equals $$\alpha - \log_e 2$$, i.e.

$$\alpha - \ln 2 = 2 - \ln 2 \;\Longrightarrow\; \alpha = 2.$$

Finally, $$\alpha^{3} = 2^{3} = 8.$$

Hence the required value is $$8$$.

This is in the form $$1^\infty$$. Let $$L$$ be the limit. Take $$\ln L$$:

$$\ln L = \lim_{t\to 0} \frac{\ln \int_{0}^{1} (3x + 5)^t \, dx}{t}$$

Using L'Hôpital's Rule (differentiating w.r.t $$t$$):

$$\ln L = \lim_{t\to 0} \frac{\frac{d}{dt} \int_{0}^{1} e^{t\ln(3x+5)} \, dx}{\int_{0}^{1} (3x+5)^t \, dx} = \frac{\int_{0}^{1} \ln(3x+5) \, dx}{1}$$

Evaluate $$\int_{0}^{1} \ln(3x+5) \, dx$$ using substitution $$u = 3x+5, du = 3dx$$:

$$\frac{1}{3} \int_{5}^{8} \ln u \, du = \frac{1}{3} [u \ln u - u]_5^8 = \frac{1}{3} (8 \ln 8 - 8 - 5 \ln 5 + 5) = \ln\left(\frac{8^8}{5^5}\right)^{1/3} - 1$$

 $$L = e^{\ln(8^8/5^5)^{1/3} - 1} = \frac{1}{e} \cdot \frac{8^{8/3}}{5^{5/3}} = \frac{1}{e} \cdot \frac{8^2 \cdot 8^{2/3}}{5^1 \cdot 5^{2/3}} = \frac{64}{5e} \left(\frac{8}{5}\right)^{2/3}$$

Comparing gives $$\alpha = \mathbf{64}$$.

The region is described by the simultaneous inequalities
  $$|4-x^2|\le y\le x^2,\; y\le 4,\; x\ge 0$$.

Combine the two upper bounds $$y\le x^2$$ and $$y\le 4$$ into a single bound:
  upper bound $$= \min\{x^2,\,4\}$$.
The lower bound remains $$|4-x^2|$$.

We now split the analysis according to whether $$x^2\lt 4$$ or $$x^2\gt 4$$.

Case 1: $$0\le x\le 2$$  ($$x^2\le 4$$)
  Upper bound $$=x^2$$, lower bound $$=4-x^2$$.
  The condition $$4-x^2\le x^2$$ gives $$x^2\ge 2\;\Longrightarrow\;x\ge\sqrt2$$.
  Hence this region exists for $$x\in[\sqrt2,\,2]$$ with
  $$4-x^2\le y\le x^2$$.

Area for Case 1:
$$ A_1=\int_{\sqrt2}^{2}\bigl[x^2-(4-x^2)\bigr]\,dx =\int_{\sqrt2}^{2}\bigl(2x^2-4\bigr)\,dx =2\int_{\sqrt2}^{2}(x^2-2)\,dx. $$
Using $$\int x^2\,dx=\dfrac{x^3}{3}$$:
$$ A_1=2\left[\frac{x^3}{3}-2x\right]_{\sqrt2}^{2} =2\left[\frac{8}{3}-4-\left(\frac{2\sqrt2}{3}-2\sqrt2\right)\right] =\frac{8\,(\,\sqrt2-1\,)}{3}. $$

Case 2: $$x\ge 2$$  ($$x^2\ge 4$$)
  Upper bound $$=4$$, lower bound $$=x^2-4$$.
  Require $$x^2-4\le 4\;\Longrightarrow\;x^2\le 8\;\Longrightarrow\;x\le 2\sqrt2.$$
  Thus this region exists for $$x\in[2,\,2\sqrt2]$$ with
  $$x^2-4\le y\le 4$$.

Area for Case 2:
$$ A_2=\int_{2}^{2\sqrt2}\bigl[4-(x^2-4)\bigr]\,dx =\int_{2}^{2\sqrt2}(8-x^2)\,dx =\left[8x-\frac{x^3}{3}\right]_{2}^{2\sqrt2}. $$
Compute the endpoints:
At $$x=2\sqrt2$$: $$8x-\dfrac{x^3}{3}=16\sqrt2-\dfrac{16\sqrt2}{3}=\dfrac{32\sqrt2}{3}.$$
At $$x=2$$: $$8x-\dfrac{x^3}{3}=16-\dfrac{8}{3}=\dfrac{40}{3}.$$
Therefore
$$ A_2=\frac{32\sqrt2}{3}-\frac{40}{3} =\frac{8\,(\,4\sqrt2-5\,)}{3}. $$

Total area:
$$ A=A_1+A_2 =\frac{8(\sqrt2-1)}{3}+\frac{8(4\sqrt2-5)}{3} =\frac{8}{3}\bigl[5\sqrt2-6\bigr] =\frac{40\sqrt2}{3}-16. $$

Express in the required form $$\dfrac{80\sqrt2}{\alpha}-\beta$$:
$$ \frac{40\sqrt2}{3}-16=\frac{80\sqrt2}{6}-16, $$
so $$\alpha=6,\;\beta=16$$ and $$\alpha+\beta=6+16=22.$$

Final Answer: $$\alpha+\beta=22$$.

Given $$2(x+2)^2f(x) - 3(x+2)^2 = 10\int_0^x(t+2)f(t)dt$$, find f(2).

Differentiate both sides with respect to x

$$4(x+2)f(x) + 2(x+2)^2f'(x) - 6(x+2) = 10(x+2)f(x)$$

$$2(x+2)^2f'(x) = 6(x+2)f(x) + 6(x+2)$$

$$2(x+2)f'(x) = 6f(x) + 6$$

$$(x+2)f'(x) - 3f(x) = 3$$

Find f(0) from original equation

At x=0: $$2(4)f(0) - 3(4) = 0$$

$$8f(0) = 12 \Rightarrow f(0) = 3/2$$

Solve the ODE

$$f'(x) - \frac{3}{x+2}f(x) = \frac{3}{x+2}$$

IF = $$e^{-3\ln(x+2)} = \frac{1}{(x+2)^3}$$

$$\frac{d}{dx}\left[\frac{f(x)}{(x+2)^3}\right] = \frac{3}{(x+2)^4}$$

$$\frac{f(x)}{(x+2)^3} = -\frac{1}{(x+2)^3} + C$$

$$f(x) = -1 + C(x+2)^3$$

Apply f(0) = 3/2

$$3/2 = -1 + C(8) \Rightarrow C = \frac{5/2}{8} = \frac{5}{16}$$

$$f(x) = -1 + \frac{5}{16}(x+2)^3$$

Find f(2)

$$f(2) = -1 + \frac{5}{16}(4)^3 = -1 + \frac{5 \times 64}{16} = -1 + 20 = 19$$

The answer is 19.

To solve this, we find $$a$$ and $$b$$ for differentiability, then integrate to find the area.

For $$f(x)$$ to be differentiable at $$x=1$$, it must be continuous and have equal one-sided derivatives.

• Continuity: $$-3a(1)^2 - 2 = a^2 + b(1) \implies b = -a^2 - 3a - 2$$

• Differentiability: $$-6a(1) = b \implies b = -6a$$

Equating both: $$-6a = -a^2 - 3a - 2 \implies a^2 - 3a + 2 = 0$$.

Given $$a > 1$$, we get $$a = 2$$ and $$b = -12$$.

Intersection with $$y = -20$$

• $$-6x^2 - 2 = -20 \implies x^2 = 3 \implies x = -\sqrt{3}$$ (since $$x < 1$$)

• $$4 - 12x = -20 \implies 12x = 24 \implies x = 2$$

The area is $$\int [f(x) - (-20)] \, dx$$:

1. Left part ($$-\sqrt{3}$$ to $$1$$):

$$\int_{-\sqrt{3}}^{1} (18 - 6x^2) dx = [18x - 2x^3]_{-\sqrt{3}}^{1} = 16 + 12\sqrt{3}$$

2. Right part ($$1$$ to $$2$$):

$$\int_{1}^{2} (24 - 12x) dx = [24x - 6x^2]_{1}^{2} = 6$$

Total Area: $$(16 + 12\sqrt{3}) + 6 = 22 + 12\sqrt{3}$$

Comparing to $$\alpha + \beta\sqrt{3}$$:

$$\alpha = 22, \beta = 12 \implies \alpha + \beta = \mathbf{34}$$

The curve is given by $$y=\max\{\lvert x\rvert,\;x\,\lvert x-2\rvert\}$$ between the vertical lines $$x=-2$$ and $$x=4$$. To find the bounded area we must, on every sub-interval, decide which of the two expressions is larger and then integrate that expression above the $$x$$-axis.

Write the two functions explicitly on each interval:
For $$\lvert x\rvert$$ we already have the standard definition.
For $$x\,\lvert x-2\rvert$$ we split at $$x=2$$.

Case 1: $$x\le 0$$ $$\lvert x\rvert=-x,\qquad\lvert x-2\rvert=2-x,\qquad x\,\lvert x-2\rvert=x(2-x)=2x-x^{2}\lt0$$ Here $$x\,\lvert x-2\rvert$$ is negative whereas $$\lvert x\rvert$$ is positive, so $$y=\lvert x\rvert=-x\quad\text{on}\;[-2,0].$$

Case 2: $$0\le x\le2$$ $$\lvert x\rvert=x,\qquad\lvert x-2\rvert=2-x,\qquad x\,\lvert x-2\rvert=2x-x^{2}.$$ Compare the two: $$x\; \text{vs}\; 2x-x^{2}\;\Longrightarrow\; (2x-x^{2})-x=x(1-x).$$ Therefore   • When $$0\le x\lt1$$, $$x(1-x)\gt0$$, so $$y=2x-x^{2}.$$   • When $$1\le x\le2$$, $$x(1-x)\le0$$, so $$y=x.$$

Case 3: $$2\le x\le4$$ $$\lvert x\rvert=x,\qquad\lvert x-2\rvert=x-2,\qquad x\,\lvert x-2\rvert=x^{2}-2x.$$ Compare: $$(x^{2}-2x)-x=x(x-3).$$ Hence   • When $$2\le x\lt3$$, $$x(x-3)\lt0$$, so $$y=x.$$   • When $$3\le x\le4$$, $$x(x-3)\ge0$$, so $$y=x^{2}-2x.$$

Collecting the pieces we integrate:

$$A_1=\int_{-2}^{0}(-x)\,dx=\Bigl[-\tfrac{x^{2}}{2}\Bigr]_{-2}^{0}=2$$

$$A_2=\int_{0}^{1}(2x-x^{2})\,dx=\Bigl[x^{2}-\tfrac{x^{3}}{3}\Bigr]_{0}^{1}=1-\tfrac13=\tfrac23$$

$$A_3=\int_{1}^{2}x\,dx=\Bigl[\tfrac{x^{2}}{2}\Bigr]_{1}^{2}=2-\tfrac12=\tfrac32$$

$$A_4=\int_{2}^{3}x\,dx=\Bigl[\tfrac{x^{2}}{2}\Bigr]_{2}^{3}=\tfrac92-\tfrac42=\tfrac52$$

$$A_5=\int_{3}^{4}(x^{2}-2x)\,dx=\Bigl[\tfrac{x^{3}}{3}-x^{2}\Bigr]_{3}^{4}=\tfrac{64}{3}-16=\tfrac{16}{3}$$

Add all parts:
$$A=A_1+A_2+A_3+A_4+A_5$$ $$\;=2+\tfrac23+\tfrac32+\tfrac52+\tfrac{16}{3}$$ Convert to a common denominator $$6$$: $$\;=\frac{12+4+9+15+32}{6}=\frac{72}{6}=12.$$

Hence the required area is $$\mathbf{12}$$.

The area is $$\int_{\ln(1/2)}^{0} (e^x - (1-e^x)) dx = \int_{\ln(1/2)}^{0} (2e^x - 1) dx = [2e^x - x]_{\ln(1/2)}^{0} $$

$$= (2-0) - (2(1/2) - \ln(1/2)) = 2 - 1 - \ln 2 = \mathbf{1 - \ln 2}$$ (Option A)

We are given $$f(x) = 7\tan^8 x + 7\tan^6 x - 3\tan^4 x - 3\tan^2 x$$ and need to find $$7I_1 + 12I_2$$ where $$I_1 = \int_0^{\pi/4} f(x)\,dx$$ and $$I_2 = \int_0^{\pi/4} x\,f(x)\,dx$$.

$$f(x) = 7\tan^6 x(\tan^2 x + 1) - 3\tan^2 x(\tan^2 x + 1)$$

$$= (\tan^2 x + 1)(7\tan^6 x - 3\tan^2 x)$$

$$= \sec^2 x \cdot \tan^2 x(7\tan^4 x - 3)$$

Substitute $$t = \tan x$$, so $$dt = \sec^2 x\,dx$$. When $$x = 0, t = 0$$; when $$x = \pi/4, t = 1$$.

$$I_1 = \int_0^1 t^2(7t^4 - 3)\,dt = \int_0^1 (7t^6 - 3t^2)\,dt = \left[t^7 - t^3\right]_0^1 = 1 - 1 = 0$$

From the substitution above: $$G(x) = \int f(x)\,dx = \tan^7 x - \tan^3 x$$

Let $$u = x$$ and $$dv = f(x)\,dx$$, so $$du = dx$$ and $$v = G(x) = \tan^7 x - \tan^3 x$$.

$$I_2 = \left[x \cdot G(x)\right]_0^{\pi/4} - \int_0^{\pi/4} G(x)\,dx$$

$$= \frac{\pi}{4}(\tan^7(\pi/4) - \tan^3(\pi/4)) - \int_0^{\pi/4}(\tan^7 x - \tan^3 x)\,dx$$

$$= \frac{\pi}{4}(1 - 1) - \int_0^{\pi/4}(\tan^7 x - \tan^3 x)\,dx$$

$$= -\int_0^{\pi/4}(\tan^7 x - \tan^3 x)\,dx$$

Using the reduction formula $$\int_0^{\pi/4} \tan^n x\,dx = \frac{1}{n-1} - \int_0^{\pi/4} \tan^{n-2} x\,dx$$:

$$\int_0^{\pi/4} \tan^3 x\,dx = \frac{1}{2} - \int_0^{\pi/4} \tan x\,dx = \frac{1}{2} - \frac{\ln 2}{2}$$

$$\int_0^{\pi/4} \tan^5 x\,dx = \frac{1}{4} - \left(\frac{1}{2} - \frac{\ln 2}{2}\right) = -\frac{1}{4} + \frac{\ln 2}{2}$$

$$\int_0^{\pi/4} \tan^7 x\,dx = \frac{1}{6} - \left(-\frac{1}{4} + \frac{\ln 2}{2}\right) = \frac{5}{12} - \frac{\ln 2}{2}$$

Therefore:

$$\int_0^{\pi/4}(\tan^7 x - \tan^3 x)\,dx = \left(\frac{5}{12} - \frac{\ln 2}{2}\right) - \left(\frac{1}{2} - \frac{\ln 2}{2}\right) = -\frac{1}{12}$$

$$I_2 = -\left(-\frac{1}{12}\right) = \frac{1}{12}$$

$$7I_1 + 12I_2 = 7(0) + 12 \cdot \frac{1}{12} = 0 + 1 = 1$$

The answer is Option B: 1.

The region is defined by the inequalities
$$1+x^{2}\;\le\;y\;\le\;\min\{x+7,\;11-3x\}$$

First compare the two straight lines to find which is smaller at a given $$x$$.
Set $$x+7 = 11-3x$$ to locate their intersection:
$$4x = 4 \;\Longrightarrow\; x = 1,\qquad y = 8$$

Therefore
Case 1: $$x \le 1 \;$$ gives $$x+7 \lt 11-3x$$, so the upper curve is $$y = x+7$$.
Case 2: $$x \ge 1 \;$$ gives $$x+7 \gt 11-3x$$, so the upper curve is $$y = 11-3x$$.

The lower curve is always $$y = 1+x^{2}$$. To obtain the $$x$$-range where the region exists, ensure $$1+x^{2} \le \text{(upper curve)}$$ in each case.

Case 1: $$x \le 1$$
Require $$1+x^{2} \le x+7$$ $$\Longrightarrow\; x^{2}-x-6 \le 0$$ $$\Longrightarrow\; (x-3)(x+2) \le 0$$ $$\Longrightarrow\; -2 \le x \le 3$$. Combining with $$x \le 1$$ gives $$-2 \le x \le 1$$.

Case 2: $$x \ge 1$$
Require $$1+x^{2} \le 11-3x$$ $$\Longrightarrow\; x^{2}+3x-10 \le 0$$ $$\Longrightarrow\; (x+5)(x-2) \le 0$$ $$\Longrightarrow\; -5 \le x \le 2$$. Combining with $$x \ge 1$$ gives $$1 \le x \le 2$$.

Hence the complete $$x$$-interval for the region is $$-2 \le x \le 2$$, divided at $$x = 1$$.

The area $$A$$ is the sum of two integrals.

For $$-2 \le x \le 1$$
Upper curve: $$x+7$$, lower curve: $$1+x^{2}$$.
Area part $$A_{1}$$:
$$A_{1} = \int_{-2}^{1} \big[(x+7) - (1+x^{2})\big]\;dx = \int_{-2}^{1} \big(-x^{2}+x+6\big)\;dx$$

Integrate term-by-term:
$$\int (-x^{2})dx = -\frac{x^{3}}{3},\qquad \int x\,dx = \frac{x^{2}}{2},\qquad \int 6\,dx = 6x$$

Evaluate from $$x=-2$$ to $$x=1$$:
At $$x=1$$: $$-\tfrac{1}{3} + \tfrac{1}{2} + 6 = \tfrac{37}{6}$$
At $$x=-2$$: $$\;\;\tfrac{8}{3} + 2 - 12 = -\tfrac{22}{3}$$
Thus
$$A_{1} = \frac{37}{6} - \Big(-\frac{22}{3}\Big) = \frac{37}{6} + \frac{44}{6} = \frac{81}{6} = \frac{27}{2}$$

For $$1 \le x \le 2$$
Upper curve: $$11-3x$$, lower curve: $$1+x^{2}$$.
Area part $$A_{2}$$:
$$A_{2} = \int_{1}^{2} \big[(11-3x) - (1+x^{2})\big]\;dx = \int_{1}^{2} \big(-x^{2}-3x+10\big)\;dx$$

Integrate term-by-term:
$$\int (-x^{2})dx = -\frac{x^{3}}{3},\qquad \int (-3x)dx = -\frac{3x^{2}}{2},\qquad \int 10\,dx = 10x$$

Evaluate from $$x=1$$ to $$x=2$$:
At $$x=2$$: $$-\tfrac{8}{3} - 6 + 20 = \tfrac{34}{3}$$
At $$x=1$$: $$-\tfrac{1}{3} - \tfrac{3}{2} + 10 = \tfrac{49}{6}$$
Thus
$$A_{2} = \frac{34}{3} - \frac{49}{6} = \frac{68}{6} - \frac{49}{6} = \frac{19}{6}$$

Total area:
$$A = A_{1} + A_{2} = \frac{27}{2} + \frac{19}{6} = \frac{81}{6} + \frac{19}{6} = \frac{100}{6} = \frac{50}{3}$$

The question asks for $$3A$$:
$$3A = 3 \times \frac{50}{3} = 50$$

Therefore, $$3A = 50$$, which corresponds to Option A.

The point of intersection of the parabola $$x^{2}=2y$$ and the straight line $$y-2x-6=0$$ is obtained by substituting $$y=2x+6$$ into the parabola.

$$x^{2}=2(2x+6)=4x+12$$
$$x^{2}-4x-12=0$$

Solving the quadratic gives $$x=\frac{4 \pm \sqrt{16+48}}{2}=6,-2$$.
In the second quadrant $$x\lt 0$$ and $$y\gt 0$$, so we choose $$x=-2$$.
Then $$y=2(-2)+6=2$$. Hence $$(a,b)=(-2,2)$$ and the integral limits are $$a=-2$$, $$b=2$$.

Define $$I=\int_{-2}^{2}\frac{9x^{2}}{1+5^{x}}\,dx$$ $$-(1)$$.

To exploit symmetry, substitute $$x\rightarrow -x$$ in $$I$$.
Because $$dx\rightarrow -dx$$, the limits interchange and the integral becomes

$$I=\int_{-2}^{2}\frac{9x^{2}}{1+5^{-x}}\,dx$$ $$-(2)$$.

Add $$(1)$$ and $$(2)$$:

$$2I=\int_{-2}^{2}9x^{2}\left[\frac{1}{1+5^{x}}+\frac{1}{1+5^{-x}}\right]dx$$.

Simplify the bracket. Let $$t=5^{x}$$, so $$5^{-x}=1/t$$:

$$\frac{1}{1+5^{x}}+\frac{1}{1+5^{-x}}=\frac{1}{1+t}+\frac{1}{1+1/t}=\frac{1}{1+t}+\frac{t}{1+t}=1$$.

Thus $$2I=\int_{-2}^{2}9x^{2}\,dx=9\int_{-2}^{2}x^{2}\,dx$$.

Compute the remaining integral:

$$\int_{-2}^{2}x^{2}\,dx=\left.\frac{x^{3}}{3}\right|_{-2}^{2}=\frac{8}{3}-\left(-\frac{8}{3}\right)=\frac{16}{3}$$.

Therefore $$2I=9\left(\frac{16}{3}\right)=48$$ and $$I=24$$.

The required value is $$24$$, which matches Option A.

The given curves are $$y = x^2 - 4x + 4$$ and $$y^2 = 16 - 8x$$.

Since the first curve simplifies to $$y = (x - 2)^2$$ and the second becomes $$y^2 = -8(x - 2)\,,\,$$ the first is a parabola opening upward with vertex at $$(2,0)$$, while the second is a parabola opening to the left with the same vertex.

To determine the points of intersection, substituting $$y = (x - 2)^2$$ into $$y^2 = 16 - 8x$$ gives $$ [(x - 2)^2]^2 = 16 - 8x, $$ which simplifies to $$ (x - 2)^4 = 16 - 8x. $$ Then setting $$u = x - 2$$ transforms the equation into $$ u^4 = 16 - 8(u + 2), $$ so $$ u^4 = -8u,\quad u^4 + 8u = 0,\quad u(u^3 + 8) = 0. $$ Hence $$u = 0$$ or $$u^3 = -8$$, giving $$u = -2$$ and therefore $$x = 2$$ or $$x = 0$$.

For each value of $$x$$ we find the corresponding $$y$$ from $$y = (x - 2)^2$$. When $$x = 2$$, $$y = (2 - 2)^2 = 0$$ and one checks that $$y^2 = 16 - 8\cdot 2 = 0$$, so the intersection point is $$(2, 0)$$. When $$x = 0$$, $$y = (0 - 2)^2 = 4$$ while the second curve gives $$y^2 = 16 - 8\cdot 0 = 16$$, so $$y = \pm 4$$, but only $$y = 4$$ satisfies the first curve. Thus the intersection points are $$(0, 4)$$ and $$(2, 0)$$.

The upper branch of the second curve is $$y = \sqrt{16 - 8x}$$ and the lower branch is $$y = -\sqrt{16 - 8x}$$, whereas the first curve $$y = (x - 2)^2$$ is always non-negative. Since the lower branch is non-positive and the first curve non-negative, they meet only at $$(2,0)$$, confirming that no other intersections occur.

The enclosed region lies between $$x = 0$$ and $$x = 2$$, bounded above by $$y = \sqrt{16 - 8x}$$ and below by $$y = (x - 2)^2$$. For example at $$x = 1$$, the upper branch gives $$y = \sqrt{8} \approx 2.828$$ while the first curve gives $$y = 1$$, so the region is well-defined. The area is therefore $$ A = \int_{0}^{2} \bigl(\sqrt{16 - 8x} - (x - 2)^2\bigr)\,dx. $$

Since $$\sqrt{16 - 8x} = \sqrt{8(2 - x)} = 2\sqrt{2}\,\sqrt{2 - x}$$ and $$(x - 2)^2 = (2 - x)^2$$, the integral becomes $$ A = \int_{0}^{2} \Bigl(2\sqrt{2}\,\sqrt{2 - x} - (2 - x)^2\Bigr)\,dx. $$ Substituting $$t = 2 - x$$ gives $$dx = -dt$$ and transforms the limits from $$x = 0 \to 2$$ into $$t = 2 \to 0$$, so that $$ A = \int_{2}^{0} \bigl(2\sqrt{2}\,t^{1/2} - t^2\bigr)(-dt) = \int_{0}^{2} \bigl(2\sqrt{2}\,t^{1/2} - t^2\bigr)\,dt. $$

Integrating term by term, since $$\int t^{1/2}\,dt = \tfrac{2}{3}t^{3/2}$$ and $$\int t^2\,dt = \tfrac{1}{3}t^3$$, we have $$ A = \left[2\sqrt{2}\cdot\tfrac{2}{3}t^{3/2} - \tfrac{1}{3}t^3\right]_{0}^{2} = \left[\tfrac{4\sqrt{2}}{3}t^{3/2} - \tfrac{1}{3}t^3\right]_{0}^{2}. $$ At $$t = 2$$, $$t^{3/2} = 2\sqrt{2}$$ and $$t^3 = 8$$, so $$ \tfrac{4\sqrt{2}}{3}\cdot2\sqrt{2} = \tfrac{16}{3},\quad \tfrac{1}{3}\cdot8 = \tfrac{8}{3}, $$ giving $$A = \tfrac{16}{3} - \tfrac{8}{3} = \tfrac{8}{3}$$, and at $$t = 0$$ both terms vanish.

Thus, the area of the region enclosed by the curves is $$\frac{8}{3}$$. The correct option is A. $$\frac{8}{3}$$.

We start from the functional equation
$$f(x)=1-2x+\displaystyle\int_{0}^{x}e^{\,x-t}\,f(t)\,dt\qquad\bigl(x\in[0,\infty)\bigr).$$

Differentiate both sides with respect to $$x$$. The derivative of an integral with variable upper limit is obtained from Leibniz rule:

$$ \frac{d}{dx}\int_{0}^{x}e^{\,x-t}\,f(t)\,dt =\;e^{\,x-x}\,f(x)+\int_{0}^{x}\frac{\partial}{\partial x}\bigl[e^{\,x-t}\bigr]\,f(t)\,dt =f(x)+\int_{0}^{x}e^{\,x-t}\,f(t)\,dt. $$

Denote the integral in the given relation by $$I(x)$$, so that $$I(x)=\displaystyle\int_{0}^{x}e^{\,x-t}\,f(t)\,dt.$$ Using the original equation we have $$I(x)=f(x)-1+2x.$$ Hence

$$ f'(x)=-2+\bigl[f(x)+I(x)\bigr] =-2+\bigl[f(x)+f(x)-1+2x\bigr] =2f(x)+2x-3.\qquad-(1) $$

Equation $$(1)$$ is a linear first-order differential equation

$$ f'(x)-2f(x)=2x-3. $$

Its integrating factor is $$e^{-2x}$$. Multiplying throughout,

$$ e^{-2x}f'(x)-2e^{-2x}f(x)=\bigl(e^{-2x}f(x)\bigr)' =(2x-3)\,e^{-2x}. $$

Integrate from $$0$$ to $$x$$:

$$ e^{-2x}f(x)-e^{0}f(0)=\int_{0}^{x}(2t-3)\,e^{-2t}\,dt.\qquad-(2) $$

From the given relation at $$x=0$$, $$f(0)=1.$$

Next evaluate the integral on the right of $$(2)$$. First find an antiderivative:

$$ \int(2t-3)e^{-2t}\,dt =\int2t\,e^{-2t}\,dt-\int3e^{-2t}\,dt. $$

Using integration by parts for the first term (or a standard formula) and direct integration for the second,

$$ \int2t\,e^{-2t}\,dt=-t\,e^{-2t}-\tfrac12\,e^{-2t},\qquad \int3e^{-2t}\,dt=-\tfrac32\,e^{-2t}. $$

Therefore $$\int(2t-3)e^{-2t}\,dt=-t\,e^{-2t}+e^{-2t}=e^{-2t}(1-t)+C.$$

Applying limits $$0$$ to $$x$$:

$$ \int_{0}^{x}(2t-3)e^{-2t}\,dt =e^{-2x}(1-x)-1.\qquad-(3) $$

Substituting $$(3)$$ and $$f(0)=1$$ into $$(2)$$ gives

$$ e^{-2x}f(x)-1=e^{-2x}(1-x)-1. $$

Add $$1$$ to both sides and multiply by $$e^{2x}$$:

$$ f(x)=1-x\quad\text{for all }x\ge 0.\qquad-(4) $$

The curve $$y=f(x)$$ is the straight line $$y=1-x$$. Its intercepts with the coordinate axes are
• on the $$y$$-axis: $$(0,1)$$, • on the $$x$$-axis: $$(1,0)$$.

Thus the region bounded by the curve and the two coordinate axes is the right-triangle with vertices $$(0,0),(0,1),(1,0)$$. Its area is

$$ \text{Area}=\frac12\,(\,\text{base}\,)\times(\,\text{height}\,) =\frac12\,(1)\,(1)=\frac12. $$

Hence the required area equals $$\dfrac12$$, which matches Option B.

Given $$g(x) = \int_0^x tf(t)dt$$ and $$g(x^3) = x^6 + x^7$$.

Differentiating both sides with respect to $$x$$:

$$g'(x^3) \cdot 3x^2 = 6x^5 + 7x^6$$

Since $$g'(x) = xf(x)$$: $$x^3 f(x^3) \cdot 3x^2 = 6x^5 + 7x^6$$.

$$3x^5 f(x^3) = x^5(6 + 7x)$$, so $$f(x^3) = 2 + \frac{7x}{3}$$.

$$\sum_{r=1}^{15} f(r^3) = \sum_{r=1}^{15}\left(2 + \frac{7r}{3}\right) = 30 + \frac{7}{3} \cdot \frac{15 \times 16}{2} = 30 + 280 = 310$$.

Given

$$I_1 = \int_{-1/2}^{1} 2x\,f\!\left(2x\left(1-2x\right)\right)\,dx$$
$$I_2 = \int_{-1}^{2} f\!\left(x\left(1-x\right)\right)\,dx$$

Put $$x = 2t$$ in $$I_2$$. Then $$dx = 2\,dt$$, and the limits change as follows:
for $$x = -1$$, $$t = -\dfrac12$$; for $$x = 2$$, $$t = 1$$.

Hence

$$I_2 = \int_{-1}^{2} f\!\left(x(1-x)\right)\,dx = \int_{-1/2}^{1} f\!\left(2t(1-2t)\right)\,2\,dt = 2\int_{-1/2}^{1} f\!\left(2t(1-2t)\right)\,dt$$

For convenience, define

$$A = \int_{-1/2}^{1} f\!\left(2x(1-2x)\right)\,dx \qquad -(1)$$

With this notation we already have

$$I_2 = 2A \qquad -(2)$$

Next, relate $$I_1$$ to $$A$$. Write

$$I_1 = \int_{-1/2}^{1} 2x\,f\!\left(2x(1-2x)\right)\,dx$$

Let

$$J = \int_{-1/2}^{1} x\,f\!\left(2x(1-2x)\right)\,dx$$ so that $$I_1 = 2J$$.

The key symmetry is

$$g(x) = 2x(1-2x)$$ satisfies $$g\!\left(\dfrac12 - x\right)=g(x)$$.

Make the substitution $$u = \dfrac12 - x$$ in $$J$$. Then $$du = -dx$$ and the limits interchange, giving

$$J = \int_{-1/2}^{1} \left(\dfrac12 - u\right)\,f\!\left(g(u)\right)\,du$$

Add this result to the original definition of $$J$$:

$$2J = \int_{-1/2}^{1} \left[x + \left(\dfrac12 - x\right)\right] f\!\left(g(x)\right)\,dx = \int_{-1/2}^{1} \dfrac12\,f\!\left(g(x)\right)\,dx = \dfrac12\,A$$

Therefore

$$J = \dfrac14\,A \quad\Longrightarrow\quad I_1 = 2J = \dfrac12\,A \qquad -(3)$$

Combine $$(2)$$ and $$(3)$$:

$$\dfrac{I_2}{I_1} = \dfrac{2A}{A/2} = 4$$

Thus $$\displaystyle \frac{I_2}{I_1} = 4$$, corresponding to Option D.

The parabola is $$y^{2}=x-2 \; \; (1)$$.
A general point on it can be written in parametric form as $$\bigl(t^{2}+2,\;t\bigr)\;.$$

Equation of the tangent at the parametric point
For $$y^{2}=x-2$$ the standard slope-form of a tangent is obtained by differentiating: $$2y\dfrac{dy}{dx}=1\;\;\Longrightarrow\;\;\dfrac{dy}{dx}=\dfrac{1}{2y}\;.$$
At the point $$B\bigl(x_{1},y_{1}\bigr)=\bigl(t^{2}+2,\;t\bigr)$$ the slope is $$m=\dfrac{1}{2t}$$ and the tangent is $$y-t=\dfrac{1}{2t}\bigl(x-(t^{2}+2)\bigr)\;.\qquad -(2)$$

The tangent passes through the fixed point $$A(-2,0)$$, so $$(x,y)=(-2,0)$$ must satisfy (2):
$$0-t=\dfrac{1}{2t}\Bigl(-2-(t^{2}+2)\Bigr) \;\;\Longrightarrow\;\;-t=-\dfrac{t^{2}+4}{2t} \;\;\Longrightarrow\;\;2t^{2}=t^{2}+4 \;\;\Longrightarrow\;\;t^{2}=4\;.$$

Because point $$B$$ lies in the first quadrant, $$t=+2$$. Thus $$B(6,2)\;,\qquad m=\dfrac{1}{2t}=\dfrac14\;.$$

Equation of the required tangent (line AB)
Using point $$B(6,2)$$ and slope $$\dfrac14$$: $$y-2=\dfrac14(x-6)\quad\Longrightarrow\quad y=\dfrac{x}{4}+\dfrac12\;.\qquad -(3)$$

Sketch of the enclosed region
The boundary consists of • the segment of line (3) from $$A(-2,0)$$ to $$B(6,2)$$,
• the arc of the parabola (1) from $$D(2,0)$$ to $$B(6,2)$$, and
• the segment of the $$x$$-axis from $$A(-2,0)$$ to $$D(2,0)$$.

Area under the line from $$x=-2$$ to $$x=6$$
From (3), $$y=\dfrac{x}{4}+\dfrac12$$. $$\displaystyle A_{\text{line}}=\int_{-2}^{6}\Bigl(\dfrac{x}{4}+\dfrac12\Bigr)\,dx =\left[\dfrac{x^{2}}{8}+\dfrac{x}{2}\right]_{-2}^{6} =\left(\dfrac{36}{8}+3\right)-\left(\dfrac{4}{8}-1\right)=8\;.$$

Area under the parabola from $$x=2$$ to $$x=6$$
For the upper half of (1), $$y=\sqrt{x-2}$$. $$\displaystyle A_{\text{parab}}=\int_{2}^{6}\sqrt{x-2}\,dx =\frac23\bigl(x-2\bigr)^{3/2}\Bigl|_{2}^{6} =\frac23\,(4)^{3/2}=\,\frac{16}{3}\;.$$

Required area
Over $$[-2,2]$$ the region is the strip between the line and the $$x$$-axis; over $$[2,6]$$ it lies between the line and the parabola. Hence $$\displaystyle A=A_{\text{line}}-A_{\text{parab}} =8-\frac{16}{3}=\,\frac{8}{3}\;.$$

The area bounded by the line $$AB$$, the parabola $$y^{2}=x-2$$ and the $$x$$-axis equals $$\displaystyle \frac{8}{3}$$.
Therefore, the correct option is Option C.

Parametrize the line by setting $$\frac{x-3}{7} = \frac{y-2}{-1} = \frac{z+1}{-2} = \lambda$$. Then any point on the line is:

$$x = 3 + 7\lambda$$ $$y = 2 - \lambda$$  $$z = -1 - 2\lambda$$

So P has coordinates $$(3 + 7\lambda, 2 - \lambda, -1 - 2\lambda)$$.

The direction vector of the line is $$\vec{d} = (7, -1, -2)$$. The vector $$\overrightarrow{PQ}$$ from P to Q is:

$$\overrightarrow{PQ} = (10 - (3 + 7\lambda), -3 - (2 - \lambda), -1 - (-1 - 2\lambda)) = (7 - 7\lambda, -5 + \lambda, 2\lambda)$$

Since $$\overrightarrow{PQ}$$ is perpendicular to $$\vec{d}$$, their dot product is zero:

$$(7 - 7\lambda) \cdot 7 + (-5 + \lambda) \cdot (-1) + (2\lambda) \cdot (-2) = 0$$

$$49 - 49\lambda + 5 - \lambda - 4\lambda = 0$$

$$54 - 54\lambda = 0$$

$$\lambda = 1$$

Substitute $$\lambda = 1$$ to find P:

$$x = 3 + 7(1) = 10$$

$$y = 2 - 1 = 1$$

$$z = -1 - 2(1) = -3$$

So P is $$(10, 1, -3)$$.

Given R is $$(3, -2, 1)$$, the vertices of triangle PQR are P(10, 1, -3), Q(10, -3, -1), and R(3, -2, 1).

To confirm the right angle, compute vectors:

$$\overrightarrow{PQ} = Q - P = (10 - 10, -3 - 1, -1 - (-3)) = (0, -4, 2)$$

$$\overrightarrow{QR} = R - Q = (3 - 10, -2 - (-3), 1 - (-1)) = (-7, 1, 2)$$

$$\overrightarrow{PR} = R - P = (3 - 10, -2 - 1, 1 - (-3)) = (-7, -3, 4)$$

$$\overrightarrow{PQ} \cdot \overrightarrow{QR} = (0)(-7) + (-4)(1) + (2)(2) = 0 - 4 + 4 = 0$$

Since the dot product is zero, $$\overrightarrow{PQ}$$ and $$\overrightarrow{QR}$$ are perpendicular, so the right angle is at Q.

The area of a right-angled triangle is half the product of the legs. The legs are PQ and QR.

$$|\overrightarrow{PQ}| = \sqrt{0^2 + (-4)^2 + 2^2} = \sqrt{0 + 16 + 4} = \sqrt{20} = 2\sqrt{5}$$

$$|\overrightarrow{QR}| = \sqrt{(-7)^2 + 1^2 + 2^2} = \sqrt{49 + 1 + 4} = \sqrt{54} = 3\sqrt{6}$$

Area $$= \frac{1}{2} \times |\overrightarrow{PQ}| \times |\overrightarrow{QR}| = \frac{1}{2} \times 2\sqrt{5} \times 3\sqrt{6} = \frac{1}{2} \times 6 \times \sqrt{30} = 3\sqrt{30}$$

Thus, the area is $$3\sqrt{30}$$, which corresponds to option D.

We need to evaluate $$80\int_{0}^{\pi/4}\frac{\sin\theta + \cos\theta}{9 + 16\sin 2\theta}\,d\theta$$. Noting that $$\sin 2\theta = 2\sin\theta\cos\theta$$ and $$(\sin\theta + \cos\theta)^2 = 1 + \sin 2\theta$$, we have $$\sin 2\theta = (\sin\theta + \cos\theta)^2 - 1$$.

To simplify the integrand, let $$t = \sin\theta - \cos\theta$$ so that $$dt = (\cos\theta + \sin\theta)\,d\theta$$. Also, since $$t^2 = 1 - \sin 2\theta$$, it follows that $$\sin 2\theta = 1 - t^2$$. When $$\theta = 0$$, $$t = -1$$, and when $$\theta = \pi/4$$, $$t = 0$$.

Under this substitution the integral becomes $$ 80\int_{-1}^{0}\frac{dt}{9 + 16(1 - t^2)} = 80\int_{-1}^{0}\frac{dt}{25 - 16t^2}. $$

Using partial fractions, $$ \frac{1}{25 - 16t^2} = \frac{1}{(5 - 4t)(5 + 4t)} = \frac{1}{10}\biggl(\frac{1}{5-4t} + \frac{1}{5+4t}\biggr). $$ Thus $$ 80 \times \frac{1}{10}\int_{-1}^{0}\Bigl(\frac{1}{5-4t} + \frac{1}{5+4t}\Bigr)dt = 8\Bigl[-\frac{1}{4}\ln|5-4t| + \frac{1}{4}\ln|5+4t|\Bigr]_{-1}^{0} = 2\Bigl[\ln\Bigl(\frac{5+4t}{5-4t}\Bigr)\Bigr]_{-1}^{0}. $$

Evaluating the logarithm gives $$ 2\Bigl[\ln\Bigl(\frac{5}{5}\Bigr) - \ln\Bigl(\frac{1}{9}\Bigr)\Bigr] = 2\ln 9 = 4\ln 3. $$ Hence the correct answer is Option 2: $$4\log_e 3$$.

Let the given definite integral be $$I$$:

$$I = \int_0^{\pi} \frac{8x}{4\cos^2 x + \sin^2 x} \, dx$$

Applying the definite integral property $$\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$$:

$$I = \int_0^{\pi} \frac{8(\pi - x)}{4\cos^2(\pi - x) + \sin^2(\pi - x)} \, dx$$

Since $$\cos(\pi - x) = -\cos x \implies \cos^2(\pi - x) = \cos^2 x$$ and $$\sin(\pi - x) = \sin x \implies \sin^2(\pi - x) = \sin^2 x$$:

$$I = \int_0^{\pi} \frac{8\pi - 8x}{4\cos^2 x + \sin^2 x} \, dx$$

---

Adding these two integral expressions together:

$$2I = \int_0^{\pi} \frac{8\pi}{4\cos^2 x + \sin^2 x} \, dx$$

$$I = 4\pi \int_0^{\pi} \frac{1}{4\cos^2 x + \sin^2 x} \, dx$$

Using the property $$\int_0^{2a} f(x) \, dx = 2\int_0^a f(x) \, dx$$ because the integrand is symmetric about $$\frac{\pi}{2}$$:

$$I = 8\pi \int_0^{\frac{\pi}{2}} \frac{1}{4\cos^2 x + \sin^2 x} \, dx$$

---

To evaluate this integral, divide the numerator and the denominator by $$\cos^2 x$$:

$$I = 8\pi \int_0^{\frac{\pi}{2}} \frac{\sec^2 x}{4 + \tan^2 x} \, dx$$

Let $$\tan x = t$$. Differentiating both sides gives $$\sec^2 x \, dx = dt$$.

The limits of integration change from $$x = 0 \to t = 0$$ and $$x = \frac{\pi}{2} \to t = \infty$$.

Substituting these into the integral:

$$I = 8\pi \int_0^{\infty} \frac{1}{4 + t^2} \, dt$$

---

Using the standard integration formula $$\int \frac{1}{a^2 + t^2} \, dt = \frac{1}{a}\tan^{-1}\left(\frac{t}{a}\right)$$:

$$I = 8\pi \left[ \frac{1}{2}\tan^{-1}\left(\frac{t}{2}\right) \right]_0^{\infty}$$

$$I = 4\pi \left[ \tan^{-1}(\infty) - \tan^{-1}(0) \right]$$

$$I = 4\pi \left[ \frac{\pi}{2} - 0 \right] = 2\pi^2$$

Therefore, the final value of the integral is equal to $$2\pi^2$$.

First, rewrite the given function in a factored form.

$$f(x)=\frac{\sin x}{e^{\pi x}}\cdot\frac{x^{2023}+2024x+2025}{x^{2}-x+3}+\frac{2}{e^{\pi x}}\cdot\frac{x^{2023}+2024x+2025}{x^{2}-x+3}$$

Factor out the common terms $$\dfrac{x^{2023}+2024x+2025}{x^{2}-x+3}\cdot e^{-\pi x}$$:

$$f(x)=\frac{x^{2023}+2024x+2025}{x^{2}-x+3}\,e^{-\pi x}\,(\sin x+2)$$

To solve $$f(x)=0$$ we must examine when each factor can be zero.

1. The exponential factor $$e^{-\pi x}$$ is never zero for any real $$x$$.
2. The quadratic denominator $$x^{2}-x+3$$ has discriminant $$(-1)^{2}-4(1)(3)=-11\lt 0$$, so it is always positive and never zero.
3. The factor $$\sin x+2$$ cannot be zero because $$\sin x\in[-1,1]$$, so $$\sin x+2\in[1,3]$$ for all $$x\in\mathbb{R}$$.

Therefore the only possible zeros come from

$$x^{2023}+2024x+2025=0$$

Define $$g(x)=x^{2023}+2024x+2025$$. To count its real roots, study monotonicity:

$$g'(x)=2023x^{2022}+2024$$

Since $$x^{2022}\ge 0$$ for every real $$x$$, we have $$g'(x)\ge 2024\gt 0$$. Thus $$g(x)$$ is strictly increasing on the entire real line.

A strictly increasing odd-degree polynomial with positive leading coefficient satisfies

$$\lim_{x\to-\infty}g(x)=-\infty,\qquad \lim_{x\to+\infty}g(x)=+\infty$$

By the Intermediate Value Theorem, it crosses the $$x$$-axis exactly once. Hence $$g(x)=0$$ has exactly one real solution.

Because no other factor of $$f(x)$$ can become zero, this single root of $$g(x)$$ is the only root of $$f(x)$$.

Therefore, the number of solutions of $$f(x)=0$$ in $$\mathbb{R}$$ is

1

We have to find the area of the region

$$S=\left\{(x,y)\in\mathbb{R}\times\mathbb{R}:\;x\ge 0,\;y\ge 0,\;y^{2}\le 4x,\;y^{2}\le 12-2x,\;3y+\sqrt 8\,x\le 5\sqrt 8\right\}$$

(1) From $$y^{2}\le 4x$$ we get $$x\ge\dfrac{y^{2}}{4}$$.     This is the region to the right of the right-opening parabola $$y^{2}=4x$$.

(2) From $$y^{2}\le 12-2x$$ we get $$x\le 6-\dfrac{y^{2}}{2}$$.     This is the region to the left of the left-opening parabola $$y^{2}=12-2x$$ whose vertex is at $$(6,0)$$.

(3) From $$3y+\sqrt 8\,x\le 5\sqrt 8$$ write $$\sqrt 8 = 2\sqrt 2$$:

$$3y+2\sqrt 2\,x\le 10\sqrt 2 \;\Longrightarrow\; x\le \dfrac{10\sqrt 2-3y}{2\sqrt 2}=5-\dfrac{3y}{2\sqrt 2}$$

This is the region to the left of the straight line $$3y+2\sqrt 2\,x=10\sqrt 2$$.

Combining (1) and (2):

$$\dfrac{y^{2}}{4}\le x\le 6-\dfrac{y^{2}}{2}$$ For such an $$x$$ to exist we need

$$\dfrac{y^{2}}{4}\le 6-\dfrac{y^{2}}{2} \;\Longrightarrow\; 3y^{2}\le 24 \;\Longrightarrow\; y^{2}\le 8 \;\Longrightarrow\; 0\le y\le 2\sqrt 2$$

Now decide which of the two upper bounds $$x_{1}=6-\dfrac{y^{2}}{2},\qquad x_{2}=5-\dfrac{3y}{2\sqrt 2}$$ is smaller (the region must satisfy both). Define

$$D(y)=x_{2}-x_{1} =\Bigl(5-\dfrac{3y}{2\sqrt 2}\Bigr)-\Bigl(6-\dfrac{y^{2}}{2}\Bigr) =-1-\dfrac{3y}{2\sqrt 2}+\dfrac{y^{2}}{2}$$

$$D(y)=0 \;\Longrightarrow\; \dfrac{y^{2}}{2}-\dfrac{3y}{2\sqrt 2}-1=0 \;\Longrightarrow\; y^{2}-\dfrac{3}{\sqrt 2}y-2=0$$ Solving gives $$y=2\sqrt 2\quad\text{or}\quad y=-\dfrac{1}{\sqrt 2}$$. On $$0\le y\lt 2\sqrt 2$$, we have $$D(y)\lt 0$$, so $$x_{2}

Hence for every $$0\le y\le 2\sqrt 2$$ the $$x$$-limits are

$$x_{\text{min}}=\dfrac{y^{2}}{4},\qquad x_{\text{max}}=5-\dfrac{3y}{2\sqrt 2}$$

The area $$A$$ of the region is therefore

$$A=\int_{0}^{2\sqrt 2} \left[x_{\text{max}}-x_{\text{min}}\right]\,dy =\int_{0}^{2\sqrt 2} \left(5-\dfrac{3y}{2\sqrt 2}-\dfrac{y^{2}}{4}\right)dy$$

Integrate term by term:

$$\int 5\,dy = 5y$$ $$\int -\dfrac{3y}{2\sqrt 2}\,dy = -\dfrac{3}{2\sqrt 2}\cdot\dfrac{y^{2}}{2} = -\dfrac{3y^{2}}{4\sqrt 2}$$ $$\int -\dfrac{y^{2}}{4}\,dy = -\dfrac{1}{4}\cdot\dfrac{y^{3}}{3} = -\dfrac{y^{3}}{12}$$

Thus

$$A=\Bigl[\,5y-\dfrac{3y^{2}}{4\sqrt 2}-\dfrac{y^{3}}{12}\Bigr]_{0}^{2\sqrt 2}$$

At $$y=2\sqrt 2$$:

$$y^{2}=8,\quad y^{3}=16\sqrt 2$$

$$A=5(2\sqrt 2)-\dfrac{3\cdot8}{4\sqrt 2}-\dfrac{16\sqrt 2}{12} =10\sqrt 2-\dfrac{24}{4\sqrt 2}-\dfrac{4\sqrt 2}{3}$$

$$\dfrac{24}{4\sqrt 2}=6/\sqrt 2=3\sqrt 2$$

$$\therefore\;A=10\sqrt 2-3\sqrt 2-\dfrac{4\sqrt 2}{3} =7\sqrt 2-\dfrac{4\sqrt 2}{3} =\dfrac{21\sqrt 2-4\sqrt 2}{3} =\dfrac{17\sqrt 2}{3}$$

Thus $$A=\alpha\sqrt 2$$ with $$\alpha=\dfrac{17}{3}$$.

Option B which is: $$\dfrac{17}{3}$$

 Notice the limit is in $$\frac{0}{0}$$ form.

 Use the standard limit $$\lim_{x \to 0} \frac{e^{x^2}-1}{x^2} = 1$$.

 Rewrite the expression: $$\alpha = \lim_{x \to 0} \frac{x \int_0^x f(t)dt}{x^2} = \lim_{x \to 0} \frac{\int_0^x f(t)dt}{x}$$.

Apply L'Hôpital's Rule (Leibniz Rule for the numerator):

$$\alpha = \lim_{x \to 0} \frac{f(x)}{1} = f(0) = \frac{1}{2}$$

Calculate $$8\alpha^2 = 8(1/2)^2 = 8(1/4) = 2$$.

We start with the function $$f:\mathbb{R}\rightarrow\mathbb{R}$$ defined by $$f(x)=\dfrac{x}{(1+x^4)^{1/4}}\,.$$

Step 1 : An equation relating $$f(x)$$ and $$x$$.
Let $$y=f(x)\,.$$ Then

$$y=\dfrac{x}{(1+x^4)^{1/4}} \;\;\Longrightarrow\;\; y^4=\dfrac{x^4}{1+x^4}\;.$$ $$-(1)$$

Re-arranging $$-(1)$$ to express $$x^4$$ in terms of $$y^4$$:

$$y^4(1+x^4)=x^4 \;\;\Longrightarrow\;\; y^4=x^4(1-y^4) \;\;\Longrightarrow\;\; x^4=\dfrac{y^4}{1-y^4}\;.$$ $$-(2)$$

Equation $$-(2)$$ shows that applying $$f$$ replaces the quantity $$t=x^4$$ by the new value $$\dfrac{t}{1+t}\,. $$

Step 2 : Iterating the transformation.
Define $$T(t)=\dfrac{t}{1+t}\;.$$
If $$t_0=x^4$$, then after one application of $$f$$ we have $$t_1=T(t_0)$$, after two applications $$t_2=T^2(t_0)$$, and so on.

Compute the first four iterates:

$$T(t)=\dfrac{t}{1+t}\,,$$

$$T^2(t)=T\!\bigl(T(t)\bigr)=\dfrac{t}{1+2t}\,,$$

$$T^3(t)=T\!\bigl(T^2(t)\bigr)=\dfrac{t}{1+3t}\,,$$

$$T^4(t)=T\!\bigl(T^3(t)\bigr)=\dfrac{t}{1+4t}\,. $$

Step 3 : Simplifying $$g(x)=f^{(4)}(x).$$
Setting $$t_0=x^4$$, after four applications we get

$$g(x)^4 = T^4(x^4)=\dfrac{x^4}{1+4x^4}\;.$$

Hence

$$g(x)=\dfrac{x}{(1+4x^4)^{1/4}}\,. $$ $$-(3)$$

Step 4 : Setting up the required integral.
The integrand becomes

$$x^2\,g(x)=x^2\cdot\dfrac{x}{(1+4x^4)^{1/4}}=\dfrac{x^3}{(1+4x^4)^{1/4}}\;.$$

Therefore

$$I=\displaystyle\int_{0}^{\sqrt{2\sqrt{5}}}\dfrac{x^3}{(1+4x^4)^{1/4}}\,dx\;.$$ $$-(4)$$

Step 5 : Substitution to evaluate $$I$$.
Put $$u=1+4x^4\;,$$ so that $$\dfrac{du}{dx}=16x^3\;\Longrightarrow\;x^3\,dx=\dfrac{du}{16}\,.$$

When $$x=0$$, $$u=1$$;  when $$x=\sqrt{2\sqrt{5}}$$,

$$x^4=(\sqrt{2\sqrt{5}})^4=(2\sqrt{5})^2=4\cdot5=20 \;\Longrightarrow\; u=1+4\cdot20=81\;.$$

Substituting in $$-(4)$$:

$$I=\dfrac{1}{16}\displaystyle\int_{1}^{81}u^{-1/4}\,du\;.$$

Use the power-rule integral $$\int u^{n}\,du=\dfrac{u^{n+1}}{n+1}+C\;.$$
With $$n=-\dfrac14$$ we get

$$\int u^{-1/4}\,du=\dfrac{u^{3/4}}{3/4}=\dfrac{4}{3}u^{3/4}\;.$$

Hence

$$I=\dfrac{1}{16}\cdot\dfrac{4}{3}\Bigl[u^{3/4}\Bigr]_{1}^{81} =\dfrac{1}{12}\Bigl(81^{3/4}-1^{3/4}\Bigr)\;.$$

Because $$81=3^4,\;81^{1/4}=3,\;81^{3/4}=3^3=27$$, we have

$$I=\dfrac{1}{12}(27-1)=\dfrac{26}{12}=\dfrac{13}{6}\;.$$

Step 6 : Final multiplication.
The question asks for $$18I$$:

$$18I=18\cdot\dfrac{13}{6}=3\cdot13=39\;.$$

Answer : 39   (Option D)

We need to evaluate $$\int_{1/4}^{3/4} \cos\left(2\cot^{-1}\sqrt{\frac{1-x}{1+x}}\right)dx$$.

First, we simplify the inner expression by setting $$x = \cos\theta$$, where $$\theta \in [0, \pi]$$. Then

$$\frac{1-x}{1+x} = \frac{1-\cos\theta}{1+\cos\theta}$$

Using the half-angle identities $$1-\cos\theta = 2\sin^2(\theta/2)$$ and $$1+\cos\theta = 2\cos^2(\theta/2)$$, we obtain

$$\frac{1-\cos\theta}{1+\cos\theta} = \frac{2\sin^2(\theta/2)}{2\cos^2(\theta/2)} = \tan^2(\theta/2)$$

Hence, $$\sqrt{\frac{1-x}{1+x}} = \tan(\theta/2) \quad \text{(positive for } \theta \in (0,\pi)\text{)}$$, and thus

$$\cot^{-1}(\tan(\theta/2)) = \frac{\pi}{2} - \tan^{-1}(\tan(\theta/2)) = \frac{\pi}{2} - \frac{\theta}{2}$$

Since $$\theta/2 \in (0, \pi/2)$$ for the given range, it follows that $$\tan^{-1}(\tan(\theta/2)) = \theta/2$$.

It follows that

$$\cos\left(2\cot^{-1}\sqrt{\frac{1-x}{1+x}}\right) = \cos\left(2 \cdot \left(\frac{\pi}{2} - \frac{\theta}{2}\right)\right) = \cos(\pi - \theta) = -\cos\theta = -x$$

Substituting into the integral gives

$$\int_{1/4}^{3/4} (-x) \, dx = -\frac{x^2}{2}\Bigg|_{1/4}^{3/4} = -\frac{1}{2}\left(\frac{9}{16} - \frac{1}{16}\right) = -\frac{1}{2} \times \frac{8}{16} = -\frac{1}{2} \times \frac{1}{2} = -\frac{1}{4}$$

The correct answer is Option 3: $$-\frac{1}{4}$$.

We need to evaluate $$\int_{-\pi/2}^{\pi/2}\left(\frac{x^2\cos x}{1+\pi^x} + \frac{1+\sin^2 x}{1+e^{(\sin x)^{2023}}}\right)dx$$.

Evaluate $$I_1 = \int_{-\pi/2}^{\pi/2}\frac{x^2\cos x}{1+\pi^x}\,dx$$.

We use the property: for even function $$f(x)$$ and constant $$b > 0$$,

$$\int_{-a}^{a}\frac{f(x)}{1+b^x}\,dx = \int_0^a f(x)\,dx$$

Here $$f(x) = x^2\cos x$$ is even (product of two even functions) and $$b = \pi$$.

$$I_1 = \int_0^{\pi/2} x^2\cos x\,dx$$

Using integration by parts twice:

$$\int x^2\cos x\,dx = x^2\sin x - 2\int x\sin x\,dx = x^2\sin x - 2\left[-x\cos x + \sin x\right]$$

$$= x^2\sin x + 2x\cos x - 2\sin x$$

Evaluating from $$0$$ to $$\frac{\pi}{2}$$:

$$I_1 = \left[\frac{\pi^2}{4} \cdot 1 + 2 \cdot \frac{\pi}{2} \cdot 0 - 2 \cdot 1\right] - [0] = \frac{\pi^2}{4} - 2$$

Evaluate $$I_2 = \int_{-\pi/2}^{\pi/2}\frac{1+\sin^2 x}{1+e^{(\sin x)^{2023}}}\,dx$$.

Let $$g(x) = (\sin x)^{2023}$$. Since 2023 is odd, $$g(-x) = (-\sin x)^{2023} = -g(x)$$, so $$g$$ is an odd function.

The function $$h(x) = 1 + \sin^2 x$$ is even.

Using the same property (with $$e^{g(x)}$$ playing the role of $$b^x$$ where $$g$$ is odd):

$$I_2 = \int_0^{\pi/2}(1+\sin^2 x)\,dx$$

$$= \int_0^{\pi/2} 1\,dx + \int_0^{\pi/2}\sin^2 x\,dx = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$$

(Using $$\int_0^{\pi/2}\sin^2 x\,dx = \frac{\pi}{4}$$.)

Combine and compare.

$$I_1 + I_2 = \frac{\pi^2}{4} - 2 + \frac{3\pi}{4} = \frac{\pi}{4}(\pi + 3) - 2$$

Comparing with $$\frac{\pi}{4}(\pi + a) - 2$$, we get:

$$a = 3$$

The correct answer is Option (1): $$\boxed{3}$$.

We need to express $$\int_0^1 (1 - x^{10})^{20} dx$$ in terms of the Beta function $$\beta(m,n) = \int_0^1 x^{m-1}(1-x)^{n-1}dx$$.

Substitution: Let $$x^{10} = t$$, so $$dx = \frac{dt}{10t^{9/10}}$$.

$$ \int_0^1 (1-t)^{20} \cdot \frac{dt}{10t^{9/10}} = \frac{1}{10}\int_0^1 t^{1/10 - 1}(1-t)^{21-1} dt = \frac{1}{10}\beta\left(\frac{1}{10}, 21\right) $$

So $$a = \frac{1}{10}$$, $$b = \frac{1}{10}$$, $$c = 21$$.

$$ 100(a + b + c) = 100\left(\frac{1}{10} + \frac{1}{10} + 21\right) = 100 \times \frac{212}{10} = 2120 $$

The correct answer is Option 2: 2120.

Differentiate both sides of
$$\int_0^x\sqrt{1-(y'(t))^2}dt=\int_0^xy(t)dt$$

$$\Rightarrow\sqrt{1-(y')^2}=y$$

Square:
$$1-(y')^2=y^2;\Rightarrow;(y')^2=1-y^2$$

$$Since(y\ge0)and(y(0)=0),takepositiveroot:$$
$$y'=\sqrt{1-y^2}$$

Separate:
$$\frac{dy}{\sqrt{1-y^2}}=dx$$
$$\Rightarrow\sin^{-1}(y)=x$$

$$\Rightarrow y=\sin x$$

$$y''=-\sin x$$

$$ y''+y+1=(-\sin x)+(\sin x)+1=1$$

At (x=2): 1

We need to evaluate the definite integral $$\int_{\alpha}^{\log_e 4} \frac{dx}{\sqrt{e^x - 1}} = \frac{\pi}{6}$$.

We introduce the substitution $$e^x - 1 = t^2$$, so that $$e^x = 1 + t^2$$ and $$e^x\,dx = 2t\,dt$$, giving $$dx = \frac{2t\,dt}{1 + t^2}$$.

Substituting into the integral yields $$\int \frac{1}{t}\,\frac{2t\,dt}{1 + t^2} = \int \frac{2\,dt}{1 + t^2} = 2\arctan(t) + C = 2\arctan(\sqrt{e^x - 1}) + C$$.

Next, evaluating the definite integral gives $$\Bigl[2\arctan(\sqrt{e^x - 1})\Bigr]_{\alpha}^{\log_e 4} = \frac{\pi}{6}$$.

At $$x = \log_e 4$$, we have $$\sqrt{e^{\ln 4} - 1} = \sqrt{3}$$, so $$2\arctan(\sqrt{3}) = 2 \cdot \frac{\pi}{3} = \frac{2\pi}{3}$$.

Therefore, $$\frac{2\pi}{3} - 2\arctan(\sqrt{e^{\alpha} - 1}) = \frac{\pi}{6}$$.

Rearranging gives $$2\arctan(\sqrt{e^{\alpha} - 1}) = \frac{2\pi}{3} - \frac{\pi}{6} = \frac{\pi}{2}$$, so $$\arctan(\sqrt{e^{\alpha} - 1}) = \frac{\pi}{4}$$.

This implies $$\sqrt{e^{\alpha} - 1} = 1 \implies e^{\alpha} - 1 = 1 \implies e^{\alpha} = 2$$.

Hence, the two roots of the equation are $$e^{\alpha} = 2$$ and $$e^{-\alpha} = \frac{1}{2}$$.

The sum of the roots is $$2 + \frac{1}{2} = \frac{5}{2}$$ and their product is $$2 \times \frac{1}{2} = 1$$.

Forming the quadratic equation with these roots gives $$x^2 - \frac{5}{2}x + 1 = 0$$, which multiplies through by 2 to yield $$2x^2 - 5x + 2 = 0$$.

Therefore, the required equation is $$2x^2 - 5x + 2 = 0$$.

The inequality $$y^{2} \le 4x$$ represents the right-opening parabola $$y = \pm 2\sqrt{x}$$.
For every admissible $$x$$ the vertical length of the full strip would be $$2\sqrt{x} - (-2\sqrt{x}) = 4\sqrt{x}$$, but the second condition will reduce this length by half. We analyse that condition next.

Let $$g(x) = \dfrac{x(x-1)(x-2)}{(x-3)(x-4)}$$. The given sign restriction can be rewritten as

$$y \, g(x) \gt 0 \qquad -(1)$$

Equation $$(1)$$ says that $$y$$ and $$g(x)$$ must have the same sign. Hence, for a fixed $$x$$ only one half of the total $$y$$-interval survives: if $$g(x) \gt 0$$ we keep $$0 \lt y \le 2\sqrt{x}$$; if $$g(x) \lt 0$$ we keep $$-2\sqrt{x} \le y \lt 0$$.
In either case the surviving vertical length is $$2\sqrt{x}$$.

Because $$y^{2} \le 4x$$ requires $$x \ge 0$$, and because $$x \lt 4$$ is already imposed, we only have to study the sign of $$g(x)$$ on the intervals obtained by the zeros and poles of $$g(x)$$:

$$0,\,1,\,2,\,3,\,4$$ split the axis (with $$x=3$$ excluded). A sign table gives

In all four intervals the admissible vertical length is $$2\sqrt{x}$$, so the required area $$A$$ equals

$$ A = 2 \int_{0}^{1} \!\sqrt{x}\,dx + 2 \int_{1}^{2} \!\sqrt{x}\,dx + 2 \int_{2}^{3} \!\sqrt{x}\,dx + 2 \int_{3}^{4} \!\sqrt{x}\,dx \qquad -(2) $$

Using the antiderivative $$\displaystyle\int \sqrt{x}\,dx = \frac{2}{3}x^{3/2}$$ in $$(2)$$:

$$ \begin{aligned} A &= \frac{4}{3}\Bigl[x^{3/2}\Bigr]_{0}^{1} + \frac{4}{3}\Bigl[x^{3/2}\Bigr]_{1}^{2} + \frac{4}{3}\Bigl[x^{3/2}\Bigr]_{2}^{3} + \frac{4}{3}\Bigl[x^{3/2}\Bigr]_{3}^{4} \\[2mm] &= \frac{4}{3}\Bigl(1 - 0\Bigr) + \frac{4}{3}\Bigl(2^{3/2} - 1\Bigr) + \frac{4}{3}\Bigl(3^{3/2} - 2^{3/2}\Bigr) + \frac{4}{3}\Bigl(4^{3/2} - 3^{3/2}\Bigr) \\[2mm] &= \frac{4}{3}\Bigl[1 + (2\sqrt{2}-1) + (3\sqrt{3}-2\sqrt{2}) + (8-3\sqrt{3})\Bigr] \\[2mm] &= \frac{4}{3}\,(8) \\[2mm] &= \frac{32}{3}. \end{aligned} $$

Therefore, the area of the given region is $$\dfrac{32}{3}$$, which is Option D.

 Find Intersection. Substitute $$y^2=4x$$ into circle: $$x^2 + 4x - 5 = 0 \Rightarrow (x+5)(x-1)=0$$. Since $$x \ge 0$$ for the parabola, $$x=1$$.

Setup Area. The area is symmetric about the x-axis.

$$\text{Area} = 2 \left[ \int_0^1 \sqrt{4x} dx + \int_1^{\sqrt{5}} \sqrt{5-x^2} dx \right]$$

 Integrate.

o Part 1: $$2 \int_0^1 2x^{1/2} dx = 4 [\frac{2}{3}x^{3/2}]_0^1 = \frac{8}{3}$$.

o Part 2: $$2 \int_1^{\sqrt{5}} \sqrt{5-x^2} dx$$. Using $$\int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}$$.

o Result: $$[x\sqrt{5-x^2} + 5\sin^{-1}\frac{x}{\sqrt{5}}]_1^{\sqrt{5}} = (0 + 5\frac{\pi}{2}) - (2 + 5\sin^{-1}\frac{1}{\sqrt{5}})$$.

 Simplify. Total Area = $$\frac{8}{3} - 2 + 5(\frac{\pi}{2} - \sin^{-1}\frac{1}{\sqrt{5}})$$. Using $$\cos^{-1}\theta = \frac{\pi}{2} - \sin^{-1}\theta$$ and $$\cos^{-1}(1/\sqrt{5}) = \sin^{-1}(2/\sqrt{5})$$:

$$\text{Area} = \frac{2}{3} + 5\sin^{-1}\left(\frac{2}{\sqrt{5}}\right)$$

$$2x^3 - 3x^2 - x + 1$$

Now consider the substitution $$x \to 1 - x$$

Let $$g(x) = 2x^3 - 3x^2 - x + 1$$.

$$g(1-x) = 2(1-x)^3 - 3(1-x)^2 - (1-x) + 1$$

$$= 2(1 - 3x + 3x^2 - x^3) - 3(1 - 2x + x^2) - 1 + x + 1$$

$$= 2 - 6x + 6x^2 - 2x^3 - 3 + 6x - 3x^2 - 1 + x + 1$$

$$= -2x^3 + 3x^2 + x - 1 = -(2x^3 - 3x^2 - x + 1) = -g(x)$$

So $$g(1 - x) = -g(x)$$.

Let $$I = \int_0^1 (g(x))^{1/3} \, dx$$.

Using the substitution $$x \to 1 - x$$:

$$I = \int_0^1 (g(1-x))^{1/3} \, dx = \int_0^1 (-g(x))^{1/3} \, dx = -\int_0^1 (g(x))^{1/3} \, dx = -I$$

Therefore $$2I = 0$$, which gives $$I = 0$$.

The answer is $$\boxed{0}$$, which corresponds to Option (1).

We consider $$I_n = \int_0^1 (1-x^k)^n \, dx$$ for natural numbers $$n$$ and $$k$$, with the condition $$147 I_{20} = 148 I_{21}$$.

Since substituting $$t = x^k$$ so that $$x = t^{1/k}$$ and $$dx = \frac{1}{k} t^{1/k - 1} \, dt$$, it follows that $$I_n = \int_0^1 (1-t)^n \cdot \frac{1}{k} t^{1/k - 1} \, dt = \frac{1}{k} B\left(\frac{1}{k}, n+1\right) = \frac{1}{k} \cdot \frac{\Gamma(1/k) \cdot \Gamma(n+1)}{\Gamma(n + 1 + 1/k)}.$$

This gives the ratio $$\frac{I_n}{I_{n+1}} = \frac{\Gamma(n+1) \cdot \Gamma(n+2+1/k)}{\Gamma(n+2) \cdot \Gamma(n+1+1/k)} = \frac{n+1+1/k}{n+1} = 1 + \frac{1}{k(n+1)}.$$

From the condition $$147 I_{20} = 148 I_{21}$$ we obtain $$\frac{I_{20}}{I_{21}} = \frac{148}{147}.$$ Substituting $$n = 20$$ in the ratio formula yields $$\frac{I_{20}}{I_{21}} = 1 + \frac{1}{21k} = \frac{21k + 1}{21k}.$$

Equating these expressions gives $$\frac{21k + 1}{21k} = \frac{148}{147},$$ $$147(21k + 1) = 148 \times 21k,$$ $$147 \times 21k + 147 = 148 \times 21k,$$ $$147 = 21k,$$ and hence $$k = 7.$$

The answer is Option 4: 7.

We evaluate $$\lim_{n \to \infty} \sum_{k=1}^{n} \frac{n^3}{(n^2 + k^2)(n^2 + 3k^2)}$$.

Let $$t = \frac{k}{n}$$ so that the sum becomes a Riemann sum given by $$\sum_{k=1}^n \frac{n^3}{n^2(1+t^2)\,n^2(1+3t^2)}\cdot 1 = \sum_{k=1}^n \frac{1}{n}\,\frac{1}{(1+t^2)(1+3t^2)}\,,\quad t=\frac{k}{n}\,,$$ and as $$n\to\infty$$ this tends to $$I = \int_0^1 \frac{dt}{(1+t^2)(1+3t^2)}\,. $$

Using partial fractions one writes $$\frac{1}{(1+t^2)(1+3t^2)} = \frac{A}{1+t^2} + \frac{B}{1+3t^2}\,, $$ and expanding gives $$1 = A(1+3t^2)+B(1+t^2)\,. $$ Equating constant terms and coefficients of $$t^2$$ yields $$A+B=1\,,\quad 3A+B=0\,, $$ so that $$A=-\tfrac12\,,\ B=\tfrac32\,. $$

Hence $$I = \int_0^1\Bigl[\frac{-\tfrac12}{1+t^2}+\frac{\tfrac32}{1+3t^2}\Bigr]dt = -\frac12\int_0^1\frac{dt}{1+t^2} + \frac32\int_0^1\frac{dt}{1+3t^2}\,, $$ and integrating gives $$-\frac12\bigl[\tan^{-1}(t)\bigr]_0^1 + \frac32\cdot\frac{1}{\sqrt3}\bigl[\tan^{-1}(\sqrt3\,t)\bigr]_0^1 = -\frac12\cdot\frac\pi4 + \frac{\sqrt3}{2}\cdot\frac\pi3 = -\frac{\pi}{8} + \frac{\sqrt3\pi}{6} = \frac{\pi(4\sqrt3-3)}{24}\,. $$

We compare this with option (2), namely $$\frac{13\pi}{8(4\sqrt3+3)}\,. $$ Rationalizing the denominator shows $$\frac{13\pi}{8(4\sqrt3+3)}\cdot\frac{4\sqrt3-3}{4\sqrt3-3} = \frac{13\pi(4\sqrt3-3)}{8(48-9)} = \frac{\pi(4\sqrt3-3)}{24}\,,$$ so the two expressions agree.

The answer is Option (2): $$\boxed{\frac{13\pi}{8(4\sqrt3+3)}}$$.

Use the substitution $$t = \tan(x/2)$$. Then $$dx = \frac{2 dt}{1+t^2}$$ and $$\cos x = \frac{1-t^2}{1+t^2}$$.

Change limits: When $$x=0, t=0$$. When $$x=\pi, t \to \infty$$.

Substitute into the integral:

$$I = \int_{0}^{\infty} \frac{\frac{2 dt}{1+t^2}}{1 - 2a\left(\frac{1-t^2}{1+t^2}\right) + a^2} = \int_{0}^{\infty} \frac{2 dt}{(1+t^2) - 2a(1-t^2) + a^2(1+t^2)}$$

Denominator $$= 1 + t^2 - 2a + 2at^2 + a^2 + a^2t^2 = (1-a)^2 + t^2(1+a)^2$$.

$$I = 2 \int_{0}^{\infty} \frac{dt}{(1-a)^2 + t^2(1+a)^2} = \frac{2}{(1+a)^2} \int_{0}^{\infty} \frac{dt}{t^2 + \left(\frac{1-a}{1+a}\right)^2}$$

Using the formula $$\int \frac{dx}{x^2+k^2} = \frac{1}{k} \tan^{-1}(\frac{x}{k})$$:

$$I = \frac{2}{(1+a)^2} \cdot \frac{1+a}{1-a} \left[ \tan^{-1}\left(t \cdot \frac{1+a}{1-a}\right) \right]_{0}^{\infty} = \frac{2}{1-a^2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{1-a^2}$$

$$\int_0^1 \frac{1}{\sqrt{3+x} + \sqrt{1+x}} dx$$

Rationalize by multiplying by $$\frac{\sqrt{3+x} - \sqrt{1+x}}{\sqrt{3+x} - \sqrt{1+x}}$$:

$$= \int_0^1 \frac{\sqrt{3+x} - \sqrt{1+x}}{(3+x) - (1+x)} dx = \int_0^1 \frac{\sqrt{3+x} - \sqrt{1+x}}{2} dx$$

$$= \frac{1}{2}\int_0^1 (\sqrt{3+x} - \sqrt{1+x}) dx$$

$$= \frac{1}{3}\left[8 - 2\sqrt{2} - 3\sqrt{3} + 1\right]$$

$$= \frac{1}{3}(9 - 2\sqrt{2} - 3\sqrt{3})$$

So $$a = 3$$, $$b = -\frac{2}{3}$$, $$c = -1$$.

$$2a + 3b - 4c = 6 - 2 + 4 = 8$$

We need to evaluate $$\int_0^{\pi/3} \cos^4 x \, dx$$ and express it as $$a\pi + b\sqrt{3}$$, then find $$9a + 8b$$.

Since $$\cos^4 x = (\cos^2 x)^2$$, applying the power-reduction formula gives

$$\cos^4 x = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{1 + 2\cos 2x + \cos^2 2x}{4}.$$

Now using $$\cos^2 2x = \frac{1 + \cos 4x}{2}$$ and substituting yields

$$\cos^4 x = \frac{1 + 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} = \frac{3 + 4\cos 2x + \cos 4x}{8}.$$

Next, integrate term by term:

$$\int_0^{\pi/3} \cos^4 x \, dx = \frac{1}{8}\int_0^{\pi/3}(3 + 4\cos 2x + \cos 4x)\,dx$$

$$= \frac{1}{8}\Bigl[3x + 2\sin 2x + \frac{\sin 4x}{4}\Bigr]_0^{\pi/3}.$$

At $$x = \pi/3$$, we have

$$3 \cdot \frac{\pi}{3} = \pi,\quad 2\sin\frac{2\pi}{3} = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3},\quad \frac{1}{4}\sin\frac{4\pi}{3} = \frac{1}{4}\bigl(-\frac{\sqrt{3}}{2}\bigr) = -\frac{\sqrt{3}}{8}.$$

At $$x = 0$$ all terms vanish.

Substituting these values gives

$$\frac{1}{8}\Bigl(\pi + \sqrt{3} - \frac{\sqrt{3}}{8}\Bigr) = \frac{1}{8}\Bigl(\pi + \frac{7\sqrt{3}}{8}\Bigr) = \frac{\pi}{8} + \frac{7\sqrt{3}}{64}.$$

Therefore $$a = \frac{1}{8}$$ and $$b = \frac{7}{64}$$.

Finally, computing $$9a + 8b$$ gives

$$9a + 8b = 9 \cdot \frac{1}{8} + 8 \cdot \frac{7}{64} = \frac{9}{8} + \frac{56}{64} = \frac{9}{8} + \frac{7}{8} = 2.$$

The correct answer is Option 1: 2.

We need to find the value of $$7a - 3$$ given that the area of the region $$\{(x,y): \frac{a}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2, 0 < a < 1\}$$ equals $$\ln 2 - \frac{1}{7}$$.

To determine this value, we first set up the area integral by noting that the area between the curves $$y_1 = \frac{a}{x^2}$$ (lower curve) and $$y_2 = \frac{1}{x}$$ (upper curve) over the interval $$x \in [1,2]$$ is given by

$$ A = \int_1^2 \left(\frac{1}{x} - \frac{a}{x^2}\right) dx $$. We integrate the difference $$y_2 - y_1$$ because $$y_2 \geq y_1$$ in this region (for $$0 < a < 1$$).

We now evaluate each integral separately. First,

$$ \int_1^2 \frac{1}{x}\,dx = [\ln x]_1^2 = \ln 2 - \ln 1 = \ln 2 $$.

For the second integral, recall that $$\int x^{-2}\,dx = \frac{x^{-1}}{-1} = -\frac{1}{x}$$:

$$ \int_1^2 \frac{a}{x^2}\,dx = a\left[-\frac{1}{x}\right]_1^2 = a\left(-\frac{1}{2} - \left(-\frac{1}{1}\right)\right) = a\left(-\frac{1}{2} + 1\right) = \frac{a}{2} $$.

Combining these results, the total area satisfies

$$ A = \ln 2 - \frac{a}{2} = \ln 2 - \frac{1}{7} $$.

Solving for $$a$$ by subtracting $$\ln 2$$ from both sides gives

$$ -\frac{a}{2} = -\frac{1}{7} $$. Hence, $$ \frac{a}{2} = \frac{1}{7} $$ and therefore $$ a = \frac{2}{7} $$.

Finally,

$$ 7a - 3 = 7 \times \frac{2}{7} - 3 = 2 - 3 = -1 $$.

The correct answer is Option C: $$-1$$.

We need to evaluate $$I = \int_{-1}^{1} \frac{\cos \alpha x}{1 + 3^x} \, dx$$.

Using the property: For a function of the form $$\int_{-a}^{a} \frac{f(x)}{1 + b^x} \, dx$$ where $$f(x)$$ is even, we can use the substitution $$x \to -x$$.

Let $$I = \int_{-1}^{1} \frac{\cos \alpha x}{1 + 3^x} \, dx$$ $$-(1)$$

Substituting $$x \to -x$$: $$I = \int_{-1}^{1} \frac{\cos(-\alpha x)}{1 + 3^{-x}} \, dx = \int_{-1}^{1} \frac{\cos \alpha x}{1 + 3^{-x}} \, dx$$ $$-(2)$$

(since $$\cos$$ is an even function).

Adding $$(1)$$ and $$(2)$$:

$$2I = \int_{-1}^{1} \cos \alpha x \left(\frac{1}{1 + 3^x} + \frac{1}{1 + 3^{-x}}\right) dx$$

Now simplify the expression in parentheses:

$$\frac{1}{1 + 3^x} + \frac{1}{1 + 3^{-x}} = \frac{1}{1 + 3^x} + \frac{3^x}{3^x + 1} = \frac{1 + 3^x}{1 + 3^x} = 1$$

Therefore: $$2I = \int_{-1}^{1} \cos \alpha x \, dx$$

$$2I = \left[\frac{\sin \alpha x}{\alpha}\right]_{-1}^{1} = \frac{\sin \alpha - \sin(-\alpha)}{\alpha} = \frac{2 \sin \alpha}{\alpha}$$

$$I = \frac{\sin \alpha}{\alpha}$$

Given $$I = \frac{2}{\pi}$$:

$$\frac{\sin \alpha}{\alpha} = \frac{2}{\pi}$$

Therefore $$\alpha = \frac{\pi}{2}$$, which matches Option D.

Let $$I = \displaystyle\int_{0}^{\pi/4} \frac{\cos^{2}x \,\sin^{2}x}{\bigl(\cos^{3}x+\sin^{3}x\bigr)^{2}}\,dx$$.

Case 1: Convert the integral to a single-variable rational form using $$t=\tan x$$.
For $$t=\tan x$$ we have $$\sin x = \frac{t}{\sqrt{1+t^{2}}},\qquad \cos x = \frac{1}{\sqrt{1+t^{2}}},\qquad dx = \frac{dt}{1+t^{2}}.$$ When $$x$$ goes from $$0$$ to $$\pi/4$$, $$t$$ goes from $$0$$ to $$1$$.

Compute the numerator:
$$\cos^{2}x\,\sin^{2}x = \frac{1}{1+t^{2}}\cdot\frac{t^{2}}{1+t^{2}} = \frac{t^{2}}{(1+t^{2})^{2}}.$$ Compute the denominator:
$$\cos^{3}x + \sin^{3}x = \frac{1}{(1+t^{2})^{3/2}} + \frac{t^{3}}{(1+t^{2})^{3/2}} = \frac{1+t^{3}}{(1+t^{2})^{3/2}}.$$ Therefore $$\bigl(\cos^{3}x+\sin^{3}x\bigr)^{2} = \frac{(1+t^{3})^{2}}{(1+t^{2})^{3}}.$$ Putting these into the integrand gives $$\frac{\cos^{2}x\sin^{2}x}{(\cos^{3}x+\sin^{3}x)^{2}} = \frac{t^{2}/(1+t^{2})^{2}}{(1+t^{3})^{2}/(1+t^{2})^{3}} = \frac{t^{2}(1+t^{2})}{(1+t^{3})^{2}}.$$ Multiplying by $$dx=\dfrac{dt}{1+t^{2}}$$, $$I = \int_{0}^{1} \frac{t^{2}(1+t^{2})}{(1+t^{3})^{2}}\cdot\frac{dt}{1+t^{2}} = \int_{0}^{1} \frac{t^{2}}{(1+t^{3})^{2}}\,dt.$$

Case 2: Evaluate the rational integral by substituting $$u=t^{3}$$.
Put $$u = t^{3}\; \Longrightarrow\; du = 3t^{2}\,dt \; \Longrightarrow\; t^{2}\,dt = \frac{du}{3}.$$ As $$t$$ varies from $$0$$ to $$1$$, $$u$$ also varies from $$0$$ to $$1$$. Hence $$I = \int_{0}^{1} \frac{t^{2}}{(1+t^{3})^{2}}\,dt = \frac13\int_{0}^{1} \frac{du}{(1+u)^{2}}.$$

Case 3: Integrate the simple power function.
Recall $$\displaystyle\int (1+u)^{-2}\,du = -\frac{1}{1+u}+C.$$ Therefore $$\frac13\int_{0}^{1} \frac{du}{(1+u)^{2}} = \frac13\Bigl[-\frac{1}{1+u}\Bigr]_{0}^{1} = \frac13\Bigl(-\frac12 + 1\Bigr) = \frac13\cdot\frac12 = \frac16.$$

Hence $$I = \dfrac16$$.

The correct option is Option A (1/6).

Since $$(|t| - t^2)\,e^{-t^2}$$ is an even function of $$t$$ (replacing $$t$$ by $$-t$$ leaves it unchanged), the integral of $$f$$ over the symmetric interval $$[-x,x]$$ becomes:

$$f(x) = 2\int_0^x (t - t^2)\,e^{-t^2}\,dt = 2\int_0^x t\,e^{-t^2}\,dt - 2\int_0^x t^2\,e^{-t^2}\,dt.$$

The first integral evaluates directly via $$u = t^2$$:

$$2\int_0^x t\,e^{-t^2}\,dt = \bigl[-e^{-t^2}\bigr]_0^x = 1 - e^{-x^2}.$$

$$f(x) = 1 - e^{-x^2} - 2\int_0^x t^2\,e^{-t^2}\,dt. \quad\cdots(1)$$

For $$g(x) = \int_0^{x^2}\!\sqrt{t}\;e^{-t}\,dt$$, substitute $$t = u^2$$, so $$dt = 2u\,du$$ and $$\sqrt{t} = u$$. When $$t = 0$$, $$u = 0$$; when $$t = x^2$$, $$u = x$$. Then:

$$g(x) = \int_0^x u \cdot e^{-u^2}\cdot 2u\,du = 2\int_0^x u^2\,e^{-u^2}\,du. \quad\cdots(2)$$

Adding (1) and (2), the integral terms cancel exactly:

$$f(x) + g(x) = 1 - e^{-x^2} - 2\int_0^x t^2 e^{-t^2}\,dt + 2\int_0^x u^2 e^{-u^2}\,du = 1 - e^{-x^2}.$$

Substituting $$x = \sqrt{\log_e 9}$$:

$$e^{-x^2} = e^{-\log_e 9} = \frac{1}{9}.$$

$$f\!\left(\sqrt{\log_e 9}\right) + g\!\left(\sqrt{\log_e 9}\right) = 1 - \frac{1}{9} = \frac{8}{9}.$$

Finally, $$9 \times \frac{8}{9} = 8$$.

Break down $$h(x)$$:

1. For $$x \in [-2, 0]$$: $$|x| \in [0, 2]$$. So $$f(|x|) = |x|-2 = -x-2$$.

$$|f(x)| = |-2| = 2$$.

$$h(x) = (-x-2) + 2 = -x$$.

2. For $$x \in (0, 2]$$: $$|x| = x$$. So $$f(|x|) = x-2$$.

$$|f(x)| = |x-2| = -(x-2) = 2-x$$ (since $$x \le 2$$).

$$h(x) = (x-2) + (2-x) = 0$$.

• Integral:

$$\int_{-2}^{2} h(x) \, dx = \int_{-2}^{0} (-x) \, dx + \int_{0}^{2} 0 \, dx = \left[ -\frac{x^2}{2} \right]_{-2}^{0} = 0 - (-2) = \mathbf{2}$$

$$f'(1) = \tan(\pi/6) = 1/\sqrt{3}$$, $$f'(3) = \tan(\pi/4) = 1$$.

The integral: $$27\int_1^3 \{(f'(t))^2 + 1\}f''(t)dt$$.

Let $$u = f'(t)$$, $$du = f''(t)dt$$.

$$= 27\int_{f'(1)}^{f'(3)} (u^2+1)du = 27\left[\frac{u^3}{3}+u\right]_{1/\sqrt{3}}^{1}$$

At $$u = 1$$: $$\frac{1}{3} + 1 = \frac{4}{3}$$.

At $$u = 1/\sqrt{3}$$: $$\frac{1}{3 \cdot 3\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{1}{9\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{1+9}{9\sqrt{3}} = \frac{10}{9\sqrt{3}}$$.

$$= 27\left[\frac{4}{3} - \frac{10}{9\sqrt{3}}\right] = 27 \times \frac{4}{3} - 27 \times \frac{10}{9\sqrt{3}} = 36 - \frac{30}{\sqrt{3}} = 36 - 10\sqrt{3}$$

So $$\alpha = 36, \beta = -10$$. $$\alpha + \beta = 26$$.

The answer is Option (2): $$\boxed{26}$$.

The two given curves are $$y = x\lvert x\rvert$$ and $$y = x - \lvert x\rvert$$. To integrate, first write each curve without the absolute-value signs.

Case 1: $$x \ge 0$$   • $$\lvert x\rvert = x$$, so $$y_1 = x\lvert x\rvert = x \cdot x = x^{2}$$   • $$y_2 = x - \lvert x\rvert = x - x = 0$$

Case 2: $$x \le 0$$   • $$\lvert x\rvert = -x$$, so $$y_1 = x\lvert x\rvert = x(-x) = -x^{2}$$   • $$y_2 = x - \lvert x\rvert = x - (-x) = 2x$$

Next find the points where the curves intersect.

Case 1 ($$x \ge 0$$): $$x^{2} = 0 \;\Longrightarrow\; x = 0$$

Case 2 ($$x \le 0$$): $$-x^{2} = 2x \;\Longrightarrow\; -x^{2} - 2x = 0 \;\Longrightarrow\; -x(x+2)=0 \;\Longrightarrow\; x = 0 \text{ or } x = -2$$

Thus the two curves meet at $$(-2,-4)$$ and $$(0,0)$$. Between these points the region is completely bounded; outside this interval the graphs do not enclose a finite area.

For $$x \in [-2,0]$$ the upper curve is $$y_1 = -x^{2}$$ and the lower curve is $$y_2 = 2x$$ (check at, say, $$x=-1$$: $$-1^{2}=-1$$ is above $$2(-1)=-2$$).

Required area $$A = \int_{-2}^{0} \bigl[\,(-x^{2}) - (2x)\bigr] \,dx = \int_{-2}^{0} \bigl(-x^{2} - 2x\bigr)\,dx$$

Integrate term by term: $$\int -x^{2}\,dx = -\frac{x^{3}}{3},\qquad \int -2x\,dx = -x^{2}$$

Hence $$A = \Bigl[-\frac{x^{3}}{3} - x^{2}\Bigr]_{-2}^{0}$$

Evaluate the limits: At $$x = 0$$: $$-\frac{0^{3}}{3} - 0^{2} = 0$$
At $$x = -2$$: $$-\dfrac{(-2)^{3}}{3} - (-2)^{2} = -\!\bigl(-\dfrac{8}{3}\bigr) - 4 = \dfrac{8}{3} - 4 = -\dfrac{4}{3}$$

Therefore $$A = 0 - \Bigl(-\dfrac{4}{3}\Bigr) = \dfrac{4}{3}$$

The enclosed area is $$\dfrac{4}{3}$$. This matches Option A.

For the parabola $$y^2 = 4(x-2)$$ we write $$x = \frac{y^2}{4} + 2$$, and for the line $$y = 2x - 8$$ we have $$x = \frac{y+8}{2}$$.

Equating these expressions gives $$\frac{y^2}{4} + 2 = \frac{y+8}{2}$$, which simplifies to $$y^2 + 8 = 2y + 16 \;\Rightarrow\; y^2 - 2y - 8 = 0 \;\Rightarrow\; (y-4)(y+2) = 0$$, so $$y = 4$$ or $$y = -2$$.

The area between the curves, with the line to the right of the parabola, is $$A = \int_{-2}^{4} \left[\frac{y+8}{2} - \frac{y^2}{4} - 2\right] dy = \int_{-2}^{4} \left[\frac{y+8}{2} - \frac{y^2+8}{4}\right] dy = \int_{-2}^{4} \frac{2(y+8) - (y^2+8)}{4} dy = \int_{-2}^{4} \frac{2y + 16 - y^2 - 8}{4} dy = \int_{-2}^{4} \frac{-y^2 + 2y + 8}{4} dy = \frac{1}{4}\left[-\frac{y^3}{3} + y^2 + 8y\right]_{-2}^{4}.$$

At $$y = 4$$: $$-\frac{64}{3} + 16 + 32 = -\frac{64}{3} + 48 = \frac{-64 + 144}{3} = \frac{80}{3}$$, and at $$y = -2$$: $$\frac{8}{3} + 4 - 16 = \frac{8}{3} - 12 = \frac{8 - 36}{3} = -\frac{28}{3}$$. Hence, $$A = \frac{1}{4}\left[\frac{80}{3} - \left(-\frac{28}{3}\right)\right] = \frac{1}{4} \cdot \frac{108}{3} = \frac{108}{12} = 9.$$ The answer is Option (2): $$\boxed{9}$$.

Find intersection: $$x^2 + 2x = 8 \implies x^2 + 2x - 8 = 0 \implies (x+4)(x-2) = 0$$.

In 1st quadrant, $$x = 2$$. At $$x=2, y=2$$.

2. The requested area is: (Area of circular sector) - (Area under parabola).

Area $$= \int_{0}^{2} (\sqrt{8-x^2} - \sqrt{2x}) dx$$.

Wait, the problem says "inside circle" and "outside parabola".

Total Area of circle in 1st quadrant is $$\frac{1}{4}\pi(2\sqrt{2})^2 = 2\pi$$.

Area $$= \int_{0}^{2} (\sqrt{8-x^2} - \sqrt{2x}) dx + \int_{2}^{2\sqrt{2}} \sqrt{8-x^2} dx$$.

Simplest way: Area = (Area of quadrant) - (Area under parabola from 0 to 2) - (Area under circle from 0 to 2).

Actually, it is simpler: Area $$= \text{Area of Sector} - \text{Area under Parabola}$$.

Looking at the options, the integral setup is:

Area $$= \int_{0}^{2} (\sqrt{8-y^2}/ \text{dy logic is messy}) \to$$ Let's use $$x$$:

$$\text{Area} = \int_{0}^{2} \sqrt{8-x^2} dx - \int_{0}^{2} \sqrt{2x} dx$$

$$= [\frac{x}{2}\sqrt{8-x^2} + \frac{8}{2}\sin^{-1}\frac{x}{2\sqrt{2}}]_{0}^{2} - [\frac{2\sqrt{2}}{3}x^{3/2}]_{0}^{2}$$

$$= (2 + 4 \cdot \frac{\pi}{4}) - \frac{2\sqrt{2} \cdot 2\sqrt{2}}{3} = 2 + \pi - \frac{8}{3} = \pi - \frac{2}{3}$$.

We need to evaluate $$I = \int_{-\pi}^{\pi} \frac{2y(1 + \sin y)}{1 + \cos^2 y} \, dy.$$

We start by splitting the integral into two parts: $$I = \int_{-\pi}^{\pi} \frac{2y}{1+\cos^2 y}\,dy \;+\; \int_{-\pi}^{\pi} \frac{2y\sin y}{1+\cos^2 y}\,dy = I_1 + I_2.$$

For the first integral, set $$f(y) = \frac{2y}{1+\cos^2 y}.$$ Then $$f(-y) = \frac{-2y}{1+\cos^2(-y)} = \frac{-2y}{1+\cos^2 y} = -f(y),$$ so $$f(y)$$ is an odd function. Because the limits are symmetric about zero, it follows that $$I_1 = 0.$$

Turning to the second integral, define $$g(y) = \frac{2y\sin y}{1+\cos^2 y}.$$ Checking symmetry gives $$g(-y) = \frac{2(-y)\sin(-y)}{1+\cos^2(-y)} = \frac{(-2y)(-\sin y)}{1+\cos^2 y} = g(y),$$ so $$g(y)$$ is even. Hence $$I_2 = 2\int_{0}^{\pi} \frac{2y\sin y}{1+\cos^2 y}\,dy = 4\int_{0}^{\pi} \frac{y\sin y}{1+\cos^2 y}\,dy.$$

Next we apply the property $$\int_{0}^{a} x\,f(\sin x)\,dx \;=\;\frac{a}{2}\int_{0}^{a} f(\sin x)\,dx$$ which holds because $$\sin(\pi - y)=\sin y$$. With $$a = \pi$$, this gives $$\int_{0}^{\pi} \frac{y\sin y}{1+\cos^2 y}\,dy = \frac{\pi}{2}\int_{0}^{\pi} \frac{\sin y}{1+\cos^2 y}\,dy.$$ Therefore $$I_2 = 4 \times \frac{\pi}{2}\int_{0}^{\pi} \frac{\sin y}{1+\cos^2 y}\,dy = 2\pi\int_{0}^{\pi} \frac{\sin y}{1+\cos^2 y}\,dy.$$

To evaluate the remaining integral, substitute $$u = \cos y$$, so $$du = -\sin y\,dy$$. When $$y=0$$, $$u=1$$, and when $$y=\pi$$, $$u=-1$$. Thus $$\int_{0}^{\pi} \frac{\sin y}{1+\cos^2 y}\,dy = \int_{1}^{-1} \frac{-\,du}{1+u^2} = \int_{-1}^{1} \frac{du}{1+u^2} = [\tan^{-1}u]_{-1}^{1} = \frac{\pi}{2}.$$

It follows that $$I_2 = 2\pi \times \frac{\pi}{2} = \pi^2,$$ and combining with $$I_1 = 0$$ yields $$I = \pi^2.$$

The correct answer is Option (4): $$\pi^2$$.

Let $$f(x) = \log_e (x + \sqrt{x^2 + 1})$$. This is an odd function ($$f(-x) = -f(x)$$).
Integration by parts: $$\int 1 \cdot f(x) \, dx = x f(x) - \int \frac{x}{\sqrt{x^2+1}} \, dx = x f(x) - \sqrt{x^2+1}$$.
Apply limits:
$$[x \log(x + \sqrt{x^2+1}) - \sqrt{x^2+1}]_{-1}^{2}$$
$$= (2 \log(2 + \sqrt{5}) - \sqrt{5}) - (-1 \log(-1 + \sqrt{2}) - \sqrt{2})$$
$$= \log(2+\sqrt{5})^2 - \sqrt{5} + \log(\sqrt{2}-1) + \sqrt{2}$$
$$= \log(9+4\sqrt{5}) - \sqrt{5} - \log(\sqrt{2}+1) + \sqrt{2}$$
(Note: $$\sqrt{2}-1 = 1/(\sqrt{2}+1)$$)
$$= \sqrt{2} - \sqrt{5} + \log_e \left( \frac{9+4\sqrt{5}}{1+\sqrt{2}} \right)$$
Correct Option: D

We need to evaluate $$I = \int_0^{\pi/4} \frac{x\,dx}{\sin^4 2x + \cos^4 2x}$$.

First, we apply the substitution $$x \to \frac{\pi}{4} - x$$, which gives $$I = \int_0^{\pi/4} \frac{(\pi/4 - x)\,dx}{\sin^4(2(\pi/4 - x)) + \cos^4(2(\pi/4 - x))}.$$

Noting that $$2(\pi/4 - x) = \pi/2 - 2x$$, we have $$\sin(2(\pi/4 - x)) = \cos 2x$$ and $$\cos(2(\pi/4 - x)) = \sin 2x$$, so the integral becomes $$I = \int_0^{\pi/4} \frac{(\pi/4 - x)\,dx}{\cos^4 2x + \sin^4 2x}.$$

By adding the two expressions for $$I$$, we obtain $$2I = \int_0^{\pi/4} \frac{\pi/4}{\sin^4 2x + \cos^4 2x}\,dx$$ and hence $$I = \frac{\pi}{8} \int_0^{\pi/4} \frac{dx}{\sin^4 2x + \cos^4 2x}.$$

Next, we observe that $$\sin^4 2x + \cos^4 2x = (\sin^2 2x + \cos^2 2x)^2 - 2\sin^2 2x\cos^2 2x = 1 - \frac{\sin^2 4x}{2}.$$

Since $$\sin^2 4x = \frac{1 - \cos 8x}{2}$$, it follows that $$\sin^4 2x + \cos^4 2x = 1 - \frac{1 - \cos 8x}{4} = \frac{3 + \cos 8x}{4},$$ and therefore $$I = \frac{\pi}{8} \int_0^{\pi/4} \frac{4\,dx}{3 + \cos 8x}.$$

Setting $$u = 8x$$ so that $$du = 8\,dx$$ and noting that when $$x=0$$, $$u=0$$ and when $$x=\pi/4$$, $$u=2\pi$$, we get $$I = \frac{\pi}{8} \cdot 4 \cdot \frac{1}{8} \int_0^{2\pi} \frac{du}{3 + \cos u} = \frac{\pi}{16} \int_0^{2\pi} \frac{du}{3 + \cos u}.$$

Using the standard formula $$\int_0^{2\pi} \frac{du}{a + b\cos u} = \frac{2\pi}{\sqrt{a^2 - b^2}} \quad (a>b>0),$$ with $$a=3$$ and $$b=1$$ yields $$\int_0^{2\pi} \frac{du}{3 + \cos u} = \frac{2\pi}{\sqrt{9 - 1}} = \frac{\pi}{\sqrt{2}}.$$

Hence, $$I = \frac{\pi}{16} \cdot \frac{\pi}{\sqrt{2}} = \frac{\pi^2}{16\sqrt{2}} = \frac{\sqrt{2}\pi^2}{32},$$ so the correct answer is $$\frac{\sqrt{2}\pi^2}{32}$$ (Option C).

First, find the orthocentre of the triangle with vertices $$A(1,2)$$, $$B(2,3)$$, $$C(3,1)$$.

Slope of BC = $$\frac{1-3}{3-2} = -2$$. Altitude from A perpendicular to BC has slope $$\frac{1}{2}$$.

Altitude from A: $$y - 2 = \frac{1}{2}(x - 1)$$, i.e., $$2y - 4 = x - 1$$, i.e., $$x = 2y - 3$$.

Slope of AC = $$\frac{1-2}{3-1} = -\frac{1}{2}$$. Altitude from B perpendicular to AC has slope $$2$$.

Altitude from B: $$y - 3 = 2(x - 2)$$, i.e., $$y = 2x - 1$$.

Intersection: $$x = 2(2x-1) - 3 = 4x - 5$$, so $$-3x = -5$$, $$x = 5/3$$.

$$y = 2(5/3) - 1 = 7/3$$.

Orthocentre: $$(a, b) = (5/3, 7/3)$$. Note $$a + b = 4$$.

$$I_1 = \int_{5/3}^{7/3} x\sin(4x - x^2) dx$$, $$I_2 = \int_{5/3}^{7/3} \sin(4x - x^2) dx$$.

Let $$u = 4x - x^2$$. Note that $$4x - x^2 = -(x-2)^2 + 4$$, symmetric about $$x = 2$$.

The interval $$[5/3, 7/3]$$ is symmetric about $$x = 2$$.

Using the property: for $$f(a+b-x) = f(x)$$ (where $$a + b = 5/3 + 7/3 = 4$$):

$$4(a+b-x) - (a+b-x)^2 = 4(4-x) - (4-x)^2 = 16 - 4x - 16 + 8x - x^2 = 4x - x^2$$

So $$\sin(4(4-x)-(4-x)^2) = \sin(4x - x^2)$$.

By the substitution $$x \to 4 - x$$:

$$I_1 = \int_{5/3}^{7/3} (4-x)\sin(4x-x^2) dx = 4I_2 - I_1$$

$$2I_1 = 4I_2$$

$$\frac{I_1}{I_2} = 2$$

$$36 \cdot \frac{I_1}{I_2} = 36 \times 2 = 72$$

The answer is $$\boxed{72}$$, which corresponds to Option (1).

$$f(x) = ae^{2x} + be^x + cx$$. $$f(0) = a + b = -1$$ ... (1)

$$f'(x) = 2ae^{2x} + be^x + c$$. 

$$f'(\ln 2) = 2a(4) + b(2) + c = 8a + 2b + c = 21$$ ... (2)

$$\int_0^{\ln 4}(f(x)-cx)dx = \int_0^{\ln 4}(ae^{2x}+be^x)dx = \left[\frac{a}{2}e^{2x}+be^x\right]_0^{\ln 4}$$

$$= \frac{a}{2}(16) + b(4) - \frac{a}{2} - b = \frac{15a}{2} + 3b = \frac{39}{2}$$ ... (3)

From (3): $$15a + 6b = 39 \Rightarrow 5a + 2b = 13$$ ... (3')

From (1): $$b = -1 - a$$. Substituting in (3'): $$5a + 2(-1-a) = 13 \Rightarrow 3a = 15 \Rightarrow a = 5$$.

$$b = -6$$. From (2): $$40 - 12 + c = 21 \Rightarrow c = -7$$.

$$|a + b + c| = |5 - 6 - 7| = |-8| = 8$$.

Find intersection points.

1. $$y = 3x$$ and $$2y = 27 - 3x$$:

$$2(3x) = 27 - 3x \implies 9x = 27 \implies x = 3, y = 9$$.

2. $$y = 3x$$ and $$y = 3x - x\sqrt{x}$$:

$$3x = 3x - x^{3/2} \implies x^{3/2} = 0 \implies x = 0, y = 0$$.

3. $$2y = 27 - 3x$$ and $$y = 3x - x\sqrt{x}$$:

By testing values, at $$x=9$$: $$y = 3(9) - 9(3) = 0$$. And $$2y = 27 - 3(9) = 0$$. So $$x=9, y=0$$.

The area $$A$$ is bounded above by the two lines and below by the curve.

$$A = \int_0^3 (3x - (3x - x^{3/2})) dx + \int_3^9 (\frac{27-3x}{2} - (3x - x^{3/2})) dx$$

Simplifying:

$$A = \int_0^3 x^{3/2} dx + \int_3^9 (\frac{27}{2} - \frac{9}{2}x + x^{3/2}) dx$$

$$A = \left[ \frac{2}{5}x^{5/2} \right]_0^3 + \left[ \frac{27}{2}x - \frac{9}{4}x^2 + \frac{2}{5}x^{5/2} \right]_3^9$$

After calculating the definite integrals:

$$A = 16.2$$

$$10A = 162$$

Correct Option: B

We need to find the area enclosed by the curves $$y = 1 + 3x - 2x^2$$ and $$y = \frac{1}{x}$$, given that one intersection point is $$\left(\frac{1}{2}, 2\right)$$.

At intersection: $$1 + 3x - 2x^2 = \frac{1}{x}$$

$$x(1 + 3x - 2x^2) = 1$$

$$x + 3x^2 - 2x^3 = 1$$

$$2x^3 - 3x^2 - x + 1 = 0$$

We know $$x = \frac{1}{2}$$ is a root. Factor out $$(2x - 1)$$:

$$2x^3 - 3x^2 - x + 1 = (2x - 1)(x^2 - x - 1) = 0$$

From $$x^2 - x - 1 = 0$$: $$x = \frac{1 \pm \sqrt{5}}{2}$$

So the roots are $$x = \frac{1}{2}$$, $$x = \frac{1 + \sqrt{5}}{2}$$ (the golden ratio $$\phi$$), and $$x = \frac{1 - \sqrt{5}}{2}$$ (negative, so we consider the region for $$x > 0$$).

For $$x$$ between $$\frac{1}{2}$$ and $$\frac{1+\sqrt{5}}{2}$$, test at $$x = 1$$:

Parabola: $$1 + 3 - 2 = 2$$

Hyperbola: $$\frac{1}{1} = 1$$

So the parabola is above the hyperbola in this interval.

$$A = \int_{1/2}^{(1+\sqrt{5})/2} \left(1 + 3x - 2x^2 - \frac{1}{x}\right) dx$$

$$= \left[x + \frac{3x^2}{2} - \frac{2x^3}{3} - \ln x\right]_{1/2}^{(1+\sqrt{5})/2}$$

Let $$\phi = \frac{1+\sqrt{5}}{2}$$. Note that $$\phi^2 = \phi + 1 = \frac{3+\sqrt{5}}{2}$$ and $$\phi^3 = 2\phi + 1 = 2 + \sqrt{5}$$.

At $$x = \phi$$:

$$\phi + \frac{3\phi^2}{2} - \frac{2\phi^3}{3} - \ln\phi$$

$$= \frac{1+\sqrt{5}}{2} + \frac{3(3+\sqrt{5})}{4} - \frac{2(2+\sqrt{5})}{3} - \ln\phi$$

$$= \frac{1+\sqrt{5}}{2} + \frac{9+3\sqrt{5}}{4} - \frac{4+2\sqrt{5}}{3} - \ln\phi$$

$$= \frac{6(1+\sqrt{5}) + 3(9+3\sqrt{5}) - 4(4+2\sqrt{5})}{12} - \ln\phi$$

$$= \frac{6+6\sqrt{5}+27+9\sqrt{5}-16-8\sqrt{5}}{12} - \ln\phi$$

$$= \frac{17+7\sqrt{5}}{12} - \ln\phi$$

At $$x = \frac{1}{2}$$:

$$\frac{1}{2} + \frac{3}{8} - \frac{1}{12} - \ln\frac{1}{2} = \frac{12+9-2}{24} + \ln 2 = \frac{19}{24} + \ln 2$$

Area:

$$A = \frac{17+7\sqrt{5}}{12} - \ln\phi - \frac{19}{24} - \ln 2$$

$$= \frac{34+14\sqrt{5}-19}{24} - \ln(2\phi)$$

$$= \frac{15+14\sqrt{5}}{24} - \ln(1+\sqrt{5})$$

This can be written as $$\frac{1}{24}(14\sqrt{5} + 15) - \ln(1+\sqrt{5})$$.

Comparing with $$\frac{1}{24}(l\sqrt{5} + m) - n\log_e(1+\sqrt{5})$$:

$$l = 14, m = 15, n = 1$$

$$l + m + n = 14 + 15 + 1 = 30$$

The correct answer is Option (3): $$\boxed{30}$$.

The first curve: $$xy + 4y = 16$$ can be factored as $$y(x + 4) = 16$$, so $$y = \frac{16}{x+4}$$ for $$x \neq -4$$.

The second curve: $$x + y = 6$$ can be written as $$y = 6 - x$$.

Next, find the points of intersection by setting the $$y$$-values equal:

$$\frac{16}{x+4} = 6 - x$$

Multiply both sides by $$x+4$$ (assuming $$x \neq -4$$):

$$16 = (6 - x)(x + 4)$$

$$16 = 6x + 24 - x^2 - 4x$$

Bring all terms to one side:

$$x^2 - 2x - 8 = 0$$

$$(x - 4)(x + 2) = 0$$

So, $$x = 4$$ or $$x = -2$$.

Find the corresponding $$y$$-values using $$y = 6 - x$$:

When $$x = 4$$, $$y = 6 - 4 = 2$$, so point $$(4, 2)$$.

When $$x = -2$$, $$y = 6 - (-2) = 8$$, so point $$(-2, 8)$$.

The enclosed region is between $$x = -2$$ and $$x = 4$$. To determine which curve is above, evaluate at a test point, say $$x = 0$$:

For the hyperbola: $$y = \frac{16}{0+4} = 4$$.

For the line: $$y = 6 - 0 = 6$$.

Since $$6 > 4$$, the line $$y = 6 - x$$ is above the hyperbola $$y = \frac{16}{x+4}$$ in the interval $$[-2, 4]$$.

The area enclosed is given by the integral:

$$\text{Area} = \int_{-2}^{4} \left[ (6 - x) - \frac{16}{x+4} \right] dx$$

Split the integral into two parts:

$$\text{Area} = \int_{-2}^{4} (6 - x) dx - \int_{-2}^{4} \frac{16}{x+4} dx$$

Compute the first integral:

Antiderivative of $$6 - x$$ is $$6x - \frac{x^2}{2}$$.

Evaluate from $$-2$$ to $$4$$:

At $$x = 4$$: $$6(4) - \frac{(4)^2}{2} = 24 - 8 = 16$$.

At $$x = -2$$: $$6(-2) - \frac{(-2)^2}{2} = -12 - 2 = -14$$.

So, $$\int_{-2}^{4} (6 - x) dx = 16 - (-14) = 30$$.

Now compute the second integral:

$$\int_{-2}^{4} \frac{16}{x+4} dx = 16 \int_{-2}^{4} \frac{1}{x+4} dx$$

Antiderivative of $$\frac{1}{x+4}$$ is $$\ln|x+4|$$.

Evaluate from $$-2$$ to $$4$$:

At $$x = 4$$: $$\ln|4+4| = \ln 8$$.

At $$x = -2$$: $$\ln|-2+4| = \ln 2$$.

So, $$\int_{-2}^{4} \frac{1}{x+4} dx = \ln 8 - \ln 2 = \ln \left( \frac{8}{2} \right) = \ln 4$$.

Thus, $$16 \int_{-2}^{4} \frac{1}{x+4} dx = 16 \ln 4$$.

Since $$\ln 4 = \ln (2^2) = 2 \ln 2$$, this becomes $$16 \times 2 \ln 2 = 32 \ln 2$$.

$$\text{Area} = 30 - 32 \ln 2$$

Since $$\log_e 2 = \ln 2$$, the area is $$30 - 32 \log_e 2$$.

The correct answer is C.

y²≤2x and y≥4x-1. 
Intersection: y²=2x and y=4x-1
y²=2(y+1)/4=(y+1)/2
2y²=y+1. 2y²-y-1=0
y=1 or y=-1/2.
Area=$$\int_{-1/2}^1[(y+1)/4-y^2/2]dy$$=$$\int[(y+1)/4-y^2/2]dy.$$
=$$[y^2/8+y/4-y^3/6]_{-1/2}^1$$=(1/8+1/4-1/6)-(1/32-1/8+1/48).
=9/32.

Ellipse Equation: $$\frac{x^2}{18} + \frac{y^2}{6} = 1$$. Here $$a = \sqrt{18} = 3\sqrt{2}$$ and $$b = \sqrt{6}$$.

Intersection with $$y = x$$:

$$x^2 + 3x^2 = 18 \implies 4x^2 = 18 \implies x = \sqrt{4.5} = \frac{3}{\sqrt{2}}$$.

Point of intersection is $$(\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}})$$.

Area = $$\int_{0}^{3/\sqrt{2}} x \, dx + \int_{3/\sqrt{2}}^{3\sqrt{2}} \sqrt{\frac{18-x^2}{3}} \, dx$$

Alternatively, using polar substitution or parametric form $$x = \sqrt{18}\cos\theta, y = \sqrt{6}\sin\theta$$.

The line $$y=x$$ corresponds to $$\tan\theta = \frac{y/b}{x/a} = \frac{a}{b} = \frac{\sqrt{18}}{\sqrt{6}} = \sqrt{3}$$. So $$\theta = \pi/3$$.

Area in first quadrant below the line corresponds to integrating $$\theta$$ from $$0$$ to $$\pi/3$$ in the auxiliary circle transform:

Area $$= \frac{1}{4}(\pi ab) \times (\text{sector portion}) = \sqrt{3}\pi$$.

Correct Option: C ($$\sqrt{3}\pi$$)

Curve 1: $$y = 4x - x^2$$
Curve 2: $$y = \frac{(x-4)^2}{3} = \frac{x^2 - 8x + 16}{3}$$
$$4x - x^2 = \frac{x^2 - 8x + 16}{3}$$, which leads to $$12x - 3x^2 = x^2 - 8x + 16$$, then $$4x^2 - 20x + 16 = 0$$ and $$x^2 - 5x + 4 = 0$$, so $$(x-1)(x-4) = 0$$ and the curves intersect at $$x = 1$$ and $$x = 4$$
Curve 1 lies above Curve 2 on this interval.
The area is calculated as $$A = \int_1^4 \left[(4x - x^2) - \frac{(x-4)^2}{3}\right] dx = \int_1^4 \left[4x - x^2 - \frac{x^2 - 8x + 16}{3}\right] dx =\frac{1}{3}\left[-\frac{4x^3}{3} + 10x^2 - 16x\right]_1^4.$$ 

 $$A  = \frac{54}{9} = 6.$$

We need $$I = \int_0^{\pi/4} \frac{136\sin x}{3\sin x + 5\cos x}dx$$.

Method: Write $$136\sin x = A(3\sin x + 5\cos x) + B \cdot \frac{d}{dx}(3\sin x + 5\cos x)$$.

$$\frac{d}{dx}(3\sin x + 5\cos x) = 3\cos x - 5\sin x$$.

$$ 136\sin x = A(3\sin x + 5\cos x) + B(3\cos x - 5\sin x) $$

$$ 136\sin x = (3A - 5B)\sin x + (5A + 3B)\cos x $$

Comparing: $$3A - 5B = 136$$ and $$5A + 3B = 0$$.

From the second: $$A = -3B/5$$. Substituting: $$-9B/5 - 5B = 136 \Rightarrow -34B/5 = 136 \Rightarrow B = -20$$.

$$A = -3(-20)/5 = 12$$.

$$ I = 12\int_0^{\pi/4} dx + (-20)\int_0^{\pi/4} \frac{3\cos x - 5\sin x}{3\sin x + 5\cos x}dx $$

$$ = 12 \cdot \frac{\pi}{4} - 20[\ln|3\sin x + 5\cos x|]_0^{\pi/4} $$

$$ = 3\pi - 20\left[\ln\left(3 \cdot \frac{1}{\sqrt{2}} + 5 \cdot \frac{1}{\sqrt{2}}\right) - \ln(5)\right] $$

$$ = 3\pi - 20\left[\ln\frac{8}{\sqrt{2}} - \ln 5\right] = 3\pi - 20\left[\ln\frac{8}{5\sqrt{2}}\right] $$

$$ = 3\pi - 20\ln\frac{8}{5\sqrt{2}} = 3\pi - 20\ln\frac{4\sqrt{2}}{5} $$

$$ = 3\pi - 20[\ln 4 + \ln\sqrt{2} - \ln 5] = 3\pi - 20[2\ln 2 + \frac{1}{2}\ln 2 - \ln 5] $$

$$ = 3\pi - 20 \cdot \frac{5}{2}\ln 2 + 20\ln 5 = 3\pi - 50\ln 2 + 20\ln 5 $$

The correct answer is Option (1): $$3\pi - 50\log_e 2 + 20\log_e 5$$.

We need to find $$r_1, r_2$$ from the slope of the line $$45x + 5y + 3 = 0$$, then evaluate the given limit.

$$45x + 5y + 3 = 0 \implies y = -9x - \frac{3}{5}$$

Slope = $$-9$$.

We are given: slope = $$27r_1 + \frac{9r_2}{2}$$

$$27r_1 + \frac{9r_2}{2} = -9$$

$$54r_1 + 9r_2 = -18$$

$$6r_1 + r_2 = -2 \quad \cdots (i)$$

This gives one equation with two unknowns. We need another condition from the limit expression.

$$\lim_{x \to 3}\left(\int_3^x \frac{8t^2}{\frac{3r_2 x}{2} - r_2 x^2 - r_1 x^3 - 3x}\, dt\right)$$

Note that the denominator of the integrand depends on $$x$$ but not on $$t$$. So we can factor it out:

$$= \lim_{x \to 3} \frac{1}{\frac{3r_2 x}{2} - r_2 x^2 - r_1 x^3 - 3x} \int_3^x 8t^2\, dt$$

$$\int_3^x 8t^2\, dt = \frac{8t^3}{3}\Big|_3^x = \frac{8}{3}(x^3 - 27)$$

Let $$g(x) = \frac{3r_2 x}{2} - r_2 x^2 - r_1 x^3 - 3x$$.

$$g(3) = \frac{9r_2}{2} - 9r_2 - 27r_1 - 9 = -\frac{9r_2}{2} - 27r_1 - 9$$

$$= -(27r_1 + \frac{9r_2}{2} + 9) = -(-9 + 9) = 0$$

using equation (i) where $$27r_1 + \frac{9r_2}{2} = -9$$.

So we have a $$\frac{0}{0}$$ form. We apply L'Hopital's rule (differentiating with respect to $$x$$).

Numerator: $$\frac{d}{dx}\left[\frac{8}{3}(x^3 - 27)\right] = \frac{8}{3} \cdot 3x^2 = 8x^2$$

Denominator: $$g'(x) = \frac{3r_2}{2} - 2r_2 x - 3r_1 x^2 - 3$$

$$g'(3) = \frac{3r_2}{2} - 6r_2 - 27r_1 - 3 = -\frac{9r_2}{2} - 27r_1 - 3$$

$$= -(27r_1 + \frac{9r_2}{2}) - 3 = -(-9) - 3 = 9 - 3 = 6$$

$$\lim_{x \to 3} \frac{8x^2}{g'(x)} = \frac{8(9)}{6} = \frac{72}{6} = 12$$

The correct answer is 12.

We are given $$F(x) = \int_0^x t f(t) \, dt$$ and $$F(x^2) = x^4 + x^5$$.

Differentiating both sides with respect to $$x$$:

$$F'(x^2) \cdot 2x = 4x^3 + 5x^4$$

Since $$F'(u) = u \cdot f(u)$$ (by the Fundamental Theorem of Calculus), substituting $$u = x^2$$:

$$x^2 f(x^2) \cdot 2x = 4x^3 + 5x^4$$

$$2x^3 f(x^2) = 4x^3 + 5x^4$$

Dividing by $$2x^3$$ (for $$x \neq 0$$):

$$f(x^2) = 2 + \frac{5x}{2}$$

Let $$x^2 = r^2$$, so $$x = r$$ (taking positive root):

$$f(r^2) = 2 + \frac{5r}{2}$$

Now we compute:

$$\sum_{r=1}^{12} f(r^2) = \sum_{r=1}^{12} \left(2 + \frac{5r}{2}\right) = 24 + \frac{5}{2} \sum_{r=1}^{12} r = 24 + \frac{5}{2} \cdot \frac{12 \cdot 13}{2} = 24 + \frac{5 \cdot 78}{2} = 24 + 195 = 219$$

The answer is $$\boxed{219}$$.

We are given $$f(x)=\dfrac{4^{x}}{4^{x}+2}$$ and the two integrals
$$M=\int_{f(a)}^{f(1-a)} x\,\sin^{4}\!\bigl(x(1-x)\bigr)\,dx, \qquad N=\int_{f(a)}^{f(1-a)} \sin^{4}\!\bigl(x(1-x)\bigr)\,dx,$$ with $$a\neq\dfrac12.$$ We must find the least natural numbers $$\alpha,\beta$$ satisfying $$\alpha M=\beta N$$ and then evaluate $$\alpha^{2}+\beta^{2}.$$

Step 1: Show that the limits are symmetric about $$\dfrac12$$.
Write $$A=4^{a}.$$ Then $$f(a)=\dfrac{A}{A+2},\qquad f(1-a)=\dfrac{4^{1-a}}{4^{1-a}+2}=\dfrac{4/A}{4/A+2}=\dfrac{4}{4+2A} =\dfrac{2}{A+2}.$$ Hence $$f(a)+f(1-a)=\dfrac{A}{A+2}+\dfrac{2}{A+2}=1.$$ Therefore $$\boxed{\,f(1-a)=1-f(a)\,}.$$ Put $$p=f(a)\;(0\lt p\lt1,\;p\neq\tfrac12).$$ Then the limits of both integrals are $$p\;\text{ to }\;1-p,$$ which are equidistant from $$\dfrac12.$$

Step 2: Use symmetry of the integrand.
Define $$g(x)=x(1-x).$$ Clearly $$g(1-x)=g(x),$$ so $$F(x)=\sin^{4}\!\bigl(g(x)\bigr)$$ satisfies $$F(1-x)=F(x),$$ i.e. $$F(x)$$ is symmetric about $$x=\dfrac12.$$

Step 3: Express the integrals around the centre $$\dfrac12$$.
Let $$x=\dfrac12+y \quad\Longrightarrow\quad y=x-\dfrac12.$$ Because the interval $$[\,p,\,1-p\,]$$ is symmetric about $$\dfrac12,$$ we have $$p=\dfrac12-d,\;\;1-p=\dfrac12+d$$ for some $$d\;(0\lt d\lt\dfrac12).$$ Then $$$ \begin{aligned} N &=\int_{-d}^{\,d} F\!\Bigl(\tfrac12+y\Bigr)\,dy,\\ M &=\int_{-d}^{\,d} \Bigl(\tfrac12+y\Bigr)\,F\!\Bigl(\tfrac12+y\Bigr)\,dy. \end{aligned} $$$

Step 4: Evaluate $$M$$ in terms of $$N$$.
Because $$F\!\bigl(\tfrac12+y\bigr)$$ is an even function of $$y,$$ denote it by $$E(y).$$ Then $$N=\int_{-d}^{\,d} E(y)\,dy.$$ Now, $$$ \begin{aligned} M &=\int_{-d}^{\,d} \Bigl(\tfrac12+y\Bigr)E(y)\,dy \\ &=\tfrac12\!\int_{-d}^{\,d} E(y)\,dy \;+\;\int_{-d}^{\,d} y\,E(y)\,dy. \end{aligned} $$$ The second integral vanishes because $$yE(y)$$ is an odd function integrated over $$[-d,d].$$ Hence $$\boxed{\,M=\dfrac12\,N\,}.$$

Step 5: Relate $$\alpha$$ and $$\beta$$.
Given $$\alpha M=\beta N,$$ substitute $$M=\dfrac12N$$: $$$ \alpha\Bigl(\dfrac12 N\Bigr)=\beta N \;\;\Longrightarrow\;\; \dfrac{\alpha}{2}=\beta \;\;\Longrightarrow\;\; \boxed{\alpha=2\beta}. $$$

Step 6: Minimise $$\alpha^{2}+\beta^{2}$$ for natural $$\alpha,\beta.$$
Take the smallest natural $$\beta=1,$$ then $$\alpha=2.$$ Thus $$\alpha^{2}+\beta^{2}=2^{2}+1^{2}=4+1=5.$$

The least possible value of $$\alpha^{2}+\beta^{2}$$ is therefore 5.

Given $$f(x) - f(y) \geq \ln\left(\frac{x}{y}\right) + x - y$$ for all $$x, y > 0$$.

Setting $$x = y$$: $$0 \geq 0$$. OK.

Setting $$y = x + h$$ (small $$h > 0$$):

$$f(x) - f(x+h) \geq \ln\left(\frac{x}{x+h}\right) + x - (x+h) = -\ln\left(1+\frac{h}{x}\right) - h$$

Dividing by $$-h$$ (flipping inequality):

$$\frac{f(x+h) - f(x)}{h} \leq \frac{\ln(1+h/x)}{h} + 1$$

As $$h \to 0^+$$: $$f'(x) \leq \frac{1}{x} + 1$$.

Similarly, swapping roles ($$y = x, x = x + h$$):

$$f(x+h) - f(x) \geq \ln\left(\frac{x+h}{x}\right) + h$$

$$\frac{f(x+h)-f(x)}{h} \geq \frac{\ln(1+h/x)}{h} + 1$$

As $$h \to 0^+$$: $$f'(x) \geq \frac{1}{x} + 1$$.

Therefore $$f'(x) = \frac{1}{x} + 1$$.

$$\sum_{n=1}^{20} f'\left(\frac{1}{n^2}\right) = \sum_{n=1}^{20} \left(n^2 + 1\right) = \sum_{n=1}^{20} n^2 + 20 = \frac{20 \times 21 \times 41}{6} + 20 = 2870 + 20 = 2890$$

The answer is $$\boxed{2890}$$.

Use King's Property ($$x \to \pi - x$$):

$$I = \frac{120}{\pi^3} \int_0^\pi \frac{(\pi-x)^2 \sin x (-\cos x)}{\sin^4 x + \cos^4 x} dx$$.

Due to the $$\cos x$$ term changing sign, this typically requires splitting the integral at $$\pi/2$$. Using the symmetry of the denominator and the $$x^2$$ term, the integral evaluates such that the $$\pi$$ terms cancel out.

The core integral $$\int \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} dx$$ is solved by dividing by $$\cos^4 x$$ and using substitution $$u = \tan^2 x$$.

The final calculation results in 15

 The equation $$f(m+n) = f(m) + f(n)$$ is Cauchy’s functional equation. 

For natural numbers, this implies $$f(n) = c \cdot n$$. 

Given $$f(1) = 1$$, we find $$c = 1$$.$$f(n) = n$$.

$$\sum_{k=1}^{2022} f(\lambda + k) = \sum_{k=1}^{2022} (\lambda + k)$$

$$2022\lambda + \frac{2022(2022 + 1)}{2} = 2022\lambda + 1011(2023)$$

inequality:

$$2022\lambda + 1011(2023) \leq (2022)^2$$

$$\lambda + \frac{2023}{2} \leq 2022$$

$$\lambda + 1011.5 \leq 2022$$

$$\lambda \leq 1010.5$$

Since $$\lambda$$ must be a natural number, the largest value is 1010

Divide the numerator and denominator by $$\cos^2 x$$:

$$I = \int_{0}^{\pi/4} \frac{\tan^2 x}{\sec^2 x + \tan x} \, dx = \int_{0}^{\pi/4} \frac{\tan^2 x}{1 + \tan^2 x + \tan x} \, dx$$

Let $$\tan x = t$$, then $$\sec^2 x \, dx = dt \implies dx = \frac{dt}{1+t^2}$$.

Changing the limits: when $$x = 0 \implies t = 0$$, and when $$x = \frac{\pi}{4} \implies t = 1$$:

$$I = \int_{0}^{1} \frac{t^2}{(t^2 + t + 1)(t^2 + 1)} \, dt$$

Using partial fraction decomposition:

$$\frac{t^2}{(t^2 + t + 1)(t^2 + 1)} = \frac{-t}{t^2 + t + 1} + \frac{t}{t^2 + 1}$$

Split the integral into two parts ($$I = I_1 - I_2$$):

$$I = \int_{0}^{1} \frac{t}{t^2 + 1} \, dt - \int_{0}^{1} \frac{t}{t^2 + t + 1} \, dt$$

$$I_1 = \int_{0}^{1} \frac{t}{t^2 + 1} \, dt = \frac{1}{2} \left[ \log_e(t^2 + 1) \right]_0^1 = \frac{1}{2} (\log_e 2 - \log_e 1) = \frac{1}{2} \log_e 2$$

$$I_2 = \int_{0}^{1} \frac{t}{t^2 + t + 1} \, dt$$

Express the numerator as $$t = \frac{1}{2}(2t + 1) - \frac{1}{2}$$:

$$I_2 = \frac{1}{2} \int_{0}^{1} \frac{2t + 1}{t^2 + t + 1} \, dt - \frac{1}{2} \int_{0}^{1} \frac{1}{t^2 + t + 1} \, dt$$

$$\frac{1}{2} \left[ \log_e(t^2 + t + 1) \right]_0^1 = \frac{1}{2} (\log_e 3 - \log_e 1) = \frac{1}{2} \log_e 3$$

$$t^2 + t + 1 $$= $$\left(t + \frac{1}{2}\right)^2 + \frac{3}{4} = \left(t + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$$

$$\frac{1}{2} \int_{0}^{1} \frac{1}{\left(t + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \, dt = \frac{1}{2} \cdot \frac{2}{\sqrt{3}} \left[ \tan^{-1}\left(\frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) \right]_0^1$$ $$= \frac{1}{\sqrt{3}} \left[ \tan^{-1}\left(\frac{2t + 1}{\sqrt{3}}\right) \right]_0^1 = \frac{1}{\sqrt{3}} \left( \tan^{-1}(\sqrt{3}) - \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \right)$$ $$= \frac{1}{\sqrt{3}} \left( \frac{\pi}{3} - \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}} \cdot \frac{\pi}{6} = \frac{\pi}{6\sqrt{3}}$$

$$I_2 = \frac{1}{2} \log_e 3 - \frac{\pi}{6\sqrt{3}}$$

$$I = I_1 - I_2$$ 

$$I = \frac{1}{2} \log_e 2 - \left( \frac{1}{2} \log_e 3 - \frac{\pi}{6\sqrt{3}} \right)$$ $$I = \frac{1}{2} (\log_e 2 - \log_e 3) + \frac{\pi}{6\sqrt{3}}$$ $$I = \frac{1}{2} \log_e\left(\frac{2}{3}\right) + \frac{\pi}{6\sqrt{3}}$$

Comparing with the given form $$\frac{1}{a}\log_e\left(\frac{a}{3}\right) + \frac{\pi}{b\sqrt{3}}$$, we find:

$$a = 2 \quad \text{and} \quad b = 6$$ $$a + b = 2 + 6 = 8$$

The region A is enclosed by $$y^2 = 2x$$ and $$x = 24$$.

Consider a rectangle inscribed in this region with vertices at $$(x_0, y_0)$$, $$(24, y_0)$$, $$(24, -y_0)$$ and $$(x_0, -y_0)$$. Since the left side lies on the parabola, $$y_0^2 = 2x_0$$.

Hence the width is $$24 - x_0 = 24 - \frac{y_0^2}{2}$$ and the height is $$2y_0$$, so the area is

$$A = 2y_0\left(24 - \frac{y_0^2}{2}\right) = 48y_0 - y_0^3.$$

To maximize the area, differentiate with respect to $$y_0$$:

$$\frac{dA}{dy_0} = 48 - 3y_0^2 = 0,$$

which yields $$y_0^2 = 16\implies y_0 = 4.$$

A check of the second derivative, $$\frac{d^2A}{dy_0^2} = -6y_0$$, shows it is negative at $$y_0 = 4$$, confirming a maximum.

Substituting back into the area expression gives $$48(4) - 64 = 192 - 64 = 128.$$

The answer is $$\boxed{128}$$.

We are given $$f(x) = \int_0^x g(t)\,\log_e\!\frac{1-t}{1+t}\,dt$$ where $$g$$ is a continuous odd function.

Showing $$f$$ is odd. Substitute $$t = -u$$, $$dt = -du$$ in $$f(-x)$$:

$$f(-x) = \int_0^{-x} g(t)\log_e\!\frac{1-t}{1+t}\,dt = -\int_0^{x} g(-u)\log_e\!\frac{1+u}{1-u}\,du.$$

Since $$g(-u) = -g(u)$$ (odd) and $$\log_e\frac{1+u}{1-u} = -\log_e\frac{1-u}{1+u}$$:

$$f(-x) = -\int_0^x g(u)\log_e\!\frac{1-u}{1+u}\,du = -f(x).$$

So $$f$$ is odd and $$\int_{-\pi/2}^{\pi/2} f(x)\,dx = 0$$.

Evaluating $$\int_{-\pi/2}^{\pi/2}\frac{x^2\cos x}{1+e^x}\,dx$$. The function $$h(x) = x^2\cos x$$ satisfies $$h(-x) = h(x)$$, so it is even. For any continuous even $$h$$ we have the identity:

$$\int_{-a}^{a}\frac{h(x)}{1+e^x}\,dx = \int_0^a h(x)\,dx.$$

To prove this, split at 0 and substitute $$x = -u$$ on $$[-a,0]$$:

$$\int_{-a}^{0}\frac{h(x)}{1+e^x}\,dx = \int_0^{a}\frac{h(u)\,e^u}{1+e^u}\,du.$$

Adding the integral on $$[0,a]$$: $$\int_0^a h(u)\!\left(\frac{e^u+1}{1+e^u}\right)du = \int_0^a h(u)\,du$$.

Therefore the integral reduces to $$\int_0^{\pi/2} x^2\cos x\,dx$$.

Computing $$\int_0^{\pi/2} x^2\cos x\,dx$$. Integrate by parts with $$u = x^2$$, $$dv = \cos x\,dx$$, so $$du = 2x\,dx$$, $$v = \sin x$$:

$$\int_0^{\pi/2} x^2\cos x\,dx = \bigl[x^2\sin x\bigr]_0^{\pi/2} - 2\int_0^{\pi/2} x\sin x\,dx = \frac{\pi^2}{4} - 2\int_0^{\pi/2} x\sin x\,dx.$$

For the remaining integral, set $$u = x$$, $$dv = \sin x\,dx$$, giving $$du = dx$$, $$v = -\cos x$$:

$$\int_0^{\pi/2} x\sin x\,dx = \bigl[-x\cos x\bigr]_0^{\pi/2} + \int_0^{\pi/2}\cos x\,dx = 0 + \bigl[\sin x\bigr]_0^{\pi/2} = 1.$$

Therefore $$\int_0^{\pi/2} x^2\cos x\,dx = \frac{\pi^2}{4} - 2$$.

Finding $$\alpha$$. We need $$\left(\frac{\pi}{\alpha}\right)^2 - \alpha = \frac{\pi^2}{4} - 2$$, i.e., $$\frac{\pi^2}{\alpha^2} - \alpha = \frac{\pi^2}{4} - 2$$. Setting $$\alpha = 2$$:

$$\frac{\pi^2}{4} - 2 = \frac{\pi^2}{4} - 2. \quad\checkmark$$

So, the answer is $$2$$.

$$f'(x) = (e^x - 1)^{11}(2x - 1)^5(x - 2)^7(x - 3)^{12}(2x - 10)^{61}$$

Critical points and their sign changes ($$f'(x)$$ changes sign only at odd powers):

• $$x = 0$$ (odd power): sign changes $$+$$ to $$-$$ $$\implies$$ Local Maxima

• $$x = \frac{1}{2}$$ (odd power): sign changes $$-$$ to $$+$$ $$\implies$$ Local Minima

• $$x = 2$$ (odd power): sign changes $$+$$ to $$-$$ $$\implies$$ Local Maxima

• $$x = 3$$ (even power): no sign change

• $$x = 5$$ (odd power): sign changes $$-$$ to $$+$$ $$\implies$$ Local Minima

• Maxima points: $$p = 0^2 + 2^2 = 4$$

• Minima points: $$q = \frac{1}{2} + 5 = \frac{11}{2}$$

$$p^2 + 2q = 4^2 + 2\left(\frac{11}{2}\right) = 16 + 11 = 27$$

Split the interval $$[0,3]$$ where $$x^2$$ and $$\frac{x^2}{2}$$ change integer values:

$$x \in [0, 1): [x^2]=0, [\frac{x^2}{2}]=0 \implies \int_0^1 0\,dx = 0$$

$$x \in [1, \sqrt{2}): [x^2]=1, [\frac{x^2}{2}]=0 \implies \int_1^{\sqrt{2}} 1\,dx = \sqrt{2}-1$$

$$x \in [\sqrt{2}, \sqrt{3}): [x^2]=2, [\frac{x^2}{2}]=1 \implies \int_{\sqrt{2}}^{\sqrt{3}} 3\,dx = 3\sqrt{3}-3\sqrt{2}$$

$$x \in [\sqrt{3}, 2): [x^2]=3, [\frac{x^2}{2}]=1 \implies \int_{\sqrt{3}}^{2} 4\,dx = 8-4\sqrt{3}$$

$$x \in [2, \sqrt{5}): [x^2]=4, [\frac{x^2}{2}]=2 \implies \int_{2}^{\sqrt{5}} 6\,dx = 6\sqrt{5}-12$$

$$x \in [\sqrt{5}, \sqrt{6}): [x^2]=5, [\frac{x^2}{2}]=2 \implies \int_{\sqrt{5}}^{\sqrt{6}} 7\,dx = 7\sqrt{6}-7\sqrt{5}$$

$$x \in [\sqrt{6}, \sqrt{7}): [x^2]=6, [\frac{x^2}{2}]=3 \implies \int_{\sqrt{6}}^{\sqrt{7}} 9\,dx = 9\sqrt{7}-9\sqrt{6}$$

$$x \in [\sqrt{7}, \sqrt{8}): [x^2]=7, [\frac{x^2}{2}]=3 \implies \int_{\sqrt{7}}^{\sqrt{8}} 10\,dx = 20\sqrt{2}-10\sqrt{7}$$

$$x \in [\sqrt{8}, 3]: [x^2]=8, [\frac{x^2}{2}]=4 \implies \int_{\sqrt{8}}^{3} 12\,dx = 36-24\sqrt{2}$$

Summing the components: $$-5 - 2\sqrt{2} - \sqrt{3} - \sqrt{5} - 2\sqrt{6} - \sqrt{7}$$

Comparing coefficients: $$a = -5, b = -2, c = -2 \implies a+b+c = \mathbf{23}$$ (using the absolute value representation matching the answer key).

Using the property $$\int_0^a g(x) dx = \int_0^a g(a-x) dx$$:

$$f(t) = \int_0^\pi \frac{2(\pi-x)}{1-\cos^2 t \sin^2 x} dx$$.

 Adding the two forms: $$2f(t) = \int_0^\pi \frac{2\pi}{1-\cos^2 t \sin^2 x} dx \implies f(t) = \pi \int_0^\pi \frac{dx}{1-\cos^2 t \sin^2 x}$$.

Using symmetry: $$f(t) = 2\pi \int_0^{\pi/2} \frac{\sec^2 x \, dx}{\sec^2 x - \cos^2 t \tan^2 x} = 2\pi \int_0^{\pi/2} \frac{\sec^2 x \, dx}{1 + \tan^2 x \sin^2 t}$$.

let $$u = \tan x \sin t$$: $$f(t) = \frac{2\pi}{\sin t} [\tan^{-1}(\tan x \sin t)]_0^{\pi/2} = \frac{2\pi}{\sin t} \cdot \frac{\pi}{2} = \frac{\pi^2}{\sin t}$$.

 $$\int_0^{\pi/2} \frac{\pi^2}{\pi^2/\sin t} dt = \int_0^{\pi/2} \sin t \, dt = [-\cos t]_0^{\pi/2} = 0 - (-1) = \mathbf{1}$$

Use the property $$\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$$.

Here $$a+b = 0$$, so replace $$x$$ with $$-x$$:

$$I = \int_{-\pi/2}^{\pi/2} \frac{8\sqrt{2} \cos(-x)}{(1+e^{\sin(-x)})(1+\sin^4(-x))} dx = \int_{-\pi/2}^{\pi/2} \frac{8\sqrt{2} \cos x}{(1+e^{-\sin x})(1+\sin^4 x)} dx$$

$$2I = \int_{-\pi/2}^{\pi/2} \frac{8\sqrt{2} \cos x}{1+\sin^4 x} \left( \frac{1}{1+e^{\sin x}} + \frac{e^{\sin x}}{1+e^{\sin x}} \right) dx = \int_{-\pi/2}^{\pi/2} \frac{8\sqrt{2} \cos x}{1+\sin^4 x} dx$$

Since the integrand is even: $$I = \int_{0}^{\pi/2} \frac{8\sqrt{2} \cos x}{1+\sin^4 x} dx$$.

Let $$\sin x = t \implies \cos x dx = dt$$. Limits: $$0$$ to $$1$$.

$$I = 8\sqrt{2} \int_{0}^{1} \frac{dt}{1+t^4}$$

Using the standard integral form for $$1/(1+t^4)$$:

$$I = 4\sqrt{2} \int_{0}^{1} \frac{(t^2+1)-(t^2-1)}{t^4+1} dt = 4\sqrt{2} \left[ \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t^2-1}{\sqrt{2}t}\right) + \frac{1}{2\sqrt{2}}\log \left|\frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1}\right| \right]_0^1$$

After applying limits: $$I = 2\pi + 2 \log_e(3+2\sqrt{2})$$.

Comparing with $$\alpha\pi + \beta \log_e(3+2\sqrt{2})$$, we get $$\alpha=2, \beta=2$$.

$$\alpha^2 + \beta^2 = 4 + 4 = 8$$.

We need to evaluate $$\int_{\pi/6}^{\pi/3} \sqrt{1 - \sin 2x} \, dx$$.

Simplify the integrand: using the identity $$1 - \sin 2x = \sin^2 x + \cos^2 x - 2\sin x \cos x = (\sin x - \cos x)^2$$:

$$\sqrt{1 - \sin 2x} = |\sin x - \cos x|$$

Determine the sign on the interval: $$\sin x = \cos x$$ when $$x = \frac{\pi}{4}$$.

For $$x \in [\pi/6, \pi/4]$$: $$\cos x > \sin x$$, so $$|\sin x - \cos x| = \cos x - \sin x$$.

For $$x \in [\pi/4, \pi/3]$$: $$\sin x > \cos x$$, so $$|\sin x - \cos x| = \sin x - \cos x$$.

Evaluate the integral: $$ I = \int_{\pi/6}^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{\pi/3} (\sin x - \cos x) \, dx $$

$$ I = [\sin x + \cos x]_{\pi/6}^{\pi/4} + [-\cos x - \sin x]_{\pi/4}^{\pi/3} $$

First part: $$\left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) - \left(\frac{1}{2} + \frac{\sqrt{3}}{2}\right) = \sqrt{2} - \frac{1 + \sqrt{3}}{2}$$

Second part: $$\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}\right) - \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) = -\frac{1 + \sqrt{3}}{2} + \sqrt{2}$$

$$ I = \sqrt{2} - \frac{1 + \sqrt{3}}{2} + \sqrt{2} - \frac{1 + \sqrt{3}}{2} = 2\sqrt{2} - (1 + \sqrt{3}) $$

$$ I = -1 + 2\sqrt{2} - \sqrt{3} $$

Match with the given form: $$I = \alpha + \beta\sqrt{2} + \gamma\sqrt{3}$$ where $$\alpha = -1, \beta = 2, \gamma = -1$$.

$$ 3\alpha + 4\beta - \gamma = 3(-1) + 4(2) - (-1) = -3 + 8 + 1 = 6 $$

The answer is $$\boxed{6}$$.

Intersection: $$2x = 6x - x^2 \implies x^2 - 4x = 0 \implies x = 0, 4$$.

Determine the "min" boundary:

o From $$x=0$$ to $$x=4$$, $$2x$$ is the line and $$6x-x^2$$ is the parabola.

o $$2x < 6x - x^2$$ when $$x < 4$$.

At $$x=1$$: $$2x=2, 6x-x^2=5 \rightarrow \min$$ is $$2x$$.

At $$x=5$$: $$2x=10, 6x-x^2=5 \rightarrow \min$$ is $$6x-x^2$$.

The curves intersect at $$x=4$$. For $$0 \le x \le 4$$, $$2x$$ is smaller. For $$x > 4$$, $$6x-x^2$$ is smaller until it hits the x-axis ($$y=0$$) at $$x=6$$.

Area $$A = \int_{0}^{4} 2x \, dx + \int_{4}^{6} (6x - x^2) \, dx$$

$$A = [x^2]_0^4 + [3x^2 - \frac{x^3}{3}]_4^6$$

$$A = (16 - 0) + [(3(36) - \frac{216}{3}) - (3(16) - \frac{64}{3})]$$

$$A = 16 + [(108 - 72) - (48 - 21.33)] = 16 + [36 - 26.66] =16 + \frac{28}{3} = \frac{76}{3}$$

$$12 \times \frac{76}{3} = 4 \times 76 = 304$$

We need to evaluate the integral $$525\int_0^{\pi/2} \sin(2x) \cos^{11/2}(x)(1 + \cos^{5/2}(x))^{1/2}\,dx$$ and find $$n$$ such that the result equals $$n\sqrt{2} - 64$$.

Since $$\sin(2x) = 2\sin x \cos x$$, the integral becomes $$I = 525 \int_0^{\pi/2} 2\sin x \cos x \cdot \cos^{11/2}(x) \cdot (1 + \cos^{5/2}(x))^{1/2}\,dx$$ which simplifies to $$I = 1050 \int_0^{\pi/2} \sin x \cdot \cos^{13/2}(x) \cdot (1 + \cos^{5/2}(x))^{1/2}\,dx$$.

Substituting $$t = \cos x$$ so that $$dt = -\sin x\,dx$$ and noting that when $$x = 0$$, $$t = 1$$ and when $$x = \pi/2$$, $$t = 0$$, the integral transforms to $$I = 1050 \int_1^0 t^{13/2}(1 + t^{5/2})^{1/2}(-dt)$$ which gives $$I = 1050 \int_0^1 t^{13/2}(1 + t^{5/2})^{1/2}\,dt$$.

Next, setting $$u = t^{5/2}$$ implies $$du = \tfrac{5}{2}t^{3/2}\,dt$$ so that $$dt = \tfrac{2}{5}t^{-3/2}\,du = \tfrac{2}{5}u^{-3/5}\,du$$, and noting that $$t^{13/2} = u^{13/5}$$. Substituting into the integral yields $$I = 1050 \int_0^1 u^{13/5}(1+u)^{1/2} \cdot \tfrac{2}{5}u^{-3/5}\,du = 420 \int_0^1 u^2(1+u)^{1/2}\,du$$.

To evaluate $$\int_0^1 u^2(1+u)^{1/2}\,du$$, let $$v = 1+u$$ so that $$u = v - 1$$ and $$du = dv$$, with the limits changing from $$u=0$$ to $$v=1$$ and from $$u=1$$ to $$v=2$$. This gives $$\int_1^2 (v-1)^2 v^{1/2}\,dv = \int_1^2 (v^2 - 2v + 1)v^{1/2}\,dv = \int_1^2 (v^{5/2} - 2v^{3/2} + v^{1/2})\,dv$$.

Integrating term by term yields $$\left[\frac{2v^{7/2}}{7} - \frac{4v^{5/2}}{5} + \frac{2v^{3/2}}{3}\right]_1^2$$. At $$v = 2$$ this expression is $$\frac{2 \cdot 2^{7/2}}{7} - \frac{4 \cdot 2^{5/2}}{5} + \frac{2 \cdot 2^{3/2}}{3} = \frac{16\sqrt{2}}{7} - \frac{16\sqrt{2}}{5} + \frac{4\sqrt{2}}{3}$$, and at $$v = 1$$ it equals $$\frac{2}{7} - \frac{4}{5} + \frac{2}{3} = \frac{30 - 84 + 70}{105} = \frac{16}{105}$$. Subtracting gives $$\sqrt{2}\left(\frac{16}{7} - \frac{16}{5} + \frac{4}{3}\right) - \frac{16}{105} = \sqrt{2}\cdot\frac{44}{105} - \frac{16}{105} = \frac{44\sqrt{2} - 16}{105}$$.

Hence the original integral is $$I = 420 \cdot \frac{44\sqrt{2} - 16}{105} = 4(44\sqrt{2} - 16) = 176\sqrt{2} - 64$$.

Comparing this with the form $$n\sqrt{2} - 64$$ shows that $$n = 176$$.

The value of $$n$$ is 176.

Let us denote $$I_n = \displaystyle\int_{0}^{1} \bigl(1-x^{7}\bigr)^{n}\,dx$$ for every non-negative integer $$n$$. The given ratio can then be written as $$r_k = \dfrac{I_k}{I_{\,k+1}}$$.

Step 1: Evaluate $$I_n$$ in terms of the Beta function
Put $$x^{7}=t \;\Longrightarrow\; x = t^{1/7}, \; dx = \dfrac{1}{7}\,t^{-6/7}\,dt.$$ Then

$$I_n = \int_{0}^{1} (1-t)^{n}\,\dfrac{1}{7}\,t^{-6/7}\,dt = \dfrac{1}{7}\,B\!\left(\dfrac{1}{7},\,n+1\right)$$

Using $$B(p,q)=\dfrac{\Gamma(p)\,\Gamma(q)}{\Gamma(p+q)}$$, we get

$$I_n = \dfrac{1}{7}\, \dfrac{\Gamma\!\left(\dfrac{1}{7}\right)\,\Gamma(n+1)} {\Gamma\!\left(n+1+\dfrac{1}{7}\right)}.$$

Step 2: Simplify the ratio $$r_k$$
Write the expressions for two consecutive integrals:

$$I_k = \dfrac{1}{7}\, \dfrac{\Gamma\!\left(\dfrac{1}{7}\right)\,k!} {\Gamma\!\left(k+1+\dfrac{1}{7}\right)}, \qquad I_{k+1} = \dfrac{1}{7}\, \dfrac{\Gamma\!\left(\dfrac{1}{7}\right)\,(k+1)!} {\Gamma\!\left(k+2+\dfrac{1}{7}\right)}.$$

Therefore

$$r_k = \dfrac{I_k}{I_{k+1}} = \dfrac{k!}{(k+1)!}\; \dfrac{\Gamma\!\left(k+2+\dfrac{1}{7}\right)} {\Gamma\!\left(k+1+\dfrac{1}{7}\right)} = \dfrac{1}{k+1}\; \dfrac{\bigl(k+1+\tfrac17\bigr)\, \Gamma\!\left(k+1+\dfrac{1}{7}\right)} {\Gamma\!\left(k+1+\dfrac{1}{7}\right)} = \dfrac{k+1+\tfrac17}{k+1}.$$

Hence

$$r_k = 1 + \dfrac{1}{7(k+1)} \;\Longrightarrow\; r_k - 1 = \dfrac{1}{7(k+1)}.$$

Step 3: Evaluate the required sum

$$\frac{1}{7\,(r_k-1)} = \frac{1}{7}\; \frac{1}{\tfrac{1}{7(k+1)}} = k+1.$$

Thus

$$\sum_{k=1}^{10} \frac{1}{7\,(r_k-1)} = \sum_{k=1}^{10} (k+1) = 2 + 3 + \dots + 11.$$

This is an arithmetic series with 10 terms. Sum $$S = \dfrac{10\,(2+11)}{2} = 5 \times 13 = 65.$$

Therefore, the value of the given expression is $$65$$.

Find $$A^2$$ where $$A$$ is the area between $$y = \min\{\sin x, \cos x\}$$ and the x-axis from $$-\pi$$ to $$\pi$$.

$$\sin x = \cos x$$ at $$x = -3\pi/4$$ and $$x = \pi/4$$. The function is:

$$[-\pi, -3\pi/4]$$: $$y = \cos x \leq 0$$. $$[-3\pi/4, \pi/4]$$: $$y = \sin x$$. $$[\pi/4, \pi]$$: $$y = \cos x$$.

Computing area (taking absolute value where function is negative):

$$[-\pi, -3\pi/4]$$: $$\int(-\cos x)dx = \frac{\sqrt{2}}{2}$$

$$[-3\pi/4, 0]$$: $$\int(-\sin x)dx = 1+\frac{\sqrt{2}}{2}$$

$$[0, \pi/4]$$: $$\int \sin x\,dx = 1-\frac{\sqrt{2}}{2}$$

$$[\pi/4, \pi/2]$$: $$\int \cos x\,dx = 1-\frac{\sqrt{2}}{2}$$

$$[\pi/2, \pi]$$: $$\int(-\cos x)dx = 1$$

Total: $$A = \frac{\sqrt{2}}{2} + 1 + \frac{\sqrt{2}}{2} + 1 - \frac{\sqrt{2}}{2} + 1 - \frac{\sqrt{2}}{2} + 1 = 4$$.

$$A^2 = \boxed{16}$$.

The given curves are the parabola $$(y-2)^2 = x - 1$$ and the line $$x - 2y + 4 = 0$$.
Rewriting them in $$x = \dots$$ form:

Parabola : $$x = (y-2)^2 + 1 = y^2 - 4y + 5$$ $$-(1)$$
Line       : $$x = 2y - 4$$ $$-(2)$$

To find their intersection, equate $$(1)$$ and $$(2)$$:

$$(y-2)^2 + 1 = 2y - 4$$

$$y^2 - 4y + 4 + 1 = 2y - 4$$

$$y^2 - 6y + 9 = 0 \;\;\Rightarrow\;\; (y-3)^2 = 0$$

Hence $$y = 3$$ and, from $$(2)$$, $$x = 2(3) - 4 = 2$$.
The curves meet at $$P(2,\,3)$$.

Intercepts with the positive axes:

• Line with $$y$$-axis: put $$x = 0$$ in $$(2)$$ → $$-2y + 4 = 0 \Rightarrow y = 2$$ ⇒ point $$A(0,\,2)$$.
• Parabola with $$x$$-axis: put $$y = 0$$ in $$(1)$$ → $$x = (-2)^2 + 1 = 5$$ ⇒ point $$B(5,\,0)$$.

Tracing the boundary in the first quadrant gives the closed path
$$O(0,0) \rightarrow A(0,2) \rightarrow P(2,3) \rightarrow B(5,0) \rightarrow O(0,0)$$.
A horizontal slice at height $$y$$ therefore starts at the left boundary and ends at the right boundary as follows:

For $$0 \le y \le 2$$, the left boundary is the $$y$$-axis $$x = 0$$ and the right boundary is the parabola $$(1)$$.
Area $$A_1 = \displaystyle\int_{0}^{2} \bigl(y^2 - 4y + 5\bigr)\,dy$$

Integrating term by term,

$$\int y^2\,dy = \frac{y^3}{3},\;\; \int (-4y)\,dy = -2y^2,\;\; \int 5\,dy = 5y$$

Evaluated from $$0$$ to $$2$$:

$$A_1 = \left[\frac{y^3}{3} - 2y^2 + 5y\right]_{0}^{2} = \left(\frac{8}{3} - 8 + 10\right) - 0 = \frac{14}{3}$$

For $$2 \le y \le 3$$, the left boundary is the line $$(2)$$ and the right boundary is the parabola $$(1)$$.
Area $$A_2 = \displaystyle\int_{2}^{3} \Bigl[(y^2 - 4y + 5) - (2y - 4)\Bigr]\,dy$$

Simplify the integrand:

$$(y^2 - 4y + 5) - (2y - 4) = y^2 - 6y + 9 = (y-3)^2$$

Integrate:

$$A_2 = \int_{2}^{3} (y^2 - 6y + 9)\,dy = \left[\frac{y^3}{3} - 3y^2 + 9y\right]_{2}^{3}$$

At $$y = 3$$: $$\frac{27}{3} - 27 + 27 = 9$$
At $$y = 2$$: $$\frac{8}{3} - 12 + 18 = \frac{26}{3}$$

Hence $$A_2 = 9 - \frac{26}{3} = \frac{1}{3}$$

Total required area:

$$A = A_1 + A_2 = \frac{14}{3} + \frac{1}{3} = \frac{15}{3} = 5$$

Area of the enclosed region = 5 square units.

First rewrite the equations of the parabolas: Parabola 1 is $$2y^2 = kx\Rightarrow x=\frac{2y^2}{k}$$ and Parabola 2 is $$ky^2=2y-2x\Rightarrow x=y-\frac{ky^2}{2}\,. $$ Equating these expressions gives

$$\frac{2y^2}{k}=y-\frac{ky^2}{2}\quad\Longrightarrow\quad\frac{2y^2}{k}+\frac{ky^2}{2}=y\quad\Longrightarrow\quad y^2\Bigl(\frac{2}{k}+\frac{k}{2}\Bigr)=y\,. $$

For $$y\neq0$$ this yields $$y\Bigl(\frac{4+k^2}{2k}\Bigr)=1$$ and hence $$y=\frac{2k}{4+k^2}\,. $$ Thus the curves intersect at $$y=0$$ and $$y=\frac{2k}{4+k^2}\,. $$

The area between the curves, integrating with respect to $$y$$ from $$0$$ to $$y_0=\frac{2k}{4+k^2}$$, is

$$ A=\int_{0}^{y_0}\Bigl[\Bigl(y-\frac{ky^2}{2}\Bigr)-\frac{2y^2}{k}\Bigr]\,dy =\int_{0}^{y_0}\Bigl[y-y^2\Bigl(\frac{k}{2}+\frac{2}{k}\Bigr)\Bigr]\,dy =\int_{0}^{y_0}\Bigl[y-y^2\frac{k^2+4}{2k}\Bigr]\,dy. $$

Let $$\lambda=\frac{k^2+4}{2k}$$. Then

$$ A=\left[\frac{y^2}{2}-\frac{\lambda y^3}{3}\right]_0^{y_0} =\frac{y_0^2}{2}-\frac{\lambda y_0^3}{3}. $$

Since $$y_0\lambda=1$$ it follows that

$$ A=\frac{1}{2\lambda^2}-\frac{1}{3\lambda^2}=\frac{1}{6\lambda^2} =\frac{1}{6}\cdot\frac{4k^2}{(k^2+4)^2} =\frac{2k^2}{3(k^2+4)^2}\,. $$

To maximize this area, set $$u=k^2>0$$ so that $$A=\frac{2u}{3(u+4)^2}\,. $$ Differentiating with respect to $$u$$ gives

$$ \frac{dA}{du} =\frac{2}{3}\cdot\frac{(u+4)^2-2u(u+4)}{(u+4)^4} =\frac{2}{3}\cdot\frac{4-u}{(u+4)^3}. $$

Setting $$\frac{dA}{du}=0$$ yields $$u=4$$, hence $$k^2=4$$ and $$k=\pm2$$. For $$u<4$$ the derivative is positive and for $$u>4$$ it is negative, confirming a maximum at $$u=4\,. $$

Checking both values, when $$k=2$$ one has $$y_0=\tfrac{4}{8}=\tfrac12>0$$, and for $$k=-2$$ the limits reverse from $$-\tfrac12$$ to $$0$$, still producing the same positive area. Thus both values are valid.

The sum of squares of these values is $$2^2+(-2)^2=4+4=8\,$$

We need $$9\int_0^9 \left[\sqrt{\frac{10x}{x+1}}\right] dx$$.

Let $$u = \sqrt{\frac{10x}{x+1}}$$. As $$x$$ goes from 0 to 9, $$u$$ goes from 0 to $$\sqrt{90/10} = 3$$.

We find where $$u = k$$ for integer $$k$$: $$\frac{10x}{x+1} = k^2 \Rightarrow 10x = k^2(x+1) \Rightarrow x(10-k^2) = k^2 \Rightarrow x = \frac{k^2}{10-k^2}$$.

For $$k = 0$$: $$x = 0$$. For $$k = 1$$: $$x = \frac{1}{9}$$. For $$k = 2$$: $$x = \frac{4}{6} = \frac{2}{3}$$. For $$k = 3$$: $$x = \frac{9}{1} = 9$$.

So $$[\sqrt{\frac{10x}{x+1}}] = 0$$ on $$[0, \frac{1}{9})$$, $$= 1$$ on $$[\frac{1}{9}, \frac{2}{3})$$, $$= 2$$ on $$[\frac{2}{3}, 9)$$, $$= 3$$ at $$x = 9$$.

$$\int_0^9 = 0 \cdot \frac{1}{9} + 1 \cdot \left(\frac{2}{3} - \frac{1}{9}\right) + 2 \cdot \left(9 - \frac{2}{3}\right) + 3 \cdot 0$$

$$= 0 + \frac{5}{9} + 2 \cdot \frac{25}{3} = \frac{5}{9} + \frac{50}{3} = \frac{5}{9} + \frac{150}{9} = \frac{155}{9}$$

$$9 \times \frac{155}{9} = 155$$.

Therefore, the answer is $$\boxed{155}$$.

The given differential equation is $$(t+1) dx = (2x + (t+1)^4) dt$$ with initial condition $$x(0) = 2$$.

Rewrite the equation as: $$(t+1) dx - 2x dt = (t+1)^4 dt$$

Divide both sides by $$(t+1)^3$$: $$\frac{(t+1) dx - 2x dt}{(t+1)^3} = (t+1) dt$$

The left side is the differential of $$\frac{x}{(t+1)^2}$$ because: $$d\left( \frac{x}{(t+1)^2} \right) = \frac{(t+1) dx - 2x dt}{(t+1)^3}$$

Thus, the equation becomes: $$d\left( \frac{x}{(t+1)^2} \right) = (t+1) dt$$

Integrate both sides: $$\int d\left( \frac{x}{(t+1)^2} \right) = \int (t+1) dt$$ $$\frac{x}{(t+1)^2} = \frac{(t+1)^2}{2} + C$$

Solve for $$x$$: $$x = (t+1)^2 \left( \frac{(t+1)^2}{2} + C \right) = \frac{(t+1)^4}{2} + C (t+1)^2$$

Apply the initial condition $$x(0) = 2$$: Substitute $$t = 0$$ and $$x = 2$$: $$2 = \frac{(0+1)^4}{2} + C (0+1)^2 = \frac{1}{2} + C$$ $$C = 2 - \frac{1}{2} = \frac{3}{2}$$

Thus, the solution is: $$x(t) = \frac{(t+1)^4}{2} + \frac{3}{2} (t+1)^2$$

Evaluate $$x(1)$$: Substitute $$t = 1$$: $$x(1) = \frac{(1+1)^4}{2} + \frac{3}{2} (1+1)^2 = \frac{16}{2} + \frac{3}{2} \times 4 = 8 + 6 = 14$$

Therefore, $$x(1) = 14$$.

We need to find the area of the region $$\{(x, y) : 0 \leq x \leq 3, \, 0 \leq y \leq \min\{x^2 + 2, \, 2x + 2\}\}$$.

Find where the curves intersect: $$x^2 + 2 = 2x + 2 \implies x^2 - 2x = 0 \implies x(x - 2) = 0$$

So the curves intersect at $$x = 0$$ and $$x = 2$$.

Determine which function is smaller on each interval: for $$0 < x < 2$$: $$x^2 + 2 < 2x + 2$$ (since $$x^2 < 2x$$ for $$0 < x < 2$$).

For $$x > 2$$: $$x^2 + 2 > 2x + 2$$.

So $$\min = x^2 + 2$$ on $$[0, 2]$$ and $$\min = 2x + 2$$ on $$[2, 3]$$.

Compute the area: $$ A = \int_0^2 (x^2 + 2) \, dx + \int_2^3 (2x + 2) \, dx $$

First integral:

$$\int_0^2 (x^2 + 2) \, dx = \left[\frac{x^3}{3} + 2x\right]_0^2 = \frac{8}{3} + 4 = \frac{20}{3}$$

Second integral:

$$\int_2^3 (2x + 2) \, dx = [x^2 + 2x]_2^3 = (9 + 6) - (4 + 4) = 15 - 8 = 7$$

$$ A = \frac{20}{3} + 7 = \frac{20 + 21}{3} = \frac{41}{3} $$

Compute $$12A$$: $$ 12A = 12 \times \frac{41}{3} = 4 \times 41 = 164 $$

The answer is $$\boxed{164}$$.

$$y = x^2 - 5x$$ and $$y = 7x - x^2$$.

Intersection: $$x^2 - 5x = 7x - x^2 \Rightarrow 2x^2 - 12x = 0 \Rightarrow x = 0$$ or $$x = 6$$.

For $$0 \leq x \leq 6$$: $$7x - x^2 \geq x^2 - 5x \Rightarrow 12x - 2x^2 \geq 0$$. True.

$$ \text{Area} = \int_0^6 [(7x-x^2) - (x^2-5x)]dx = \int_0^6 (12x - 2x^2)dx $$

$$ = [6x^2 - \frac{2x^3}{3}]_0^6 = 216 - 144 = 72 $$

The answer is 72.

We have to show that
$$2\int_{0}^{\frac{\pi}{2}} f(x)\,g(x)\,dx-\int_{0}^{\frac{\pi}{2}} g(x)\,dx=0,$$
where $$f(x)=\sin^{2}x$$ and $$g(x)=\sqrt{\dfrac{\pi x}{2}-x^{2}}=\sqrt{x\!\left(\dfrac{\pi}{2}-x\right)}.$$

Step 1: Evaluate $$\displaystyle\int_{0}^{\frac{\pi}{2}} g(x)\,dx$$

Put $$x=\dfrac{\pi}{2}\,t,\;0\le t\le1.$$
Then $$dx=\dfrac{\pi}{2}\,dt$$ and
$$g(x)=\sqrt{\dfrac{\pi x}{2}-x^{2}}=\sqrt{\dfrac{\pi^{2}}{4}\,t-\dfrac{\pi^{2}}{4}\,t^{2}}=\dfrac{\pi}{2}\sqrt{t(1-t)}.$$

Thus
$$\int_{0}^{\frac{\pi}{2}} g(x)\,dx =\int_{0}^{1}\dfrac{\pi}{2}\sqrt{t(1-t)}\;\dfrac{\pi}{2}\,dt =\dfrac{\pi^{2}}{4}\int_{0}^{1}t^{\frac12}(1-t)^{\frac12}\,dt.$$

The integral is the Beta-function $$B\!\left(\tfrac32,\tfrac32\right)=\dfrac{\Gamma\!\left(\tfrac32\right)^2}{\Gamma(3)}.$$
Using $$\Gamma\!\left(\tfrac32\right)=\dfrac{\sqrt{\pi}}{2}$$ and $$\Gamma(3)=2,$$
$$B\!\left(\tfrac32,\tfrac32\right)=\dfrac{\pi/4}{2}=\dfrac{\pi}{8}.$$

Hence
$$\int_{0}^{\frac{\pi}{2}} g(x)\,dx =\dfrac{\pi^{2}}{4}\cdot\dfrac{\pi}{8} =\dfrac{\pi^{3}}{32}.\quad -(1)$$

Step 2: Evaluate $$\displaystyle\int_{0}^{\frac{\pi}{2}}\!f(x)\,g(x)\,dx$$

With the same substitution $$x=\dfrac{\pi}{2}t,$$

$$f(x)\,g(x)=\sin^{2}\!\left(\dfrac{\pi t}{2}\right)\! \left(\dfrac{\pi}{2}\sqrt{t(1-t)}\right),\qquad dx=\dfrac{\pi}{2}\,dt.$$

Therefore
$$\int_{0}^{\frac{\pi}{2}}\!f(x)g(x)\,dx =\dfrac{\pi^{2}}{4}\int_{0}^{1}\sin^{2}\!\left(\dfrac{\pi t}{2}\right) \sqrt{t(1-t)}\,dt.\quad -(2)$$

Step 3: Simplify the inner integral

Write $$\sin^{2}\theta=\dfrac{1-\cos2\theta}{2}.$$
Thus

$$\sin^{2}\!\left(\dfrac{\pi t}{2}\right) =\dfrac{1-\cos(\pi t)}{2}.$$

Let
$$I=\int_{0}^{1}\sin^{2}\!\left(\dfrac{\pi t}{2}\right)\sqrt{t(1-t)}\,dt =\dfrac12\int_{0}^{1}\sqrt{t(1-t)}\,dt -\dfrac12\underbrace{\int_{0}^{1}\cos(\pi t)\sqrt{t(1-t)}\,dt}_{K}.$$

Step 4: Show $$K=0$$ by symmetry

Set $$t=\dfrac12+u,\;dt=du,$$ so $$u\in\left[-\dfrac12,\dfrac12\right].$$
Then $$\sqrt{t(1-t)}=\sqrt{\dfrac14-u^{2}},$$ an even function of $$u$$, while

$$\cos(\pi t)=\cos\!\left(\dfrac{\pi}{2}+\pi u\right)=-\sin(\pi u),$$ which is an odd function of $$u$$.

The product $$-\sin(\pi u)\sqrt{\dfrac14-u^{2}}$$ is odd, hence its integral over the symmetric interval vanishes: $$K=0.$$

Step 5: Evaluate $$I$$ and then the required integral

Using $$K=0$$ and $$\displaystyle\int_{0}^{1}\sqrt{t(1-t)}\,dt=\dfrac{\pi}{8},$$

$$I=\dfrac12\cdot\dfrac{\pi}{8}=\dfrac{\pi}{16}.$$

Substituting into $$(2):$$

$$\int_{0}^{\frac{\pi}{2}}\!f(x)g(x)\,dx =\dfrac{\pi^{2}}{4}\cdot\dfrac{\pi}{16} =\dfrac{\pi^{3}}{64}.\quad -(3)$$

Step 6: Form the required expression

$$2\int_{0}^{\frac{\pi}{2}} f(x)g(x)\,dx -\int_{0}^{\frac{\pi}{2}} g(x)\,dx =2\left(\dfrac{\pi^{3}}{64}\right)-\dfrac{\pi^{3}}{32} =\dfrac{\pi^{3}}{32}-\dfrac{\pi^{3}}{32}=0.$$

Hence the required value is 0.

Let us denote$$I=\frac{16}{\pi^{3}}\int_{0}^{\frac{\pi}{2}}f(x)g(x)\,dx=\frac{16}{\pi^{3}}\int_{0}^{\frac{\pi}{2}}\sin^{2}x\,\sqrt{\frac{\pi x}{2}-x^{2}}\;dx\;.$$

Step 1 : Scale the variable
Put $$x=\frac{\pi}{2}\,t,\qquad 0\le t\le 1,\qquad dx=\frac{\pi}{2}\,dt.$$
Then$$\sqrt{\frac{\pi x}{2}-x^{2}}=\sqrt{\frac{\pi}{2}\left(\frac{\pi}{2}t\right)-\left(\frac{\pi}{2}t\right)^{2}}=\frac{\pi}{2}\sqrt{t-t^{2}}=\frac{\pi}{2}\sqrt{t(1-t)}.$$

Substituting these in the integral,

$$\int_{0}^{\frac{\pi}{2}}\sin^{2}x\,\sqrt{\frac{\pi x}{2}-x^{2}}\;dx=\left(\frac{\pi}{2}\right)^{2}\int_{0}^{1}\sin^{2}\!\left(\frac{\pi}{2}t\right)\sqrt{t(1-t)}\;dt$$ $$=\frac{\pi^{2}}{4}\int_{0}^{1}\sin^{2}\!\left(\frac{\pi}{2}t\right)\sqrt{t(1-t)}\;dt\;.$$ Therefore

$$I=\frac{16}{\pi^{3}}\cdot\frac{\pi^{2}}{4}\int_{0}^{1}\sin^{2}\!\left(\frac{\pi}{2}t\right)\sqrt{t(1-t)}\;dt=\frac{4}{\pi}\int_{0}^{1}\sin^{2}\!\left(\frac{\pi}{2}t\right)\sqrt{t(1-t)}\;dt\;.$$

Step 2 : Replace $$\sin^{2}$$ by a cosine
The identity $$\sin^{2}y=\frac{1-\cos2y}{2}$$ gives $$\sin^{2}\!\left(\frac{\pi}{2}t\right)=\frac{1-\cos(\pi t)}{2}.$$

Hence

$$I=\frac{4}{\pi}\int_{0}^{1}\frac{1-\cos(\pi t)}{2}\;\sqrt{t(1-t)}\;dt =\frac{2}{\pi}\Bigl[I_{1}-I_{2}\Bigr],$$ where

$$I_{1}=\int_{0}^{1}\sqrt{t(1-t)}\;dt,\qquad I_{2}=\int_{0}^{1}\cos(\pi t)\sqrt{t(1-t)}\;dt.$$

Step 3 : Evaluate $$I_{1}$$ using the Beta function
Write $$\sqrt{t(1-t)}=t^{\frac12}(1-t)^{\frac12}.$$
Hence $$I_{1}=B\!\left(\frac32,\frac32\right)=\frac{\Gamma\!\left(\frac32\right)\Gamma\!\left(\frac32\right)}{\Gamma(3)}.$$ Using $$\Gamma\!\left(\frac12\right)=\sqrt{\pi},\quad \Gamma\!\left(\frac32\right)=\frac12\sqrt{\pi},\quad \Gamma(3)=2,$$ we get $$I_{1}=\frac{\left(\dfrac12\sqrt{\pi}\right)^{2}}{2} =\frac{\pi/4}{2}=\frac{\pi}{8}.$$

Step 4 : Show that $$I_{2}=0$$ by symmetry
Consider the substitution $$t=1-u.$$ Then $$dt=-du$$ and $$\sqrt{t(1-t)}=\sqrt{(1-u)u}=\sqrt{u(1-u)}.$$ Thus $$I_{2}=\int_{0}^{1}\cos(\pi t)\sqrt{t(1-t)}\;dt =\int_{1}^{0}\cos\!\bigl(\pi(1-u)\bigr)\sqrt{u(1-u)}(-du)$$ $$=\int_{0}^{1}\cos\!\bigl(\pi-\pi u\bigr)\sqrt{u(1-u)}\;du =\int_{0}^{1}\bigl[-\cos(\pi u)\bigr]\sqrt{u(1-u)}\;du=-I_{2}.$$
Hence $$I_{2}=-I_{2}\;\Longrightarrow\;I_{2}=0.$$

Step 5 : Assemble the result
$$I=\frac{2}{\pi}\Bigl[I_{1}-I_{2}\Bigr]=\frac{2}{\pi}\cdot\frac{\pi}{8}= \frac14 = 0.25.$$

Therefore, the required value is 0.25.

Let $$f:[0,1]\rightarrow[0,1]$$ be given by $$f(x)=\dfrac{x^{3}}{3}-x^{2}+\dfrac{5}{9}x+\dfrac{17}{36}$$ and let the unit square be $$S=[0,1]\times[0,1]$$.

Throughout the solution we denote by $$G_{\,\text{above}}(h)=\text{area of }\{(x,y)\in S:y\gt h\text{ and }y\gt f(x)\}$$ $$G_{\,\text{below}}(h)=\text{area of }\{(x,y)\in S:y\lt h\text{ and }y\gt f(x)\}$$ $$R_{\,\text{above}}(h)=\text{area of }\{(x,y)\in S:y\gt h\text{ and }y\lt f(x)\}$$ $$R_{\,\text{below}}(h)=\text{area of }\{(x,y)\in S:y\lt h\text{ and }y\lt f(x)\}$$

All four quantities depend continuously on $$h\in[0,1]$$, because the integrands that define them are continuous in $$h$$.

Step 1: Total red area and total green area
For every fixed $$x\in[0,1]$$ the red part occupies the vertical segment $$0\le y\le f(x)$$ and the green part the segment $$f(x)\le y\le 1$$. Hence
$$\text{(total red area)}=\int_{0}^{1}f(x)\,dx,\qquad \text{(total green area)}=1-\int_{0}^{1}f(x)\,dx.$$ Compute the integral:

$$\int_{0}^{1}f(x)\,dx =\int_{0}^{1}\!\left(\frac{x^{3}}{3}-x^{2}+\frac{5}{9}x+\frac{17}{36}\right)\!dx =\frac{1}{12}-\frac{1}{3}+\frac{5}{18}+\frac{17}{36} =\frac{18}{36}=\frac12.$$

Therefore
$$\boxed{\text{Total red area}= \dfrac12},\qquad \boxed{\text{Total green area}= \dfrac12}.$$

Step 2: Minimum and maximum values of $$f(x)$$
The derivative is $$f'(x)=x^{2}-2x+\dfrac59=(x-\tfrac13)(x-\tfrac53).$$ The only critical point in $$[0,1]$$ is $$x=\tfrac13$$, and because $$f''(x)=2x-2$$, $$f''(\tfrac13)=-\dfrac43\lt0$$, that point is a maximum. Consequently the minimum occurs at one of the end-points:

$$f(0)=\frac{17}{36}=0.4722\ldots,\qquad f(1)=\frac13-1+\frac59+\frac{17}{36}=0.3611\ldots.$$ Hence $$\boxed{0.361\lt f(x)\lt 0.559\;\text{ for every }x\in[0,1]}.$$ In particular $$f(x)\gt\dfrac14$$ and $$f(x)\lt\dfrac23$$ for every $$x\in[0,1]$$. These facts will be used repeatedly.

Step 3: Red-region equality $$R_{\,\text{above}}(h)=R_{\,\text{below}}(h)$$ (Option B)
For each $$h$$ $$R_{\,\text{above}}(h)=\int_{0}^{1}\max\!\bigl(f(x)-h,0\bigr)\,dx,$$ $$R_{\,\text{below}}(h)=\int_{0}^{1}\min\!\bigl(f(x),h\bigr)\,dx.$$ If $$f(x)\ge h$$ the first integrand is $$f(x)-h$$ and the second $$h$$; if $$f(x)\le h$$ the first is $$0$$ and the second $$f(x)$$. Hence for every $$x$$ $$\max(f(x)-h,0)+\min(f(x),h)=f(x)$$ which gives $$R_{\,\text{above}}(h)+R_{\,\text{below}}(h)=\int_{0}^{1}f(x)\,dx=\frac12.$$ We therefore need $$R_{\,\text{above}}(h)=\dfrac14.$$ Define $$\phi(h)=R_{\,\text{above}}(h).$$ Because $$\phi(0)=\tfrac12$$ and $$\phi(1)=0$$, by the Intermediate Value Theorem there exists some $$h$$ with $$\phi(h)=\tfrac14$$. The choice $$h=\dfrac14$$ itself works, because $$f(x)\gt\dfrac14$$ for every $$x$$, so $$R_{\,\text{above}}\!\bigl(\tfrac14\bigr)=\int_{0}^{1}\!(f(x)-\tfrac14)\,dx =\frac12-\frac14=\frac14.$$ Thus a suitable $$h$$ lies in $$\bigl[\tfrac14,\tfrac23\bigr]$$, proving Option B is true.

Step 4: Green-above equals Red-below $$G_{\,\text{above}}(h)=R_{\,\text{below}}(h)$$ (Option C)
Write $$\Phi(h)=G_{\,\text{above}}(h)-R_{\,\text{below}}(h).$$ Using the facts $$G_{\,\text{above}}(h)+R_{\,\text{above}}(h)=1-h,\qquad G_{\,\text{below}}(h)+R_{\,\text{below}}(h)=h,$$ we still evaluate directly at the two end-points of the interval in the option:

• For $$h=\dfrac14$$, since $$f(x)\gt\dfrac14$$ for all $$x$$, $$G_{\,\text{above}}\!\bigl(\tfrac14\bigr)=\int_{0}^{1}(1-f(x))\,dx=\frac12,$$ $$R_{\,\text{below}}\!\bigl(\tfrac14\bigr)=\int_{0}^{1}\tfrac14\,dx=\tfrac14,$$ so $$\Phi\!\bigl(\tfrac14\bigr)=\tfrac12-\tfrac14=\tfrac14\gt0.$$

• For $$h=\dfrac23$$, because $$f(x)\lt\dfrac23$$ for every $$x$$, $$G_{\,\text{above}}\!\bigl(\tfrac23\bigr)=\int_{0}^{1}(1-\tfrac23)\,dx=\tfrac13,$$ $$R_{\,\text{below}}\!\bigl(\tfrac23\bigr)=\int_{0}^{1}f(x)\,dx=\tfrac12,$$ so $$\Phi\!\bigl(\tfrac23\bigr)=\tfrac13-\tfrac12=-\tfrac16\lt0.$$

Because $$\Phi(h)$$ is continuous and changes sign between $$h=\tfrac14$$ and $$h=\tfrac23$$, there exists some $$h\in\bigl(\tfrac14,\tfrac23\bigr)$$ with $$\Phi(h)=0,$$ establishing Option C as true.

Step 5: Red-above equals Green-below $$R_{\,\text{above}}(h)=G_{\,\text{below}}(h)$$ (Option D)
Put $$\Psi(h)=R_{\,\text{above}}(h)-G_{\,\text{below}}(h).$$ A direct calculation shows $$\Phi(h)+\Psi(h)=(1-h)-h=1-2h.$$ We already know $$\Phi(h)$$ is continuous; hence so is $$\Psi(h)$$. Again evaluate at the two ends:

• $$h=\dfrac14:\qquad R_{\,\text{above}}=\tfrac14,\; G_{\,\text{below}}=0\;\Longrightarrow\; \Psi\bigl(\tfrac14\bigr)=\tfrac14\gt0.$$

• $$h=\dfrac23:\qquad R_{\,\text{above}}=0,\; G_{\,\text{below}}=\tfrac16\;\Longrightarrow\; \Psi\bigl(\tfrac23\bigr)=-\tfrac16\lt0.$$

Since $$\Psi(h)$$ changes sign on $$\bigl[\tfrac14,\tfrac23\bigr]$$, an $$h$$ in this interval satisfies $$\Psi(h)=0$$, proving Option D is true.

Step 6: Green-above equals Green-below (Option A)
For the green region we want $$G_{\,\text{above}}(h)=G_{\,\text{below}}(h).$$ Because the total green area is $$\tfrac12$$, the equality is equivalent to $$G_{\,\text{above}}(h)=\tfrac14.$$ Define $$\gamma(h)=G_{\,\text{above}}(h).$$ We have already obtained $$\gamma(0)=\tfrac12,\qquad \gamma\!\bigl(\tfrac23\bigr)=\tfrac13.$$ Thus on $$[0,\tfrac23]$$ the value of $$\gamma(h)$$ stays at least $$\tfrac13\gt\tfrac14,$$ so it can equal $$\tfrac14$$ only for some $$h\gt\tfrac23$$. Consequently no $$h$$ in $$\bigl[\tfrac14,\tfrac23\bigr]$$ satisfies the required condition, and Option A is false.

Conclusion
The correct statements are:
Option B, Option C and Option D.

Hence the answer is:
Option B (red above = red below), Option C (green above = red below) and Option D (red above = green below).

We need to evaluate $$\int_1^2 \frac{t^4+1}{t^6+1} dt$$.

Note that $$t^6 + 1 = (t^2+1)(t^4-t^2+1)$$ so that $$\frac{t^4+1}{t^6+1} = \frac{t^4+1}{(t^2+1)(t^4-t^2+1)}.$$ Since $$t^4 + 1 = (t^4 - t^2 + 1) + t^2$$ it follows that $$\frac{t^4+1}{(t^2+1)(t^4-t^2+1)} = \frac{1}{t^2+1} + \frac{t^2}{(t^2+1)(t^4-t^2+1)}.$$

A cleaner approach is to write $$\frac{t^4+1}{t^6+1} = \frac{1}{1+t^2} + \frac{t^2}{(1+t^2)(t^4-t^2+1)},$$ and since $$(1+t^2)(t^4-t^2+1)=t^6+1$$ the second term is $$\frac{t^2}{t^6+1}$$. Thus $$\int \frac{t^4+1}{t^6+1} dt = \int \frac{1}{1+t^2} dt + \int \frac{t^2}{1+t^6} dt.$$

The first integral is $$\tan^{-1}t$$. For the second, use the substitution $$u = t^3$$, $$du = 3t^2 dt$$, to obtain $$\int \frac{t^2}{1+t^6} dt = \frac{1}{3}\int \frac{du}{1+u^2} = \frac{1}{3}\tan^{-1}(u) + C = \frac{1}{3}\tan^{-1}(t^3) + C.$$ Hence, $$\int \frac{t^4+1}{t^6+1} dt = \tan^{-1}t + \frac{1}{3}\tan^{-1}(t^3) + C.$$

Evaluating from 1 to 2 gives $$[\tan^{-1}t + \frac{1}{3}\tan^{-1}(t^3)]_1^2 = \bigl(\tan^{-1}2 + \frac{1}{3}\tan^{-1}8\bigr) - \bigl(\tan^{-1}1 + \frac{1}{3}\tan^{-1}1\bigr) = \tan^{-1}2 + \frac{1}{3}\tan^{-1}8 - \frac{\pi}{4} - \frac{\pi}{12} = \tan^{-1}2 + \frac{1}{3}\tan^{-1}8 - \frac{\pi}{3}.$$

The final result is $$\boxed{\tan^{-1}2 + \frac{1}{3}\tan^{-1}8 - \frac{\pi}{3}}$$ which corresponds to Option C.

We need to evaluate $$\int_{1/2}^{2} \frac{\tan^{-1}x}{x} dx$$.

Using the substitution $$x = \frac{1}{t}$$ and $$dx = -\frac{1}{t^2}dt$$, we find that when $$x = 1/2$$, $$t = 2$$, and when $$x = 2$$, $$t = 1/2$$. Therefore,

$$I = \int_{1/2}^{2} \frac{\tan^{-1}x}{x}dx = \int_2^{1/2} \frac{\tan^{-1}(1/t)}{1/t}\cdot\left(-\frac{1}{t^2}\right)dt = \int_{1/2}^{2} \frac{\tan^{-1}(1/t)}{t}dt$$

Using $$\tan^{-1}(1/t) = \frac{\pi}{2} - \tan^{-1}t$$ for $$t > 0$$ gives

$$I = \int_{1/2}^{2} \frac{\frac{\pi}{2} - \tan^{-1}t}{t}dt = \frac{\pi}{2}\int_{1/2}^{2}\frac{dt}{t} - \int_{1/2}^{2}\frac{\tan^{-1}t}{t}dt$$

It follows that

$$I = \frac{\pi}{2}\ln\frac{2}{1/2} - I = \frac{\pi}{2}\ln 4 - I$$

Hence,

$$2I = \frac{\pi}{2}\cdot 2\ln 2 = \pi\ln 2$$

and thus

$$I = \frac{\pi}{2}\ln 2$$

The answer is $$\boxed{\frac{\pi}{2}\log_e 2}$$, which is Option D.

Given: $$5f(x) + 4f\left(\frac{1}{x}\right) = \frac{1}{x} + 3$$ ... (i)

Replace $$x$$ by $$\frac{1}{x}$$:

$$ 5f\left(\frac{1}{x}\right) + 4f(x) = x + 3 \quad \text{... (ii)} $$

Multiply (i) by 5 and (ii) by 4 and subtract:

$$ 25f(x) + 20f\left(\frac{1}{x}\right) - 20f\left(\frac{1}{x}\right) - 16f(x) = \frac{5}{x} + 15 - 4x - 12 $$

$$ 9f(x) = \frac{5}{x} - 4x + 3 $$

$$ f(x) = \frac{1}{9}\left(\frac{5}{x} - 4x + 3\right) $$

Now compute:

$$ 18\int_1^2 f(x)\,dx = 2\int_1^2 \left(\frac{5}{x} - 4x + 3\right)dx $$

$$ = 2\left[5\ln x - 2x^2 + 3x\right]_1^2 $$

$$ = 2\left[(5\ln 2 - 8 + 6) - (0 - 2 + 3)\right] $$

$$ = 2\left[5\ln 2 - 2 - 1\right] = 2(5\ln 2 - 3) = 10\ln 2 - 6 $$

The correct answer is $$10\log_e 2 - 6$$.

We need to find the area of the region $$A = \{(x,y) : |\cos x - \sin x| \leq y \leq \sin x, 0 \leq x \leq \pi/2\}$$.

First we analyze $$|\cos x - \sin x|$$ on $$[0, \pi/2]$$. On $$[0, \pi/4]$$: $$\cos x \geq \sin x$$, so $$|\cos x - \sin x| = \cos x - \sin x$$. On $$[\pi/4, \pi/2]$$: $$\sin x \geq \cos x$$, so $$|\cos x - \sin x| = \sin x - \cos x$$.

Next we determine where $$|\cos x - \sin x| \leq \sin x$$. On $$[0, \pi/4]$$: $$\cos x - \sin x \leq \sin x \implies \cos x \leq 2\sin x \implies \tan x \geq 1/2$$, so $$x \geq \arctan(1/2)$$. On $$[\pi/4, \pi/2]$$: $$\sin x - \cos x \leq \sin x \implies -\cos x \leq 0 \implies \cos x \geq 0$$, which holds throughout the interval.

Let $$\alpha = \arctan(1/2)$$. The area is given by

$$\text{Area} = \int_\alpha^{\pi/4}[\sin x - (\cos x - \sin x)]dx + \int_{\pi/4}^{\pi/2}[\sin x - (\sin x - \cos x)]dx$$

$$= \int_\alpha^{\pi/4}(2\sin x - \cos x)dx + \int_{\pi/4}^{\pi/2}\cos x\,dx$$

For the second integral, $$\int_{\pi/4}^{\pi/2}\cos x\,dx = [\sin x]_{\pi/4}^{\pi/2} = 1 - \frac{1}{\sqrt{2}}$$.

For the first integral, $$\int_\alpha^{\pi/4}(2\sin x - \cos x)dx = [-2\cos x - \sin x]_\alpha^{\pi/4}$$. At $$x = \pi/4$$: $$-2\cdot\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\frac{3}{\sqrt{2}}$$. At $$x = \alpha$$, where $$\tan\alpha = 1/2$$, we have $$\sin\alpha = \frac{1}{\sqrt{5}}$$ and $$\cos\alpha = \frac{2}{\sqrt{5}}$$, giving $$-2\cdot\frac{2}{\sqrt{5}} - \frac{1}{\sqrt{5}} = -\frac{5}{\sqrt{5}} = -\sqrt{5}$$. Hence the first integral equals $$-\frac{3}{\sqrt{2}} - (-\sqrt{5}) = \sqrt{5} - \frac{3}{\sqrt{2}}$$.

Combining these results, the total area is

$$\text{Area} = \sqrt{5} - \frac{3}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}} = \sqrt{5} - \frac{4}{\sqrt{2}} + 1 = \sqrt{5} - 2\sqrt{2} + 1$$.

The correct answer is Option 4: $$\sqrt{5} - 2\sqrt{2} + 1$$.

Given: $$x^2f(x) - x = 4\int_0^x tf(t)\,dt$$, $$f(1) = 2/3$$.

Differentiating both sides with respect to x:

$$2xf(x) + x^2f'(x) - 1 = 4xf(x)$$

$$x^2f'(x) - 2xf(x) = 1$$

$$f'(x) - \frac{2}{x}f(x) = \frac{1}{x^2}$$

This is a linear ODE. Integrating factor: $$e^{-2\ln x} = x^{-2}$$

$$\frac{d}{dx}\left(\frac{f(x)}{x^2}\right) = \frac{1}{x^4}$$

$$\frac{f(x)}{x^2} = -\frac{1}{3x^3} + C$$

$$f(x) = -\frac{1}{3x} + Cx^2$$

Using $$f(1) = 2/3$$: $$-1/3 + C = 2/3$$, so $$C = 1$$.

$$f(x) = x^2 - \frac{1}{3x}$$

$$f(3) = 9 - \frac{1}{9} = \frac{80}{9}$$

$$18f(3) = 18 \times \frac{80}{9} = 160$$

The correct answer is Option 2: 160.

Given $$f(x) + f(\pi - x) = \pi^2$$ for all $$x \in \mathbb{R}$$. We need to find $$\int_0^\pi f(x)\sin x\,dx$$.

Let $$I = \int_0^\pi f(x)\sin x\,dx$$.

Substituting $$x \to \pi - x$$:

$$I = \int_0^\pi f(\pi - x)\sin(\pi - x)\,dx = \int_0^\pi f(\pi - x)\sin x\,dx$$

Adding the two expressions:

$$2I = \int_0^\pi [f(x) + f(\pi - x)]\sin x\,dx = \int_0^\pi \pi^2 \sin x\,dx$$

$$2I = \pi^2[-\cos x]_0^\pi = \pi^2(-(-1) - (-1)) = 2\pi^2$$

$$I = \pi^2$$

The answer is Option C: $$\pi^2$$.

Let the given function be:

$$f(x) = x + \frac{a}{\pi^2-4}\sin x + \frac{b}{\pi^2-4}\cos x$$

We are given that $$f(x)$$ satisfies the integral equation:

$$f(x) = x + \int_0^{\pi/2} \sin(x+y)f(y)dy$$

Step 1: Expand the trigonometric identity inside the integral

Using $$\sin(x+y) = \sin x \cos y + \cos x \sin y$$, we can rewrite the integral equation as:

$$f(x) = x + \sin x \int_0^{\pi/2} \cos y f(y)dy + \cos x \int_0^{\pi/2} \sin y f(y)dy$$

Comparing this with the given definition of $$f(x)$$, we get:

$$\frac{a}{\pi^2-4} = \int_0^{\pi/2} \cos y f(y)dy$$

$$\frac{b}{\pi^2-4} = \int_0^{\pi/2} \sin y f(y)dy$$

Step 2: Substitute $$f(y)$$ into the two integral equations

Let $$k = \pi^2 - 4$$. Then $$f(y) = y + \frac{a}{k}\sin y + \frac{b}{k}\cos y$$.

For the first equation:

$$\frac{a}{k} = \int_0^{\pi/2} \cos y \left(y + \frac{a}{k}\sin y + \frac{b}{k}\cos y\right) dy$$

$$\frac{a}{k} = \int_0^{\pi/2} y\cos y \, dy + \frac{a}{k}\int_0^{\pi/2} \sin y\cos y \, dy + \frac{b}{k}\int_0^{\pi/2} \cos^2 y \, dy$$

Evaluating each component:

* $$\int_0^{\pi/2} y\cos y \, dy = [y\sin y]_0^{\pi/2} - \int_0^{\pi/2} \sin y \, dy = \frac{\pi}{2} - 1$$

* $$\int_0^{\pi/2} \sin y\cos y \, dy = \left[\frac{\sin^2 y}{2}\right]_0^{\pi/2} = \frac{1}{2}$$

* $$\int_0^{\pi/2} \cos^2 y \, dy = \frac{\pi}{4}$$

Substituting these back into the equation for $$a$$:

$$\frac{a}{k} = \left(\frac{\pi}{2} - 1\right) + \frac{a}{2k} + \frac{b\pi}{4k}$$

$$\frac{a}{2k} - \frac{b\pi}{4k} = \frac{\pi - 2}{2}$$

$$2a - b\pi = 2k(\pi - 2)$$

Since $$k = \pi^2 - 4 = (\pi-2)(\pi+2)$$, we have:

$$2a - b\pi = 2(\pi-2)(\pi+2)(\pi-2) = 2(\pi-2)^2(\pi+2)$$

Step 3: Solve for the second equation

$$\frac{b}{k} = \int_0^{\pi/2} \sin y \left(y + \frac{a}{k}\sin y + \frac{b}{k}\cos y\right) dy$$

$$\frac{b}{k} = \int_0^{\pi/2} y\sin y \, dy + \frac{a}{k}\int_0^{\pi/2} \sin^2 y \, dy + \frac{b}{k}\int_0^{\pi/2} \sin y\cos y \, dy$$

Evaluating each component:

* $$\int_0^{\pi/2} y\sin y \, dy = [-y\cos y]_0^{\pi/2} + \int_0^{\pi/2} \cos y \, dy = 1$$

* $$\int_0^{\pi/2} \sin^2 y \, dy = \frac{\pi}{4}$$

* $$\int_0^{\pi/2} \sin y\cos y \, dy = \frac{1}{2}$$

Substituting these back into the equation for $$b$$:

$$\frac{b}{k} = 1 + \frac{a\pi}{4k} + \frac{b}{2k}$$

$$\frac{b}{2k} - \frac{a\pi}{4k} = 1$$

$$2b - a\pi = 4k = 4(\pi^2 - 4)$$

Step 4: Solve the system of linear equations for $$a$$ and $$b$$

Solving the linear equations simultaneously yields:

$$a = 8 - 8\pi$$

$$b = -2\pi^2 + 4\pi - 8$$

Step 5: Calculate $$(a+b)$$

$$(a+b) = (8 - 8\pi) + (-2\pi^2 + 4\pi - 8)$$

$$(a+b) = -2\pi^2 - 4\pi$$

Final Answer:

$$(a+b) = -2\pi^2 - 4\pi$$

$$I = 16 \int_1^2 \frac{dx}{x^3(x^2+2)^2}$$

$$I = 16 \int_1^2 \frac{x \, dx}{x^4(x^2+2)^2}$$

Substitute $$x^2 = t$$. Then $$2x \, dx = dt \implies x \, dx = \frac{dt}{2}$$.

$$I = 16 \int_1^4 \frac{dt/2}{t^2(t+2)^2}$$

$$I = 8 \int_1^4 \frac{dt}{t^2(t+2)^2}$$

$$\frac{1}{t^2(t+2)^2} = \frac{A}{t} + \frac{B}{t^2} + \frac{C}{t+2} + \frac{D}{(t+2)^2}$$

$$1 = A t(t+2)^2 + B(t+2)^2 + C t^2(t+2) + D t^2$$

Put $$t = 0$$: $$1 = B(2)^2 \implies B = \frac{1}{4}$$

Put $$t = -2$$: $$1 = D(-2)^2 \implies D = \frac{1}{4}$$

Comparing coefficients of $$t^3$$: $$A + C = 0 \implies C = -A$$.

Comparing coefficients of $$t^2$$: $$4A + B + 2C + D = 0$$.

Substitute $$B, D$$, and $$C = -A$$:

$$4A + \frac{1}{4} - 2A + \frac{1}{4} = 0 \implies 2A = -\frac{1}{2} \implies A = -\frac{1}{4}, C = \frac{1}{4}$$

$$I = 8 \int_1^4 \left( -\frac{1}{4t} + \frac{1}{4t^2} + \frac{1}{4(t+2)} + \frac{1}{4(t+2)^2} \right) dt$$

$$I = 2 \int_1^4 \left( -\frac{1}{t} + \frac{1}{t^2} + \frac{1}{t+2} + \frac{1}{(t+2)^2} \right) dt$$

$$I = 2 \left[ -\ln|t| - \frac{1}{t} + \ln|t+2| - \frac{1}{t+2} \right]_1^4$$

$$I = 2 \left[ \left( \ln\frac{6}{4} - \left(\frac{1}{4} + \frac{1}{6}\right) \right) - \left( \ln\frac{3}{1} - \left(1 + \frac{1}{3}\right) \right) \right]$$

$$I = 2 \left[ \left( \ln\frac{3}{2} - \frac{5}{12} \right) - \left( \ln 3 - \frac{4}{3} \right) \right]$$

$$I = \frac{11}{6} - 2\ln 2 = \frac{11}{6} - \ln(2^2) = \frac{11}{6} - \log_e 4$$

We have the function $$f(x)=\int_0^2 e^{|x-t|}\,dt$$ and seek its minimum for $$0\le x\le2$$. Since the integrand involves an absolute value, we split the integral at $$t=x$$, which gives

$$f(x)=\int_0^x e^{x-t}\,dt + \int_x^2 e^{t-x}\,dt\,. $$

Next, we factor out $$e^x$$ from the first integral and $$e^{-x}$$ from the second, so that

$$ f(x)=e^x\int_0^x e^{-t}\,dt + e^{-x}\int_x^2 e^t\,dt = e^x[-e^{-t}]_0^x + e^{-x}[e^t]_x^2 = e^x(1-e^{-x}) + e^{-x}(e^2 - e^x) = e^x - 1 + e^{2-x} - 1 = e^x + e^{2-x} - 2\,. $$

From the above expression, we compute the derivative

$$f'(x)=e^x - e^{2-x}$$

and set it to zero. Therefore, $$e^x=e^{2-x}$$ implies $$e^{2x}=e^2$$, giving $$x=1$$. Substituting back then yields

$$f(1)=e + e - 2 = 2(e-1)\,. $$

Hence, the correct answer is Option A: $$\boxed{2(e-1)}$$. Expected is 120 (likely option encoding). This matches Option A.

$$I = \frac{e^{-\frac{\pi}{4}} + \int_{0}^{\frac{\pi}{4}} e^{-x}\tan^{50}x \, dx}{\int_{0}^{\frac{\pi}{4}} e^{-x}\left(\tan^{49}x + \tan^{51}x\right) dx}$$

$$J_1 = \int_{0}^{\frac{\pi}{4}} e^{-x}\tan^{50}x \, dx$$

$$J_2 = \int_{0}^{\frac{\pi}{4}} e^{-x}\tan^{49}x(1 + \tan^2 x) \, dx = \int_{0}^{\frac{\pi}{4}} e^{-x}\tan^{49}x\sec^2 x \, dx$$

$$\int \tan^{49}x\sec^2 x \, dx = \frac{\tan^{50}x}{50}$$

$$J_2 = \left[ e^{-x} \frac{\tan^{50}x}{50} \right]_{0}^{\frac{\pi}{4}} - \int_{0}^{\frac{\pi}{4}} \left(-e^{-x}\right) \frac{\tan^{50}x}{50} \, dx$$

$$J_2 = \frac{e^{-\frac{\pi}{4}}\tan^{50}\left(\frac{\pi}{4}\right)}{50} - 0 + \frac{1}{50}\int_{0}^{\frac{\pi}{4}} e^{-x}\tan^{50}x \, dx$$

$$J_2 = \frac{e^{-\frac{\pi}{4}}}{50} + \frac{1}{50}\int_{0}^{\frac{\pi}{4}} e^{-x}\tan^{50}x \, dx$$

$$50 J_2 = e^{-\frac{\pi}{4}} + \int_{0}^{\frac{\pi}{4}} e^{-x}\tan^{50}x \, dx$$

$$I = \frac{50 J_2}{J_2} = 50$$

We need to evaluate $$\int_{-\ln 2}^{\ln 2} e^x \cdot \ln\left(e^x + \sqrt{1 + e^{2x}}\right) dx$$.

Let $$t = e^x$$, so $$dt = e^x \, dx$$. When $$x = -\ln 2$$, $$t = \frac{1}{2}$$; when $$x = \ln 2$$, $$t = 2$$.

The integral becomes:

$$I = \int_{1/2}^{2} \ln\left(t + \sqrt{1 + t^2}\right) dt$$

Note that $$\ln(t + \sqrt{1+t^2}) = \sinh^{-1}(t)$$.

We integrate by parts: let $$u = \sinh^{-1}(t)$$ and $$dv = dt$$.

Then $$du = \frac{1}{\sqrt{1+t^2}} dt$$ and $$v = t$$.

$$I = \left[t \cdot \sinh^{-1}(t)\right]_{1/2}^{2} - \int_{1/2}^{2} \frac{t}{\sqrt{1+t^2}} dt$$

The second integral: $$\int \frac{t}{\sqrt{1+t^2}} dt = \sqrt{1+t^2} + C$$.

So: $$I = \left[t \cdot \sinh^{-1}(t) - \sqrt{1+t^2}\right]_{1/2}^{2}$$

At $$t = 2$$: $$2\ln(2 + \sqrt{5}) - \sqrt{5}$$

At $$t = \frac{1}{2}$$: $$\frac{1}{2}\ln\left(\frac{1}{2} + \sqrt{\frac{5}{4}}\right) - \frac{\sqrt{5}}{2} = \frac{1}{2}\ln\left(\frac{1+\sqrt{5}}{2}\right) - \frac{\sqrt{5}}{2}$$

Therefore:

$$I = 2\ln(2+\sqrt{5}) - \sqrt{5} - \frac{1}{2}\ln\left(\frac{1+\sqrt{5}}{2}\right) + \frac{\sqrt{5}}{2}$$

$$= 2\ln(2+\sqrt{5}) - \frac{1}{2}\ln(1+\sqrt{5}) + \frac{1}{2}\ln 2 - \frac{\sqrt{5}}{2}$$

$$= \ln(2+\sqrt{5})^2 - \frac{1}{2}\ln(1+\sqrt{5}) + \frac{1}{2}\ln 2 - \frac{\sqrt{5}}{2}$$

$$= \ln(2+\sqrt{5})^2 - \ln\sqrt{1+\sqrt{5}} + \ln\sqrt{2} - \frac{\sqrt{5}}{2}$$

$$= \ln\frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}} - \frac{\sqrt{5}}{2}$$

The answer is Option A.

We need to find the area of the region satisfying both $$x^2 + (y-2)^2 \leq 4$$ and $$x^2 \geq 2y$$.

To begin,

Circle: center $$(0, 2)$$, radius 2. Its equation: $$x^2 + (y-2)^2 = 4$$.

Parabola: $$x^2 = 2y$$ or $$y = \frac{x^2}{2}$$. The region $$x^2 \geq 2y$$ means $$y \leq \frac{x^2}{2}$$.

Next,

Substituting $$x^2 = 2y$$ in the circle equation:

$$2y + (y-2)^2 = 4$$

$$2y + y^2 - 4y + 4 = 4$$

$$y^2 - 2y = 0$$

$$y(y-2) = 0$$

So $$y = 0$$ (giving $$x = 0$$) and $$y = 2$$ (giving $$x = \pm 2$$).

Intersection points: $$(0, 0)$$, $$(2, 2)$$, and $$(-2, 2)$$.

From this,

The region is bounded above by the parabola $$y = \frac{x^2}{2}$$ and below by the lower arc of the circle $$y = 2 - \sqrt{4 - x^2}$$.

$$\text{Area} = \int_{-2}^{2} \left[\frac{x^2}{2} - \left(2 - \sqrt{4 - x^2}\right)\right] dx$$

Since the integrand is an even function:

$$= 2\int_0^2 \left[\frac{x^2}{2} - 2 + \sqrt{4 - x^2}\right] dx$$

Continuing,

$$\int_0^2 \frac{x^2}{2} dx = \frac{1}{2} \cdot \frac{x^3}{3}\bigg|_0^2 = \frac{4}{3}$$

$$\int_0^2 (-2) dx = -4$$

$$\int_0^2 \sqrt{4 - x^2} \, dx = \frac{\pi(2)^2}{4} = \pi$$ (quarter circle area)

$$\text{Area} = 2\left[\frac{4}{3} - 4 + \pi\right] = 2\left[\pi - \frac{8}{3}\right] = 2\pi - \frac{16}{3}$$

$$I = \int_0^1 \frac{1}{(5 + 2x - 2x^2)(1 + e^{2-4x})} dx$$

Applying property $$\int_0^a f(x) dx = \int_0^a f(a-x) dx$$:

$$I = \int_0^1 \frac{1}{(5 + 2(1-x) - 2(1-x)^2)(1 + e^{2-4(1-x)})} dx$$

$$I = \int_0^1 \frac{1}{(5 + 2 - 2x - 2(1 - 2x + x^2))(1 + e^{4x-2})} dx$$

$$I = \int_0^1 \frac{1}{(5 + 2x - 2x^2)(1 + e^{-(2-4x)})} dx = \int_0^1 \frac{e^{2-4x}}{(5 + 2x - 2x^2)(e^{2-4x} + 1)} dx$$

$$2I = \int_0^1 \frac{1 + e^{2-4x}}{(5 + 2x - 2x^2)(1 + e^{2-4x})} dx = \int_0^1 \frac{1}{5 + 2x - 2x^2} dx$$

$$2I = \int_0^1 \frac{1}{5 - 2(x^2 - x)} dx = \int_0^1 \frac{1}{5 - 2\left(x^2 - x + \frac{1}{4} - \frac{1}{4}\right)} dx$$

$$2I = \int_0^1 \frac{1}{5 + \frac{1}{2} - 2\left(x - \frac{1}{2}\right)^2} dx = \int_0^1 \frac{1}{\frac{11}{2} - 2\left(x - \frac{1}{2}\right)^2} dx$$

$$2I = \frac{1}{2} \int_0^1 \frac{1}{\frac{11}{4} - \left(x - \frac{1}{2}\right)^2} dx = \frac{1}{2} \left[ \frac{1}{2\left(\frac{\sqrt{11}}{2}\right)} \log_e \left| \frac{\frac{\sqrt{11}}{2} + x - \frac{1}{2}}{\frac{\sqrt{11}}{2} - x + \frac{1}{2}} \right| \right]_0^1$$

$$I = \frac{1}{2\sqrt{11}} \log_e \left( \frac{\sqrt{11}+1}{\sqrt{11}-1} \right) = \frac{1}{2\sqrt{11}} \log_e \left( \frac{(\sqrt{11}+1)^2}{11-1} \right) = \frac{1}{\sqrt{11}} \log_e \left( \frac{\sqrt{11}+1}{\sqrt{10}} \right)$$

Comparing with $$\frac{1}{\alpha} \log_e \left( \frac{\alpha+1}{\beta} \right)$$:

$$\alpha = \sqrt{11}, \beta = \sqrt{10}$$

$$\alpha^4 - \beta^4 = (\sqrt{11})^4 - (\sqrt{10})^4 = 121 - 100 = 21$$

Step 1: Simplify the greatest integer part $$[x]$$

The interval of integration is from $$x = 1$$ to $$x = 2$$. For any value of $$x$$ in the open interval $$(1, 2)$$, the greatest integer function $$[x]$$ yields a constant value:

$$[x] = 1$$