Chapter 8: Application of Integrals

The circle $$x^2 + y^2 = a^2$$ has its centre at the origin and radius $$a$$. It is symmetrical about both the $$x$$-axis and the $$y$$-axis, so the total enclosed area is four times the area of the portion lying in the first quadrant.

In the first quadrant the boundary of the circle is the upper curve $$y = \sqrt{a^2 - x^2}$$, and $$x$$ runs from $$0$$ to $$a$$. Hence

$$\text{Area} = 4\int_{0}^{a} y \, dx = 4\int_{0}^{a} \sqrt{a^2 - x^2} \, dx.$$

Using the standard result $$\int \sqrt{a^2 - x^2}\, dx = \dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + C$$,

$$\text{Area} = 4\left[\dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}\right]_{0}^{a}.$$

At $$x = a$$: $$\dfrac{a}{2}\sqrt{a^2 - a^2} + \dfrac{a^2}{2}\sin^{-1}(1) = 0 + \dfrac{a^2}{2}\cdot\dfrac{\pi}{2} = \dfrac{\pi a^2}{4}.$$ At $$x = 0$$ the expression is $$0$$.

$$\text{Area} = 4\left(\dfrac{\pi a^2}{4} - 0\right) = \pi a^2.$$

$$\text{Area enclosed by the circle} = \pi a^2 \text{ square units}$$

The ellipse $$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$ is symmetrical about both axes, so the enclosed area equals four times the area lying in the first quadrant.

Solving for $$y$$ in the first quadrant (where $$y \ge 0$$): $$\dfrac{y^2}{b^2} = 1 - \dfrac{x^2}{a^2}$$, so $$y = \dfrac{b}{a}\sqrt{a^2 - x^2}$$. Here $$x$$ varies from $$0$$ to $$a$$.

$$\text{Area} = 4\int_{0}^{a} y\, dx = 4\int_{0}^{a}\dfrac{b}{a}\sqrt{a^2 - x^2}\, dx = \dfrac{4b}{a}\int_{0}^{a}\sqrt{a^2 - x^2}\, dx.$$

Using $$\int \sqrt{a^2 - x^2}\, dx = \dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + C$$,

$$\text{Area} = \dfrac{4b}{a}\left[\dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}\right]_{0}^{a}.$$

At $$x = a$$ the bracket equals $$0 + \dfrac{a^2}{2}\cdot\dfrac{\pi}{2} = \dfrac{\pi a^2}{4}$$; at $$x = 0$$ it equals $$0$$.

$$\text{Area} = \dfrac{4b}{a}\cdot\dfrac{\pi a^2}{4} = \pi a b.$$

$$\text{Area enclosed by the ellipse} = \pi a b \text{ square units}$$

Comparing $$\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$$ with the standard form $$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$ gives $$a^2 = 16$$ and $$b^2 = 9$$, so $$a = 4$$ and $$b = 3$$.

The ellipse is symmetrical about both axes, so its area is four times the area in the first quadrant. There the curve is $$y = \dfrac{3}{4}\sqrt{16 - x^2}$$ with $$x$$ running from $$0$$ to $$4$$.

$$\text{Area} = 4\int_{0}^{4}\dfrac{3}{4}\sqrt{16 - x^2}\, dx = 3\int_{0}^{4}\sqrt{16 - x^2}\, dx.$$

Using $$\int \sqrt{a^2 - x^2}\, dx = \dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + C$$ with $$a = 4$$,

$$\text{Area} = 3\left[\dfrac{x}{2}\sqrt{16 - x^2} + \dfrac{16}{2}\sin^{-1}\dfrac{x}{4}\right]_{0}^{4}.$$

At $$x = 4$$: $$\dfrac{4}{2}\sqrt{16 - 16} + 8\sin^{-1}(1) = 0 + 8\cdot\dfrac{\pi}{2} = 4\pi.$$ At $$x = 0$$ the bracket is $$0$$.

$$\text{Area} = 3(4\pi - 0) = 12\pi.$$

Comparing $$\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$$ with the standard form $$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$ gives $$a^2 = 4$$ and $$b^2 = 9$$, so $$a = 2$$ and $$b = 3$$.

By symmetry about both axes, the area is four times the area in the first quadrant. There the curve is $$y = \dfrac{3}{2}\sqrt{4 - x^2}$$ with $$x$$ running from $$0$$ to $$2$$.

$$\text{Area} = 4\int_{0}^{2}\dfrac{3}{2}\sqrt{4 - x^2}\, dx = 6\int_{0}^{2}\sqrt{4 - x^2}\, dx.$$

Using $$\int \sqrt{a^2 - x^2}\, dx = \dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + C$$ with $$a = 2$$,

$$\text{Area} = 6\left[\dfrac{x}{2}\sqrt{4 - x^2} + \dfrac{4}{2}\sin^{-1}\dfrac{x}{2}\right]_{0}^{2}.$$

At $$x = 2$$: $$\dfrac{2}{2}\sqrt{4 - 4} + 2\sin^{-1}(1) = 0 + 2\cdot\dfrac{\pi}{2} = \pi.$$ At $$x = 0$$ the bracket is $$0$$.

$$\text{Area} = 6(\pi - 0) = 6\pi.$$

The required region lies in the first quadrant, bounded by the circle $$x^2 + y^2 = 4$$ (centre origin, radius $$2$$), the line $$x = 0$$ (the $$y$$-axis) and the line $$x = 2$$.

In the first quadrant the circle gives $$y = \sqrt{4 - x^2}$$, and $$x$$ runs from $$0$$ to $$2$$.

$$\text{Area} = \int_{0}^{2}\sqrt{4 - x^2}\, dx = \left[\dfrac{x}{2}\sqrt{4 - x^2} + \dfrac{4}{2}\sin^{-1}\dfrac{x}{2}\right]_{0}^{2}.$$

At $$x = 2$$: $$\dfrac{2}{2}\sqrt{4 - 4} + 2\sin^{-1}(1) = 0 + 2\cdot\dfrac{\pi}{2} = \pi.$$ At $$x = 0$$ the bracket is $$0$$.

$$\text{Area} = \pi - 0 = \pi.$$

Hence the correct option is (A).

Step 1: Identify the region
The parabola $$y^2 = 4x$$ opens to the right with vertex at the origin. The y-axis is $$x = 0$$ and the horizontal line is $$y = 3$$. Hence the required region is enclosed between $$x = 0$$ and $$x = \frac{y^2}{4}$$ for $$0 \le y \le 3$$.

Step 2: Width of a horizontal strip
For a fixed ordinate $$y$$, the left boundary is the y-axis (so $$x = 0$$) and the right boundary is the parabola $$x = \frac{y^2}{4}$$. Thus the strip’s width is $$\frac{y^2}{4}$$.

Step 3: Area integral
Taking an element of height $$dy$$, the differential area is $$dA = \left(\frac{y^2}{4}\right)dy$$. Therefore
$$\text{Area} = \int_{0}^{3} \frac{y^2}{4}\,dy$$

Step 4: Evaluate the integral
$$\text{Area} = \frac{1}{4}\int_{0}^{3} y^2\,dy = \frac{1}{4}\Bigl[\frac{y^3}{3}\Bigr]_{0}^{3} = \frac{1}{4}\left(\frac{27}{3}\right) = \frac{1}{4}\times9 = \frac{9}{4}.$$

Step 5: Choose the correct option
The required area is $$\dfrac{9}{4}$$ square units. Hence, option (B) is correct.

Step 1 : Points of intersection with the coordinate axes

• With the x-axis (put $$y = 0$$): $$3x + 2 = 0 \;\Longrightarrow\; x = -\dfrac{2}{3}$$. Hence the line cuts the x-axis at $$\bigl(-\dfrac{2}{3}, 0\bigr)$$.
• With the given ordinates:
  • At $$x = -1$$: $$y = 3(-1) + 2 = -1<0$$ (point lies below the x-axis).
  • At $$x = 1$$: $$y = 3(1) + 2 = 5>0$$ (point lies above the x-axis).

Thus, between $$x = -1$$ and $$x = 1$$ the line crosses the x-axis at $$x = -\dfrac{2}{3}$$. The bounded region therefore has two parts, one below and one above the x-axis.

Step 2 : Writing the required area as an integral

The area $$A$$ is obtained by integrating the absolute value of the ordinate:

$$A = \int_{-1}^{-2/3} |3x+2|\,dx + \int_{-2/3}^{1} |3x+2|\,dx$$

Because $$3x+2<0$$ on $$[-1,-\tfrac{2}{3}]$$ and $$3x+2>0$$ on $$[-\tfrac{2}{3},1]$$, we rewrite

$$A = \int_{-1}^{-2/3} \bigl(-(3x+2)\bigr)\,dx + \int_{-2/3}^{1} (3x+2)\,dx.$$

Step 3 : Evaluating the integrals

First part:

\(\displaystyle \int \!\bigl(-3x-2\bigr)\,dx = -\dfrac{3x^{2}}{2}-2x\).

Evaluate from $$x=-1$$ to $$x=-\dfrac{2}{3}$$:

\(\displaystyle\Bigl[-\dfrac{3x^{2}}{2}-2x\Bigr]_{-1}^{-2/3}=\Bigl(-\dfrac{3(\tfrac{4}{9})}{2}+\dfrac{4}{3}\Bigr)-\Bigl(-\dfrac{3(1)}{2}+2\Bigr)=\dfrac{2}{3}-\dfrac{1}{2}=\dfrac{1}{6}.\)

Second part:

\(\displaystyle \int (3x+2)\,dx = \dfrac{3x^{2}}{2}+2x\).

Evaluate from $$x=-\dfrac{2}{3}$$ to $$x=1$$:

\(\displaystyle\Bigl[\dfrac{3x^{2}}{2}+2x\Bigr]_{-2/3}^{1}=\Bigl(\dfrac{3(1)^{2}}{2}+2\Bigr)-\Bigl(\dfrac{3(\tfrac{4}{9})}{2}-\dfrac{4}{3}\Bigr)=\dfrac{7}{2}+\dfrac{2}{3}=\dfrac{25}{6}.\)

Step 4 : Total area

$$A = \dfrac{1}{6} + \dfrac{25}{6} = \dfrac{26}{6} = \dfrac{13}{3}.$$

Hence the required area is $$\dfrac{13}{3}$$ square units.

Objective: Find the area enclosed between the curve $$y = \cos x$$ and the $$x$$-axis from $$x = 0$$ to $$x = 2\pi$$.

1. Locating the points where the curve meets the $$x$$-axis

$$\cos x = 0 \;\Longrightarrow\; x = \dfrac{\pi}{2},\; \dfrac{3\pi}{2}$$ (within the interval $$0 \le x \le 2\pi$$).

Thus the interval $$[0,2\pi]$$ is split into three sub-intervals:

• $$0 \le x \le \dfrac{\pi}{2}$$, where $$\cos x \ge 0$$.
• $$\dfrac{\pi}{2} \le x \le \dfrac{3\pi}{2}$$, where $$\cos x \le 0$$.
• $$\dfrac{3\pi}{2} \le x \le 2\pi$$, where $$\cos x \ge 0$$.

2. Setting up the integral for the (non-negative) area

Where $$\cos x$$ is negative we integrate its absolute value (i.e. change the sign):

$$A = \int_{0}^{\pi/2} \cos x\,dx \; + \; \int_{\pi/2}^{3\pi/2} (-\cos x)\,dx \; + \; \int_{3\pi/2}^{2\pi} \cos x\,dx.$$

3. Evaluating each part

• First part: $$\int_{0}^{\pi/2} \cos x\,dx = \big[\sin x\big]_{0}^{\pi/2} = 1.$$
• Second part: $$\int_{\pi/2}^{3\pi/2} (-\cos x)\,dx = \big[-\sin x\big]_{\pi/2}^{3\pi/2} = -\sin(3\pi/2) - \big(-\sin(\pi/2)\big) = 1 - (-1) = 2.$$
• Third part: $$\int_{3\pi/2}^{2\pi} \cos x\,dx = \big[\sin x\big]_{3\pi/2}^{2\pi} = 0 - (-1) = 1.$$

4. Total area

$$A = 1 + 2 + 1 = 4.$$

Result: The required area is $$4$$ square units.

Step 1: Understand the function

The given function is $$y = |x + 3|$$. An absolute–value expression changes its sign when the quantity inside the bars becomes zero. Solve $$x + 3 = 0$$ to find the critical point:

$$x = -3$$.

Thus the definition in piece-wise form is

$$|x + 3| = \begin{cases}-(x + 3), & x \lt -3,\\ x + 3, & x \ge -3.\end{cases}$$

Step 2: Sketch the graph

• Plot the vertex at the point $$(-3,0)$$, because $$|x+3|=0$$ there.
• For $$x \ge -3$$ the formula is $$y = x + 3$$, a straight line of slope $$+1$$ passing through $$(-3,0)$$.
• For $$x \lt -3$$ the formula is $$y = -(x + 3) = -x - 3$$, a straight line of slope $$-1$$ passing through $$(-3,0)$$.
• Combine the two rays to obtain the familiar V-shape of an absolute-value graph.

Diagram to draw: On a pair of perpendicular axes mark the vertex at $$(-3,0)$$. Draw one ray to the right of the vertex with slope $$+1$$ (the line $$y = x + 3$$) and another ray to the left of the vertex with slope $$-1$$ (the line $$y = -x - 3$$). Label the important points $$(-6,3)$$, $$(-3,0)$$ and $$(0,3)$$, and shade the region under the curve between $$x=-6$$ and $$x=0$$.

Step 3: Set up the integral

The sign change at $$x=-3$$ breaks the interval $$[-6,0]$$ into two sub-intervals:

$$\int_{-6}^{0}|x+3|\,dx = \int_{-6}^{-3}|x+3|\,dx + \int_{-3}^{0}|x+3|\,dx.$$

Using the piece-wise definition, rewrite each part with an ordinary algebraic expression.

First part ($$-6 \le x \lt -3$$): $$|x+3| = -(x+3) = -x-3$$.

$$\int_{-6}^{-3}(-x-3)\,dx.$$

Second part ($$-3 \le x \le 0$$): $$|x+3| = x+3$$.

$$\int_{-3}^{0}(x+3)\,dx.$$

Step 4: Evaluate each integral

1. Integrate the first part:

$$\int(-x-3)\,dx = -\frac{x^{2}}{2}-3x.$$

Evaluate from $$x=-6$$ to $$x=-3$$:

At $$x=-3$$: $$-\dfrac{(-3)^{2}}{2}-3(-3) = -\dfrac{9}{2}+9 = \dfrac{9}{2}.$$

At $$x=-6$$: $$-\dfrac{(-6)^{2}}{2}-3(-6) = -\dfrac{36}{2}+18 = -18+18 = 0.$$

Hence $$\int_{-6}^{-3}(-x-3)\,dx = \dfrac{9}{2}-0 = \dfrac{9}{2}.$$

2. Integrate the second part:

$$\int (x+3)\,dx = \frac{x^{2}}{2}+3x.$$

Evaluate from $$x=-3$$ to $$x=0$$:

At $$x=0$$: $$0+0=0.$$

At $$x=-3$$: $$\dfrac{(-3)^{2}}{2}+3(-3) = \dfrac{9}{2}-9 = -\dfrac{9}{2}.$$

Therefore $$\int_{-3}^{0}(x+3)\,dx = 0 - \left(-\dfrac{9}{2}\right) = \dfrac{9}{2}.$$

Step 5: Add the two results

$$\int_{-6}^{0}|x+3|\,dx = \dfrac{9}{2}+\dfrac{9}{2} = 9.$$

Geometric check – The region under each straight segment is a right-triangle of base $$3$$ and height $$3$$, area $$\tfrac{1}{2}\times 3\times 3 = \tfrac{9}{2}$$. Two such equal triangles give total area $$9$$, confirming the integral.

Conclusion: The required definite integral equals $$9$$.

Step 1: Locate the points where the curve meets the x-axis

For $$y = \sin x$$, we have $$\sin x = 0$$ when $$x = 0,\; \pi,\; 2\pi,\;\ldots$$ Between $$x = 0$$ and $$x = 2\pi$$ the curve meets the x-axis at the three points $$x = 0$$, $$x = \pi$$ and $$x = 2\pi$$.

Step 2: Understand the sign of $$\sin x$$ in each sub-interval

• On $$0 \le x \le \pi$$, $$\sin x \ge 0$$ (curve lies above the x-axis).
• On $$\pi \le x \le 2\pi$$, $$\sin x \le 0$$ (curve lies below the x-axis).

The required area is the sum of the absolute areas in these two parts.

Step 3: Write the definite integrals for the two parts

Required area $$A$$ is given by

$$A = \int_{0}^{\pi} \sin x\,dx \; - \int_{\pi}^{2\pi} \sin x\,dx.$$

The minus sign before the second integral converts the (negative) signed area on $$[\pi,2\pi]$$ into its positive magnitude.

Step 4: Evaluate the first integral

Since $$\int \sin x\,dx = -\cos x$$,

$$\int_{0}^{\pi} \sin x\,dx = \big[-\cos x\big]_{0}^{\pi} = -\cos\pi - (-\cos 0) = -(-1) - (-1) = 1 + 1 = 2.$$

Step 5: Evaluate the second integral

$$\int_{\pi}^{2\pi} \sin x\,dx = \big[-\cos x\big]_{\pi}^{2\pi} = -\cos(2\pi) - (-\cos\pi) = -1 - 1 = -2.$$

The signed value is negative because $$\sin x \le 0$$ on $$[\pi, 2\pi]$$. Its magnitude (the geometrical area) is

$$-\int_{\pi}^{2\pi} \sin x\,dx = -(-2) = 2.$$

Step 6: Add the two magnitudes

$$A = 2 + 2 = 4.$$

Conclusion: The area enclosed by the curve $$y = \sin x$$ and the x-axis between $$x = 0$$ and $$x = 2\pi$$ is $$4$$ square units.

Given the curve $$y = x^3$$, the $$x$$-axis (that is, $$y = 0$$) and the two ordinates $$x = -2$$ and $$x = 1$$. We have to calculate the area enclosed by these lines.

1. Sign of the function on the interval

$$y = x^3$$ crosses the $$x$$-axis at $$x = 0$$.

• On $$-2 \le x \le 0$$, $$x^3 \le 0$$ (curve is on or below the $$x$$-axis).
• On $$0 \le x \le 1$$, $$x^3 \ge 0$$ (curve is on or above the $$x$$-axis).

2. Express the required area

Since area is always positive we integrate separately on the two sub-intervals and take the absolute value where the curve is below the axis:

$$\text{Area}=\left|\int_{-2}^{0} x^{3}\,dx\right|+\int_{0}^{1} x^{3}\,dx.$$

3. Evaluate the integrals

Because $$\int x^{3}\,dx = \dfrac{x^{4}}{4}+C$$, we get

$$\int_{-2}^{0} x^{3}\,dx = \left[\dfrac{x^{4}}{4}\right]_{-2}^{0}=0-\dfrac{16}{4}=-4,$$

so its signed value is $$-4$$ and its area contribution is $$4$$.

Similarly,

$$\int_{0}^{1} x^{3}\,dx = \left[\dfrac{x^{4}}{4}\right]_{0}^{1}=\dfrac{1}{4}-0=\dfrac{1}{4}.$$

4. Add the positive contributions

$$\text{Total area}=4+\dfrac{1}{4}=\dfrac{17}{4}.$$

Hence, the required area is $$\dfrac{17}{4}$$ square units, which corresponds to option (D).

Given curve: $$y = x|x|$$.

Because

$$|x| = \begin{cases}-x, & x \lt 0\\ x, & x \ge 0\end{cases}$$

the curve can be written piece-wise as

$$y = x|x| = \begin{cases}-x^{2}, & x \lt 0\\ x^{2}, & x \ge 0\end{cases}$$

We must find the area enclosed by this curve, the $$x$$-axis and the lines $$x = -1$$, $$x = 1$$.

1. Identify the parts above and below the $$x$$-axis

• For $$-1 \le x \lt 0$$, $$y = -x^{2} \le 0$$ (curve lies on or below the axis).
• For $$0 \le x \le 1$$, $$y = x^{2} \ge 0$$ (curve lies on or above the axis).

When the curve is below the $$x$$-axis we take the absolute value of $$y$$ while integrating, so the required area is

$$\text{Area}= \int_{-1}^{0} \bigl(-y\bigr)\,dx + \int_{0}^{1} y\,dx.$$

2. Write the integrals explicitly

Bottom branch (below axis): $$y = -x^{2}\;\Rightarrow\;-y = x^{2}.$$
Top branch (above axis): $$y = x^{2}.$$
Therefore,

$$\text{Area}= \int_{-1}^{0} x^{2}\,dx + \int_{0}^{1} x^{2}\,dx.$$

3. Evaluate

Both integrals are the same in magnitude, so

$$\text{Area}= 2\int_{0}^{1} x^{2}\,dx = 2\left[\frac{x^{3}}{3}\right]_{0}^{1}=2\left(\frac{1^{3}}{3}-\frac{0^{3}}{3}\right)=2\times\frac{1}{3}=\frac{2}{3}.$$

4. Conclusion

The required bounded area equals $$\dfrac{2}{3}$$. Hence option (C) is correct.