JEE Areas Questions

Question 1 JEE Mains

What is the area bounded by $$y=x^3-7x^2+14x-8$$ and the x-axis in the interval $$[0,2]$$?

$$2$$

$$5/2$$

$$3$$

$$7/2$$

Solution

Factor the polynomial to find where the curve meets the x axis:

$$y = x^3 - 7x^2 + 14x - 8$$

$$y = (x - 1)(x - 2)(x - 4)$$

The required interval goes from zero to two.

The roots of the polynomial inside this range are at $$x = 1$$ and $$x = 2$$.

The curve changes sign at $$x = 1$$:

From zero to one the function is negative so we take the negative of the integral
From one to two the function is positive so we take the standard integral

Set up the area calculation:

$$\text{Area} = -\int_0^1 (x^3 - 7x^2 + 14x - 8) \, dx + \int_1^2 (x^3 - 7x^2 + 14x - 8) \, dx$$

Find the antiderivative function:

$$F(x) = \frac{x^4}{4} - \frac{7x^3}{3} + 7x^2 - 8x$$

Evaluate the first integral from zero to one:

$$-\left( \frac{1}{4} - \frac{7}{3} + 7 - 8 \right) = -\left( \frac{3 - 28 - 12}{12} \right) = \frac{37}{12}$$

Evaluate the second integral from one to two:

$$\left( 4 - \frac{56}{3} + 28 - 16 \right) - \left( \frac{1}{4} - \frac{7}{3} + 7 - 8 \right) = \left( 16 - \frac{56}{3} \right) - \left( -\frac{37}{12} \right)$$

$$= -\frac{8}{3} + \frac{37}{12} = \frac{-32 + 37}{12} = \frac{5}{12}$$

Sum both portions to get the total area:

$$\text{Total Area} = \frac{37}{12} + \frac{5}{12} = \frac{42}{12} = \frac{7}{2}$$

The correct option is D.

Question 2 JEE Mains

Let $$A_{1}$$ be the bounded area enclosed by the curves $$y=x^{2}+2,x+y=8$$ and y-axis that lies in the first quadrant. Let $$A_{2}$$ be the bounded area enclosed by the curves $$y=x^{2}+2,y^{2}=x,x=2$$ and y-axis that lies in the first quadrant. Then $$A_{1}-A_{2}$$ is equal to

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

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

$$\frac{2}{3}(4\sqrt{2}+1)$$

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

Solution

Given curves for $$A_1$$:

$$y = x^2 + 2, \quad x + y = 8 \implies y = 8 - x, \quad x = 0 \quad (\text{y-axis})$$

Finding the point of intersection of $$y = x^2 + 2$$ and $$y = 8 - x$$:

$$x^2 + 2 = 8 - x \implies x^2 + x - 6 = 0 \implies (x+3)(x-2) = 0 \implies x = 2 \quad (\text{since } x \ge 0)$$

$$A_1 = \int_{0}^{2} \left[ (8 - x) - (x^2 + 2) \right] dx = \int_{0}^{2} (6 - x - x^2) dx$$

$$A_1 = \left[ 6x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{2} = 12 - 2 - \frac{8}{3} = 10 - \frac{8}{3} = \frac{22}{3}$$

Given curves for $$A_2$$:

$$y = x^2 + 2, \quad y^2 = x \implies y = \sqrt{x}, \quad x = 2, \quad x = 0$$

$$A_2 = \int_{0}^{2} \left[ (x^2 + 2) - \sqrt{x} \right] dx$$

$$A_2 = \left[ \frac{x^3}{3} + 2x - \frac{2}{3}x^{3/2} \right]_{0}^{2} = \frac{8}{3} + 4 - \frac{2}{3}(2\sqrt{2}) = \frac{20}{3} - \frac{4\sqrt{2}}{3}$$

$$A_1 - A_2 = \frac{22}{3} - \left( \frac{20}{3} - \frac{4\sqrt{2}}{3} \right) = \frac{2}{3} + \frac{4\sqrt{2}}{3} = \frac{2}{3}(2\sqrt{2} + 1)$$

Question 1 JEE Mains

Let the area of the region $$\{(x, y) : x - 2y + 4 \geq 0, \; x + 2y^2 \geq 0, \; x + 4y^2 \leq 8, \; y \geq 0\}$$ be $$\frac{m}{n}$$, where $$m$$ and $$n$$ are coprime numbers. Then $$m + n$$ is equal to _______.

Solution

We need the area of the region defined by:

1) $$x - 2y + 4 \geq 0$$ (i.e., $$x \geq 2y - 4$$)

2) $$x + 2y^2 \geq 0$$ (i.e., $$x \geq -2y^2$$)

3) $$x + 4y^2 \leq 8$$ (i.e., $$x \leq 8 - 4y^2$$)

4) $$y \geq 0$$

Let us integrate with respect to $$y$$. From condition 3: $$x \leq 8 - 4y^2$$, so $$y \leq \sqrt{2}$$ (for $$x \geq 0$$, max).

The right boundary is $$x = 8 - 4y^2$$.

The left boundary is $$x = \max(2y - 4, -2y^2)$$.

For $$y \geq 0$$: When is $$2y - 4 \geq -2y^2$$?

$$2y^2 + 2y - 4 \geq 0$$, i.e., $$y^2 + y - 2 \geq 0$$, i.e., $$(y+2)(y-1) \geq 0$$. Since $$y \geq 0$$: $$y \geq 1$$.

So left boundary = $$-2y^2$$ for $$0 \leq y \leq 1$$ and $$2y - 4$$ for $$y \geq 1$$.

Upper limit from right boundary: $$8 - 4y^2 \geq 2y - 4$$ gives $$4y^2 + 2y - 12 \leq 0$$, $$2y^2 + y - 6 \leq 0$$, $$(2y - 3)(y + 2) \leq 0$$, so $$y \leq 3/2$$.

Also $$8 - 4y^2 \geq -2y^2$$ gives $$2y^2 \leq 8$$, $$y \leq 2$$.

Area = $$\int_0^1 (8 - 4y^2 - (-2y^2)) dy + \int_1^{3/2} (8 - 4y^2 - (2y - 4)) dy$$

$$= \int_0^1 (8 - 2y^2) dy + \int_1^{3/2} (12 - 4y^2 - 2y) dy$$

First integral: $$\left[8y - \frac{2y^3}{3}\right]_0^1 = 8 - \frac{2}{3} = \frac{22}{3}$$

Second integral: $$\left[12y - \frac{4y^3}{3} - y^2\right]_1^{3/2}$$

At $$y = 3/2$$: $$18 - \frac{4 \cdot 27/8}{3} - \frac{9}{4} = 18 - \frac{9}{2} - \frac{9}{4} = 18 - \frac{27}{4} = \frac{45}{4}$$

At $$y = 1$$: $$12 - \frac{4}{3} - 1 = 11 - \frac{4}{3} = \frac{29}{3}$$

Second integral = $$\frac{45}{4} - \frac{29}{3} = \frac{135 - 116}{12} = \frac{19}{12}$$

Total area = $$\frac{22}{3} + \frac{19}{12} = \frac{88 + 19}{12} = \frac{107}{12}$$

$$m + n = 107 + 12 = 119$$.

The answer is $$\boxed{119}$$.

JEE Areas Questions

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