Class 12 Ch 8: Application of Integrals - Miscellaneous Exercise

Q1

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Find the area under the given curves and given lines:
(i) $y = x^2, x = 1, x = 2$ and x-axis
(ii) $y = x^4, x = 1, x = 5$ and x-axis

(i) Area = $\int_1^2 x^2 dx = [\frac{x^3}{3}]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$ sq. units.

(ii) Area = $\int_1^5 x^4 dx = [\frac{x^5}{5}]_1^5 = \frac{3125}{5} - \frac{1}{5} = \frac{3124}{5}$ sq. units.

$\boxed{\text{(i) } \frac{7}{3}, \text{ (ii) } \frac{3124}{5}}$

Q2

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Find the area between the curves $y = x$ and $y = x^2$.

Intersection points: $x^2 = x \Rightarrow x(x-1) = 0 \Rightarrow x = 0, 1$.

Area = $\int_0^1 (x - x^2) dx = [\frac{x^2}{2} - \frac{x^3}{3}]_0^1$.

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

$\boxed{\frac{1}{6} \text{ sq. units}}$

Q3

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Find the area of the region lying in the first quadrant and bounded by $y = 4x^2$, $x = 0$, $y = 1$ and $y = 4$.

From $y = 4x^2$, we get $x = \frac{\sqrt{y}}{2}$ (in 1st quadrant).

Area = $\int_1^4 x dy = \int_1^4 \frac{\sqrt{y}}{2} dy = \frac{1}{2} [\frac{2}{3}y^{3/2}]_1^4$.

$= \frac{1}{3} [4^{3/2} - 1^{3/2}] = \frac{1}{3} [8 - 1] = \frac{7}{3}$.

$\boxed{\frac{7}{3} \text{ sq. units}}$

Q4

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Sketch the graph of $y = |x+3|$ and evaluate $\int_{-6}^0 |x+3| dx$.

$|x+3| = -(x+3)$ for $x < -3$ and $(x+3)$ for $x \ge -3$.

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

$= [-\frac{x^2}{2}-3x]_{-6}^{-3} + [\frac{x^2}{2}+3x]_{-3}^0$.

$= [(\frac{-9}{2}+9) - (-18+18)] + [(0) - (\frac{9}{2}-9)] = 4.5 + 4.5 = 9$.

$\boxed{9}$

Q5

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Find the area bounded by the curve $y = \sin x$ between $x = 0$ and $x = 2\pi$.

Area = $\int_0^{2\pi} |\sin x| dx = \int_0^\pi \sin x dx + \int_\pi^{2\pi} -\sin x dx$.

$= [-\cos x]_0^\pi - [-\cos x]_\pi^{2\pi} = -(-1-1) - (-1 - (-(-1))) = 2 + 2 = 4$.

$\boxed{4 \text{ sq. units}}$

Q6

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Find the area enclosed between the parabola $y^2 = 4ax$ and the line $y = mx$.

Intersection: $(mx)^2 = 4ax \Rightarrow m^2x^2 - 4ax = 0 \Rightarrow x(m^2x - 4a) = 0$.

$x = 0$ and $x = \frac{4a}{m^2}$.

Area = $\int_0^{4a/m^2} (\sqrt{4ax} - mx) dx = [2\sqrt{a}\frac{2}{3}x^{3/2} - \frac{mx^2}{2}]_0^{4a/m^2}$.

$= \frac{4\sqrt{a}}{3}(\frac{4a}{m^2})^{3/2} - \frac{m}{2}(\frac{16a^2}{m^4}) = \frac{32a^2}{3m^3} - \frac{8a^2}{m^3} = \frac{8a^2}{3m^3}$.

$\boxed{\frac{8a^2}{3m^3} \text{ sq. units}}$

Q7

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Find the area enclosed by the parabola $4y = 3x^2$ and the line $2y = 3x + 12$.

Parabola: $y = \frac{3x^2}{4}$. Line: $y = \frac{3x+12}{2}$.

Intersection: $\frac{3x^2}{4} = \frac{3x+12}{2} \Rightarrow 3x^2 = 6x+24 \Rightarrow x^2-2x-8=0$.

$(x-4)(x+2)=0 \Rightarrow x = -2, 4$.

Area = $\int_{-2}^4 (\frac{3x+12}{2} - \frac{3x^2}{4}) dx = [\frac{3x^2}{4} + 6x - \frac{x^3}{4}]_{-2}^4$.

$= (12+24-16) - (3-12+2) = 20 - (-7) = 27$.

$\boxed{27 \text{ sq. units}}$

Q8

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Find the area of the smaller region bounded by the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ and the line $\frac{x}{3} + \frac{y}{2} = 1$.

Area = Area of quadrant of ellipse - Area of triangle formed by line.

Area = $\frac{\pi ab}{4} - \frac{1}{2}ab$. Here $a=3, b=2$.

$= \frac{\pi(3)(2)}{4} - \frac{1}{2}(3)(2) = \frac{3\pi}{2} - 3$.

$\boxed{\frac{3}{2}(\pi-2) \text{ sq. units}}$

Q9

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Find the area of the smaller region bounded by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and the line $\frac{x}{a} + \frac{y}{b} = 1$.

Similar to Q8. Area = $\frac{\pi ab}{4} - \frac{1}{2}ab$.

$\boxed{\frac{ab}{4}(\pi-2) \text{ sq. units}}$

Q10

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Find the area of the region enclosed by the parabola $x^2 = y$, the line $y = x + 2$ and the x-axis.

Intersection: $x^2 = x+2 \Rightarrow x^2-x-2=0 \Rightarrow x=-1, 2$.

Area = $\int_{-1}^2 (x+2 - x^2) dx = [\frac{x^2}{2} + 2x - \frac{x^3}{3}]_{-1}^2$.

$= (2+4-\frac{8}{3}) - (\frac{1}{2}-2+\frac{1}{3}) = \frac{10}{3} - (-\frac{7}{6}) = \frac{27}{6} = \frac{9}{2}$.

$\boxed{\frac{9}{2} \text{ sq. units}}$

Q11

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Using the method of integration find the area bounded by the curve $|x| + |y| = 1$.

The curve is a square with vertices $(\pm 1, 0)$ and $(0, \pm 1)$.

Area = $4 \times \int_0^1 (1-x) dx = 4 [x - \frac{x^2}{2}]_0^1$.

$= 4(1 - \frac{1}{2}) = 2$.

$\boxed{2 \text{ sq. units}}$

Q12

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Find the area bounded by curves $\{(x, y) : y \ge x^2 \text{ and } y = |x|\}$.

Intersection of $y=x^2$ and $y=|x|$: $x^2=|x| \Rightarrow x=0, \pm 1$.

Area = $2 \int_0^1 (x - x^2) dx = 2 [\frac{x^2}{2} - \frac{x^3}{3}]_0^1$.

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

$\boxed{\frac{1}{3} \text{ sq. units}}$

Q13

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Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A(2, 0), B(4, 5) and C(6, 3).

Eq AB: $y = \frac{5}{2}(x-2)$. Eq BC: $y = -x+9$. Eq AC: $y = \frac{3}{4}(x-2)$.

Area = $\int_2^4 \frac{5}{2}(x-2) dx + \int_4^6 (-x+9) dx - \int_2^6 \frac{3}{4}(x-2) dx$.

$= 5 + 8 - 6 = 7$.

$\boxed{7 \text{ sq. units}}$

Q14

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Using the method of integration find the area of the region bounded by lines: $2x + y = 4$, $3x - 2y = 6$ and $x - 3y + 5 = 0$.

Vertices: (1,2), (2,0), (4,3).

Area = $\int_1^4 \frac{x+5}{3} dx - \int_1^2 (4-2x) dx - \int_2^4 \frac{3x-6}{2} dx$.

$= 7.5 - 1 - 3 = 3.5$.

$\boxed{3.5 \text{ sq. units}}$

Q15

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Find the area of the region $\{(x, y) : y^2 \le 4x, 4x^2 + 4y^2 \le 9\}$.

Intersection of $y^2=4x$ and $x^2+y^2=9/4$: $x=1/2$.

Area = $2 [\int_0^{1/2} \sqrt{4x} dx + \int_{1/2}^{3/2} \sqrt{9/4-x^2} dx]$.

$= \frac{\sqrt{2}}{6} + \frac{9\pi}{8} - \frac{9}{4}\sin^{-1}\frac{1}{3}$.

$\boxed{\frac{\sqrt{2}}{6} + \frac{9\pi}{8} - \frac{9}{4}\sin^{-1}\frac{1}{3} \text{ sq. units}}$

Q16

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Area bounded by the curve $y = x^3$, the x-axis and the ordinates $x = -2$ and $x = 1$ is: (A) -9 (B) -15/4 (C) 15/4 (D) 17/4

Area = $|\int_{-2}^0 x^3 dx| + \int_0^1 x^3 dx$.

$= |[\frac{x^4}{4}]_{-2}^0| + [\frac{x^4}{4}]_0^1 = |-4| + \frac{1}{4} = 4.25 = \frac{17}{4}$.

$\boxed{\text{(D) } \frac{17}{4}}$

Q17

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The area bounded by the curve $y = x|x|$, x-axis and the ordinates $x = -1$ and $x = 1$ is given by: (A) 0 (B) 1/3 (C) 2/3 (D) 4/3

Area = $\int_{-1}^1 |x|x|| dx = \int_{-1}^0 x^2 dx + \int_0^1 x^2 dx$.

$= [\frac{x^3}{3}]_{-1}^0 + [\frac{x^3}{3}]_0^1$ (taking absolute area).

$= \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$.

$\boxed{\text{(C) } \frac{2}{3}}$

Q18

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The area of the circle $x^2 + y^2 = 16$ exterior to the parabola $y^2 = 6x$ is: (A) $\frac{4}{3}(4\pi - \sqrt{3})$ (B) $\frac{4}{3}(4\pi + \sqrt{3})$ (C) $\frac{4}{3}(8\pi - \sqrt{3})$ (D) $\frac{4}{3}(8\pi + \sqrt{3})$

Total Area = $16\pi$. Interior Area = $\frac{4}{3}(4\pi + \sqrt{3})$.

Exterior Area = $16\pi - (\frac{16\pi}{3} + \frac{4\sqrt{3}}{3}) = \frac{32\pi}{3} - \frac{4\sqrt{3}}{3}$.

$\boxed{\text{(C) } \frac{4}{3}(8\pi - \sqrt{3})}$

Q19

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The area bounded by the y-axis, $y = \cos x$ and $y = \sin x$ when $0 \le x \le \frac{\pi}{2}$ is: (A) $2(\sqrt{2}-1)$ (B) $\sqrt{2}-1$ (C) $\sqrt{2}+1$ (D) $\sqrt{2}$

Intersection at $x=\pi/4$. Bounded by y-axis means $x=0$ to $x=\pi/4$.

Area = $\int_0^{\pi/4} (\cos x - \sin x) dx = [\sin x + \cos x]_0^{\pi/4}$.

$= (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0+1) = \sqrt{2}-1$.

$\boxed{\text{(B) } \sqrt{2}-1}$

Miscellaneous Exercise

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