Exercise 8.1 Practice
Chapter 8: Application of Integrals – NCERT Solutions
Q1
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Find the area of the region bounded by the curve $y^2 = x$ and the lines $x = 1$, $x = 4$ and the x-axis in the first quadrant.
The area is given by $\int_1^4 y \, dx$.
From $y^2 = x$, we get $y = \sqrt{x}$ for the first quadrant.
Area = $\int_1^4 \sqrt{x} \, dx = \int_1^4 x^{1/2} \, dx = \left[\frac{x^{3/2}}{3/2}\right]_1^4 = \frac{2}{3}[4^{3/2} - 1^{3/2}]$.
$= \frac{2}{3}[8 - 1] = \frac{14}{3}$.
$\boxed{\frac{14}{3} \text{ sq. units}}$
Q2
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Find the area of the region bounded by $y^2 = 9x$, $x = 2$, $x = 4$ and the x-axis in the first quadrant.
Area = $\int_2^4 y \, dx$. Here $y = \sqrt{9x} = 3\sqrt{x}$.
Area = $\int_2^4 3\sqrt{x} \, dx = 3 \left[\frac{2}{3}x^{3/2}\right]_2^4 = 2[4^{3/2} - 2^{3/2}]$.
$= 2[8 - 2\sqrt{2}] = 16 - 4\sqrt{2}$.
$\boxed{(16 - 4\sqrt{2}) \text{ sq. units}}$
Q3
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Find the area of the region bounded by $x^2 = 4y$, $y = 2$, $y = 4$ and the y-axis in the first quadrant.
Area = $\int_2^4 x \, dy$. Here $x = \sqrt{4y} = 2\sqrt{y}$.
Area = $\int_2^4 2\sqrt{y} \, dy = 2 \left[\frac{2}{3}y^{3/2}\right]_2^4 = \frac{4}{3}[4^{3/2} - 2^{3/2}]$.
$= \frac{4}{3}[8 - 2\sqrt{2}] = \frac{32 - 8\sqrt{2}}{3}$.
$\boxed{\frac{32 - 8\sqrt{2}}{3} \text{ sq. units}}$
Q4
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Find the area of the region bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$.
The area of the ellipse is $4 \times$ (Area in first quadrant).
$y = \frac{3}{4}\sqrt{16-x^2}$. Area = $4 \int_0^4 \frac{3}{4}\sqrt{16-x^2} \, dx = 3 \int_0^4 \sqrt{4^2-x^2} \, dx$.
Using the standard integral: $3 \left[\frac{x}{2}\sqrt{16-x^2} + \frac{16}{2}\sin^{-1}\frac{x}{4}\right]_0^4$.
$= 3 [ (0 + 8\sin^{-1}(1)) - (0) ] = 3 \cdot 8 \cdot \frac{\pi}{2} = 12\pi$.
$\boxed{12\pi \text{ sq. units}}$
Q5
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Find the area of the region bounded by the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$.
Area is $4 \times$ (Area in first quadrant). Here $a=2, b=3$.
$y = \frac{3}{2}\sqrt{4-x^2}$. Area = $4 \int_0^2 \frac{3}{2}\sqrt{4-x^2} \, dx = 6 \int_0^2 \sqrt{2^2-x^2} \, dx$.
$= 6 \left[\frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}\frac{x}{2}\right]_0^2 = 6 [ (0 + 2\sin^{-1}(1)) - 0 ] = 6 \cdot 2 \cdot \frac{\pi}{2} = 6\pi$.
$\boxed{6\pi \text{ sq. units}}$
Q6
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Find the area of the region in the first quadrant enclosed by x-axis, line $x = \sqrt{3}y$ and the circle $x^2 + y^2 = 4$.
The line is $y = \frac{x}{\sqrt{3}}$ (i.e., $\tan\theta = 1/\sqrt{3} \Rightarrow \theta = \pi/6$). The circle has radius 2.
Intersection point: $(\sqrt{3}y)^2 + y^2 = 4 \Rightarrow 4y^2=4 \Rightarrow y=1, x=\sqrt{3}$.
Area = $\int_0^{\sqrt{3}} \frac{x}{\sqrt{3}} \, dx + \int_{\sqrt{3}}^2 \sqrt{4-x^2} \, dx$.
First integral: $\frac{1}{\sqrt{3}}[\frac{x^2}{2}]_0^{\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.
Second integral: $[\frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\frac{x}{2}]_{\sqrt{3}}^2 = (2\frac{\pi}{2}) - (\frac{\sqrt{3}}{2} + 2\frac{\pi}{3}) = \pi - \frac{\sqrt{3}}{2} - \frac{2\pi}{3} = \frac{\pi}{3} - \frac{\sqrt{3}}{2}$.
Total Area = $\frac{\sqrt{3}}{2} + \frac{\pi}{3} - \frac{\sqrt{3}}{2} = \frac{\pi}{3}$.
$\boxed{\frac{\pi}{3} \text{ sq. units}}$
Q7
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Find the area of the smaller part of the circle $x^2 + y^2 = a^2$ cut off by the line $x = \frac{a}{\sqrt{2}}$.
Area = $2 \int_{a/\sqrt{2}}^a \sqrt{a^2-x^2} \, dx$.
$= 2 \left[\frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]_{a/\sqrt{2}}^a$.
$= 2 \left[ (0 + \frac{a^2}{2}\frac{\pi}{2}) - (\frac{a}{2\sqrt{2}}\sqrt{a^2-\frac{a^2}{2}} + \frac{a^2}{2}\sin^{-1}\frac{1}{\sqrt{2}}) \right]$.
$= 2 \left[ \frac{\pi a^2}{4} - \frac{a^2}{4} - \frac{\pi a^2}{8} \right] = \frac{\pi a^2}{4} - \frac{a^2}{2}$.
$\boxed{\frac{a^2}{2}(\frac{\pi}{2} - 1) \text{ sq. units}}$
Q8
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The area between $x = y^2$ and $x = 4$ is divided into two equal parts by the line $x = a$, find the value of $a$.
Total area = $\int_0^4 \sqrt{x} \, dx = [\frac{2}{3}x^{3/2}]_0^4 = \frac{16}{3}$.
Area from $0$ to $a$ is half of total: $\int_0^a \sqrt{x} \, dx = \frac{1}{2} \cdot \frac{16}{3} = \frac{8}{3}$.
$\frac{2}{3}[x^{3/2}]_0^a = \frac{8}{3} \Rightarrow \frac{2}{3}a^{3/2} = \frac{8}{3} \Rightarrow a^{3/2} = 4$.
$a = 4^{2/3}$.
$\boxed{a = 4^{2/3}}$
Q9
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Find the area of the region bounded by the parabola $y = x^2$ and $y = |x|$.
The region is symmetric. We find area in first quadrant and multiply by 2.
In first quadrant, $y=x$. Intersection: $x^2=x \Rightarrow x=1$.
Area = $2 \int_0^1 (x - x^2) \, dx = 2[\frac{x^2}{2} - \frac{x^3}{3}]_0^1$.
$= 2[\frac{1}{2} - \frac{1}{3}] = 2[\frac{1}{6}] = \frac{1}{3}$.
$\boxed{\frac{1}{3} \text{ sq. units}}$
Q10
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Find the area bounded by the curve $x^2 = 4y$ and the line $x = 4y - 2$.
Solve for intersection points: $x = x^2 - 2 \Rightarrow x^2-x-2=0 \Rightarrow (x-2)(x+1)=0$. So $x=-1, 2$.
Area = $\int_{-1}^2 (\text{line} - \text{parabola}) \, dx = \int_{-1}^2 (\frac{x+2}{4} - \frac{x^2}{4}) \, dx$.
$= \frac{1}{4} [\frac{x^2}{2} + 2x - \frac{x^3}{3}]_{-1}^2 = \frac{1}{4} [ (2+4-\frac{8}{3}) - (\frac{1}{2}-2+\frac{1}{3}) ]$.
$= \frac{1}{4} [ \frac{10}{3} - (-\frac{7}{6}) ] = \frac{1}{4}[\frac{27}{6}] = \frac{9}{8}$.
$\boxed{\frac{9}{8} \text{ sq. units}}$
Q11
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Find the area of the region bounded by the curve $y^2 = 4x$ and the line $x = 3$.
Area is symmetric about x-axis. Area = $2 \int_0^3 y \, dx$.
$y = \sqrt{4x} = 2\sqrt{x}$.
Area = $2 \int_0^3 2\sqrt{x} \, dx = 4 [\frac{2}{3}x^{3/2}]_0^3 = \frac{8}{3}[3\sqrt{3}] = 8\sqrt{3}$.
$\boxed{8\sqrt{3} \text{ sq. units}}$
Q12
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Area lying in the first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the lines $x = 0$ and $x = 2$ is: (A) $\pi$ (B) $\pi/2$ (C) $\pi/3$ (D) $\pi/4$
This is the area of a quarter circle of radius 2.
Area = $\frac{1}{4}\pi r^2 = \frac{1}{4}\pi (2^2) = \pi$.
$\boxed{\text{(A) } \pi}$
Q13
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Area of the region bounded by the curve $y^2 = 4x$, y-axis and the line $y = 3$ is: (A) 2 (B) 9/4 (C) 9/3 (D) 9/2
Area = $\int_0^3 x \, dy$. From $y^2=4x$, we have $x = \frac{y^2}{4}$.
Area = $\int_0^3 \frac{y^2}{4} \, dy = \frac{1}{4}[\frac{y^3}{3}]_0^3 = \frac{1}{12}[3^3 - 0] = \frac{27}{12} = \frac{9}{4}$.
$\boxed{\text{(B) } 9/4}$