Exercise 8.1 ncert math solution class 12

Exercise 8.1(Application of Integrals)

Chapter 8 Exercise 8.1 ncert math solution class 12

Question1: Find the area of the region bounded by the curve $y^{2}=x$ and the lines $x=1, x=4$ and the x-axis.

Solution : The area of the region bounded by the curve, $y^{2}=x$ , the lines, x = 1 and x = 4, and the X-axis is the area E B C F.

Area of $\mathrm{E B C F}=\int_{1}^{4} y d x$

$=\int_{1}^{4} \sqrt{x} d x$

$=\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{4}$

$=\frac{2}{3}\left[(4)^{\frac{3}{2}}-(1)^{\frac{3}{2}}\right]$

$=\frac{2}{3}[8-1]=\frac{14}{3}$ units

Question 2: Find the area of the region bounded by $y^{2}=9 x, x=2, x=4$ and the x – axis in the first quadrant.

Solution : The area of the region bounded by the curve, $y^{2}=9 x, x=2$ , and x = 4, and the x-axis, is the area EBCF.

Area of $\mathrm{E B C F}=\int_{2}^{4} y d x$

$=\int_{2}^{4} 3 \sqrt{x} d x$

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

$=2\left[(4)^{\frac{3}{2}}-(2)^{\frac{3}{2}}\right]$

$=2[8-2 \sqrt{2}]$

$=16-4 \sqrt{2}$ units

Question 3: Find the area of the region bounded by $x^{2}=4 y, y=2, y=4$ and the y-axis in the first quadrant.

Solution : The area of the region bounded by the curve, $x^{2}=4 y, y=2$ , and y = 4, and the y-axis is the area $E B C F$ .

Area of $\mathrm{E B C F}=\int_{2}^{4} x d y$

$=\int_{2}^{4} 2 \sqrt{y} d x$

$=2 \int_{2}^{4} \sqrt{y} d x=2\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{2}^{4}$

$=\frac{4}{3}\left[(4)^{\frac{3}{2}}-(2)^{\frac{3}{2}}\right]=\frac{4}{3}[8-2 \sqrt{2}]$

$=\left(\frac{32-8 \sqrt{2}}{3}\right)$ units

Question 4: Find the area of the region bounded by the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$

Solution : The given equation of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$

It can be observed that the ellipse is symmetrical about x-axis and y-axis.

Area bounded by ellipse $=4 \times$ Area of OAB

Area of $\mathrm{OAB}=\int_{0}^{4} y d x$

$=\int_{0}^{4} 3 \sqrt{1-\frac{x^{2}}{16} d x}=\frac{3}{4} \int_{0}^{4} \sqrt{16-x^{2}} d x$

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

$=\frac{3}{4}\left[2 \sqrt{16-16}+8 \sin ^{-1}(1)-0-8 \sin ^{-1}(0)\right]$

$=\frac{3}{4}\left[\frac{8 \pi}{2}\right]$

$\Rightarrow \frac{3}{4}\left[\frac{8 \pi}{2}\right]=3 \pi$

Therefore, area bounded by the ellipse =4 × Area of $\mathrm{OAB}=4 \times 3 \pi=12 \pi$ units

Question 5: Find the area of the region bounded by the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$

Solution : The given equation of the ellipse can be represented as $\frac{x^2}{4}+\frac{y^2}{9}=1$ It can be observed that the ellipse is symmetrical about x-axis and y-axis.

Area bounded by ellipse =4 × Area of OAB

$\text { Area of } \mathrm{OAB}=\int_0^2 y d x=\int_0^2 3 \sqrt{1-\frac{x^2}{4}} d x$

$=\frac{3}{2} \int_0^2 \sqrt{4-x^2} d x$

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

$=\frac{3}{4}\left[\frac{2 \pi}{2}\right]=\frac{3 \pi}{2}$

Therefore, area bounded by the ellipse $=4 \times$ Area of $O A B=4 \times \frac{3 \pi}{2}=6 \pi$ units

Question 6: Find the area of the region in the first quadrant enclosed by x-axis, line $x=\sqrt{3} y$ and the $\operatorname{circle} x^2+y^2=4$

Solution : The area of the region bounded by the circle, $x^2+y^2=4$ ,line $x=\sqrt{3}y$ and the x-axis is the area OAB.

Solving $x^2+y^2=4$ and $x=\sqrt{3}y$

$(\sqrt{3}y)^2+y^2=4$

$\Rightarrow 3y^2+y^2=4$

$\Rightarrow 4y^2 = 4$

$\Rightarrow y^2 = 1$

$\Rightarrow y = 1$

and $x= \sqrt{3}$

The point of intersection of the line and the circle in the first quadrant is $(\sqrt{3}, 1)$ .

Area $O A C=$ Area $\triangle O C B+$ Area $A C B$

Area of $\mathrm{OAB}= \int_0^{\sqrt{3}}y_1dx$

$=\int_0^{\sqrt{3}}\frac{x}{\sqrt{3}}dx$

$=\frac{1}{\sqrt{3}}\int_0^{\sqrt{3}} xdx$

$= \frac{1}{\sqrt{3}}\left[\frac{x^2}{2}\right]_0^{\sqrt{3}}$

$=\frac{1}{2\sqrt{3}}\left[(\sqrt{3})^2-0\right]$

$=\frac{1}{2\sqrt{3}}\times 3$

$=\frac{\sqrt{3}}{2}$ ….(i)

$\text { Area of } \mathrm{ABC}=\int_{\sqrt{3}}^2 y d x=\int_{\sqrt{3}}^2 \sqrt{4-x^2} d x$

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

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

$=\pi-\frac{\sqrt{3} \pi}{2}-2\left(\frac{\pi}{3}\right)=\pi-\frac{\sqrt{3} \pi}{2}-\frac{2 \pi}{3}=\left[\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right]$ …..(ii)

Adding the equation (1) and (2), we get

The total area of $\mathrm{OAC}=\frac{\sqrt{3}}{2}+\frac{\pi}{3}-\frac{\sqrt{3}}{2}=\frac{\pi}{3}$ units

Therefore, area enclosed by x – axis, the line $x=\sqrt{3}$ and the circle $x^2+y^2=4$ in the first quadrant is $\frac{\pi}{3}$ square units.

Question 7: Find the area of the smaller part of the circle $\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2$ cut off by the line $x=\frac{a}{\sqrt{2}}$ .

Solution : It can be observed that the area $A B C D$ is symmetrical about x -axis.

Given, $x^2+y^2=a^21$

$\Rightarrow y = \sqrt{a^2-x^2}$

Area of ABC $= \int_{\frac{a}{\sqrt{2}}}^a y dx$

$= \int_{\frac{a}{\sqrt{2}}}^a \sqrt{a^2-x^2}dx$

$= \left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]_{\frac{a}{\sqrt{2}}}^a$

$=\left[\left(\frac{a}{2}\sqrt{a^2-a^2}+\frac{a^2}{2}\sin^{-1}\frac{a}{a}\right)-\left(\frac{1}{2}\frac{a}{\sqrt{2}}\sqrt{a^2-\frac{a^2}{2}}+\frac{a^2}{2}\sin^{-1}\frac{a}{\sqrt{2}a}\right) \right]$

$=\left[\frac{a^2}{2}\times \frac{\pi}{2}-\left(\frac{a^2}{4}+\frac{a^2}{2}\times \frac{\pi}{4}\right)\right]$

$=\left[\frac{\pi a^2}{4}-\frac{a^2}{4}-\frac{\pi a^2}{8}\right]$

$\left[\frac{\pi a^2}{8}-\frac{a^2}{4}\right]$

$=\frac{a^2}{4}\left[\frac{\pi}{2}-1\right]$

Therefore, the area of smaller part of the circle, $\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2$ , cut off by the line $x=\frac{a}{\sqrt{2}}$ is $\frac{a^2}{4}\left(\frac{\pi}{2}-1\right)$ square units.

Question 8: 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.

Solution : The line, x = a, divides the area bounded by the parabola and x = 4 into two equal parts.

⇒ Area $\mathrm{OBA}=$ Area $\mathrm{ABCD}$

It can be observed that the given area is symmetrical about x-axis.

⇒ Area $2\times O B E=$ Area $2\times E B C F$

Area $O B E=\int_0^a y d x=\int_0^a \sqrt{x} d x$

$=\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^a$

$=\frac{2}{3}(a)^{\frac{3}{2}}$ …..(i)

Area of $E B C F=\int_0^4 \sqrt{x} d x$

$=\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^4$

$=\frac{2}{3}\left[8-(a)^{\frac{3}{2}}\right]$ …..(ii)

From (1) and (2), we have

$\frac{2}{3}(a)^{\frac{3}{2}}=\frac{2}{3}\left[8-(a)^{\frac{3}{2}}\right]$

$\Rightarrow 2(a)^{\frac{3}{2}}=8$

$\Rightarrow(a)^{\frac{3}{2}}=4$

$\Rightarrow a=(4)^{\frac{2}{3}}$

Therefore, the value of a is $a=(4)^{\frac{2}{3}}$ .

Question 9: Find the area of the region bounded by the parabola $y=x^{2}$ and $y=|x|$

Solution: The area bounded by the parabola, $\mathrm{x}^{2}=\mathrm{y}$ , and the line, $y=|x|$ , can be represented as The given area is symmetrical about y-axis.

⇒ Area OACO = Area ODBO

Solving $y=x^{2}$ and y = x

$\Rightarrow x = x^2$

$\Rightarrow x = 1 \text{ and } x = 0$

and y = 1 and y = 0

The point of intersection of parabola, $x^{2}=y$ , and line, y = x, is A(1, 1) and (0, 0).

Area of $O A C O=$ Area $\triangle O A M - \text{ Area of OCAMO }$

Area of $\triangle O A M=\int_0^1 y_1 dx$

$=\int_0^1 x dx$

$= \left[\frac{x^2}{2}\right]_0^1$

$=\frac{1}{2}\left[1-0\right]$

$=\frac{1}{2}$

Area of $O B A C O=\int_{0}^{1} y_2 d x$

$=\int_{0}^{1} x^{2} d x$

$=\left[\frac{x^{3}}{3}\right]_{0}^{1}=\frac{1}{3}$

⇒ Area of $O A C O=$ Area of $\triangle O A M - \text{ Area of OCAMO }$

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

Therefore, required area $=2\left[\frac{1}{6}\right]=\frac{1}{3}$ units

Question 10: Find the area bounded by the curve $x^{2}=4 y$ and the line $x=4 y-2$ .

Solution : The area bounded by the curve $x^{2}=4 y$ and the line $x=4 y-2$ is represented by the shaded area OBAO.

Solving $x^{2}=4 y$ and $x=4 y-2$

$x^2=x+2$

$\Rightarrow x^-x-2=0$

$\Rightarrow x^2-2x+x-2=0$

$\Rightarrow x(x-2)+1(x-2)=0$

$\Rightarrow (x-2)(x+1)=0$

$\Rightarrow x = 2$ and $x=-1$

aand y=1 and $y=\frac{1}{4}$

Let A and B be the points of intersection of the line and parabola. Coordinates of point A are $\left(-1, \frac{1}{4}\right)$ and point B are (2, 1).

We draw AL and BM perpendicular to x-axis.

It can be observed that,

Area OBAO $=\int_{-1}^{2} \frac{x+2}{4} d x-\int_{-1}^{2} \frac{x^{2}}{4} d x$

$=\frac{1}{4}\left[\frac{x^{2}}{2}+2 x\right]_{-1}^{2}-\frac{1}{4}\left[\frac{x^{3}}{3}\right]_{-1}^{2}$

$=\frac{1}{4}[(2+4)-(\frac{1}{2}-2)]-\frac{1}{4}\left[\frac{8}{3}-\frac{-1}{3}\right]$

$=\frac{1}{4}\left[\frac{15}{2}\right]-\frac{1}{4}\left[\frac{9}{3}\right]$

$=\frac{1}{4}\left[\frac{45-18}{6}\right]$

$=\frac{1}{4}\times \frac{27}{6}$

$= \frac{9}{8}$

Question 11:Find the area of the region bounded by the curve $y^{2}=4 x$ and the line x=3.

Solution : The region bounded by the parabola, $y^{2}=4 x$ , and the line, x=3, is the area OABO.

The area OABO is symmetrical about x-axis.

Area of $\mathrm{OABO}=2$ (Area of $\mathrm{OAC})$

Area $O A B O=2\left[\int_{0}^{3} y d x\right]$

$=2\left[\int_{0}^{3} 2 \sqrt{x} d x\right]$

$=4\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{3}$

$=\frac{8}{3}\left[(3)^{\frac{3}{2}}\right]$

$=8 \sqrt{3}$

Therefore, the required area is $8 \sqrt{3}$ units

Question 12: 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) $\frac{\pi}{2}$

(D) $\frac{\pi}{4}$

Solution : The correct answer is (A).

The area bounded by the circle and the lines, x = 0 and x = 2, in the first quadrant is represented as

Area of $O A B=\int_{0}^{2} y d x=\left[\int_{0}^{2} \sqrt{4-x^{2}} d x\right]$

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

$=2\left(\frac{\pi}{2}\right)$

$=\pi$ units

Thus, the correct answer is (A).

Question 13: Area of the region bounded by the curve $y^{2}=4 x, y$ -axis and the line y = 3 is

(A) 2

(B) $\frac{9}{4}$

(D) $\frac{9}{2}$

Solution : The correct answer is (B)

The area bounded by the curve, $y^{2}=4 x, y$ -axis and y = 3 is represented as

Area $O A B=\int_{0}^{3} x d y=\int_{0}^{3} \frac{y^{2}}{4} d y$

$=\frac{1}{4}\left[\frac{y^{2}}{4}\right]_{0}^{3}=\frac{1}{12}(27)=\frac{9}{4}$ units

Thus, the correct answer is (B).

Chapter 8: Application of Integrals Class 12

Exercise 8.1 ncert math solution class 12

Exercise 8.2 ncert math solution class 12

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Chapter 8: Application of Integrals Class 12

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