Exercise 8.2(Application of Integrals)

Chapter 8 Exercise 8.2 ncert math solution class 12

Question 1: Find the area of the circle $4 x^2+4 y^2=9$ which is interior to the parabola $x^2=4 y$ .

Solution : The required area is represented by the shaded area OBCDO.

Solving the given equation of circle $4 x^2+4 y^2=9$ , and parabola $x^2=4 y$ ,

$4(4y)+4y^2=9$

$\Rightarrow 4y^2+16y-9=0$

$\Rightarrow 4y^2+18y -2y-9=0$

$\Rightarrow 2y(2y+9)-1(2y+9)=0$

$\Rightarrow (2y+9)(2y-1)=0$

$\Rightarrow y =-\frac{9}{2}, \frac{1}{2}$

Whent $y=\frac{1}{2}$ then $x = \pm\sqrt{2}$

we obtain the B Point of intersection as $B\left(\sqrt{2}, \frac{1}{2}\right)$ and $D\left(-\sqrt{2}, \frac{1}{2}\right)$ .

Area $\mathrm{OBCDO}=2 \times$ Area $\mathrm{OBCO}$
We draw BM perpendicular to OA.

Therefore, the coordinates of $M$ are $(\sqrt{2}, 0)$ .

Therefore, Area OBCO = Area OMBCO – Area OMBO

$=\int_0^{\sqrt{2}} \sqrt{\frac{\left(9-4 x^2\right)}{4}} d x-\int_0^{\sqrt{2}} \sqrt{\frac{\left(x^2\right)}{4}} d x$

$=\frac{1}{2} \int_0^{\sqrt{2}} \sqrt{9-4 x^2}-\frac{1}{4} \int_0^{\sqrt{2}} x^2 d x$

$=\frac{1}{4}\left[x \sqrt{9-4 x^2}+\frac{9}{2} \sin ^{-1} \frac{2 \sqrt{2}}{3}\right]_0^{\sqrt{2}}-\frac{1}{4}\left[\frac{x^3}{3}\right]_0^{\sqrt{2}}$

$=\frac{1}{4}\left[\sqrt{2} \sqrt{9-8}+\frac{9}{2} \sin ^{-1} \frac{2 \sqrt{2}}{3}\right]-\frac{1}{12}(\sqrt{2})^3$

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

$=\frac{\sqrt{2}}{12}+\frac{9}{8} \sin ^{-1} \frac{2 \sqrt{2}}{3}$

$=\frac{1}{2}\left(\frac{\sqrt{2}}{6}+\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}\right)$

Therefore, the required area OBCDO = $2 \times \frac{1}{2}\left(\frac{\sqrt{2}}{6}+\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}\right)$

$=\left(\frac{\sqrt{2}}{6}+\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}\right)$ units.

Question 2: Find the area bounded by curves $(x-1)^2+y^2=1$ and $x^2+y^2=1$ .

Solution : The area bounded by curves $(x-1)^2+y^2=1$ and $x^2+y^2=1$ is represented by the shaded area OACBO.

On solving the equations, $(x-1)^2+y^2=1$ and $x^2+y^2=1$ ,

$(x-1)^2+1-x^2=1$

$\Rightarrow x^2+1-2x-x^2=0$

$\Rightarrow 1-2x=0$

$\Rightarrow x = \frac{1}{2}$

When $x= \frac{1}{2}$ then $y= \pm\frac{\sqrt{3}}{2}$

we obtain the point of intersection as $B\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and $D\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$ .

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

Area $\mathrm{OBCAO}=2 \times$ Area OCAO

We join $A B$ , which intersects $O C$ at $M$ , such that $A M$ is perpendicular to OC.

The coordinates of $M$ are $\left(\frac{1}{2}, 0\right)$

Area OCAO $=$ Area OMAO + Area MCAM

$= \int_0^{\frac{1}{2}}y_1 dx+\int_{\frac{1}{2}}^1 y_2 dx$

$=\left[\int_0^{\frac{1}{2}} \sqrt{1-(x-1)^2} d x+\int_{\frac{1}{2}}^1 \sqrt{1-x^2} d x\right]$

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

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

$=\left[-\frac{\sqrt{3}}{8}+\frac{1}{2}\left(-\frac{\pi}{6}\right)-\frac{1}{2}\left(-\frac{\pi}{2}\right)\right]+\left[\frac{1}{2}\left(\frac{\pi}{2}\right)-\frac{\sqrt{3}}{8}-\frac{1}{2}\left(\frac{\pi}{6}\right)\right]$

$=\left[-\frac{\sqrt{3}}{4}-\frac{\pi}{12}+\frac{\pi}{4}+\frac{\pi}{4}-\frac{\pi}{12}\right]$

$=\left[-\frac{\sqrt{3}}{4}-\frac{\pi}{6}+\frac{\pi}{2}\right]$

$=\left[\frac{2 \pi}{6}-\frac{\sqrt{3}}{4}\right]$

Therefore, required area OBCAO = $2 \times\left(\frac{2 \pi}{6}-\frac{\sqrt{3}}{4}\right)$

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

Question 3: Find the area of the region bounded by the curves $y=x^2+2, y=x \quad x=0$ and $x=3$ .

Solution: The area bounded by the curves:

$y=x^2+2, \quad y=x, \quad x=0 \quad \text { and } \quad x=3 .$

Then, Area OCBAO = Area ODBAO – Area ODCO

$=\int_0^3\left(x^2+2\right) d x-\int_0^3 x d x$

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

$=[9+6]-\left[\frac{9}{2}\right]=15-\frac{9}{2}=\frac{21}{2} \text { units }$

Question 4: Using integration finds the area of the region bounded by the triangle whose vertices are $(-1,0),(1,3)$ and $(3,2)$ .

Solution: BL and CM are drawn perpendicular to x-axis.
It can be observed in the given figure that,

Area $(\triangle \mathrm{ACB})=\operatorname{Area}(A L B A)+$ Area (BLMCB) – Area (AMCA)

Equation of line segment $A B$ is $y-0=\frac{3-0}{1+1}(x+1)$

$\Rightarrow y=\frac{3}{2}(x+1)$

$\text { Area (ALBA) }=\int_{-1}^1 \frac{3}{2}(x+1) d x$

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

Equation of line segment BC is $y-3=\frac{2-3}{3-1}(x-1)$

$\Rightarrow y=\frac{1}{2}(-x+7)$

$\text { Area (BLMCB) }=\int_1^3 \frac{1}{2}(-x+7) d x=\frac{1}{2}\left[-\frac{x^2}{2}+7 x\right]_1^3$

$=\frac{1}{2}\left[-\frac{9}{2}+21+\frac{1}{2}-7\right]=5 \text { units }$

Equation of line segment AC is $y-0=\frac{2-0}{3+1}(x+1) \Rightarrow y=\frac{1}{2}(x+1)$

Area (AMCA) $=\frac{1}{2} \int_{-1}^3(x+1) d x$

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

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

Therefore, from equation (1), we have

Area $(\triangle A B C)=(3+5-4)=4$ units

Question 5: Using integration find the area of the triangular region whose sides have the equations $y=2 x+1, y=3 x+1$ and $x=4$

Solution: The equations of sides of the triangle are $y=2 x+1, y=3 x+1$ , and $x=4$ .

Solving $y=2 x+1$ and $y=3 x+1$

$3x+1=2x+1$

$\Rightarrow x=0$

When $x=0$ then $y=1$

Point $A(0,1)$

Again solving $y=2 x+1$ and $x=4$

$y=2\times 4+1$

$\Rightarrow y=9$

Point $C(4,9)$

Again solving $y=3x+1$ and $x=4$

$y = 3\times 4+1=13$

Point $B(4,13)$

we obtain the vertices of triangle as $A(0,1), B(4,13)$ , and $C(4,9)$

It can be observed that,

Area $(\triangle \mathrm{ACB})=\operatorname{Area}(\mathrm{OLBAO})-$ Area (OLCAO)

$=\int_1^4(3 x+1) d x-\int_0^4(2 x+1) d x$

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

$=(24+4)-(16+4)$

$=28-20=8$ units

Question 6: Smaller area enclosed by the circle $x^2+y^2=4$ and the line $\mathrm{x}+\mathrm{y}=2$ is

(A) $2(\pi-2)$

(B) $(\pi-2)$

(D) $2(\pi+2)$

Solution: The correct answer is (B)

The smaller area enclosed by the circle, $x^2+y^2=4$ and the line, $x+y=2$ , is represented by the shaded area ACBA.

Area $A C B A=$ Area $O A C B O-\operatorname{Area}(\triangle \mathrm{OAB})$

$=\int_0^{2} \sqrt{4-x^2} d x-\int_0^{2}(2-x) d x$

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

$=\left[2 \cdot \frac{\pi}{2}\right]-[4-2]=[\pi-2]$ units

Thus, the correct answer is (B).

Question 7: Area lying between the curve $y^2=4$ and $y=2 x$ is
(A) $\frac{2}{3}$

(B) $\frac{1}{3}$

(D) $\frac{3}{4}$

Solution: The correct answer is (B)

The area lying between the curve, $y^2=4$ and $y=2 x$ , is represented by the shaded area OBAO.

The points of intersection of these curves are $0(0,0)$ and $A(1,2)$ .

We draw AC perpendicular to -axis such that the coordinates of $C$ are $(1,0)$ .

Area $O B A O=\operatorname{Area}(\triangle O C A)-$ Area $(O C A B O)$

$=\int_0^1 2 x d x-\int_0^1 2 \sqrt{x}$

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

$=\left|1-\frac{4}{3}\right|=\left|-\frac{1}{3}\right|$

$=-\frac{1}{3} \text { units }$

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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Exercise 8.2(Application of Integrals)

Chapter 8 Exercise 8.2 ncert math solution class 12

Chapter 8: Application of Integrals Class 12

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