Exercise 8.2 ncert math solution class 12

Exercise 8.2(Application of Integrals)

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.

Exercise 8.2 ncert math solution class 12

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 $(x-1)^2+y^2=1[latex] and [latex]x^2+y^2=1$ is represented by the shaded area OACBO.

Exercise 8.2 ncert math solution class 12

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:

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

Exercise 8.2 ncert math solution class 12

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,

Exercise 8.2 ncert math solution class 12

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

Equation of line segment AB 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 $x=4$ .

Exercise 8.2 ncert math solution class 12

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

3x+1=2x+1

⇒ x=0

When x=0 then y=1

Point A(0, 1)

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

y=2× 4+1

⇒ y=9

Point C(4, 9)

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

y = 3× 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.

Exercise 8.2 ncert math solution class 12

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.

Exercise 8.2 ncert math solution class 12

The points of intersection of these curves are O(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 $(O C A B O)$

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

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

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

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

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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Chapter 8: Application of Integrals Class 12

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