# NCERT Solutions for Exercise 8.2 Class 12 Maths Chapter 8 - Application of Integrals

NCERT solutions for Class 12 Maths chapter 8 exercise 8.2 is similar to the exercise 8.1. Also it has linkages with the previous chapter. Exercise 8.2 Class 12 Maths consists of questions related to areas bounded by two different curves. NCERT solutions for Class 12 Maths chapter 8 exercise 8.2 can be solved easily if the concept is understood from initial few questions. All other questions have the same concept used. Similar questions which are in this exercise are asked in Physics also, hence this exercise becomes more important. Students can find solutions to other exercises apart from this exercise from below mentioned NCERT exercise list.

• Application of Integrals Exercise 8.1

• Application of Integrals Miscellaneous Exercise

## Application of Integrals Class 12 Chapter 8 Exercise: 8.2

Question: 1 Find the area of the circle $\dpi{100} \small 4x^2+4y^2=9$ which is interior to the parabola $\dpi{100} \small x^2=4y$ .

The area bounded by the circle $\dpi{100} \small 4x^2+4y^2=9$ and the parabola $\dpi{100} \small x^2=4y$ .

By solving the equation we get the intersecting point $D(-\sqrt{2},\frac{1}{2})$ and $B(\sqrt{2},\frac{1}{2})$
So, the required area (OBCDO)=2 times the area of (OBCO)
Draw a normal on the x-axis (M = $\sqrt{2},0$ )

Thus the area of OBCO = Area of OMBCO - Area of OMBO

$\\\int_{0}^{\sqrt{2}}\sqrt{\frac{(9-4x^2)}{4}}dx-\int_{0}^{\sqrt{2}}{\frac{x^2}{4}}dx\\ =\frac{1}{2}\int_{0}^{\sqrt{2}}\sqrt{9-4x^2}-\frac{1}{4}\int_{0}^{\sqrt{2}}x^2dx\\ =\frac{1}{4}[x\sqrt{9-4x^2}+\frac{9}{2}\sin^{-1}\frac{2x}{3}]_0^{\sqrt{2}}-\frac{1}{4}[\frac{x^3}{3}]_0^{\sqrt{2}}\\ =\frac{1}{4}[\sqrt{2}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}]-\frac{1}{12}(\sqrt{2})^3\\ =\frac{\sqrt{2}}{12}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}\\ =\frac{1}{2}(\frac{\sqrt{2}}{6}+\frac{9}{4}\sin^{-1}\frac{2\sqrt{2}}{3})$
S0, total area =
$\\=2\times \frac{1}{2}(\frac{\sqrt{2}}{6}+\frac{9}{4}\sin^{-1}\frac{2\sqrt{2}}{3})\\ =\frac{\sqrt{2}}{6}+\frac{9}{4}\sin^{-1}\frac{2\sqrt{2}}{3}$

Question:2 Find the area bounded by curves $\dpi{100} \small (x-1)^2+y^2=1$ and $\dpi{100} \small x^2+y^2=1$ .

Given curves are $\dpi{100} \small (x-1)^2+y^2=1$ and $\dpi{100} \small x^2+y^2=1$

Point of intersection of these two curves are

$A = \left ( \frac{1}{2},\frac{\sqrt3}{2} \right )$ and $B = \left ( \frac{1}{2},-\frac{\sqrt3}{2} \right )$

We can clearly see that the required area is symmetrical about the x-axis

Therefore,

Area of OBCAO = 2 $\times$ Area of OCAO

Now, join AB such that it intersects the x-axis at M and AM is perpendicular to OC

Coordinates of M = $\left ( \frac{1}{2},0 \right )$

Now,

Area OCAO = Area OMAO + Area CMAC

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

$=\left [- \frac{1}{4}\sqrt{1-(-\frac{1}{2})^2}+\frac{1}{2}\sin^{-1}(\frac{1}{2}-1)-0-\frac{1}{2}\sin^{-1}(-1) \right ]+\left [ 0+\frac{1}{2}\sin^{-1}(1)- \frac{1}{4}\sqrt{1-\left ( \frac{1}{2} \right )^2}-\frac{1}{2}\sin^{-1}\left ( \frac{1}{2} \right )\right ]$
$=\left [ -\frac{\sqrt3}{8}+\frac{1}{2}\left ( -\frac{\pi}{6} \right )-\frac{1}{2}\left ( -\frac{\pi}{2} \right ) \right ]+\left [ \frac{1}{2}\left ( \frac{\pi}{2} \right ) -\frac{\sqrt3}{8}-\frac{1}{2}\left ( \frac{\pi}{6} \right )\right ]$
$= \left [ -\frac{\sqrt3}{8}+\frac{\pi}{6} \right ]+\left [ \frac{\pi}{6}-\frac{\sqrt3}{8} \right ]$
$=2 \left [ -\frac{\sqrt3}{8}+\frac{\pi}{6} \right ]$
Now,
Area of OBCAO = 2 $\times$ Area of OCAO

$=2\times 2 \left [ -\frac{\sqrt3}{8}+\frac{\pi}{6} \right ]$
$=\frac{2\pi}{3}-\frac{\sqrt3}{2}$

Therefore, the answer is $\frac{2\pi}{3}-\frac{\sqrt3}{2}$

Question: 3 Find the area of the region bounded by the curves $\dpi{100} \small y=x^2+2,y=x,x=0$ and $\dpi{100} \small x=3$ .

The area of the region bounded by the curves,

$\dpi{100} \small y=x^2+2,y=x,x=0$ and $\dpi{100} \small x=3$ is represented by the shaded area OCBAO as

Then, Area OCBAO will be = Area of ODBAO - Area of ODCO

which is equal to

$\int_0^3(x^2+2)dx - \int_0^3x dx$

$= \left ( \frac{x^3}{3}+2x \right )_0^3 -\left ( \frac{x^3}{2} \right )_0^3$

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

Question: 4 Using integration find the area of region bounded by the triangle whose vertices are $\dpi{100} \small (-1,0),(1,3)$ and $\dpi{100} \small (3,2)$ .

So, we draw BL and CM perpendicular to x-axis.

Then it can be observed in the following figure that,

$Area(\triangle ACB) = Area (ALBA)+Area(BLMCB) - Area (AMCA)$

We have the graph as follows:

Equation of the line segment AB is:

$y-0 = \frac{3-0}{1+1}(x+1)$ or $y = \frac{3}{2}(x+1)$

Therefore we have Area of $ALBA$

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

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

So, the equation of line segment BC is

$y-3 = \frac{2-3}{3-1}(x-1)$ or $y= \frac{1}{2}(-x+7)$

Therefore the area of BLMCB will be,

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

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

Equation of the line segment AC is,

$y-0 = \frac{2-0}{3+1}(x+1)$ or $y = \frac{1}{2}(x+1)$

Therefore the area of AMCA will be,

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

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

Therefore, from equations (1), we get

The area of the triangle $\triangle ABC =3+5-4 =4units.$

Question:5 Using integration find the area of the triangular region whose sides have the equations $\dpi{100} \small y=2x+1,y=3x+1$ and $\dpi{100} \small x=4$ .

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

ON solving these equations, we will get the vertices of the triangle as $A(0,1),B(4,13),\ and\ C(4,9)$

Thus it can be seen that,

$Area (\triangle ACB) = Area (OLBAO) -Area (OLCAO)$

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

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

$=(24+4) - (16+4) = 28-20 =8units.$

Smaller area enclosed by the circle $\dpi{100} \small x^2+y^2=4$ and the line $\dpi{100} \small x+y=2$ is

(A) $\dpi{100} \small 2(\pi -2)$ (B) $\dpi{100} \small \pi -2$ (C) $\dpi{100} \small 2\pi -1$ (D) $\dpi{100} \small 2(\pi +2)$

So, 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 as

Thus it can be observed that,

Area of ACBA = Area OACBO - Area of $(\triangle OAB)$

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

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

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

$= (\pi -2) units.$

Thus, the correct answer is B.

Area lying between the curves $\dpi{100} \small y^2=4x$ and $\dpi{100} \small y=2x$ is

(A) $\dpi{100} \small \frac{2}{3}$ (B) $\dpi{100} \small \frac{1}{3}$ (C) $\dpi{100} \small \frac{1}{4}$ (D) $\dpi{100} \small \frac{3}{4}$

The area lying between the curve, $\dpi{100} \small y^2=4x$ and $\dpi{100} \small y=2x$ is represented by the shaded area OBAO as

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

So, we draw AC perpendicular to x-axis such that the coordinates of C are (1,0).

Therefore the Area OBAO = $Area(\triangle OCA) -Area (OCABO)$

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

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

Thus the correct answer is B.

## More About NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2

The NCERT Class 12 Maths chapter application of Integrals mainly deals with the finding out of an area bounded by two curves. Exercise 8.2 Class 12 Maths is an extension of last exercise only. Hence before doing NCERT solutions for Class 12 Maths chapter 8 exercise 8.2 one should complete the exercise 8.1.

## Benefits of NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2

• The Class 12th Maths chapter 8 exercise has 3 exercises in total maily dealing with the application of integrals.

• These Class 12 Maths chapter 8 exercise 8.2 solutions are helpful in solving the questions in the upcoming exercise also.

• NCERT solutions for Class 12 Maths chapter 8 exercise 8.2 provides mostly moderate to difficult level of questions.

## Also see-

• NCERT Exemplar Solutions Class 12 Maths Chapter 8

• NCERT Solutions for Class 12 Maths Chapter 8

## NCERT Solutions Subject Wise

• NCERT Solutions Class 12 Chemistry

• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Biology

• NCERT Solutions for Class 12 Mathematics

## Subject Wise NCERT Exemplar Solutions

• NCERT Exemplar Class 12 Maths

• NCERT Exemplar Class 12 Physics

• NCERT Exemplar Class 12 Chemistry

• NCERT Exemplar Class 12 Biology

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