# NCERT Solutions for Exercise 8.1 Class 12 Maths Chapter 8 - Application of Integrals

NCERT solutions for Class 12 Maths chapter 8 exercise 8.1 is the extension of last chapter i.e integrals. Concepts learnt in Class 12 Maths NCERT book chapter Integrals will be used heavily in this chapter. Exercise 8. Class 12 Maths deals with area finding of simple curves. NCERT solutions for Class 12 Maths chapter 8 exercise 8.1 requires only basic understanding of integrals which is learnt in the NCERT book previous chapter. Similar questions are asked in Physics also, hence this NCERT syllabus exercise becomes more important. Upcoming exercises are also linked to this exercise, hence sincerely it should be practice. Apart from this below mentioned NCERT exercise can be seen for further topics.

• Application of Integrals Exercise 8.2

• Application of Integrals Miscellaneous Exercise

## Application of IntegralsClass 12 Chapter 8 Exercise: 8.1

Question:1 Find the area of the region bounded by the curve $\dpi{100} y^2=x$ and the lines $\dpi{100} x=1,x=4$ and the $\dpi{100} x$ -axis in the first quadrant.

Area of the region bounded by the curve $\dpi{100} y^2=x$ and the lines $\dpi{100} x=1,x=4$ and the $\dpi{100} x$ -axis in the first quadrant

Area = $\int_{1}^{4}ydy = \int_{1}^{4}\sqrt{x}dx$

$\left [\frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]^4_1 = \frac{2}{3}\left [ (4)^\frac{3}{2}- (1)^\frac{3}{2} \right ]$

$= \frac{2}{3}\left [ 8 -1 \right ]$

= 14/3 units

Question:2 Find the area of the region bounded by $\dpi{100} y^2=9x,x=2,x=4$ and the $\dpi{100} x$ -axis in the first quadrant.

Area of the region bounded by the curve $\dpi{100} y^2=9x,x=2,x=4$ and the $\dpi{100} x$ -axis in the first quadrant

Area = $\int_{2}^{4}ydy = \int_{2}^{4}\sqrt{9x}dx = 3\int_{2}^{4}\sqrt{x}dx$

$3\left [\frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]^4_2 = 3.\frac{2}{3}\left [ (4)^\frac{3}{2}- (2)^\frac{3}{2} \right ]$

$= 2\left [ 8 -2\sqrt2 \right ]$

$= \left [ 16 -4\sqrt2 \right ]$ units

Question:3 Find the area of the region bounded by $\dpi{100} x^2=4y,y=2,y=4$ and the $\dpi{100} y$ -axis in the first quadrant.

The area bounded by the curves $\dpi{100} x^2=4y,y=2,y=4$ and the $\dpi{100} y$ -axis in the first quadrant is ABCD.

$= \int^4_{2} x dy$

$= \int^4_{2} 2\sqrt{y} dy$

$= 2\int^4_{2} \sqrt{y} dy$

$=2\left \{ \frac{y^{\frac{3}{2}}}{\frac{3}{2}} \right \}^4_{2}$

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

$= \frac{4}{3} \left \{ 8 -2\sqrt 2 \right \}$

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

Question:4 Find the area of the region bounded by the ellipse $\dpi{100} \frac{x^2}{16}+\frac{y^2}{9}=1.$

The area bounded by the ellipse : $\dpi{100} \frac{x^2}{16}+\frac{y^2}{9}=1.$

Area will be 4 times the area of EAB.

Therefore, $Area\ of\ EAB= \int^4_{0} y dx$

$= \int^4_{0}3\sqrt{1-\frac{x^2}{16}} dx$

$= \frac{3}{4}\int^4_{0}\sqrt{16-x^2} dx$

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

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

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

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

Therefore the area bounded by the ellipse will be $= 4\times {3\pi} = 12\pi\ units.$

Question: 5 Find the area of the region bounded by the ellipse $\dpi{100} \small \frac{x^2}{4}+\frac{y^2}{9}=1$

The area bounded by the ellipse : $\dpi{100} \small \frac{x^2}{4}+\frac{y^2}{9}=1$

The area will be 4 times the area of EAB.

Therefore, $Area\ of\ EAB= \int^2_{0} y dx$

$= \int^2_{0}3\sqrt{1-\frac{x^2}{4}} dx$

$= \frac{3}{2}\int^2_{0}\sqrt{4-x^2} dx$

$= \frac{3}{2}\left [ \frac{x}{2}\sqrt4-x^2 +\frac{4}{2}\sin^{-1}\frac{x}{2} \right ]^2_{0}$

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

$= \frac{3\pi}{2}$

Therefore the area bounded by the ellipse will be $= 4\times \frac{3\pi}{2} = 6\pi\ units.$

Question: 6 Find the area of the region in the first quadrant enclosed by $\dpi{100} \small x$ -axis, line $\dpi{100} \small x=\sqrt{3}y$ and the circle $\dpi{100} \small x^2+y^2=4$

The area of the region bounded by $\dpi{100} \small x=\sqrt{3}y$ and $\dpi{100} \small x^2+y^2=4$ is ABC shown:

The point B of the intersection of the line and the circle in the first quadrant is $(\sqrt3,1)$ .

Area ABC = Area ABM + Area BMC where, M is point in x-axis perpendicular drawn from the line.

Now,area of $ABM = \frac{1}{2}\times AM\times BM = \frac{1}{2}\times \sqrt{3}\times 1 =\frac{\sqrt3}{2}$ ............(1)

and Area of $BMC = \int^2_{\sqrt{3}} ydx$

$= \int^2_{\sqrt3} \sqrt{4-x^2} dx$

$= \left [ \frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}\frac{x}{2} \right ]^2_{\sqrt3}$

$= \left [ 2\times\frac{\pi}{2}-\frac{\sqrt3}{2}\sqrt{4-3}-2\sin^{-1}\left ( \frac{\sqrt3}{2} \right ) \right ]$

$= \left [ \pi - \frac{\sqrt3\pi}{2}-2\frac{\pi}{3} \right ]$

$= \left [ \pi-\frac{\sqrt3}{2}-\frac{2\pi}{3} \right ]$

$= \left [ \frac{\pi}{3}-\frac{\sqrt3}{2} \right ]$ ..................................(2)

then adding the area (1) and (2), we have then

The Area under ABC $= \frac{\sqrt3}{2} +\frac{\pi}{3}-\frac{\sqrt3}{2} = \frac{\pi}{3}\ units.$

Question: 7 Find the area of the smaller part of the circle $\dpi{100} \small x^2+y^2=a^2$ cut off by the line
$\dpi{100} \small x=\frac{a}{\sqrt{2}}$

we need to find the area of smaller part of the circle
Now,
Area of ABCD = 2 X Area of ABC
Area of ABC = $\int_{\frac{a}{\sqrt2}}^{a} ydx= \int_{\frac{a}{\sqrt2}}^{a} \sqrt{a^2-x^2}dx= \left [ \frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}\right ]^{a}_\frac{a}{\sqrt2}\\ \\$
$=\left [ \frac{a}{2}\sqrt{a^2-a^2}+\frac{a^2}{2}\sin^{-1}\frac{a}{a}- \frac{a}{2\sqrt2}\sqrt{a^2-(\frac{a}{\sqrt2})^2}-\frac{a^2}{2}\sin^{-1}\frac{a}{a\sqrt2}\right ]$
$=\left [ \frac{a}{2}\sqrt{0}+\frac{a^2}{2}\sin^{-1}1- \frac{a}{2\sqrt2}\sqrt{\frac{a^2}{2}}-\frac{a^2}{2}\sin^{-1}\frac{1}{\sqrt2}\right ]$
$=\left [ 0+\frac{a^2}{2}\frac{\pi}{2}- \frac{a^2}{4}-\frac{a^2}{2}\frac{\pi}{4}\right ]$
$=\frac{a^2}{4}\left ( \frac{\pi}{2}-1 \right )$
Area of ABCD = 2 X Area of ABC
$=2\times\frac{a^2}{4}\left ( \frac{\pi}{2}-1 \right )= \frac{a^2}{2}\left ( \frac{\pi}{2}-1 \right )$
Therefore, the area of the smaller part of the circle is $\frac{a^2}{2}\left ( \frac{\pi}{2}-1 \right )$

Question:8 The area between $\dpi{100} \small x=y^2$ and $\dpi{100} \small x=4$ is divided into two equal parts by the line $\dpi{100} \small x=a$ , find the value of $\dpi{100} \small a$ .

we can clearly see that given area is symmetrical about x - axis
It is given that
Area of OED = Area of EFCD
Area of OED = $\int_{0}^{a}ydx = \int_{0}^{a}\sqrt xdx = \left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]_0^{a}= \frac{a^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2a^{\frac{3}{2}}}{3}$
Area of EFCD = $\int_{a}^{4}ydx = \int_{a}^{4}\sqrt xdx = \left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]_a^{4}= \frac{4^{\frac{3}{2}}-a^\frac{3}{2}}{\frac{3}{2}} = \frac{2(8-a^\frac{3}{2})}{3}=\frac{2(8-a^\frac{3}{2})}{3}$
Area of OED = Area of EFCD
$\frac{2a^{\frac{3}{2}}}{3}= \frac{2(8-a^{\frac{3}{2}})}{3}\\ \\ 2a^\frac{3}{2} = 8\\ a^\frac{3}{2} = 4\\ a = (4)^\frac{2}{3}$
Therefore, the value of a is $a = (4)^\frac{2}{3}$

Question:9 Find the area of the region bounded by the parabola $\dpi{100} \small y=x^2$ and $\dpi{100} \small y=|x|$ .

We can clearly see that given area is symmetrical about y-axis
Therefore,
Area of OCAO = Area of OBDO
Point of intersection of $y=x^2 \ and \ y = |x|$ is (1 , 1) and (-1 , 1)
Now,
Area od OCAO = Area OAM - Area of OCMO
Area of OAM = $\frac{1}{2}.OM.AM = \frac{1}{2}.1.1 = \frac{1}{2}$
Area of OCMO = $\int_{0}^{1}ydx= \int_{0}^{1}x^2dx= \left [ \frac{x^3}{3} \right ]_{0}^{1}= \frac{1}{3}$
Therefore,
Area od OCAO $=\frac{1}{2}- \frac{1}{3}= \frac{1}{6}$
Now,
Area of the region bounded by the parabola $\dpi{100} \small y=x^2$ and $\dpi{100} \small y=|x|$ is = 2 X Area od OCAO $=2\times \frac{1}{6} = \frac{1}{3}$ Units

Question: 10 Find the area bounded by the curve $\dpi{100} \small x^2=4y$ and the line $\dpi{100} \small x=4y-2$ .

Points of intersections of $y = x^2 \ and \ x = 4y-2$ is
$A\left ( -1,\frac{1}{4} \right ) \ and \ B(2,1)$
Now,
Area of OBAO = Area of OBCO + Area of OCAO
Area of OBCO = Area of OMBCO- Area of OMBO

Area of OMBCO = $\int_{0}^{2}ydx = \int_{0}^{2}\frac{x+2}{4}dx = \left [ \frac{x^2}{8} \right ]_{0}^{2}+\left [ \frac{x}{2} \right ]_{0}^{2}= \frac{4}{8}+\frac{2}{2}=\frac{3}{2}$

Area of OMBO = $\int_{0}^{2}ydx = \int_{0}^{2}\frac{x^2}{4}dx = \left [ \frac{x^3}{12} \right ]_{0}^{2}= \frac{8}{12}= \frac{2}{3}$

Area of OBCO = Area of OMBCO- Area of OMBO
$= \frac{3}{2}-\frac{2}{3}= \frac{5}{6}$
Similarly,
Area of OCAO = Area of OCALO - Area of OALO

Area of OCALO = $\int_{-1}^{0}ydx = \int_{-1}^{0}\frac{x+2}{4}dx = \left [ \frac{x^2}{8} \right ]_{-1}^{0}+\left [ \frac{x}{2} \right ]_{-1}^{0}=- \frac{1}{8}-\frac{(-1)}{2}=-\frac{1}{8}+\frac{1}{2}=\frac{3}{8}$

Area of OALO = $\int_{-1}^{0}ydx = \int_{-1}^{0}\frac{x^2}{4}dx = \left [ \frac{x^3}{12} \right ]_{-1}^{0}= -\frac{(-1)}{12}= \frac{1}{12}$

Area of OCAO = Area of OCALO - Area of OALO
$=\frac{3}{8}- \frac{1}{12}= \frac{9-2}{24}= \frac{7}{24}$
Now,
Area of OBAO = Area of OBCO + Area of OCAO
$=\frac{5}{6}+ \frac{7}{24}= \frac{20+7}{24}= \frac{27}{24} = \frac{9}{8}$

Therefore, area bounded by the curve $\dpi{100} \small x^2=4y$ and the line $\dpi{100} \small x=4y-2$ is $\frac{9}{8} \ units$

Question: 11 Find the area of the region bounded by the curve $\dpi{100} \small y^2=4x$ and the line $\dpi{100} \small x=3$ .

The combined figure of the curve $y^2=4x$ and $x=3$

The required are is OABCO, and it is symmetrical about the horizontal axis.
Therefore, Area of OABCO = 2 $\times$ Area of OAB
$\\=2[\int_{0}^{3}ydx]\\ =2\int^3_02\sqrt{x}dx\\ =4[\frac{x^{3/2}}{3/2}]^3_0\\ =8\sqrt{3}$
therefore the required area is $8\sqrt{3}$ units.

Question: 12 Choose the correct answer in the following

Area lying in the first quadrant and bounded by the circle $\dpi{100} \small x^2+y^2=4$ and the lines $\dpi{100} \small x=0$ and $\dpi{100} \small x=2$ is

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

The area bounded by circle C(0,0,4) and the line x=2 is

The required area = area of OAB
$\int^2_0ydx = \int^2_0\sqrt{4-x^2}dx$
$\\=[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}\frac{x}{2}]^2_0\\ =2(\pi/2)\\ =\pi$

Question: 13 Choose the correct answer in the following.

Area of the region bounded by the curve $\dpi{100} \small y^2=4x$ , $\dpi{100} \small y$ -axis and the line $\dpi{100} \small y=3$ is

(A) $\dpi{100} \small 2$ (B) $\dpi{100} \small \frac{9}{4}$ (C) $\dpi{100} \small \frac{9}{3}$ (D) $\dpi{100} \small \frac{9}{2}$

The area bounded by the curve $y^2=4x$ and y =3

the required area = OAB =
$\\\int ^3_0xdy\\ =\int ^3_0\frac{y^2}{4}dy\\ =\frac{1}{4}.[\frac{y^3}{3}]^3_0\\ =\frac{9}{4}$

## More About NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1

The NCERT Class 12 Maths chapter Application of Integrals mainly deals with the finding out of area, volume etc. Exercise 8.1 Class 12 Maths is the extension and application of everything learnt in chapter 7. Hence before doing NCERT Solutions for Class 12 Maths chapter 8 exercise 8.1 one should complete chapter 7 Integrals.

## Benefits of NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1

• 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 8.1 solutions are helpful in understanding the theory also.

• NCERT solutions for Class 12 Maths chapter 8 exercise 8.1 provides some basic questions initially and then moderate to advanced level of questions later on.

## 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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