NCERT Solutions for Class 12 Maths Chapter 8 Application of integrals

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Q: What are the important topics in chapter Application of integrals ?

Some applications of integrals like finding area under simple curves, area of the region bounded by a curve and a line, and area between two curves are the important topics covered in this chapter.

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Edited By Ramraj Saini | Updated on Jun 21, 2022 - 6:24 p.m. IST | #CBSE Class 12th

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals: In geometry, you must have learned some formulas to calculate areas of simple geometrical figures like triangles, rectangles, trapeziums, circles, etc. But how do you calculate areas enclosed by curves? In this article, you will get NCERT solutions for class 12 maths chapter 8 application of integrals. This article also includes applications of integrals class 12. Important topics that are going to be discussed in this chapter 8 class 12 maths are the area under simple curves, the area between lines and arcs of circles, parabolas, and ellipses.

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In NCERT solutions for class 12 maths chapter 8 applications of integrals article, questions from all these topics are covered. In this composition of Class 12 Maths Chapter 8 NCERT solutions application of integrals, you will learn some important application of integrals class 12. If you are interested to check all NCERT solutions from class 6 to 12 in a single place, which will help you to learn CBSE maths and science. Here you will get NCERT solutions for class 12.

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Topics and sub-topics of NCERT Grade 12 Maths Chapter - 8 Application of integrals-

8.1 Introduction

8.2 Area under Simple Curves

8.2.1 The area of the region bounded by a curve and a line

8.3 Area between Two Curves

NCERT solutions for class 12 maths chapter 8 applications of integrals Exercise: 8.1

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

Answer:

Area of the region bounded by the curve $y^2=x$ and the lines $x=1,x=4$ and the $x$ -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 $y^2=9x,x=2,x=4$ and the $x$ -axis in the first quadrant.

Answer:

Area of the region bounded by the curve $y^2=9x,x=2,x=4$ and the $x$ -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 $x^2=4y,y=2,y=4$ and the $y$ -axis in the first quadrant.

Answer:

The area bounded by the curves $x^2=4y,y=2,y=4$ and the $y$ -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 $\frac{x^2}{16}+\frac{y^2}{9}=1.$

Answer:

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

1646972152541

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 $\small \frac{x^2}{4}+\frac{y^2}{9}=1$

Answer:

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

1646972197883

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 $\small x$ -axis, line $\small x=\sqrt{3}y$ and the circle $\small x^2+y^2=4$
Answer:

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

1646972234725

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 $\small x^2+y^2=a^2$ cut off by the line
$\small x=\frac{a}{\sqrt{2}}$

Answer:

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 $\small x=y^2$ and $\small x=4$ is divided into two equal parts by the line $\small x=a$ , find the value of $\small a$ .

Answer:

1646972308309 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 $\small y=x^2$ and $\small y=|x|$ .

Answer:

1646972345767 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 $\small y=x^2$ and $\small y=|x|$ is = 2 X Area od OCAO $=2\times \frac{1}{6} = \frac{1}{3}$ Units

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

Answer:

1646972388198 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 $\small x^2=4y$ and the line $\small x=4y-2$ is $\frac{9}{8} \ units$

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

Answer:
The combined figure of the curve $y^2=4x$ and $x=3$
1594726575288
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 $\small x^2+y^2=4$ and the lines $\small x=0$ and $\small x=2$ is

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

Answer:

The correct answer is A
The area bounded by circle C(0,0,4) and the line x=2 is
1594726886037
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 $\small y^2=4x$ , $\small y$ -axis and the line $\small y=3$ is

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

Answer:

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

1594727150537
the required area = OAB =
$\\\int ^3_0xdy\\ =\int ^3_0\frac{y^2}{4}dy\\ =\frac{1}{4}.[\frac{y^3}{3}]^3_0\\ =\frac{9}{4}$

NCERT solutions for class 12 maths chapter 8 application of integrals Exercise: 8.2

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

Answer:

The area bounded by the circle $\small 4x^2+4y^2=9$ and the parabola $\small x^2=4y$ .
1594727818345
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 $\small (x-1)^2+y^2=1$ and $\small x^2+y^2=1$ .

Answer:

1646972457138 Given curves are $\small (x-1)^2+y^2=1$ and $\small x^2+y^2=1$

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 $\small y=x^2+2,y=x,x=0$ and $\small x=3$ .

Answer:

The area of the region bounded by the curves,

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

1646972497382

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 $\small (-1,0),(1,3)$ and $\small (3,2)$ .

Answer:

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:

1646972546438

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 $\small y=2x+1,y=3x+1$ and $\small x=4$ .

Answer:

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 $A(0,1),B(4,13),\ and\ C(4,9)$

1646972586017

Thus it can be seen that,

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

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

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

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

Question:6 Choose the correct answer.

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

(A) $\small 2(\pi -2)$ (B) $\small \pi -2$ (C) $\small 2\pi -1$ (D) $\small 2(\pi +2)$

Answer:

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

1646972621266

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.

Question:7 Choose the correct answer.

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

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

Answer:

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

1646972657802

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.

NCERT solutions for class 12 maths chapter 8 application of integrals Miscellaneous: Exercise

Question:1 Find the area under the given curves and given lines:

(i) $\small y=x^2,x=1,x=2$ and $\small x$ -axis

Answer:

The area bounded by the curve $\small y=x^2,x=1,x=2$ and $\small x$ -axis
1594728126741
The area of the required region = area of ABCD
$\\=\int_{1}^{2}ydx\\ =\int_{1}^{2}x^2dx\\ =[\frac{x^3}{3}]_1^2\\ =\frac{7}{3}$
Hence the area of shaded region is 7/3 units

Question:1 Find the area under the given curves and given lines:

(ii) $\small y=x^4,x=1,x=5$ and $\small x$ -axis

Answer:

The area bounded by the curev $\small y=x^4,x=1,x=5$ and $\small x$ -axis

1594728286834
The area of the required region = area of ABCD
$\\=\int_{1}^{5}ydx\\ =\int_{1}^{2}x^4dx\\ =[\frac{x^5}{5}]_1^2\\ =625-\frac{1}{5}\\ =624.8$
Hence the area of the shaded region is 624.8 units

Question:2 Find the area between the curves $\small y=x$ and $\small y=x^2$ .

Answer:

the area between the curves $\small y=x$ and $\small y=x^2$ .
AOI graph
The curves intersect at A(1,1)
Draw a normal to AC to OC(x-axis)
therefore, the required area (OBAO)= area of (OCAO) - area of (OCABO)
$\\=\int_{0}^{1}xdx-\int_{0}^{1}x^2dx\\ =[\frac{x^2}{2}]_0^1-[\frac{x^3}{3}]_0^1\\ =1/2-1/3\\ =\frac{1}{6}$
Thus the area of shaded region is 1/6 units

Question:3 Find the area of the region lying in the first quadrant and bounded by $\small y=4x^2,x=0,y=1$ and $\small y=4$ .

Answer:

the area of the region lying in the first quadrant and bounded by $\small y=4x^2,x=0,y=1$ and $\small y=4$ .

1594728678865
The required area (ABCD) =
$\\=\int_{1}^{4}xdy\\ =\int_{1}^{4}\frac{\sqrt{y}}{2}dy\\ =\frac{1}{2}.\frac{2}{3}[y^{3/2}]_1^4\\ =\frac{1}{3}[8-1]\\ =\frac{7}{3}$
The area of the shaded region is 7/3 units

Question:4 Sketch the graph of $\small y=|x+3|$ and evaluate $\small \int_{-6}^{0}|x+3|dx.$

Answer:

y=|x+3|

the given modulus function can be written as

x+3>0

x>-3

for x>-3

y=|x+3|=x+3

x+3<0

x<-3

For x<-3

y=|x+3|=-(x+3)

1646972694635

Integral to be evaluated is

$\\\int_{-6}^{0}|x+3|dx\\ =\int_{-6}^{-3}(-x-3)dx+\int_{-3}^{0}(x+3)dx\\ =[-\frac{x^{2}}{2}-3x]_{-6}^{-3}+[\frac{x^{2}}{2}+3x]_{-3}^{0}\\ =(-\frac{9}{2}+9)-(-18+18)+0-(\frac{9}{2}-9)\\ =9$

Question:5 Find the area bounded by the curve $\small y=\sin x$ between $\small x=0$ and $\small x=2\pi$ .

Answer:

The graph of y=sinx is as follows

1646972720677

We need to find the area of the shaded region

ar(OAB)+ar(BCD)

=2ar(OAB)

$\\=2\times \int_{0}^{\pi }sinxdx\\ =2\times [-cosx]_{0}^{\pi }\\ =2\times [-(-1)-(-1)]\\ =4$

The bounded area is 4 units.

Question:6 Find the area enclosed between the parabola $\small y^2=4ax$ and the line $\small y=mx$ .

Answer:

1646972749429

We have to find the area of the shaded region OBA

The curves y=mx and y ² =4ax intersect at the following points

$\left ( 0,0 \right )and\left ( \frac{4a}{m^{2}},\frac{4a}{m} \right )$

$\\y^{2}=4ax\\ \Rightarrow y=2\sqrt{ax}$

The required area is

$\\\int_{0}^{\frac{4a}{m^{2}}}(2\sqrt{ax}-mx)\\ =2\sqrt{a}[\frac{2x^{\frac{3}{2}}}{3}]_{0}^{\frac{4a}{m^{2}}}-m[\frac{x^{2}}{2}]_{0}^{\frac{4a}{m^{2}}}\\ =\frac{32a^{2}}{3m^{3}}-\frac{8a^{2}}{m^{3}}\\ =\frac{8a^{2}}{3m^{3}}units$

Question:7 Find the area enclosed by the parabola $\small 4y=3x^2$ and the line $\small 2y=3x+12$ .

Answer:

1646972775838

We have to find the area of the shaded region COB

$\\2y=3x+12\\ \Rightarrow y=\frac{3}{2}x+6\\ 4y=3x^{2}\\ \Rightarrow y=\frac{3x^{2}}{4}$

The two curves intersect at points (2,3) and (4,12)

Required area is

$\\\int_{-2}^{4}(\frac{3}{2}x+6-\frac{3x^{2}}{4})dx\\ =[\frac{3x^{2}}{4}+6x-\frac{x^{3}}{4}]{_{-2}}^{4}\\ =[12+24-16]-[3-12+2]\\ =20-(-7)\\ =27\ units$

Question:8 Find the area of the smaller region bounded by the ellipse $\small \frac{x^2}{9}+\frac{y^2}{4}=1$ and the line $\small \frac{x}{3}+\frac{y}{2}=1$ .

Answer:

1646972803949

We have to find the area of the shaded region

The given ellipse and the given line intersect at following points

$\left ( 0,2 \right )and \left ( 3,0 \right )$

$\\\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\\ y=\frac{2}{3}\sqrt{9-x^{2}}$

Since the shaded region lies above x axis we take y to be positive

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

The required area is

$\\\frac{2}{3}\int_{0}^{3}\left ( \sqrt{9-x^{2}}-(3-x) \right )dx\\ =\frac{2}{3}[\frac{x}{2}(\sqrt{9-x^{2}})+\frac{9}{2}sin^{-1}\frac{x}{3}-3x+\frac{x^{2}}{2}]_{0}^{3}\\ =\frac{2}{3}\left ( \left [ \frac{9}{2}\times \frac{\pi }{2}-9+\frac{9}{2} \right ]-0 \right )\\ =\frac{2}{3}(\frac{9\pi }{4}-\frac{9}{2})\\ =\frac{3}{2}(\pi -2)units$

Question:9 Find the area of the smaller region bounded by the ellipse $\small \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the line $\small \frac{x}{a}+\frac{y}{b}=1$ .

Answer:

1646972828912

The area of the shaded region ACB is to be found

The given ellipse and the line intersect at following points

$\left ( 0,b \right )and\left ( a,0 \right )$

$\\\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\\ \Rightarrow y=\frac{b}{a}\sqrt{a^{2}-x^{2}}$

Y will always be positive since the shaded region lies above x axis

$\\\frac{x}{a}+\frac{y}{b}=1\\ \Rightarrow y=\frac{b}{a}(a-x)$

The required area is

$\\\frac{b}{a}\int_{0}^{a}(\sqrt{a^{2}-x^{2}}-(a-x))dx\\ =\frac{b}{a}[\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\frac{x}{a}-ax+\frac{x^{2}}{2}]_{0}^{a}\\ =\frac{b}{a}[(\frac{a^{2}}{2}\times \frac{\pi }{2}-a^{2}+\frac{a^{2}}{2})]\\ =\frac{b}{a}(\frac{\pi a^{2}}{4}-\frac{a^{2}}{2})\\ =\frac{ab}{4}(\pi -2)units$

Question:10 Find the area of the region enclosed by the parabola $\small x^2=y,$ the line $\small y=x+2$ and the $\small x$ -axis.

Answer:

1646972855917

We have to find the area of the shaded region BAOB

O is(0,0)

The line and the parabola intersect in the second quadrant at (-1,1)

The line y=x+2 intersects the x axis at (-2,0)

$\\ar(BAOB)=ar(BAC)+ar(ACO)\\ =\int_{-2}^{-1}(x+2)dx+\int_{-1}^{0}(x^{2})dx\\ =[\frac{x^{2}}{2}+2x]_{-2}^{-1}+[\frac{x^{3}}{3}]_{-1}^{0}\\ =(\frac{1}{2}-2)-(2-4)+0-(-\frac{1}{3})\\ =\frac{5}{6}\ units$

The area of the region enclosed by the parabola $\small x^2=y,$ the line $\small y=x+2$ and the $\small x$ -axis is 5/6 units.

Question:11 Using the method of integration find the area bounded by the curve $\small |x|+|y|=1.$

[ Hint: The required region is bounded by lines $\small x+y=1,x-y=1,-x+y=1$ and $\small -x-y=1$ ]

Answer:

1646972880742

We need to find the area of the shaded region ABCD

ar(ABCD)=4ar(AOB)

Coordinates of points A and B are (0,1) and (1,0)

Equation of line through A and B is y=1-x

$\\ar(AOB)=\int_{0}^{1}(1-x)dx\\ =[x-\frac{x^{2}}{2}]_{0}^{1}\\ =(1-\frac{1}{2})-0 \\=\frac{1}{2}\ units\\ ar(ABCD)=4ar(AOB)\\ =4\times \frac{1}{2}\\ =2\ units$

The area bounded by the curve $\small |x|+|y|=1$ is 2 units.

Question:12 Find the area bounded by curves $\small \left \{ (x,y);y\geq x^2\hspace{1mm} and \hspace{1mm}y=|x| \right \}$ .

Answer:

1646972905611

We have to find the area of the shaded region

In the first quadrant

y=|x|=x

Area of the shaded region=2ar(OADO)

$\\=2\int_{0}^{1}(x-x^{2})dx\\ =2[\frac{x^{2}}{2}-\frac{x^{3}}{3}]{_{0}}^{1} \\=2(\frac{1}{2}-\frac{1}{3})-0\\ =1-\frac{2}{3}\\ =\frac{1}{3}\ units$

The area bounded by the curves is 1/3 units.

Question:13 Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are $\small A(2,0),B(4,5)\hspace{1mm}and\hspace{1mm}C(6,3).$

Answer:

1646972934221

Equation of line joining A and B is

$\\\frac{y-0}{x-2}=\frac{5-0}{4-2}\\ y=\frac{5x}{2}-5$

Equation of line joining B and C is

$\\\frac{y-5}{x-4}=\frac{5-3}{4-6}\\ y=9-x$

Equation of line joining A and C is

$\\\frac{y-0}{x-2}=\frac{3-0}{6-2}\\ y=\frac{3x}{4}-\frac{3}{2}$

ar(ABC)=ar(ABL)+ar(LBCM)-ar(ACM)

$\\ar(ABL)=\int_{2}^{4}(\frac{5x}{2}-5)dx\\ =[\frac{5x^{2}}{4}-5x]_{2}^{4}\\ =(20-20)-(5-10)\\ =5\ units$

$\\ar(LBCM)=\int_{4}^{6}(9-x)dx\\ =[9x-\frac{x^{2}}{2}]_{4}^{6}\\ =(54-18)-(36-8)\\ =8\ units$

$\\ar(ACM)=\int_{2}^{6}(\frac{3x}{4}-\frac{3}{2})dx\\ =[\frac{3x^{2}}{8}-\frac{3x}{2}]_{2}^{6}\\ =(\frac{27}{2}-9)-(\frac{3}{2}-3)\\ =6\ units$

ar(ABC)=8+5-6=7

Therefore the area of the triangle ABC is 7 units.

Question:14 Using the method of integration find the area of the region bounded by lines:
$\small 2x+y=4,3x-2y=6\hspace{1mm}and\hspace{1mm}x-3y+5=0.$

Answer:

1646972962041

We have to find the area of the shaded region ABC

ar(ABC)=ar(ACLM)-ar(ALB)-ar(BMC)

The lines intersect at points (1,2), (4,3) and (2,0)

$\\x-3y=-5\\ y=\frac{x}{3}+\frac{5}{3}$

$\\ar(ACLM)=\int_{1}^{4}(\frac{x}{3}+\frac{5}{3})dx\\ =[\frac{x^{2}}{6}+\frac{5x}{3}]_{1}^{4}\\ =(\frac{4^{2}}{6}+\frac{5\times 4}{3})-(\frac{1}{6}+\frac{5}{3})\\ =\frac{15}{2}\ units$

$\\2x+y=4\\ y=4-2x$

$\\ar(ALB)=\int_{1}^{2}(4-2x)dx\\ =[4x-x^{2}]_{1}^{2}\\ =(8-4)-(4-1)\\ =1\ unit$

$\\3x-2y=6\\ y=\frac{3x}{2}-3$

$\\ar(BMC)=\int_{2}^{4}(\frac{3x}{2}-3)dx\\ =[\frac{3x^{2}}{4}-3x]_{2}^{4}\\ =(12-12)-(3-6)\\ =3\ units$

$\\ ar(ABC)=\frac{15}{2}-1-3\\ =\frac{7}{2}\ units$

Area of the region bounded by the lines is 3.5 units

Question:15 Find the area of the region $\small \left \{ (x,y);y^2\leq 4x,4x^2+4y^2\leq 9 \right \}$ .

Answer:

1646972988754

We have to find the area of the shaded region OCBAO

Ar(OCBAO)=2ar(OCBO)

For the fist quadrant

$\\4x^{2}+4y^{2}=9\\ y=\sqrt{\frac{9}{4}-x^{2}}$

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

In the first quadrant, the curves intersect at a point $\left ( \frac{1}{2},\sqrt{2} \right )$

Area of the unshaded region in the first quadrant is

$\\\int_{0}^{\frac{1}{2}}\left ( \sqrt{\frac{9}{4}-x^{2}} -2\sqrt{x}\right )dx\\ =[\frac{x}{2}\sqrt{\frac{9}{4}-x^{2}}+\frac{9}{8}sin^{-1}\frac{2x}{3}]{_{0}}^{\frac{1}{2}}-4[\frac{x^{3/2}}{3}]_0^{1/2}\\ =\frac{\sqrt{2}}{4}+\frac{9}{8}sin^{-1}\frac{1}{3}-\frac{\sqrt{2}}{3}$

The total area of the shaded region is-
= Area of half circle - area of the shaded region in the first quadrant

$\\\frac{\pi }{2}\times (\frac{3}{2})^{2}-2\left ( \frac{\sqrt{2}}{4}-\frac{\sqrt{2}}{3}+\frac{9}{8}sin^{-1}\frac{1}{3}\right )\\ =\frac{9 }{8}\left ( \pi-2sin^{-1}\frac{1}{3} \right )+\frac{\sqrt{2}}{6}\ units$

Question:16 Choose the correct answer.

Area bounded by the curve $\small y=x^3$ , the $\small x$ -axis and the ordinates $\small x=-2$ and $\small x=1$ is

(A) $\small -9$ (B) $\small \frac{-15}{4}$ (C) $\small \frac{15}{4}$ (D) $\small \frac{17}{4}$

Answer:

1646973022742

Hence the required area

$=\int_{-2}^1 ydx$

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

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

$= \left [ 0-\frac{(-2)^4}{4} \right ] + \left [ \frac{1}{4} - 0 \right ]$

$= -4+\frac{1}{4} = \frac{-15}{4}$

Therefore the correct answer is B.

Question:17 Choose the correct answer.

T he area bounded by the curve $\small y=x|x|$ , $\small x$ -axis and the ordinates $\small x=-1$ and $\small x=1$ is given by

(A) $\small 0$ (B) $\small \frac{1}{3}$ (C) $\small \frac{2}{3}$ (D) $\small \frac{4}{3}$

[ Hint : $y=x^2$ if $x> 0$ and $y=-x^2$ if $x<0$ . ]

Answer:

The required area is

$\\2\int_{0}^{1}x^{2}dx\\ =2\left [ \frac{x^{3}}{3} \right ]_{0}^{1}\\ =\frac{2}{3}\ units$

Question:18 Choose the correct answer.

The area of the circle $\small x^2+y^2=16$ exterior to the parabola $\small y^2=6x$ is

(A) $\small \frac{4}{3}(4\pi -\sqrt{3} )$ (B) $\small \frac{4}{3}(4\pi +\sqrt{3} )$ (C) $\small \frac{4}{3}(8\pi -\sqrt{3} )$ (D) $\small \frac{4}{3}(8\pi +\sqrt{3} )$

Answer:

1646973081315

The area of the shaded region is to be found.

Required area =ar(DOC)+ar(DOA)

The region to the left of the y-axis is half of the circle with radius 4 units and centre origin.

Area of the shaded region to the left of y axis is ar(1) = $\frac{\pi 4^{2}}{2}=8\pi\ units$

For the region to the right of y-axis and above x axis

$\\x^{2}+y^{2}=16\\ y=\sqrt{16-x^{2}}$

$\\y^{2}=6x\\ y=\sqrt{6x}$

The parabola and the circle in the first quadrant intersect at point

$\left ( 2,2\sqrt{3} \right )$

Remaining area is 2ar(2) is

$\\ar(2)=\int_{0}^{2}\left ( \sqrt{16-x^{2}}-\sqrt{6x} \right )dx\\ =[ \frac{x}{2}\sqrt{16-x^{2}}+\frac{16}{2}sin^{-1}(\frac{x}{4})-\frac{2\sqrt{6}}{3}x^{\frac{3}{2}} ]{_{0}}^{2}\\ =[\sqrt{12}+\frac{16}{2}\times \frac{\pi }{6}-\frac{4\sqrt{12}}{3}]\\ =\frac{4\pi }{3}-\frac{2\sqrt{3}}{3}$

Total area of shaded region is

$\\ar(1)+2ar(2)\\ =8\pi +\frac{8\pi}{3}-\frac{4\sqrt{3}}{3}\\ =\frac{4}{3}(8\pi -\sqrt{3})\ units$

Question:19 Choose the correct answer The area bounded by the $\small y$ -axis, $\small y=\cos x$ and $\small y=\sin x$ when $\small 0\leq x\leq \frac{\pi }{2}$ is

(A) $\small 2(\sqrt{2}-1)$ (B) $\small \sqrt{2}-1$ (C) $\small \sqrt{2}+1$ (D) $\small \sqrt{2}$

Answer:

Given : $\small y=\cos x$ and $\small y=\sin x$

1646973113069

Area of shaded region = area of BCDB + are of ADCA

$=\int_{\frac{1}{\sqrt{2}}}^{1}x dy +\int_{1}^{\frac{1}{\sqrt{2}}}x dy$

$=\int_{\frac{1}{\sqrt{2}}}^{1} cos^{-1} y .dy +\int_{1}^{\frac{1}{\sqrt{2}}} sin^{-1}x dy$

$=[y. cos^{-1}y - \sqrt{1-y^2}]_\frac{1}{\sqrt{2}}^1 + [x. sin^{-1}x + \sqrt{1-x^2}]_1^\frac{1}{\sqrt{2}}$

$= cos^{-1}(1)-\frac{1}{\sqrt{2}} cos^{-1}(\frac{1}{\sqrt{2}})+\sqrt{1-\frac{1}{2}}+\frac{1}{\sqrt{2}} sin^{-1}(\frac{1}{\sqrt{2}})+\sqrt{1-\frac{1}{2}}-1$

$=\frac{-\pi }{4\sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{\pi }{4\sqrt{2}}+\frac{1}{\sqrt{2}}-1$

$=\frac{2}{\sqrt{2}} - 1$

$=\sqrt{2} - 1$

Hence, the correct answer is B.

If you are looking for exercises solutions for chapter application of integrals class 12 then these are listed below.

Frequently Asked Question (FAQs) - NCERT Solutions for Class 12 Maths Chapter 8 Application of integrals

Question: What is the weightage of the chapter Application of integrals for CBSE board exam ?

Answer:

Generally, one question of 5 marks is asked from this chapter in the 12th board final exam but if you want to obtain the full 5 five marks that demands practice, good strategy, and command of concepts, and therefore NCERT textbooks, NCERT syllabus becomes very important. it is advised for students to precisely go through the basics to score good marks.

Question: How does the NCERT solutions are helpful in the CBSE board exam ?

Answer:

Only knowing the answer is not enough to score good marks in the exam. One should know how to write the answer to board exam in order to get good marks. NCERT solutions are provided by experts who knows how best to write the answer in the board exam in order to get good marks.

Question: Does CBSE provides the solutions of NCERT for class 12 maths ?

Answer:

No, CBSE doesn’t provide NCERT solutions for any class or subject but you can download NCERT solutions from the careers360 website. On the careers360 website, you not only find NCERT solutions but also you can read NCERT Notes, NCERT Syllabus, and solutions of NCERT exercises.

Question: Where can I find the complete solutions of NCERT for class 12 maths ?

Answer:

Here you will get the detailed and comprehensive NCERT Solutions for class 12 maths by clicking on the link. if you want to study other material like NCERT Syllabus, NCERT Notes, NCERT Exercise Solutions, etc then you can browse the careers360 official website.

Question: What are the important topics in chapter Application of integrals ?

Answer:

Some applications of integrals like finding area under simple curves, area of the region bounded by a curve and a line, and area between two curves are the important topics covered in this chapter.

Question: Which is the official website of NCERT ?

Answer:

NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12 that can be downloaded and studied offline.

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Showing 146 out of 146 Questions

7 Views

pls tell correct info about compartment and improvement exams 2022. pls o want to know all the processes. pleasw

Devyanshi Sharma 23rd Jun, 2022

Hello,if you are from cbse board then your compartment exams will be organised by the central Board of secondary education.These examinations are held for those students who will not clear the class 10th or class 12th exam in one or two subjects.Cbse has a provision for providing three chances to clear the compartment exam.If you are not able to clear your exams or secure good grades even after three attempts you are considered failed .Both private and regular students can apply for cbse compartment examination.For class 10th compartment examinations will begin from 25th August and end on 8th of September whereas for class 12th students examinations will begin from 25th August and end on 15th September.The application form is different for every individual hailing from different categories.If you are a regular student who appeared in class 12th examinations during the month of April/June and their result is declared compartment are eligible to appear for compartment examinations.if you have failed in more than one subject then you will have to appear for two subjects in which you faced compartment.If you were not able to clear one subject you can appear for compartment examinations under the improvement of performance category.Regular students may habe to go to the school and enquire about compartment examinations whereas private students can easily apply through the cbse site .

This is the criteria i have told you for the cbse board.If you are from any other board like hp board,up board or Maharashtra board,let me know and i would enlighten you with the knowledge i have regarding the compartment examinations held by these boards

42 Views

if someone got 80% in term one, passed term 2 practical but failed in term 2 theory cbse class 12th will they be considered pass or fail in 12th?

prashant kumar 11th Jun, 2022

See , frankly speaking, this year many new guidelines have been issued by the C.B.S.E . So , it will be better if you wait for the results to be declared. If , you are failing in one or two subjects, you can again give the exam this year and pass it .

40 Views

I want to take a admission in class 12 privately.. so can I???

Parvati Solanki 2nd Jun, 2022

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

70 Views

Im in class 12 CBSE right now. I feel Im going to fail in my core subjects because of my weak preparation. Please guide me on what should I do in case I end up failing? Should I take private or compartment or improvement? I might fail in all 3 main subjects (PCM) or 2 subjects. Its confusing.

Pallavi 6th Jun, 2022

Hello aspirant,

--> If you fail in 2 subjects - You will be eligible for giving compartment exams of both of them. However, if you fail in 3 exams your result will be declared as - Fail. Hence, you have to repeat a year again in-order to clear 12th.

Remember that life doesn't end, if you haven't passed your 12th. There are lot of options to get succeed in life.

In case you couldn't clear in Class 12, you may use your 10th standard certificate and get enrolled in diploma courses such as Mechanical Engineering, Electrical Engineering etc.

All the best!

58 Views

sir board exam 2022-23 me 30% syllabus reduce hoga?

Shivanshu 15th May, 2022

Dear aspirant !

Hope you are doing great ! No, why will it reduce ,there is no covid like situation as in the past years ,so why are you wasting your time by thinking such thoughts,no syllabus will be reduced onwards ,if covid situation arises then there may be possibility to reduce the syllabus , but as see the present situation ,no syllabus will deduct ,and also if syllabus reduced then it is not profitable for you because in neet and jee ,full syllabus comes , so if they reduce the syllabus then study full syllabus !

Hope it helps you

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chapter 1	NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
chapter 2	NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions
chapter 3	NCERT solutions for class 12 maths chapter 3 Matrices
chapter 4	NCERT solutions for class 12 maths chapter 4 Determinants
chapter 5	NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability
chapter 6	NCERT solutions for class 12 maths chapter 6 Application of Derivatives
chapter 7	NCERT solutions for class 12 maths chapter 7 Integrals
chapter 8	NCERT solutions for class 12 maths chapter 8 Application of Integrals
chapter 9	NCERT solutions for class 12 maths chapter 9 Differential Equations
chapter 10	NCERT solutions for class 12 maths chapter 10 Vector Algebra
chapter 11	NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry
chapter 12	NCERT solutions for class 12 maths chapter 12 Linear Programming
chapter 13	NCERT solutions for class 12 maths chapter 13 Probability

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NCERT Solutions

NCERT Solutions for Class 12 Maths Chapter 8 Application of integrals

NCERT solutions for class 12 maths chapter 8 applications of integrals Exercise: 8.1

NCERT solutions for class 12 maths chapter 8 application of integrals Exercise: 8.2

NCERT solutions for class 12 maths chapter 8 application of integrals Miscellaneous: Exercise

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