NCERT Solutions for Miscellaneous Exercise Chapter 8 Class 12 - Application of Integrals

NCERT solutions for Class 12 Maths chapter 8 Miscellaneous exercise deals with the area finding of the various curves and intersection of two curves. Class 12 Maths chapter 8 miscellaneous exercise is a good source to practice good questions before the exam. NCERT book Class 12 Maths chapter 8 miscellaneous exercise is relatively easier than the chapter 7 exercise as it provides questions based on the application part. Class 12 Maths chapter 8 miscellaneous exercise solutions with other two exercises are sufficient to score good marks. Also NCERT Exemplar questions can be referred for wide understanding of the topic.

Application of Integrals Exercise 8.1
Application of Integrals Exercise 8.2

Application of Integrals Class 12 Chapter 8 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 1594728124007
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 1594728284206
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 1594728523616
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 1594728676557
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)

1654760706138

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

1654760755958

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:

1654760804965

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:

1654764435919

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:

1654764622123

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:

1654764663614

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:

1654764778111

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:

1654764830021

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:

1654764925341

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:

1654764957923

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:

1654764994031

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:

1654765056298

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:

1654765098486

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:

1654765195847

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 $\small y=\sin x$

1654765304392

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.

Benefits of Ncert Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercises

The Class 12th Maths chapter 8 exercise has some good questions which must be done before examination.
NCERT syllabus Class 12 Maths chapter 8 miscellaneous exercises can be done along with chapter 7 as basic concepts used are the same.
These Class 12 Maths chapter 8 miscellaneous exercises are also useful in Physics and chemistry problems.

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

Happy learning!!!

NCERT Solutions for Miscellaneous Exercise Chapter 8 Class 12 - Application of Integrals

Application of Integrals Class 12 Chapter 8 Miscellaneous: Exercise

More About NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise

Also Read| Application of Integrals Class 12 Notes

Benefits of Ncert Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercises

Also see-

NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions