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) $\dpi{100} \small y=x^2,x=1,x=2$ and $\dpi{100} \small x$ -axis

The area bounded by the curve $\dpi{100} \small y=x^2,x=1,x=2$ and $\dpi{100} \small x$ -axis

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) $\dpi{100} \small y=x^4,x=1,x=5$ and $\dpi{100} \small x$ -axis

The area bounded by the curev $\dpi{100} \small y=x^4,x=1,x=5$ and $\dpi{100} \small x$ -axis

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 $\dpi{100} \small y=x$ and $\dpi{100} \small y=x^2$ .

the area between the curves $\dpi{100} \small y=x$ and $\dpi{100} \small y=x^2$ .

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 $\dpi{100} \small y=4x^2,x=0,y=1$ and $\dpi{100} \small y=4$ .

the area of the region lying in the first quadrant and bounded by $\dpi{100} \small y=4x^2,x=0,y=1$ and $\dpi{100} \small y=4$ .

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 $\dpi{100} \small y=|x+3|$ and evaluate $\dpi{100} \small \int_{-6}^{0}|x+3|dx.$

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)

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 $\dpi{100} \small y=\sin x$ between $\dpi{100} \small x=0$ and $\dpi{100} \small x=2\pi$ .

The graph of y=sinx is as follows

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 $\dpi{100} \small y^2=4ax$ and the line $\dpi{100} \small y=mx$ .

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 $\dpi{100} \small 4y=3x^2$ and the line $\dpi{100} \small 2y=3x+12$ .

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 $\dpi{100} \small \frac{x^2}{9}+\frac{y^2}{4}=1$ and the line $\dpi{100} \small \frac{x}{3}+\frac{y}{2}=1$ .

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 $\dpi{100} \small \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the line $\dpi{100} \small \frac{x}{a}+\frac{y}{b}=1$ .

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 $\dpi{100} \small x^2=y,$ the line $\dpi{100} \small y=x+2$ and the $\dpi{100} \small x$ -axis.

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 $\dpi{100} \small x^2=y,$ the line $\dpi{100} \small y=x+2$ and the $\dpi{100} \small x$ -axis is 5/6 units.

Question:11 Using the method of integration find the area bounded by the curve $\dpi{100} \small |x|+|y|=1.$

[ Hint: The required region is bounded by lines $\dpi{100} \small x+y=1,x-y=1,-x+y=1$ and $\dpi{100} \small -x-y=1$ ]

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 $\dpi{100} \small |x|+|y|=1$ is 2 units.

Question:12 Find the area bounded by curves $\dpi{100} \small \left \{ (x,y);y\geq x^2\hspace{1mm} and \hspace{1mm}y=|x| \right \}$ .

We have to find the area of the shaded region

y=|x|=x

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

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 $\dpi{100} \small A(2,0),B(4,5)\hspace{1mm}and\hspace{1mm}C(6,3).$

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:
$\dpi{100} \small 2x+y=4,3x-2y=6\hspace{1mm}and\hspace{1mm}x-3y+5=0.$

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 $\dpi{100} \small \left \{ (x,y);y^2\leq 4x,4x^2+4y^2\leq 9 \right \}$ .

We have to find the area of the shaded region OCBAO

Ar(OCBAO)=2ar(OCBO)

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

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

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

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

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

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

Area bounded by the curve $\dpi{100} \small y=x^3$ , the $\dpi{100} \small x$ -axis and the ordinates $\dpi{100} \small x=-2$ and $\dpi{100} \small x=1$ is

(A) $\dpi{100} \small -9$ (B) $\dpi{100} \small \frac{-15}{4}$ (C) $\dpi{100} \small \frac{15}{4}$ (D) $\dpi{100} \small \frac{17}{4}$

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.

T he area bounded by the curve $\dpi{100} \small y=x|x|$ , $\dpi{100} \small x$ -axis and the ordinates $\dpi{100} \small x=-1$ and $\dpi{100} \small x=1$ is given by

(A) $\dpi{100} \small 0$ (B) $\dpi{100} \small \frac{1}{3}$ (C) $\dpi{100} \small \frac{2}{3}$ (D) $\dpi{100} \small \frac{4}{3}$

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

The required area is

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

The area of the circle $\dpi{100} \small x^2+y^2=16$ exterior to the parabola $\dpi{100} \small y^2=6x$ is

(A) $\dpi{100} \small \frac{4}{3}(4\pi -\sqrt{3} )$ (B) $\dpi{100} \small \frac{4}{3}(4\pi +\sqrt{3} )$ (C) $\dpi{100} \small \frac{4}{3}(8\pi -\sqrt{3} )$ (D) $\dpi{100} \small \frac{4}{3}(8\pi +\sqrt{3} )$

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 $\dpi{100} \small y$ -axis, $\dpi{100} \small y=\cos x$ and $\dpi{100} \small y=\sin x$ when $\dpi{100} \small 0\leq x\leq \frac{\pi }{2}$ is

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

Given : $\dpi{100} \small y=\cos x$ and $\dpi{100} \small y=\sin x$

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.

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

Miscellaneous exercise chapter 8 Class 12 chapter application of integrals is quite interesting and important from exam perspective. Class 12 Maths chapter 8 miscellaneous solutions mainly deals with application based questions like area finding between two curves etc. NCERT solutions for Class 12 Maths chapter 8 miscellaneous exercise will not take much time to solve provided chapter 7 integrals are done well before chapter 8.

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!!!