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

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

Edited By Ramraj Saini | Updated on Dec 04, 2023 01:24 PM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Chapter 8 Miscellaneous Exercise

NCERT Solutions for miscellaneous exercise chapter 8 class 12 Application of Integrals are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. 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.

Miscellaneous exercise class 12 chapter 8 are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise enumerated in NCERT Book together using the link provided below.

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

(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

(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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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.

[ 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.

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.

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.

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

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

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

(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.
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## Key Features Of NCERT Solutions For Class 12 Chapter 8 Miscellaneous Exercise

• Comprehensive Coverage: The solutions encompass all the topics covered in miscellaneous exercise class 12 chapter 8, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 chapter 8 maths miscellaneous solutions, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 maths miscellaneous exercise chapter 8 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this class 12 maths ch 8 miscellaneous exercise solutions, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
• Inclusive Approach: Solutions for class 12 chapter 8 miscellaneous exercise cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
• Relevance to Curriculum: The solutions for miscellaneous exercise class 12 chapter 8 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.
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## Subject Wise NCERT Exemplar Solutions

Happy learning!!!

1. How many questions are there in Miscellaneous exercise Chapter 8 ?

There are 19 questions total in Miscellaneous exercise Chapter 8.

2. Can we find an area without using integrals ?

Simple figures like triangle, circle etc. can be tackled without integration but not the complex ones.

3. Are questions repeated in the examination from this Chapter ?

Yes, in the Board exam the questions are repeated every year.

4. What is the level of questions asked from this Chapter ?

Moderate level questions are asked from this Chapter.

5. Can one skip Miscellaneous exercise ?

No, as it has some good questions, miscellaneous exercise must be done.

6. What is the time it will take to complete for the first time ?

It will take around 5-6 hours to complete for the first time.

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### Questions related to CBSE Class 12th

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Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

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Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Hello Aspirant , Hope your doing great . As per your query , your eligible for JEE mains in the year of 2025 , Every candidate can appear for the JEE Main exam 6 times over three consecutive years . The JEE Main exam is held two times every year, in January and April.

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 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9