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NCERT Solutions for Exercise 8.2 Class 12 Maths Chapter 8 - Application of Integrals

NCERT Solutions for Exercise 8.2 Class 12 Maths Chapter 8 - Application of Integrals

Edited By Ramraj Saini | Updated on Dec 03, 2023 10:26 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 8 Exercise 8.2

NCERT Solutions for Exercise 8.2 Class 12 Maths Chapter 8 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 exercise 8.2 is similar to the exercise 8.1. Also it has linkages with the previous chapter. Exercise 8.2 Class 12 Maths consists of questions related to areas bounded by two different curves. NCERT solutions for Class 12 Maths chapter 8 exercise 8.2 can be solved easily if the concept is understood from initial few questions. All other questions have the same concept used. Similar questions which are in this exercise are asked in Physics also, hence this exercise becomes more important.

This Story also Contains
  1. NCERT Solutions For Class 12 Maths Chapter 8 Exercise 8.2
  2. Access NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2
  3. Application of Integrals Class 12 Chapter 8 Exercise: 8.2
  4. More About NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2
  5. Also Read| Application of Integrals Class 12 Notes
  6. Benefits of NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2
  7. Key Features Of NCERT Solutions for Exercise 8.2 Class 12 Maths Chapter 8
  8. Also see-
  9. NCERT Solutions Subject Wise
  10. Subject Wise NCERT Exemplar Solutions

12th class Maths exercise 8.2 answers 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 together using the link provided below.

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Application of Integrals Class 12 Chapter 8 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 .
15947278183451594727815379
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:

1654758558749 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

1654758611010

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:

1654758688421

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)

1654758730701

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

1654758775744

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

1654758937961

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.



More About NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2

The NCERT Class 12 Maths chapter application of Integrals mainly deals with the finding out of an area bounded by two curves. Exercise 8.2 Class 12 Maths is an extension of last exercise only. Hence before doing NCERT solutions for Class 12 Maths chapter 8 exercise 8.2 one should complete the exercise 8.1.

Also Read| Application of Integrals Class 12 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2

  • 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 8.2 solutions are helpful in solving the questions in the upcoming exercise also.
  • NCERT solutions for Class 12 Maths chapter 8 exercise 8.2 provides mostly moderate to difficult level of questions.
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Key Features Of NCERT Solutions for Exercise 8.2 Class 12 Maths Chapter 8

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 8.2 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 8.2, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 8.2 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 8.2 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 8.2 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 8.2 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions

Happy learning!!!


Frequently Asked Question (FAQs)

1. How many questions are there in exercise 8.2 Class 12 Maths ?

Total of 7 questions are there in exercise 8.2 Class 12 Maths.

2. Which applications are mentioned in this exercise ?

Area under two curves are discussed in this chapter mainly. 

3. Is this chapter very important for Board exams ?

Yes, this chapter holds good weightage so it is important to cover comprehensively. 

4. What is the difficulty level of this exercise 8.2 ?

Moderate to difficult level of questions are asked from this exercise. 

5. Mention some of the topics discussed in this chapter ?

Maily area and at higher level volume etc. are discussed in this. 

6. How many exercises does this chapter include ?

3 exercises are there in this chapter including miscellaneous exercise.

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