# NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals (Ex 8.2) Exercise 8.2

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**NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals (Ex 8.2) Exercise 8.2 **

The Central Board of Secondary Education (CSBE) conducts the Class 12 Board Examination annually, in which over 16 lakh candidates appear. It could be an intimidating time for students. Strengths and weaknesses are part of preparation. Mathematics is a very important subject in Class 12. Every chapter carries significant marks value in the examination, and it should be revised thoroughly. With the help of the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2, students can easily understand the exercise. It helps students prepare for their final exam. Extramarks provides NCERT Solutions to help candidates with all the relevant resources. If students can deal with these subjects tactfully then they will not face many problems. The NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 can be used during the time of exam preparation for revision purposes. To fetch more marks in the class, it is very important to have conceptual clarity of the topic. Students should not delay taking references from the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2.

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

The NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 is inclusive of the process of calculating the areas bounded by the curve. Students are also introduced to the Application of Integrals such as the Area under Simple Curves, Definite Integral as the Limit of a Sum, between Lines, and Ellipses, and Parabolas. By the Integration Method, we can calculate the Average Value of a Function.

The NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 helps students strategise their learning of this topic.

The Area under Simple Curves:

The definition of the area under simple curves is the area bounded by the curve y = f(x) using the principles. It helps to calculate the area of the space by using a line and a parabola, a line and an ellipse, a line and a circle. A few examples accompany the method of calculation in the chapter.

The Area between Two Curves:

This part explains the method of calculating the area between two curves with a solution. It denotes that the area can be calculated by dividing the region into a number of pieces of small area and then adding up the area of those tiny pieces. Calculating the area of these tiny pieces is vertical is easy.

Candidate can align their notes with the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 for taking quick references and understanding difficult topics.

The NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 deals with multiple Integrals applications. Students should not delay any more to take benefit from the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2.

**Q.1** Find the area bounded by curves (x – 1)^{2} + y^{2} = 1 and x^{2} + y^{2} = 1.

**Ans**

The required area bounded by (x – 1)^{2} + y^{2} = 1 and x^{2} + y^{2} = 1 is represented by shaded area ACBOA.

\begin{array}{l}On\text{solving the given equations}{\left(\text{x}-\text{1}\right)}^{\text{2}}{\text{+ y}}^{\text{2}}={\text{1 and x}}^{\text{2}}{\text{+y}}^{\text{2}}=1,\\ we\text{get the coordinates of point of intersection as A}\left(\frac{1}{2},\text{\hspace{0.17em}}\frac{\sqrt{3}}{2}\right)\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\\ B\left(\frac{1}{2},-\text{\hspace{0.17em}}\frac{\sqrt{3}}{2}\right).This\text{area is symmetrical about x-axis}\text{.}\\ \therefore \text{}\text{\hspace{0.17em}}AreaOBCAO=2\left(area\text{\hspace{0.17em}}OCAO\right)\\ Join\text{}\text{AB such that}\text{\hspace{0.17em}}AM\perp OC.\\ So,\text{\hspace{0.17em}}the\text{coordinates of M are}\left(\frac{1}{2},0\right).\\ Therefore,\text{\hspace{0.17em}}area\text{OCAO}=Area\text{\hspace{0.17em}}OMAO+Area\text{\hspace{0.17em}}MCAM\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\displaystyle {\int}_{0}^{\frac{1}{2}}\sqrt{1-{\left(x-1\right)}^{2}}dx}+{\displaystyle {\int}_{\frac{1}{2}}^{1}\sqrt{1-{x}^{2}}}dx\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left[\frac{\left(x-1\right)}{2}\sqrt{1-{\left(x-1\right)}^{2}}+\frac{1}{2}{\mathrm{sin}}^{-1}\left(x-1\right)\right]}_{0}^{\frac{1}{2}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}+{\left[\frac{x}{2}\sqrt{1-{x}^{2}}+\frac{1}{2}{\mathrm{sin}}^{-1}x\right]}_{\frac{1}{2}}^{1}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\frac{\left(\frac{1}{2}-1\right)}{2}\sqrt{1-{\left(\frac{1}{2}-1\right)}^{2}}+\frac{1}{2}{\mathrm{sin}}^{-1}\left(\frac{1}{2}-1\right)\right]\\ \text{}-\left[\frac{\left(0-1\right)}{2}\sqrt{1-{\left(0-1\right)}^{2}}+\frac{1}{2}{\mathrm{sin}}^{-1}\left(0-1\right)\right]+\left[\frac{1}{2}\sqrt{1-{1}^{2}}+\frac{1}{2}{\mathrm{sin}}^{-1}1\right]\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}-\left[\frac{\frac{1}{2}}{2}\sqrt{1-{\left(\frac{1}{2}\right)}^{2}}+\frac{1}{2}{\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)\right]\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[-\frac{\sqrt{3}}{8}+\frac{1}{2}\left(-\frac{\pi}{6}\right)-\frac{1}{2}\left(-\frac{\pi}{2}\right)\right]+\frac{1}{2}\left(\frac{\pi}{2}\right)-\frac{\sqrt{3}}{8}-\frac{1}{2}\left(\frac{\pi}{6}\right)\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{\sqrt{3}}{8}-\frac{\pi}{12}+\frac{\pi}{4}+\frac{\pi}{4}-\frac{\sqrt{3}}{8}-\frac{\pi}{12}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{\sqrt{3}}{4}-\frac{\pi}{6}+\frac{\pi}{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{2\pi}{6}-\frac{\sqrt{3}}{4}\\ Therefore,\text{required area of OBCAO}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\left(\frac{2\pi}{6}-\frac{\sqrt{3}}{4}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\right)\text{\hspace{0.17em}}\text{square units}\end{array}

**Q.2 **Find the area of the region bounded by the curves y = x^{2} + 2, y = x, x = 0 and x = 3.

**Ans**

The area bounded by y = x^{2 }+2, y = x, x = 0 and x = 3 is represented by the shaded area

\begin{array}{l}Then,\text{\hspace{0.17em}}area\text{\hspace{0.17em}}OCBAO=area\text{\hspace{0.17em}}ODBAO-area\text{\hspace{0.17em}}ODCO\\ \text{}\text{}\text{}\text{}={\displaystyle {\int}_{0}^{3}\left({x}^{2}+2\right)dx}-{\displaystyle {\int}_{0}^{3}x\text{\hspace{0.17em}}dx}\\ \text{}\text{}\text{}\text{}={\left[\frac{{x}^{3}}{3}+2x\right]}_{0}^{3}-{\left[\frac{{x}^{2}}{2}\right]}_{0}^{3}\\ \text{}\text{}\text{}\text{}=\left[\frac{{3}^{3}}{3}+2\left(3\right)-0-0\right]-\left[\frac{{3}^{2}}{2}-0\right]\\ \text{}\text{}\text{}\text{}=15-\frac{9}{2}=\frac{21}{2}\text{\hspace{0.17em}}units\\ Thus,\text{the required area of shaded region is}\frac{21}{2}\text{\hspace{0.17em}}sq\text{\hspace{0.17em}}units.\end{array}

**Q.3 **Using integration find the area of region bounded by the triangle whose vertices are (–1, 0), (1, 3) and (3, 2).

**Ans**

Here, BL and CM are perpendicular to x-axis.

\begin{array}{l}Equation\text{of line AB is}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{y}-0=\frac{3-0}{1+1}\left(x+1\right)\\ \text{}y=\frac{3}{2}\left(x+1\right)\\ Equation\text{of line BC is}\\ \text{y}-3=\frac{2-3}{3-1}\left(x-1\right)\\ \text{}y=\frac{-1}{2}\left(x-1\right)+3\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}\left(-x+7\right)\\ Equation\text{of line CA is}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{y}-2=\frac{2-0}{3+1}\left(x-3\right)\\ \text{}y=\frac{1}{2}\left(x-3\right)+2\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}\left(x+1\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Area\text{\hspace{0.17em}}\Delta ACB=Area\left(ALBA\right)+Area\left(BLMCB\right)-Area\left(\Delta AMCA\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\displaystyle {\int}_{-1}^{1}\frac{3}{2}\left(x+1\right)dx}+{\displaystyle {\int}_{1}^{3}\frac{1}{2}\left(-x+7\right)dx}-{\displaystyle {\int}_{-1}^{3}\frac{1}{2}\left(x+1\right)dx}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3}{2}{\left[\frac{{x}^{2}}{2}+x\right]}_{-1}^{1}+\frac{1}{2}{\left[\frac{-{x}^{2}}{2}+7x\right]}_{1}^{3}-\frac{1}{2}{\left[\frac{{x}^{2}}{2}+x\right]}_{-1}^{3}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3}{2}\left[\frac{{1}^{2}}{2}+1-\frac{{\left(-1\right)}^{2}}{2}-\left(-1\right)\right]+\frac{1}{2}\left[\frac{-{3}^{2}}{2}+7\left(3\right)+\frac{{1}^{2}}{2}-7\left(1\right)\right]\\ \text{}\text{}\text{}\text{}\text{}-\frac{1}{2}\left[\frac{{3}^{2}}{2}+3-\frac{{\left(-1\right)}^{2}}{2}-\left(-1\right)\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3}{2}\left(2\right)+\frac{1}{2}\left(-\frac{9}{2}+21+\frac{1}{2}-7\right)-\frac{1}{2}\left(\frac{9}{2}+3-\frac{1}{2}+1\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=3+5-4\\ \end{array}

**Q.4 **Using integration find the area of the triangular region whose sides have the equations y = 2x + 1, y = 3x + 1 and x = 4.

**Ans**

The coordinates of intersection points of the given lines are A(0, 1), B(4, 13) and C(4, 9).** **

\begin{array}{l}Area\text{\hspace{0.17em}}\Delta ABC=Area\text{\hspace{0.17em}}OLBAO-Area\text{\hspace{0.17em}}\text{\hspace{0.17em}}OLCAO\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\displaystyle {\int}_{0}^{4}\left(3x+1\right)}dx-{\displaystyle {\int}_{0}^{4}\left(2x+1\right)dx}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left[3\frac{{x}^{2}}{2}+x\right]}_{0}^{4}-{\left[2.\frac{{x}^{2}}{2}+x\right]}_{0}^{4}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[3\frac{{\left(4\right)}^{2}}{2}+4-3\frac{{\left(0\right)}^{2}}{2}-0\right]-\left[2.\frac{{4}^{2}}{2}+4-2.\frac{{0}^{2}}{2}+0\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(24+4\right)-\left(16+4\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=28-20\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=8\text{square units}\end{array}

**Q.5 **Choose the correct answer

Smaller area enclosed by the circle x^{2} + y^{2} = 4 and the line x + y =2 is

\begin{array}{l}(a)\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\left(\pi -2\right)\\ (b)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\pi -2\\ (c)\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\pi -1\\ (d)\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\left(\pi +2\right)\end{array}

**Ans**

The smaller area enclosed by the circle, x^{2} + y^{2} = 4 and line x + y = 2 is given by shaded region ABCA.

\begin{array}{l}Area\text{\hspace{0.17em}}ABCA=Area\text{\hspace{0.17em}}OACBO-Area\Delta AOB\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\displaystyle {\int}_{0}^{2}\sqrt{\left(4-{x}^{2}\right)}}dx-{\displaystyle {\int}_{0}^{2}\left(2-x\right)dx}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left[\frac{x}{2}\sqrt{4-{x}^{2}}+\frac{4}{2}{\mathrm{sin}}^{-1}\left(\frac{x}{2}\right)\right]}_{0}^{2}-{\left[2x-\frac{{x}^{2}}{2}\right]}_{0}^{2}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\frac{2}{2}\sqrt{4-{2}^{2}}+2{\mathrm{sin}}^{-1}\left(\frac{2}{2}\right)\right]-\left[2\left(2\right)-\frac{{2}^{2}}{2}\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(2\times \frac{\pi}{2}\right)-\left(4-2\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(\pi -2\right)\text{square units}\\ \text{Therefore, correct option is B}\text{.}\end{array}

**Q.6 **Area lying between the curves y^{2} = 4x and y = 2x is

(A) 2/3 (B) 1/3 (C) 1/4 (D) 3/4

**Ans**

The area lying between y^{2 }= 4x and y = 2x is shown by shaded area given graph.

\begin{array}{l}Area\text{\hspace{0.17em}}OBAO=area\text{\hspace{0.17em}}OCABO-area\text{\hspace{0.17em}}OCAO\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\displaystyle {\int}_{0}^{1}2\sqrt{x}dx}-{\displaystyle {\int}_{0}^{1}2xdx}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2{\left[\frac{{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_{0}^{1}-2{\left[\frac{{x}^{2}}{2}\right]}_{0}^{1}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{4}{3}\left[1-0\right]-\left[{1}^{2}-0\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{4}{3}-1\text{\hspace{0.17em}}=\frac{1}{3}\text{\hspace{0.17em}}square\text{units}\\ \end{array}

**Q.7** Find the area under the given curves and given lines:

(i) y = x^{2}, x = 1, x = 2 and x- axis

(ii) y = x^{4}, x = 1, x = 5 and x- axis

**Ans**

**(i) **The required area ADCBA is represented by shaded region in graph.

\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Area\text{\hspace{0.17em}}ABCDA={\displaystyle {\int}_{1}^{2}{x}^{2}dx}\\ \text{}\text{}\text{}={\left[\frac{{x}^{3}}{3}\right]}_{1}^{2}\\ \text{}\text{}\text{}=\frac{1}{3}\left[{2}^{3}-{1}^{3}\right]\\ \text{}\text{}\text{}=\frac{7}{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}s\text{quare units}\end{array}

**(ii) **The required area ADCBA is represented by shaded region in graph.

\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Area\text{\hspace{0.17em}}ABCDA={\displaystyle {\int}_{1}^{5}{x}^{4}dx}\\ \text{}\text{}\text{}={\left[\frac{{x}^{5}}{5}\right]}_{1}^{5}\\ \text{}\text{}\text{}=\frac{1}{5}\left[{5}^{5}-{1}^{5}\right]\\ \text{}\text{}\text{}=\frac{3124}{5}\\ \text{}\text{}\text{}=624.8\text{\hspace{0.17em}}\text{\hspace{0.17em}}s\text{quare units}\end{array}

## FAQs (Frequently Asked Questions)

### 1. Que 1.What are the Mathematical Applications of Integrals?

Lots of applications of integrals are used in many subjects like Mathematics, Physics

some of them are the following:

It is used in Physics** **to know the Centre of Gravity, to calculate the Mass and Momentum of Towers, Satellites, Inertia of Vehicles, and calculate the Centre of Mass, etc.

In Mathematics to find the Area Under a Curve, the Average Value of a Curve or Between Two Curves, etc. These definitions and concepts can also be found in the **NCERT** **Solutions for Class 12 Maths Chapter 8 Exercise 8.2**

### 2. Que 2.What are the benefits of having NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2?

To solve problems in **Ex 8.2 Class 12 **students should use the **NCERT** **Solutions for Class 12 Maths Chapter 8 Exercise 8.2** provided by Extramarks. To verify the final answer or to rectify the process students can refer to the** NCERT** **Solutions for Class 12 Maths Chapter 8 Exercise 8.2.** Using the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2, candidates can easily identify and correct their errors and simple mistakes.It is uploaded on the Extramarks’ website for the students.

### 3. How can the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 help with board exam preparation?

The **NCERT** **Solutions for Class 12 Maths Chapter 8 Exercise 8.2 **have been compiled for students to understand the topics easily. Extramarks ensures that students can clarify their doubts and obtain maximum marks in the examination. The** NCERT** **Solutions for Class 12 Maths Chapter 8 Exercise 8.2 **includes questions relating to the Areas Bounded by the Curve, Definite Integral as the Limit of a Sum, between Lines, and Ellipses, Parabolas and introduces the Application of Integrals such as the Area under Simple Curves.

### 4. What are the key features of NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2?

The **NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2** are prepared with the help of subject experts. These solutions help students to firm their core concepts on Applications of Integrals. These are well-structured and organised solutions. It will help students with their assignments and ace their competitive exams.

### 5. Are the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 sufficient to score well in the CBSE exams?

These solutions have been created by the professional faculty at Extramarks. These solutions have been curated in consideration of the current CBSE Syllabus and its guidelines. Candidates need to measure their calibre and manage their study routine accordingly. The **NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2** are exercise-wise solutions to help students obtain an overall idea about the concepts for the examination. Practising these questions regularly will improve efficiency and problem-solving techniques.