# NCERT Solutions Class 12 Maths Chapter-8 (Application Of Integrals)Exercise 8.2

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-8 (Application Of Integrals)Exercise 8.2 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

### Exercise 8.2

Q1. Find the area of the circle

Answer. The required area is represented by the shaded area OBCDO. $\begin{array}{rl}& ={\int }_{0}^{\sqrt{2}}\sqrt{\frac{\left(9-4{x}^{2}\right)}{4}}dx-\int \sqrt{\frac{{x}^{2}}{4}}dx\\ & =\frac{1}{2}{\int }_{0}^{\sqrt{2}}\sqrt{9-4{x}^{2}}dx-\frac{1}{4}{\int }_{0}^{\sqrt{2}}{x}^{2}dx\\ & =\frac{1}{4}{\left[x\sqrt{9-4{x}^{2}}+\frac{9}{2}{\mathrm{sin}}^{-1}\frac{2x}{3}\right]}_{0}^{\sqrt{2}}-\frac{1}{4}{\left[\frac{{x}^{3}}{3}\right]}_{0}^{\sqrt{2}}\end{array}$ $\begin{array}{rl}& =\frac{1}{4}\left[\sqrt{2}\sqrt{9-8}+\frac{9}{2}{\mathrm{sin}}^{-1}\frac{2\sqrt{2}}{3}\right]-\frac{1}{12}\left(\sqrt{2}{\right)}^{3}\\ & =\frac{\sqrt{2}}{4}+\frac{9}{8}{\mathrm{sin}}^{-1}\frac{2\sqrt{2}}{3}-\frac{\sqrt{2}}{6}\\ & =\frac{\sqrt{2}}{12}+\frac{9}{8}{\mathrm{sin}}^{-1}\frac{2\sqrt{2}}{3}\\ & =\frac{1}{2}\left(\frac{\sqrt{2}}{6}+\frac{9}{4}{\mathrm{sin}}^{-1}\frac{2\sqrt{2}}{3}\right)\end{array}$

Q2. Find the area bounded by curves

Answer. $\left[\frac{1}{2}{\mathrm{sin}}^{-1}\left(1\right)-\frac{1}{4}\sqrt{1-{\left(\frac{1}{2}\right)}^{2}}-\frac{1}{2}{\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)\right]$ $\begin{array}{rl}& =\left[-\frac{\sqrt{3}}{8}+\frac{1}{2}\left(-\frac{\pi }{6}\right)-\frac{1}{2}\left(-\frac{\pi }{2}\right)\right]+\left[\frac{1}{2}\left(\frac{\pi }{2}\right)-\frac{\sqrt{3}}{8}-\frac{1}{2}\left(\frac{\pi }{6}\right)\right]\\ & =\left[-\frac{\sqrt{3}}{4}-\frac{\pi }{12}+\frac{\pi }{4}+\frac{\pi }{4}-\frac{\pi }{12}\right]+\left[\frac{1}{2}\left(\frac{\pi }{2}\right)-\frac{\sqrt{3}}{8}-\frac{1}{2}\left(\frac{\pi }{6}\right)\right]\\ & =\left[-\frac{\sqrt{3}}{4}-\frac{\pi }{6}+\frac{\pi }{2}\right]\\ & =\left[\frac{2\pi }{6}-\frac{\sqrt{3}}{4}\right]\end{array}$ Therefore ,required area OBCAO

Q3. Find the area of the region bounded by the curves

Answer. Q4.

Answer. Q5.

Answer. Q6.

Answer. Q7. Area lying between the curve  is