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 $4x^2+4y^2=9\text{which is interior of the parabola}{x}^2=4y$

Answer. The required area is represented by the shaded area OBCDO.  $\begin{array}{c}\text{Solving the given equation of circle,}4x^2+4y^2=9,\text{and parabola,}{x}^2=4y,\text{we obtain the}\\ \text{point of intersection as}(\sqrt{2},\frac{1}{2})\text{and}D(-\sqrt{2},\frac{1}{2})\\ \text{It can be observed that the required area is symmetrical about y-axis.}\end{array}$ $\begin{array}{c}\therefore \text{Area OBCDO =}2\times \text{Area OBCO}\\ \text{We draw BM perpendicular to OA.}\\ \text{Therefore, the coordinates of M are}(\sqrt{2},0)\text{.}\\ \text{Therefore, Area OBCO = Area OMBCO - Area OMBO}\end{array}$ $\begin{array}{cc}& ={\int }_0^{\sqrt{2}}\sqrt{\frac{(9-4x^2)}{4}}dx-\int \sqrt{\frac{x^2}{4}}dx\\ & =\frac{1}{2}{\int }_0^{\sqrt{2}}\sqrt{9-4x^2}dx-\frac{1}{4}{\int }_0^{\sqrt{2}}{x}^{2}dx\\ & =\frac{1}{4}{[x\sqrt{9-4x^2}+\frac{9}{2}{\sin }^{-1}\frac{2x}{3}]}_0^{\sqrt{2}}-\frac{1}{4}{\left[\frac{x^3}{3}\right]}_0^{\sqrt{2}}\end{array}$ $\begin{array}{cc}& =\frac{1}{4}[\sqrt{2}\sqrt{9-8}+\frac{9}{2}{\sin }^{-1}\frac{2\sqrt{2}}{3}]-\frac{1}{12}(\sqrt{2}{)}^3\\ & =\frac{\sqrt{2}}{4}+\frac{9}{8}{\sin }^{-1}\frac{2\sqrt{2}}{3}-\frac{\sqrt{2}}{6}\\ & =\frac{\sqrt{2}}{12}+\frac{9}{8}{\sin }^{-1}\frac{2\sqrt{2}}{3}\\ & =\frac{1}{2}(\frac{\sqrt{2}}{6}+\frac{9}{4}{\sin }^{-1}\frac{2\sqrt{2}}{3})\end{array}$ $\begin{array}{c}\text{Therefore, the required area OBCDO is}\\ (2\times \frac{1}{2}[\frac{\sqrt{2}}{6}+\frac{9}{4}{\sin }^{-1}\frac{2\sqrt{2}}{3}])=[\frac{\sqrt{2}}{6}+\frac{9}{4}{\sin }^{-1}\frac{2\sqrt{2}}{3}]\text{units}\end{array}$

Q2. Find the area bounded by curves $(x-1{)}^2+y^2=1\text{and}{x}^2+y^2=1$

Answer. $\begin{array}{c}\text{The area bounded by the curves,}(x-1{)}^2+y^2=1\text{and}{x}^2+y^2=1,\text{is represented by}\\ \text{the shaded area as}\end{array}$  $\begin{array}{c}\text{On solving the equations,}(x-1{)}^2+y^2=1\text{and}{x}^2+y^2=1,\text{we obtain the point of}\\ \text{intersection as}A(\frac{1}{2},\frac{\sqrt{3}}{2})\\ \text{It can be observed that the required area is symmetrical about x-axis.}\end{array}$ $\begin{array}{c}\therefore \text{Area OBCAO}=2\times \text{Area OCAO}\\ \text{We join AB, which intersects oC at M, such that AM is perpendicular to oc.}\\ \text{The coordinates of M are}(\frac{1}{2},0)\end{array}$ $\begin{array}{cc}\Rightarrow \text{Area}OCAO& =\text{Area OMAO+Area MCAM}\\ & =[{\int }_0^{\frac{1}{2}}\sqrt{1-(x-1{)}^2}dx+{\int }_{\frac{1}{2}}^1\sqrt{1-x^2}dx]\\ & ={[\frac{x-1}{2}\sqrt{1-(x-1{)}^2}+\frac{1}{2}{\sin }^{-1}(x-1\left)\right]}_0^{\frac{1}{2}}+{[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x]}_{\frac{1}{2}}^1\\ & =[-\frac{1}{4}\sqrt{1-{(-\frac{1}{2})}^2}+\frac{1}{2}{\sin }^{-1}(\frac{1}{2}-1)-\frac{1}{2}{\sin }^{-1}(-1\left)\right]+\end{array}$ $\left[\frac{1}{2}{\sin }^{-1}\right(1)-\frac{1}{4}\sqrt{1-{\left(\frac{1}{2}\right)}^2}-\frac{1}{2}{\sin }^{-1}\left(\frac{1}{2}\right)]$ $\begin{array}{cc}& =[-\frac{\sqrt{3}}{8}+\frac{1}{2}(-\frac{\pi }{6})-\frac{1}{2}(-\frac{\pi }{2})]+[\frac{1}{2}\left(\frac{\pi }{2}\right)-\frac{\sqrt{3}}{8}-\frac{1}{2}\left(\frac{\pi }{6}\right)]\\ & =[-\frac{\sqrt{3}}{4}-\frac{\pi }{12}+\frac{\pi }{4}+\frac{\pi }{4}-\frac{\pi }{12}]+[\frac{1}{2}\left(\frac{\pi }{2}\right)-\frac{\sqrt{3}}{8}-\frac{1}{2}\left(\frac{\pi }{6}\right)]\\ & =[-\frac{\sqrt{3}}{4}-\frac{\pi }{6}+\frac{\pi }{2}]\\ & =[\frac{2\pi }{6}-\frac{\sqrt{3}}{4}]\end{array}$ Therefore ,required area OBCAO $=2\times (\frac{2\pi }{6}-\frac{\sqrt{3}}{4})=(\frac{2\pi }{3}-\frac{\sqrt{3}}{2})\text{units}$

Q3. Find the area of the region bounded by the curves $y=x^2+2,y=x,x=0\text{and}x=3$

Answer. $\begin{array}{c}\text{The area bounded by the curves,}y=x^2+2,y=x,x=0,\text{and}x=3,\text{is represented by}\\ \text{the shaded area OCBAO as}\end{array}$  $\begin{array}{c}\text{Then, Area}\mathrm{OCBAO}=\text{Area ODBAO - Area ODCO}\\ ={\int }_0^3(x^2+2)dx-{\int }_0^{3}xdx\\ ={[\frac{x^3}{3}+2x]}_0^3-{\left[\frac{x^2}{2}\right]}_0^3\end{array}$ $\begin{array}{cc}& =[9+6]-\left[\frac{9}{2}\right]\\ & =15-\frac{9}{2}\\ & =\frac{21}{2}\text{units}\end{array}$

Q4. $\begin{array}{c}\text{Using integration finds the area of the region bounded by the triangle whose vertices are}\\ (-1,0),(1,3)\text{and}(3,2)\end{array}$

Answer. $\begin{array}{c}\text{BL and}CM\text{are drawn perpendicular to}x\text{-axis.}\\ \text{It can be observed in the following figure that,}\\ \text{Area}\left(\Delta ACB\right)=\text{Area}\left(ALBA\right)+\text{Area}\left(BLMCB\right)-\text{Area}\left(AMCA\right)\dots \left(1\right)\end{array}$  $\begin{array}{c}\text{Equation of line segment AB is}\\ y-0=\frac{3-0}{1+1}(x+1)\\ y=\frac{3}{2}(x+1)\\ \therefore \text{Area}\left(ALBA\right)={\int }_{-1}^{+}\frac{3}{2}(x+1)dx=\frac{3}{2}{[\frac{x^2}{2}+x]}_{-1}^1=\frac{3}{2}[\frac{1}{2}+1-\frac{1}{2}+1]=3\text{units}\end{array}$ $\begin{array}{c}\text{Equation of line segment BC is}\\ y-3=\frac{2-3}{3-1}(x-1)\\ y=\frac{1}{2}(-x+7)\\ \therefore \text{Area}\left(BLMCB\right)={\int }_1^3\frac{1}{2}(-x+7)dx=\frac{1}{2}{[-\frac{x^2}{2}+7x]}_1^3=\frac{1}{2}[-\frac{9}{2}+21+\frac{1}{2}-7]=5\text{units}\end{array}$ $\begin{array}{c}\text{Equation of line segment}AC\text{is}\\ y-0=\frac{2-0}{3+1}(x+1)\\ y=\frac{1}{2}(x+1)\\ \therefore \text{Area}\left(AMCA\right)=\frac{1}{2}{\int }_1^3(x+1)dx=\frac{1}{2}{[\frac{x^2}{2}+x]}_{-1}^3=\frac{1}{2}[\frac{9}{2}+3-\frac{1}{2}+1]=4\text{units}\\ \text{Therefore, from equation}\left(1\right),\text{we obtain}\\ \text{Area (}\Delta \text{ABC)}=(3+5-4)=4\text{units}\end{array}$

Q5. $\begin{array}{c}\text{Using integration find the area of the triangular region whose sides have the equations}y\\ =2x+1,y=3x+1\text{and}x=4\text{.}\end{array}$

Answer. $\begin{array}{c}\text{The equations of sides of the triangle are}y=2x+1,y=3x+1,\text{and}x=4\text{.}\\ \text{On solving these equations, we obtain the vertices of triangle as}A(0,1),B(4,13),\text{and}C\\ (4,9).\end{array}$  $\begin{array}{c}\text{It can be observed that,}\\ \text{Area}\left(\Delta ACB\right)=\text{Area (OLBAO) -Area (OLCAO)}\\ ={\int }_0^4(3x+1)dx-{\int }_0^4(2x+1)dx\\ ={[\frac{3x^2}{2}+x]}_0^4-{(\frac{2x^2}{2}+x]}_0^4\\ =28-20\\ =8\text{units}\end{array}$

Q6. $\begin{array}{c}\text{Smaller area enclosed by the circle}{x}^2+y^2=4\text{and the line}x+y=2\text{is}\\ \text{A.}2(n-2)\\ \text{B.}\pi -2\\ \text{C.}2\pi -1\\ \text{D.}2(n+2)\end{array}$

Answer. $\begin{array}{c}\text{The smaller area enclosed by the circle,}{x}^2+y^2=4,\text{and the line,}x+y=2,\text{is}\\ \text{represented by the shaded area ACBA as}\end{array}$  $\begin{array}{c}\text{It can be observed that,}\\ \text{Area}ACBA=\text{Area OACBO - Area (DeltaOAB)}\\ ={\int }_0^2\sqrt{4-x^2}dx-{\int }_0^2(2-x)dx\end{array}$ $\begin{array}{cc}& ={[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\frac{x}{2}]}_0^2-{[2x-\frac{x^2}{2}]}_0^2\\ & =[2\cdot \frac{\pi }{2}]-[4-2]\\ & =(\pi -2)-[4-2]\\ & =(\pi -2)\text{units}\\ \text{Thus, the correct answer is B.}\end{array}$

Q7. Area lying between the curve $y^2=4x\text{and}y=2x$ is $\begin{array}{c}\text{A.}\frac{2}{3}\\ \text{B.}\frac{1}{3}\end{array}$ $\begin{array}{c}\text{C.}\frac{1}{4}\\ \text{D.}\frac{3}{4}\end{array}$

$\begin{array}{l}\frac{\mathrm{<br>}}{}\end{array}$

Answer. $\begin{array}{c}\text{The area lying between the curve,}{y}^2=4x\text{and}y=2x,\text{is represented by the shaded}\\ \text{area OBAO as}\end{array}$  $\begin{array}{c}\text{The points of intersection of these curves are}0(0,0)\text{and}A(1,2)\text{.}\\ \text{We draw AC perpendicular to}x\text{-axis such that the coordinates of}C\text{are}(1,0)\end{array}$ $\therefore \text{Area OBAO}=\text{Area}\left(\Delta OCA\right)-$ Area(OCABO) $\begin{array}{c}={\int }_0^{4}2xdx-{\int }_0^{2}2\sqrt{x}dx\\ =2{\left[\frac{x^2}{2}\right]}_0^1-2{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^1\\ =|1-\frac{4}{3}|\\ =|-\frac{1}{3}|\\ =\frac{1}{3}\text{units}\\ \text{Thus, the correct answer is B.}\end{array}$

Chapter-8 (integrals)

• Exercise 8.1