NCERT Solutions Class 12 Maths Chapter-8 (Application Of Integrals)Exercise 8.1

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-8 (Application Of Integrals)Exercise 8.1 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.

NCERT Question-Answer

Class 12 Mathematics

Chapter-8 (Application Of Integrals)

Questions and answers given in practice

Chapter-8 (Application Of Integrals)

 Exercise 8.1

Q1. Find the area of the region bounded by the curve $y^2=x$ and the lines x = 1, x = 4 and the x-axis in the first quadrant.

Answer.  The area of the region bounded by the curve, $y^2$ = x, the lines, x = 1 and x = 4, and the x-axis is the area ABCD. $\begin{array}{cc}\text{Area of}ABCD& ={\int }^{4}ydx\\ & =\int \sqrt{x}dx\\ & ={\left[\frac{\frac{3}{2}}{\frac{3}{2}}\right]}_1\end{array}$ $\begin{array}{cc}& =\frac{2}{3}\left[\right(4{)}^{\frac{3}{2}}-\left(1{)}^{\frac{3}{2}}\right]\\ & =\frac{2}{3}[8-1]\\ & =\frac{14}{3}\text{units}\end{array}$

Q2. Find the area of the region bounded by $y^2=9x$, x = 2, x = 4 and the x-axis in the first quadrant.

Answer.  The area of the region bounded by the curve, $y^2$ = 9x, x = 2, and x = 4, and the x-axis is the area ABCD. $\begin{array}{cc}\text{Area of}ABCD& ={\int }_2^{4}ydx\\ & ={\int }_2^{4}3\sqrt{x}dx\\ & =3{\left[\frac{3}{2}\right]}_2^4\\ & =2{\left[x^{\frac{3}{2}}\right]}_2^4\end{array}$ $\begin{array}{cc}& =2\left[\right(4{)}^{\frac{3}{2}}-\left(2{)}^{\frac{3}{2}}\right]\\ & =2[8-2\sqrt{2}]\\ & =(16-4\sqrt{2})\text{units}\end{array}$

Q3. Find the area of the region bounded by $x^2=4y$, y = 2, y = 4 and the y-axis in the first quadrant.

Answer.  the area of the region bounded by $x^2$ = 4y, y = 2, y = 4 and the y-axis is the area ABCD. $\begin{array}{cc}\text{Area of}ABCD& ={\int }_2^{4}xdy\\ & ={\int }_2^{4}2\sqrt{y}dy\\ & =2{\int }_2^4\sqrt{y}dy\\ & =2{\left[\frac{\frac{3}{2}}{2}\right]}_2\end{array}$ $\begin{array}{cc}& =\frac{4}{3}\left[\right(4{)}^{\frac{3}{2}}-\left(2{)}^{\frac{3}{2}}\right]\\ & =\frac{4}{3}[8-2\sqrt{2}]\\ & =\left(\frac{32-8\sqrt{2}}{3}\right)\text{units}\end{array}$

Q4. Find the area of the region bounded by the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$

Answer. The given equation of the ellipse, $\frac{x^2}{16}+\frac{y^2}{9}=1$, can be represented as  It can be observed that the ellipse is symmetrical about x-axis and y-axis. ∴ Area bounded by ellipse = 4 × Area of OAB $\begin{array}{cc}\text{Area of}OAB& ={\int }_0^{4}ydx\\ & ={\int }_0^4\sqrt{1-\frac{x^2}{16}}dx\\ & =\frac{3}{4}{\int }_0^4\sqrt{16-x^2}dx\end{array}$ $\begin{array}{cc}& =\frac{3}{4}{[\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}{\sin }^{-1}\frac{x}{4}]}_0^4\\ & =\frac{3}{4}[2\sqrt{16-16}+8{\sin }^{-1}(1)-0-8{\sin }^{-1}(0\left)\right]\\ & =\frac{3}{4}\left[\frac{8\pi }{2}\right]\\ & =\frac{3}{4}\left[4\pi \right]\\ & =3\pi \end{array}$ Therefore, area bounded by the ellipse = 4 × 3π = 12π units

Q5. Find the area of the region bounded by the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$

Answer. The given equation of the ellipse can be represented as  $\begin{array}{c}\frac{x^2}{4}+\frac{y^2}{9}=1\\ \Rightarrow y=3\sqrt{1-\frac{x^2}{4}}\end{array}$ It can be observed that the ellipse is symmetrical about x-axis and y-axis. ∴ Area bounded by ellipse = 4 × Area OAB $\begin{array}{cc}\therefore \text{Area of OAB}& ={\int }_0^{2}ydx\\ & ={\int }_0^2\sqrt[3]{31-\frac{x^2}{4}}dx\left[\text{Using}\right(1\left)\right]\\ & =\frac{3}{2}{\int }_0^2\sqrt{4-x^2}dx\end{array}$ $\begin{array}{cc}& =\frac{3}{2}{[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-}\frac{x}{2}]}_0^2\\ & =\frac{3}{2}\left[\frac{2\pi }{2}\right]\\ =& \frac{3\pi }{2}\\ \text{Therefore, area bounded by the ellipse}=& 4\times \frac{3\pi }{2}=6\pi \text{units}\end{array}$

Q6. Find the area of the region in the first quadrant enclosed by x-axis, line $x=\sqrt{3}y\text{and the circle}{x}^2+y^2=4$

Answer. The area of the region bounded by the circle,$x^2+y^2=4,x=\sqrt{3}y$ , and the x-axis is the area OAB.  The point of intersection of the line and the circle in the first quadrant is $(\sqrt{3},1)$ . Area OAB = Area ΔOCA + Area ACB $\begin{array}{ccc}& \text{Area of}OAC& =\frac{1}{2}\times OC\times AC=\frac{1}{2}\times \sqrt{3}\times 1=\frac{\sqrt{3}}{2}\dots \left(i\right)\\ \text{Area of}ABC& ={\int }_{\sqrt{3}}^{2}ydx& \end{array}$ $\begin{array}{cc}& ={\int }_{\sqrt{3}}^2\sqrt{4-x^2}dx\\ & ={[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\frac{x}{2}]}_{\sqrt{5}}^2\\ & =[2\times \frac{\pi }{2}-\frac{\sqrt{3}}{2}\sqrt{4-3}-2{\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)]\\ & =[\pi -\frac{\sqrt{3}}{2}-2\left(\frac{1}{3}\right)]\\ & =[\pi -\frac{\sqrt{3}}{2}-\frac{2\pi }{3}]\\ & =[\frac{\pi }{3}-\frac{\sqrt{3}}{2}]...\left(ii\right)\end{array}$ $\begin{array}{c}\text{Therefore, area enclosed by}x\text{-axis, the line}x=\sqrt{3}y,\text{and the circle}{x}^2+y^2=4\text{in the first}\\ \frac{\sqrt{3}\pi }{2}+\frac{3\sqrt{2}}{2}=\frac{\pi }{3}\text{units}\end{array}$

Q7. Find the area of the smaller part of the circle $x^2+y^2=a^2$ cut off by the line $x=\frac{a}{\sqrt{2}}$

Answer. $\begin{array}{c}\text{The area of the smaller part of the circle,}{x}^2+y^2=a^2,\text{cut off by the line,}\\ \text{area ABCDA.}\end{array}$  It can be observed that the area ABCD is symmetrical about x-axis. ∴ Area ABCD = 2 × Area ABC $\begin{array}{cc}\text{Area of}ABC& ={\int }_a^{\pi }ydx\\ & ={\int }_{\sqrt{2}}^{\pi }\sqrt{a^2-x^2}dx\\ & ={[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}]}_{\frac{a}{\sqrt{2}}}^a\\ & =[\frac{a^2}{2}\left(\frac{\pi }{2}\right)-\frac{a}{2\sqrt{2}}\sqrt{a^2-\frac{a^2}{2}}-\frac{a^2}{2}{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)]\end{array}$ $\begin{array}{cc}& =\frac{a^2\pi }{4}-\frac{a}{2\sqrt{2}}\cdot \frac{a}{\sqrt{2}}-\frac{a^2}{2}\left(\frac{\pi }{4}\right)\\ & =\frac{a^2\pi }{4}-\frac{a^2}{4}-\frac{a^2\pi }{8}\\ & =\frac{a^2}{4}[\pi -1-\frac{\pi }{2}]\\ & =\frac{a^2}{4}[\frac{\pi }{2}-1]\\ \Rightarrow \text{Area ABCD}& =2\left[\frac{a^2}{2}(\frac{\pi }{2}-1)\right]=\frac{a^2}{2}(\frac{\pi }{2}-1)\end{array}$ $\begin{array}{c}\text{Therefore, the area of smaller part of the circle,}{x}^2+y^2=a^2,\text{cut off by the line,}x=\frac{a}{\sqrt{2}}\\ \frac{a^2}{2}(\frac{\pi }{2}-1)\text{units.}\end{array}$

Q8. The area between x = $y^2$ and x = 4 is divided into two equal parts by the line x = a, find the value of a.

Answer.

 It can be observed that the area ABCD is symmetrical about x-axis. ∴ Area ABCD = 2 × Area ABC Area OED=Area EFCD $\begin{array}{cc}\text{Area}OED& ={\int }_0^{t}ydx\\ & ={\int }_0^{\pi }\sqrt{x}dx\\ & ={\left[\frac{3}{2}\right]}_0^a\\ & =\frac{2}{3}(a{)}^{\frac{3}{2}}...(i)\end{array}$ $\begin{array}{cc}\text{Area of}EFCD& ={\int }_0^4\sqrt{x}dx\\ & ={\left[\begin{array}{c}\frac{3}{2}\\ \frac{3}{2}\end{array}\right]}_0\\ & =\frac{2}{3}[8-a^{\frac{3}{2}}]\dots \left(ii\right)\end{array}$ $\begin{array}{c}\text{From}\left(1\right)\text{and}\left(2\right),\text{we obtain}\\ \frac{2}{3}(a{)}^{\frac{3}{2}}=\frac{2}{3}[8-(a{)}^{\frac{3}{2}}]\\ \Rightarrow 2\cdot (a{)}^{\frac{3}{2}}=8\\ \Rightarrow (a{)}^{\frac{3}{2}}=4\\ \Rightarrow a=(4{)}^{\frac{2}{3}}\\ \text{Therefore, the value of a is}(4{)}^{\frac{2}{3}}\end{array}$

Q9. Find the area of the region bounded by the parabola y = $x^2\text{and}y=|x|$

$\begin{array}{c}\\ \\ \text{The area bounded by the parabola,}{x}^2=y,\text{and the line,}y=|x|,\text{can be represented as}\end{array}$  $\begin{array}{c}\text{The given area is symmetrical about}y\text{-axis.}\\ \therefore \text{Area OACO = Area ODBO}\\ \text{The point of intersection of parabola,}{x}^2=y,\text{and line,}y=x,\text{is A}(1,1)\text{.}\\ \text{Area of}OACO=\text{Area}\Delta OAB-\text{Area OBACO}\end{array}$ $\therefore \text{Area of}\Delta OAB=\frac{1}{2}\times OB\times AB=\frac{1}{2}\times 1\times 1=\frac{1}{2}$ $\begin{array}{c}\text{Area of OBACO}={\int }_{0}ydx=\int x^{2}dx={\left[\frac{x^3}{3}\right]}_0^1=\frac{1}{3}\\ \Rightarrow \text{Area of OACO}=\text{Area of}\Delta OAB-\text{Area of OBACO}\\ =\frac{1}{2}-\frac{1}{3}\\ =\frac{1}{6}\end{array}$Therefore,required area $=2\left[\frac{1}{6}\right]=\frac{1}{3}$

Q10. Find the area bounded by the curve $x^2$ = 4y and the line x = 4y – 2

Answer.

$\begin{array}{c}\text{The area bounded by the curve,}{x}^2=4y,\text{and line,}x=4y-2,\text{is represented by the}\\ \text{shaded area OBAO.}\end{array}$  $\begin{array}{c}\text{Let}A\text{and}B\text{be the points of intersection of the line and parabola.}\\ \text{A are}(-1,\frac{1}{4})\\ \text{Coordinates of point}\\ \text{Coordinates of point}B\text{are}(2,1)\\ \text{We draw AL and BM perpendicular to}x\text{-axis.}\end{array}$ $\begin{array}{c}\text{It can be observed that,}\\ \text{Area}\mathrm{OBAO}=\text{Area OBCO}+\text{Area OACO}\dots \text{(i)}\end{array}$ $\begin{array}{cc}\text{Then, Area}\mathrm{OBCO}& =\text{Area OMBC - Area OMBO}\\ & ={\int }_0^2\frac{x+2}{4}dx-{\int }_0^2\frac{x^2}{4}dx\\ & =\frac{1}{4}{[\frac{x^2}{2}+2x]}_0^2-\frac{1}{4}{\left[\frac{x^3}{3}\right]}_0^2\end{array}$ $\begin{array}{c}=\frac{1}{4}[2+4]-\frac{1}{4}\left[\frac{8}{3}\right]\\ =\frac{3}{2}-\frac{2}{3}\\ =\frac{5}{6}\\ \text{Similarly, Area OACO = Area OLAC - Area OLAO}\end{array}$ $\begin{array}{c}={\int }_{-1}^{\infty }\frac{x+2}{4}dx-{\int }_{-1}\frac{x^2}{4}dx\\ =\frac{1}{4}{[\frac{x^2}{2}+2x]}_{-1}^0-\frac{1}{4}{\left[\frac{x^3}{3}\right]}^0\\ =-\frac{1}{4}[\frac{(-1{)}^2}{2}+2(-1\left)\right]-[-\frac{1}{4}\left(\frac{(-1{)}^3}{3}\right)]\\ =-\frac{1}{4}[\frac{1}{2}-2]-\frac{1}{12}\\ =\frac{1}{2}-\frac{1}{8}-\frac{1}{12}\\ =\frac{7}{24}\end{array}$ Therefore, required area $=(\frac{5}{6}+\frac{7}{24})=\frac{9}{8}$

Q11. Find the area of the region bounded by the curve $y^2$ = 4x and the line x = 3

Answer. The region bounded by the parabola, $y^2=4x,\text{and the line,}x=3,\text{is the area OACO.}$  $\begin{array}{c}\text{The area OACO is symmetrical about}x\text{-axis.}\\ \therefore \text{Area of}OACO=2\text{(Area of OAB)}\end{array}$ $\begin{array}{cc}\text{Area OACO}& =2\left[{\int }_0^{3}ydx\right]\\ & =2\int 2\sqrt{x}dx\\ & =4{\left[\frac{x^3}{2}\right]}_0^3\\ & =\frac{8}{3}\left[\right(3{)}^{\frac{3}{2}}]\\ & =8\sqrt{3}\end{array}$ Therefore,the reqiured area is $8\sqrt{3}$$\begin{array}{c}$ x^2+y^2$$ \begin{array}{c}\text{A.}\pi \\ \text{B.}\frac{\pi }{2}\\ \text{C.}\frac{\pi }{3}\\ \text{D.}\frac{\pi }{4}\end{array}$\\ \\ \\ \end{array}$

$\begin{array}{c}\text{The area bounded by the circle and the lines,}x=0\text{and}x=2,\text{in the first quadrant is}\\ \text{represented as}\end{array}$  $\begin{array}{cc}\therefore \text{Area}OAB& ={\int }_0^{2}ydx\\ & ={\int }_0^2\sqrt{4-x^2}dx\\ & ={[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\frac{x}{2}]}_0^2\\ & =2\left(\frac{\pi }{2}\right)\\ & =\pi \text{units}\end{array}$ Thus,the correct answer is A.

Q13. Area of the region bounded by the curve $y^2=4x,$y-axis and the line y = 3 is $\begin{array}{c}\text{A.}2\\ \text{B.}\frac{9}{4}\\ \text{C.}\frac{9}{3}\\ \text{D.}\frac{9}{2}\end{array}$

Answer. The bounded by the curve, $y^2=4x,y\text{-axis, and}y=3$ is represented as  $\begin{array}{cc}\therefore \text{Area}OAB& ={\int }_0^{3}xdy\\ & ={\int }_0^3\frac{y^2}{4}dy\\ & =\frac{1}{4}{\left[\frac{y^3}{3}\right]}_0^3\\ & =\frac{1}{12}\left(27\right)\\ & =\frac{9}{4}\text{units}\end{array}$ Thus the correct answer is B.

Chapter-8 (integrals)

• Exercise 8.2