NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals (Ex 8.1) Exercise 8.1

Q.1 Find the area of the region bounded by the curve y² = x and the lines x = 1, x = 4 and the x-axis.

Ans

The area of the region bounded by the curve, y² = x, the lines, x = 1 and x = 4, and the x-axis is the area ABCD.

\begin{array}{l}Area\text{of ABCD}={\displaystyle {\int }_1^{4}ydx}\\ ={\displaystyle {\int }_1^4\sqrt{x}dx}\\ ={\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_1^4=\frac{2}{3}\left[4^{\frac{3}{2}}-1^{\frac{3}{2}}\right]\\ =\frac{2}{3}\left(8-1\right)\\ =\frac{14}{3}units\end{array}

Q.2 Find the area of the region bounded by y²= 9x, x = 2, x = 4 and the x-axis in the first quadrant.

Ans

The area of the region bounded by the curve, y² = 9x, the lines, x = 2 and x = 4, and the x-axis is the area ABCD.

\begin{array}{l}Area\text{of ABCD}={\displaystyle {\int }_2^{4}ydx}\\ ={\displaystyle {\int }_2^{4}3\sqrt{x}dx}\\ =3{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_2^4=3\times \frac{2}{3}\left[4^{\frac{3}{2}}-2^{\frac{3}{2}}\right]\\ =2\left(8-2\sqrt{2}\right)\\ =\left(16-4\sqrt{2}\right)units\end{array}

Q.3 Find the area of the region bounded by x² = 4y, y = 2, y = 4 and the y-axis in the first quadrant.

Ans

The area of the region bounded by the curve, x² = 4y, the lines, y = 2 and y = 4, and the y-axis is the area ABCD.

\begin{array}{l}Area\text{of ABCD}={\displaystyle {\int }_2^{4}xdy}\\ ={\displaystyle {\int }_2^{4}2\sqrt{y}dx}\\ =2{\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]}_2^4=\frac{4}{3}\left[4^{\frac{3}{2}}-2^{\frac{3}{2}}\right]\\ =\frac{4}{3}\left(8-2\sqrt{2}\right)\\ =\left(\frac{32-8\sqrt{2}}{3}\right)units\end{array}

Q.4 Find the area of the region bounded by the ellipse (x²/16) + (y²/9) = 1.

Ans

The given ellipse (x²/16) + (y²/9)=1 can be represented as given below:

Ellipse is symmetrical about x-axis and y-axis.
So, area bounded by ellipse = 4 x Area of OAB

\begin{array}{l}Area\text{of OAB}={\displaystyle {\int }_0^{4}ydx}\\ ={\displaystyle {\int }_0^{4}3\sqrt{1-\frac{x^2}{16}}dx}\\ =\frac{3}{4}{\displaystyle {\int }_0^4\sqrt{16-x^2}dx}\\ =\frac{3}{4}{\left[\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}{\sin }^{-1}\frac{x}{4}\right]}_0^4\\ =\frac{3}{4}\left[\frac{4}{2}\sqrt{16-4^2}+\frac{16}{2}{\sin }^{-1}\frac{4}{4}\right]-\frac{3}{4}\left[\frac{0}{2}\sqrt{16-0^2}+\frac{16}{2}{\sin }^{-1}\frac{0}{4}\right]\\ =\frac{3}{4}\times 8\frac{\pi }{2}\\ =3\pi squareunits\\ Therefore,\text{area bounded by ellipse}\\ =4\times Area\text{of OAB}\\ =\text{4}\times 3\pi squareunits\\ =12\pi squareunits\end{array}

Q.5 Find the area of the region bounded by the ellipse (x²/4) + (y²/9) = 1.

Ans

The given ellipse (x²/4) + (y²/9) =1 can be represented as given below:

\begin{array}{l}The\text{given equation of ellipse is}\\ \frac{x^2}{4}+\frac{y^2}{9}=1\\ y=3\sqrt{1-\frac{x^2}{4}}\\ It\text{can be observed that ellipse is symmetrical about x-axis and}\\ \text{y-axis}\text{.}\\ \therefore \text{Area bounded by ellipse}=4\times Area\text{OAB}\\ \therefore \text{Area of OAB}={\displaystyle {\int }_0^{2}ydx}\\ ={\displaystyle {\int }_0^{2}3\sqrt{1-\frac{x^2}{4}}dx}\\ =\frac{3}{2}{\displaystyle {\int }_0^2\sqrt{\left(4-x^2\right)}dx}\\ =\frac{3}{2}{\left[\frac{x}{2}\sqrt{\left(4-x^2\right)}+\frac{4}{2}{\sin }^{-1}\frac{x}{2}\right]}_0^2\end{array} \begin{array}{l}=\frac{3}{2}\left[\frac{2}{2}\sqrt{\left(4-2^2\right)}+\frac{4}{2}{\sin }^{-1}\frac{2}{2}\right]-\frac{3}{2}\left[\frac{0}{2}\sqrt{\left(4-0^2\right)}+\frac{4}{2}{\sin }^{-1}\frac{0}{2}\right]\\ =\frac{3}{2}\times 2\frac{\pi }{2}\\ =\frac{3}{2}\text{}\pi square\text{units}\\ \text{Therefore, area bounded by the ellipse}\\ =4\times \frac{3}{2}\text{}\pi square\text{units}\\ =\text{6}square\text{unit}\end{array}

Q.6

$\begin{array}{l}\mathrm{Find\; the\; area\; of\; the\; region\; in\; the\; first}\hspace{0.17em}\hspace{0.17em}\mathrm{quadrant\; enclosed}\\ \mathrm{by\; x}–\mathrm{axis},\mathrm{line\; x}=\sqrt{3}{\mathrm{yandthecirclex}}^2+{\mathrm{y}}^2=4.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{smaller}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{circle},{\mathrm{x}}^2+{\mathrm{y}}^{\mathrm{2}}={\mathrm{a}}^{\mathrm{2}},\mathrm{\hspace{0.17em}}\mathrm{cut}\mathrm{\hspace{0.17em}}\mathrm{off}\mathrm{\hspace{0.17em}}\mathrm{by}\\ \mathrm{the}\mathrm{line},\mathrm{x}=\frac{\mathrm{a}}{\sqrt{2}},\mathrm{is}\mathrm{the}\mathrm{area}\mathrm{ABCDA}\mathrm{.}\end{array}$

\begin{array}{l}It\text{is observed that the area ABCD is symmetrical about x-axis}\text{.}\\ \therefore \text{area ABCD}=2\times Area\text{ABC}\\ Area\text{of}\text{ABC}={\displaystyle {\int }_{\frac{a}{\sqrt{2}}}^{a}ydx}\\ ={\displaystyle {\int }_{\frac{a}{\sqrt{2}}}^a\sqrt{a^2-x^2}dx}\\ ={\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}\right]}_{\frac{a}{\sqrt{2}}}^a\\ =\left[\frac{a}{2}\sqrt{a^2-a^2}+\frac{a^2}{2}{\sin }^{-1}\frac{a}{a}\right]-\left[\frac{\left(\frac{a}{\sqrt{2}}\right)}{2}\sqrt{a^2-{\left(\frac{a}{\sqrt{2}}\right)}^2}+\frac{a^2}{2}{\sin }^{-1}\frac{\left(\frac{a}{\sqrt{2}}\right)}{a}\right]\\ =\frac{a^2}{2}\times \frac{\pi }{2}-\frac{a}{2\sqrt{2}}\times \frac{a}{\sqrt{2}}-\frac{a^2}{2}\left(\frac{\pi }{4}\right)\\ =\frac{a^2}{2}\times \frac{\pi }{2}-\frac{a^2}{4}-\frac{a^2}{2}\left(\frac{\pi }{4}\right)\\ =\frac{a^2}{4}\left(\pi -1-\frac{\pi }{2}\right)\\ =\frac{a^2}{4}\left(\frac{\pi }{2}-1\right)\\ \therefore Area\text{ABCD}=2\left\{\frac{a^2}{4}\left(\frac{\pi }{2}-1\right)\right\}\\ =\frac{a^2}{2}\left(\frac{\pi }{2}-1\right)squareunits\\ Therefore,{\text{the area of the smaller part of the circle, x}}^{\text{2}}{\text{+y}}^{\text{2}}={\text{a}}^{\text{2}},\\ \text{cut off by the line, x=}\frac{a}{\sqrt{2}}\text{is}\frac{a^2}{2}\left(\frac{\pi }{2}-1\right)squareunits.\end{array}

Q.7 The area between x = y² and x = 4 is divided into two equal parts by the line x = a, find the value of a.

Ans

The line x = a divides the area bounded by the parabola and line x= 4 into two equal parts.

\begin{array}{l}AreaOED={\displaystyle {\int }_0^{a}ydx}\\ ={\displaystyle {\int }_0^a\sqrt{x}dx}\\ ={\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^a\\ =\frac{2}{3}{a}^{\frac{3}{2}}\\ Area\text{}\text{of CDEF}\\ ={\displaystyle {\int }_a^{4}ydx}\\ ={\displaystyle {\int }_a^4\sqrt{x}dx}\\ ={\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_a^4\\ =\frac{2}{3}\left(4^{\frac{3}{2}}-a^{\frac{3}{2}}\right)\\ =\frac{2}{3}\left(8-a^{\frac{3}{2}}\right)\\ According\text{to equation}\left(i\right),\text{we have}\\ \frac{2}{3}{a}^{\frac{3}{2}}=\frac{2}{3}\left(8-a^{\frac{3}{2}}\right)\\ \Rightarrow a^{\frac{3}{2}}=\left(8-a^{\frac{3}{2}}\right)\\ \Rightarrow 2a^{\frac{3}{2}}=8\\ \Rightarrow a^{\frac{3}{2}}=4\\ \Rightarrow a=4^{\frac{2}{3}}\\ Therefore,\text{the value of a is}{4}^{\frac{2}{3}}.\end{array}

Q.8 Find the area of the region bounded by the parabola y = x² and y = |x|.

Ans

The area bounded by the parabola y = x² and
y = |x| is shown as below:

Since the required area is symmetrical about y-axis.

So, area OACO = area OBDO

The points of intersection of parabola and line are A(1,1) and B(–1, 1).

\begin{array}{l}Area\text{of OACO}=Area\Delta \text{AMO}-areaOMACO\\ =\frac{1}{2}\times OM\times AM-{\displaystyle {\int }_0^{1}ydx}\\ =\frac{1}{2}\times 1\times 1-{\displaystyle {\int }_0^1{x}^{2}dx}\\ =\frac{1}{2}-{\left[\frac{x^3}{3}\right]}_0^1\\ =\frac{1}{2}-\left[\frac{1^3}{3}-\frac{0^3}{3}\right]\\ =\frac{1}{2}-\frac{1}{3}\\ =\frac{1}{6}\\ \therefore \text{Required area}\\ =2\left(Area\text{of OACO}\right)\\ =2\times \frac{1}{6}\\ =\frac{1}{3}square\text{units}\end{array}

Q.9 Find the area of the region bounded by the curve x² = 4y and the line x = 4y – 2.

Ans

The area bounded by the parabola x² = 4y and x = 4y – 2 is shown in figure.
The intersection points of line and parabola are A(–1, 1/4) and B(2,1).

\begin{array}{l}HereAL\perp X-axis\text{and BM}\perp X-axis.Then\\ AreaOBAO=AreaOBCO+AreaOACO\\ =\left(AreaOMBCO-AreaOMBO\right)+\left(AreaOLAC-AreaOLAO\right)\\ =\left({\displaystyle {\int }_0^2\frac{x+2}{4}dx}-{\displaystyle {\int }_0^2\frac{x^2}{4}dx}\right)+\left({\displaystyle {\int }_{-1}^0\frac{x+2}{4}dx}-{\displaystyle {\int }_{-1}^0\frac{x^2}{4}dx}\right)\\ =\frac{1}{4}{\left[\frac{x^2}{2}+2x\right]}_0^2-\frac{1}{4}{\left[\frac{x^3}{3}\right]}_0^2+\frac{1}{4}{\left[\frac{x^2}{2}+2x\right]}_{-1}^0-\frac{1}{4}{\left[\frac{x^3}{3}\right]}_{-1}^0\\ =\frac{1}{4}\left[\frac{2^2}{2}+2\left(2\right)\right]-\frac{1}{4}\left[\frac{2^3}{3}\right]+0-\frac{1}{4}\left[\frac{{\left(-1\right)}^2}{2}+2\left(-1\right)\right]-0+\frac{1}{4}\left[\frac{{\left(-1\right)}^3}{3}\right]\\ =\frac{1}{4}\times 6-\frac{1}{4}\times \frac{8}{3}-\frac{1}{4}\times \frac{-3}{2}+\frac{1}{4}\times \frac{-1}{3}\\ =\frac{3}{2}-\frac{2}{3}+\frac{3}{8}-\frac{1}{12}\\ =\frac{36-16+9-2}{24}\\ =\frac{27}{24}=\frac{9}{8}squnits\\ Thus,\text{the area of OBAO is}\frac{9}{8}sq.units.\end{array}

Q.10 Find the area of the region bounded by the curve y² = 4x and the line x = 3.

Ans

The region bounded by the parabola and line is given below:

The area COAC is symmetrical about x-axis. So,
Area OABO = Area OCBO
Then, Area OCAO = 2(area OABO)

\begin{array}{l}AreaOACO=2{\displaystyle {\int }_0^{3}ydx}\\ =2{\displaystyle {\int }_0^{3}2\sqrt{x}}dx\\ =4{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^3\\ =\frac{2}{3}\times 4\left[3^{\frac{3}{2}}-0\right]=8\sqrt{3}\\ Therefore,\text{the required area is 8}\sqrt{3}squareunits.\end{array}

Q.11 Choose the correct answer
Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is

(a)\pi (b)\frac{\pi }{2}(c)\frac{\pi }{3}(d)\frac{\pi }{4}

Ans

The area bounded by the circle and the lines x = 0 and x = 2 lies in first quadrant.

\begin{array}{l}\therefore AreaOABO={\displaystyle {\int }_0^{2}ydx}\\ ={\displaystyle {\int }_0^2\sqrt{4-x^2}dx}\\ ={\left[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\left(\frac{x}{2}\right)\right]}_0^2\\ =\left[\frac{2}{2}\sqrt{4-2^2}+\frac{4}{2}{\sin }^{-1}\left(\frac{2}{2}\right)\right]-\left[\frac{0}{2}\sqrt{4-0^2}+\frac{4}{2}{\sin }^{-1}\left(\frac{0}{2}\right)\right]\\ =\left[0+2\times \frac{\pi }{2}\right]-\left[0+0\right]=\pi \\ Thus,\text{}the\text{correct option is A}\text{.}\end{array}

Q.12 Choose the correct answer

Area of the region bounded by the curve y² = 4x, and y-axis and the line y = 3 is

(A) 2 (B) 9/4 (C) 9/3 (D) 9/2

Ans

The area bounded by the parabola and the lines y = 0 and y = 3 lies in first quadrant.

\begin{array}{l}\therefore AreaOAB={\displaystyle {\int }_0^{3}xdy}\\ ={\displaystyle {\int }_0^2\frac{y^2}{4}dy}\\ =\frac{1}{4}{\left[\frac{y^3}{3}\right]}_0^3\\ =\frac{1}{12}\left[3^3-0^3\right]\\ =\frac{27}{12}=\frac{9}{4}\\ Thus,\text{}the\text{correct option is B}\text{.}\end{array}

Q.13 Find the area of the circle 4x² + 4y² = 9 which is interior to the parabola x² = 4y.

Ans

The required area is represented by shaded area OBCDO.

\begin{array}{l}On{\text{solving the given equations 4x}}^{\text{2}}{\text{+ 4y}}^{\text{2}}={\text{9 and parabola x}}^2=\text{4y,}\\ we\text{get the coordinates of point of intersection as B}\left(\sqrt{2},\frac{1}{2}\right)and\\ C\left(-\sqrt{2},\frac{1}{2}\right).This\text{area is symmetrical about y-axis}\text{.}\\ \therefore \text{}AreaOBCD=2\left(areaOBCO\right)\\ Since,BM\perp X-axis.\\ So,the\text{coordinates of M are}\left(\sqrt{2},0\right).\\ Therefore,area\text{OBCO}=AreaOMBCO-AreaOMBO\\ ={\displaystyle {\int }_0^{\sqrt{2}}\sqrt{\frac{9-4x^2}{4}}dx}-{\displaystyle {\int }_0^{\sqrt{2}}\frac{x^2}{4}}dx\\ =\frac{1}{2}{\displaystyle {\int }_0^{\sqrt{2}}\sqrt{9-4x^2}dx}-{\displaystyle {\int }_0^{\sqrt{2}}\frac{x^2}{4}}dx\\ =\frac{1}{2}{\displaystyle {\int }_0^{2\sqrt{2}}\sqrt{9-t^2}\frac{dt}{2}}-\frac{1}{4}{\displaystyle {\int }_0^{\sqrt{2}}{x}^2}dx\left[\begin{array}{l}Let\text{t}=2x\Rightarrow \frac{dt}{dx}=2\\ andt=2\times \sqrt{2},\left(Whenx=\sqrt{2}\right)\\ =2\sqrt{2}\\ t=2\times 0\left(Whenx=0\right)\\ =0\end{array}\right]\\ =\frac{1}{4}{\left[\frac{t}{2}\sqrt{9-t^2}+\frac{9}{2}{\sin }^{-1}\frac{t}{3}\right]}_0^{2\sqrt{2}}-\frac{1}{4}{\left[\frac{x^3}{3}\right]}_0^{\sqrt{2}}\\ =\frac{1}{4}\left[\frac{2\sqrt{2}}{2}\sqrt{9-{\left(2\sqrt{2}\right)}^2}+\frac{9}{2}{\sin }^{-1}\frac{2\sqrt{2}}{3}\right]-\\ \frac{1}{4}\left[\left(0\right)\sqrt{9-4{\left(0\right)}^2}+\frac{9}{2}{\sin }^{-1}\frac{2\left(0\right)}{3}\right]-\frac{1}{4}\left[\frac{{\left(\sqrt{2}\right)}^3}{3}\right]\\ =\frac{1}{4}\left(\sqrt{2}+\frac{9}{2}{\sin }^{-1}\frac{2\sqrt{2}}{3}\right)-\frac{\sqrt{2}}{6}\\ =\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}\left(\frac{\sqrt{2}}{6}+\frac{9}{4}{\sin }^{-1}\frac{2\sqrt{2}}{3}\right)\\ Therefore,\text{area of shaded region is 2}\times \frac{1}{2}\left(\frac{\sqrt{2}}{6}+\frac{9}{4}{\sin }^{-1}\frac{2\sqrt{2}}{3}\right)\\ =\left(\frac{\sqrt{2}}{6}+\frac{9}{4}{\sin }^{-1}\frac{2\sqrt{2}}{3}\right)square\text{}\text{units}\text{.}\end{array}