NCERT Solutions Class 12 Mathematics Chapter 8

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

$\begin{array}{l}\mathrm{Findtheareaoftheregioninthefirst}\hspace{0.17em}\hspace{0.17em}\mathrm{quadrantenclosed}\\ \mathrm{byx}–\mathrm{axis},\mathrm{linex}=\sqrt{3}{\mathrm{yandthecirclex}}^2+{\mathrm{y}}^2=4.\end{array}$

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}

Q.14 Find the area bounded by curves (x – 1)² + y² = 1 and x² + y² = 1.

Ans.

The required area bounded by (x – 1)² + y² = 1 and x² + y² = 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},\frac{\sqrt{3}}{2}\right)and\\ B\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right).This\text{area is symmetrical about x-axis}\text{.}\\ \therefore \text{}AreaOBCAO=2\left(areaOCAO\right)\\ Join\text{}\text{AB such that}AM\perp OC.\\ So,the\text{coordinates of M are}\left(\frac{1}{2},0\right).\\ Therefore,area\text{OCAO}=AreaOMAO+AreaMCAM\\ ={\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\\ ={\left[\frac{\left(x-1\right)}{2}\sqrt{1-{\left(x-1\right)}^2}+\frac{1}{2}{\sin }^{-1}\left(x-1\right)\right]}_0^{\frac{1}{2}}\\ +{\left[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x\right]}_{\frac{1}{2}}^1\\ =\left[\frac{\left(\frac{1}{2}-1\right)}{2}\sqrt{1-{\left(\frac{1}{2}-1\right)}^2}+\frac{1}{2}{\sin }^{-1}\left(\frac{1}{2}-1\right)\right]\\ -\left[\frac{\left(0-1\right)}{2}\sqrt{1-{\left(0-1\right)}^2}+\frac{1}{2}{\sin }^{-1}\left(0-1\right)\right]+\left[\frac{1}{2}\sqrt{1-1^2}+\frac{1}{2}{\sin }^{-1}1\right]\\ -\left[\frac{\frac{1}{2}}{2}\sqrt{1-{\left(\frac{1}{2}\right)}^2}+\frac{1}{2}{\sin }^{-1}\left(\frac{1}{2}\right)\right]\\ =\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)\\ =-\frac{\sqrt{3}}{8}-\frac{\pi }{12}+\frac{\pi }{4}+\frac{\pi }{4}-\frac{\sqrt{3}}{8}-\frac{\pi }{12}\\ =-\frac{\sqrt{3}}{4}-\frac{\pi }{6}+\frac{\pi }{2}=\frac{2\pi }{6}-\frac{\sqrt{3}}{4}\\ Therefore,\text{required area of OBCAO}\\ =2\left(\frac{2\pi }{6}-\frac{\sqrt{3}}{4}\right)=\left(\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\right)\text{square units}\end{array}

Q.15 Find the area of the region bounded by the curves y = x² + 2, y = x, x = 0 and x = 3.

Ans.

The area bounded by y = x² +2, y = x, x = 0 and x = 3 is represented by the shaded area

\begin{array}{l}Then,areaOCBAO=areaODBAO-areaODCO\\ ={\displaystyle {\int }_0^3\left(x^2+2\right)dx}-{\displaystyle {\int }_0^{3}xdx}\\ ={\left[\frac{x^3}{3}+2x\right]}_0^3-{\left[\frac{x^2}{2}\right]}_0^3\\ =\left[\frac{3^3}{3}+2\left(3\right)-0-0\right]-\left[\frac{3^2}{2}-0\right]\\ =15-\frac{9}{2}=\frac{21}{2}units\\ Thus,\text{the required area of shaded region is}\frac{21}{2}squnits.\end{array}

Q.16 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{y}-0=\frac{3-0}{1+1}\left(x+1\right)\\ 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)\\ y=\frac{-1}{2}\left(x-1\right)+3\\ =\frac{1}{2}\left(-x+7\right)\\ Equation\text{of line CA is}\\ \text{y}-2=\frac{2-0}{3+1}\left(x-3\right)\\ y=\frac{1}{2}\left(x-3\right)+2\\ =\frac{1}{2}\left(x+1\right)\\ Area\Delta ACB=Area\left(ALBA\right)+Area\left(BLMCB\right)-Area\left(\Delta AMCA\right)\\ ={\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}\\ =\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\\ =\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]\\ -\frac{1}{2}\left[\frac{3^2}{2}+3-\frac{{\left(-1\right)}^2}{2}-\left(-1\right)\right]\\ =\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)\\ =3+5-4\\ \end{array}

   =4​​ square units

Q.17 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\Delta ABC=AreaOLBAO-AreaOLCAO\\ ={\displaystyle {\int }_0^4\left(3x+1\right)}dx-{\displaystyle {\int }_0^4\left(2x+1\right)dx}\\ ={\left[3\frac{x^2}{2}+x\right]}_0^4-{\left[2.\frac{x^2}{2}+x\right]}_0^4\\ =\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]\\ =\left(24+4\right)-\left(16+4\right)\\ =28-20\\ =8\text{square units}\end{array}

Q.18 Choose the correct answer

Smaller area enclosed by the circle x² + y² = 4 and the line x + y =2 is

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

Ans.

The smaller area enclosed by the circle, x² + y² = 4 and line x + y = 2 is given by shaded region ABCA.

\begin{array}{l}AreaABCA=AreaOACBO-Area\Delta AOB\\ ={\displaystyle {\int }_0^2\sqrt{\left(4-x^2\right)}}dx-{\displaystyle {\int }_0^2\left(2-x\right)dx}\\ ={\left[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\left(\frac{x}{2}\right)\right]}_0^2-{\left[2x-\frac{x^2}{2}\right]}_0^2\\ =\left[\frac{2}{2}\sqrt{4-2^2}+2{\sin }^{-1}\left(\frac{2}{2}\right)\right]-\left[2\left(2\right)-\frac{2^2}{2}\right]\\ =\left(2\times \frac{\pi }{2}\right)-\left(4-2\right)\\ =\left(\pi -2\right)\text{square units}\\ \text{Therefore, correct option is B}\text{.}\end{array}

Q.19 Area lying between the curves y² = 4x and y = 2x is
(A) 2/3 (B) 1/3 (C) 1/4 (D) 3/4

Ans.

The area lying between y² = 4x and y = 2x is shown by shaded area given graph.

\begin{array}{l}AreaOBAO=areaOCABO-areaOCAO\\ ={\displaystyle {\int }_0^{1}2\sqrt{x}dx}-{\displaystyle {\int }_0^{1}2xdx}\\ =2{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^1-2{\left[\frac{x^2}{2}\right]}_0^1\\ =\frac{4}{3}\left[1-0\right]-\left[1^2-0\right]\\ =\frac{4}{3}-1=\frac{1}{3}square\text{units}\\ \end{array}

Therefore, correct option is B.

Q.20 Find the area under the given curves and given lines:
(i) y = x², x = 1, x = 2 and x- axis

(ii) y = x⁴, x = 1, x = 5 and x- axis

Ans.

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

\begin{array}{l}AreaABCDA={\displaystyle {\int }_1^2{x}^{2}dx}\\ ={\left[\frac{x^3}{3}\right]}_1^2\\ =\frac{1}{3}\left[2^3-1^3\right]\\ =\frac{7}{3}s\text{quare units}\end{array}

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

\begin{array}{l}AreaABCDA={\displaystyle {\int }_1^5{x}^{4}dx}\\ ={\left[\frac{x^5}{5}\right]}_1^5\\ =\frac{1}{5}\left[5^5-1^5\right]\\ =\frac{3124}{5}\\ =624.8s\text{quare units}\end{array}

Q.21 Find the area between the curves y = x and y = x².

Ans.

The point of intersection of curve y = x² and line y = x is (1, 1). Here AC is drawn as perpendicular to x-axis.

\begin{array}{l}\therefore AreaOBAO=Area\left(\Delta OCA\right)-Area\left(OCABO\right)\\ ={\displaystyle {\int }_0^{1}xdx}-{\displaystyle {\int }_0^1{x}^{2}dx}\\ ={\left[\frac{x^2}{2}\right]}_0^1-{\left[\frac{x^3}{3}\right]}_0^1\\ =\left[\frac{1^2}{2}-0\right]-\left[\frac{1^3}{3}-0\right]\\ =\frac{1}{2}-\frac{1}{3}\\ =\frac{1}{6}\text{square units}\end{array}

Q.22 Find the area of the region lying in the first quadrant and bounded by y = 4x², x = 0, y = 1 and y = 4.

Ans.

The area of the region lying in the first quadrant and bounded by y = 4x², x = 0, y = 1 and y = 4 is shown by shaded region ABCDA.

\begin{array}{l}\therefore AreaABCD={\displaystyle {\int }_1^4\frac{\sqrt{y}}{2}dy}\left[\because y=4x^2\right]\\ =\frac{1}{2}{\displaystyle {\int }_1^4{y}^{\frac{1}{2}}dy}\\ =\frac{1}{2}{\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]}_1^4\\ =\frac{1}{3}\left[4^{\frac{3}{2}}-1^{\frac{3}{2}}\right]\\ =\frac{1}{3}\left(8-1\right)=\frac{7}{3}\text{square units}\end{array}

Q.23

Sketchthegraphofy=\left|x+3\right|andevaluate{\displaystyle {\int }_{-6}^0\left|x+3\right|\text{}dx.}

Ans.

The graph of y = |x + 3| is as shown below:

\begin{array}{l}\because y=\left|x+3\right|=\{\begin{array}{l}-\left(x+3\right),x+3<0\Rightarrow x<-3\\ \left(x+3\right),x+3\ge 0\Rightarrow x\ge -3\end{array}\\ \therefore {\displaystyle {\int }_{-6}^0\left|\left(x+3\right)\right|dx}={\displaystyle {\int }_{-6}^{-3}-\left(x+3\right)dx}+{\displaystyle {\int }_{-3}^0\left(x+3\right)dx}\\ =-{\left[\frac{x^2}{2}+3x\right]}_{-6}^{-3}+{\left[\frac{x^2}{2}+3x\right]}_{-3}^0\\ =-\left[\frac{{\left(-3\right)}^2}{2}+3\left(-3\right)-\frac{{\left(-6\right)}^2}{2}-3\left(-6\right)\right]+\left[0-\frac{{\left(-3\right)}^2}{2}-3\left(-3\right)\right]\\ =-\left(\frac{9}{2}-9-18+18\right)+\left(-\frac{9}{2}+9\right)\\ =\frac{9}{2}+\frac{9}{2}=9\text{square units}\end{array}

Q.24

$\begin{array}{l}\mathrm{Findtheareaboundedbythecurvey}=\mathrm{sinxbetweenx}=0\\ \mathrm{andx}=2\mathrm{\pi }.\end{array}$

Ans.

The graph of y = sin x is shown below:

$\begin{array}{l}\mathrm{Required}\mathrm{area}=\mathrm{Area}\mathrm{OABO}+\mathrm{Area}\mathrm{BCDB}\\ \hspace{0.17em}\hspace{0.17em}={\int }_0^{\mathrm{\pi }}\mathrm{sinxdx}+\left|{\int }_{\mathrm{\pi }}^{2\mathrm{\pi }}\mathrm{sinxdx}\right|\\ \hspace{0.17em}\hspace{0.17em}={[-\mathrm{cosx}]}_0^{\mathrm{\pi }}+\left|{[-\mathrm{cosx}]}_{\mathrm{\pi }}^{2\mathrm{\pi }}\right|\\ \hspace{0.17em}\hspace{0.17em}=-[\mathrm{cos\pi }+\cos 0]+|\cos 2\mathrm{\pi }-\mathrm{cos\pi }|\\ \hspace{0.17em}\hspace{0.17em}=-[-1-1]+(1+1)\\ \mathrm{\hspace{0.17em}}=4\mathrm{\hspace{0.17em}}\mathrm{square}\mathrm{units}\end{array}$

Q.25 Find the area enclosed between the parabola y² = 4ax and the line y = mx.

Ans.

Solving y² = 4ax and y = mx, we get x = (4a/m²)
and y = (4a/m).

\begin{array}{l}The\text{points of intersection of both the curves are}\left(0,0\right)\text{and}\\ \left(\frac{4a}{m^2},\frac{4a}{m}\right).\\ \therefore AreaOBAO=AreaOCABO-Area\left(\Delta OCA\right)\\ ={\displaystyle {\int }_0^{\frac{4a}{m^2}}2\sqrt{ax}}dx-{\displaystyle {\int }_0^{\frac{4a}{m^2}}mx}dx\\ =2\sqrt{a}{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^{\frac{4a}{m^2}}-m{\left[\frac{x^2}{2}\right]}_0^{\frac{4a}{m^2}}\\ =\frac{4}{3}\sqrt{a}\left\{{\left(\frac{4a}{m^2}\right)}^{\frac{3}{2}}-0\right\}-\frac{m}{2}\left\{{\left(\frac{4a}{m^2}\right)}^2-0\right\}\\ =\frac{4}{3}\sqrt{a}\left(\frac{8a\sqrt{a}}{m^3}\right)-\frac{m}{2}\left(\frac{16a^2}{m^4}\right)\\ =\frac{4}{3}\times \frac{8a^2}{m^3}-\frac{8a^2}{m^3}\\ =\frac{8a^2}{3m^3}\end{array}