Application of Integrals

Class 12 NCERT

NCERT

1   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.

Solution :

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 $ABCDA.$ Area $ABCDA =\int_1^4 \sqrt{x}dx\\ =\left[\dfrac{\dfrac{x^{\frac{3}{2}}}{3}}{2}\right]_1^4$$\\$ $ =\dfrac{2}{3}[(4)^{\frac{3}{2}}-(1)^{\frac{3}{2}}]\\ =\dfrac{2}{3}[8-1]\\ =\dfrac{14}{3}sq. units$

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

Solution :

The area of the region bounded by the curve, $y ^2 = 9 x , x = 2,$ and $x = 4 $, and the x-axis is the area $ABCDA.$ $\\$ Area $ABCDA =\int_2^4 3\sqrt{x}\mathrm{d}x$$\\$ $ =3[\dfrac{x ^{\frac{3}{2}}}{{\frac{3}{2}}}]_2^4\\ =2[x\frac{3}{2}]_2^4\\ =2[4^\frac{3}{2}-2^\frac{3}{2}]\\ =2[2^3-8^\frac{1}{2}] =2[8-2\sqrt{2}]\\ =16-4\sqrt{2} sq.units$

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

Solution :

The area of the region bounded by the curve, $x ^2 = 4 y , y = 2,$ and $ y = 4,$ and the y-axis is the area $ABCDA.$$\\$ Area of $ABCDA=\int _2^4 x \mathrm{d}y\\ x^2=4y\\ x=2\sqrt{y}\\ \int _2^4 x\mathrm{d}y=\int_2^4 2\sqrt{y}\mathrm{d}y\\ =2\int _2^4 \sqrt{y}\mathrm{d}y\\ =2\left[ \dfrac{\frac{y^\frac{3}{2}}{3}}{2}\right]_2^4\\ =\dfrac{4}{3}[(4)^\frac{3}{2}-(2)^\frac{3}{2}]\\ =\dfrac{4}{3}[8-2\sqrt{2}]\\ =(\dfrac{32-8\sqrt{2}}{3})sq.units$

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

Solution :

The given equation of the ellipse,$\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$can be represented as$\\$ It can be observed that the ellipse is symmetrical about x-axis and y-axis.$\\$ $\therefore $ Area bounded by ellipse =$ 4 *$ Area of $OABO$$\\$ Area $OABO=\int _0^4 y \mathrm{d}x\\ \dfrac{x^2}{16}+\dfrac{y^2}{9}=1$$\\$ $\implies \dfrac{y^2}{9}=1-\dfrac{x^2}{16}\\ \implies y^2-9(1-\dfrac{x^2}{16})\\ y=3\sqrt{1-\dfrac{x^2}{16}}$$\\$ Area $OABO =\int_0^4 3\sqrt{1-\dfrac{x^2}{16}\mathrm{d}x}\\ =\dfrac{3}{4}\int_0^4\sqrt{16-x^2}\mathrm{d}x$$\\$ Substitute $ x=4 \sin \theta,\theta =\sin^-1\dfrac{x}{4}\\ =\dfrac{3}{4}\int_0^\frac{\pi}{2}\sqrt{16-16\sin^2\theta.}4\cos\theta \mathrm{d}\theta \\ =12\int_0^\frac{\pi}{2}\sqrt{1-\sin^2\theta},\cos\theta \mathrm{d}\theta\\ =12\int_0^\frac{\pi}{2}\cos^2\theta \mathrm{d}\theta \\ =12\int_0^\frac{\pi}{2}\dfrac{1+\cos2\theta}{2}\mathrm{d}\theta\\ =6\int_0^\frac{\pi}{2}(1+\cos2\theta)\mathrm{d}\theta \\ =6[\theta+\dfrac{\sin2\theta}{2}]_0^\frac{\pi}{2}\\ =6[\dfrac{\pi}{2}+\dfrac{\sin \pi}{2}-0-\dfrac{\sin0}{2}]\\ =6[\dfrac{\pi}{2}]=3\pi$$\\$ Therefore, area bounded by the ellipse $= 4 * 3 \pi= 12 \pi sq. units$

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

Solution :

The given equation of the ellipse can be represented as$\\$ $\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$$\\$ $\dfrac{y^2}{9}=1-\dfrac{x^2}{4}\\ y^2=9(1-\dfrac{x^2}{4})\\ \implies y=3\sqrt{1-\dfrac{x^2}{4}} ......(1)$$\\$ It can be observed that the ellipse is symmetrical about x-axis and y-axis.$\\$ $\therefore $ Area bounded by ellipse = 4 *Area OABO$\\$ Area of $OABO =\int_0^2 y\mathrm{d}x$$\\$ $=\int_0^2 3\sqrt{1-\dfrac{x^2}{4}\mathrm{d}x} \ \ \ \ \ [\text{Using } (1)]\\ =\dfrac{3}{2}\int_0^2\sqrt{4-x^2}\mathrm{d}x$$\\$ Substitute $ x=2\sin\theta \implies \theta =\sin^-1(\dfrac{x}{2})\\ \mathrm{d}x=2 \cos\theta \mathrm{d}\theta$$\\$ when,$x=0,\theta=0 \& x=2 \ \ \theta =\dfrac{x}{2}\\ \therefore \dfrac{3}{2}\int_0^2\sqrt{4-x^2}\mathrm{d}x\\ =\dfrac{3}{2}\int_0^\frac{\pi}{2}\sqrt{4-4\sin^2\theta}.2\cos\theta \mathrm{d}\theta \\ =3\int_0^\frac{\pi}{2}\sqrt{4-4\sin^2\theta}.\cos\theta \mathrm{d}\theta \\ =6\int_0^\frac{\pi}{2}\sqrt{1-\sin^2\theta}\cos\theta \mathrm{d}\theta\\ =6\int_0^\frac{\pi}{2}\cos2\theta\mathrm{d}\theta\\ =6\int_0^\frac{\pi}{2}\dfrac{1+\cos2\theta}{2}\mathrm{d}\theta\\ =\dfrac{6}{2}\int_0^\frac{\pi}{2}(1+\cos2\theta)\mathrm{d}\theta\\ =3[0+\dfrac{\sin2\theta}{2}]_0^\frac{\pi}{2}\\ 3[\dfrac{\pi}{2}+\dfrac{\sin\pi}{2}-0]\\ =3*\dfrac{\pi}{2}=\dfrac{3\pi}{2}$$\\$ Therefore, area bounded by the ellipse$=4*\dfrac{3\pi}{2}=6\pi sq.units$

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

Solution :

The area of the region bounded by the circle, $x^2+y62=4,x=\sqrt{3}y$, and the x-axis is the area $OAB$

Substituting $ x=\sqrt{3}y$ in $x^2+y^2=4$ for finding the point of intersection.$\\$ $\therefore (\sqrt{3}y)+y^2=4\Rightarrow y^2-1 \Rightarrow y=\pm1,x=\pm\sqrt{3}$$\\$ The point of intersection of the line and the circle in the first quadrant is$(\sqrt{3},1)$$\\$ Area $OABO$=Area $\Delta OCA$ + Area $ACBA$$\\$ Area of $OAC=\dfrac{1}{2}*OC* AC=\dfrac{1}{2}*\sqrt{3}*1=\dfrac{\sqrt{3}}{2} \ \ ...(1)$$\\$ Area of$ABCA=\int_{\sqrt{3}}^2 y\mathrm dx$$\\$ $\int_{\sqrt{3}}^2 y\mathrm dx=\int_{\sqrt{3}}^2 \sqrt{4-x^2}\mathrm dx $$\\$ $x=2 \sin \theta \ \ \ \theta =\sin^{-1}(\dfrac{x}{2})$$\\$

when $ x=2 \ \ \ \theta =\dfrac{\pi}{2}\\ x=\sqrt{3} \ \ \theta =\dfrac{\pi}{3}\\ \therefore \int_{\sqrt{3}}^2 \sqrt{4-x}\mathrm dx \\ =\int_{\frac{\pi}{2}}^{\frac{\pi}{3}}\sqrt{4-4 \sin^2 \theta }(2 \cos \theta ) \mathrm d \theta \\ =4\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \cos^2 \theta \mathrm d \theta \\ =4\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} 1+\dfrac{\cos2 \theta }{2}\mathrm d \theta \\ =2\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}(1+\cos 2 \theta )\mathrm \theta \\ =2[\theta +\dfrac{\sin 2 \theta }{2}]_{\dfrac{\pi}{3}}^{\dfrac{\pi}{2}} \\ =2[\dfrac{\pi}{2}+\dfrac{1}{2}\sin \pi -\dfrac{\pi}{3}-\dfrac{1}{2}\sin \dfrac{2 \pi}{3}]\\ =2[\dfrac{\pi}{2}-\dfrac{\pi}{3}-\dfrac{1}{2}*\dfrac{\sqrt{3}}{2}]=2[\dfrac{\pi}{6}-\dfrac{\sqrt{3}}{4}]=\dfrac{\pi}{3} \ \ \ ......(2)$$\\$ From (1) & (2)$\\$ Area of $OAB=\dfrac{\sqrt{3}}{2}+2[\dfrac{\pi}{6}-\dfrac{\sqrt{3}}{4}]=\dfrac{\pi}{3}$$\\$ Therefore, area enclosed by x-axis, the line $ x=\sqrt{3}y$,and the circle $ x^2+y^2=4$ in the first quadrant =$\dfrac{\pi}{3}$ sq.units

7   Find the area of the smaller part of the circle $4x^2+y^2=a^2$ cut off by the line $x=\dfrac{a}{\sqrt{2}}.$

Solution :

The area of the smaller part of the circle $4x^2+y^2=a^2$ cut off by the line $x=\dfrac{a}{\sqrt{2}},$ is the area $ABCDA$$\\$ It can be observed that the area $ABCDA$ is symmetrical about x-axis.$\\$ Area of $ABCA =\int_{\frac{a}{\sqrt{2}}}^a y\mathrm dx\\ =\int_{\frac{a}{\sqrt{2}}}^a \sqrt{a^2-x^2}\mathrm dx \\ x= a \sin \theta \ \ \ \mathrm dx = a \cos \theta \mathrm d\theta \\ x=\dfrac{a}{\sqrt{2}} \ \ \theta =\sin^{-1}(\dfrac{x}{a})\\ =\sin^{-1}(\dfrac{1}{\sqrt{2}})=\dfrac{\pi}{4}\\ x=a, \theta =\sin^{-1}(\dfrac{a}{a})=\dfrac{\pi}{2}\\ \therefore =\int_{\frac{a}{\sqrt{2}}}^a \sqrt{a^2-x^2}\mathrm dx \\ =\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{a^2-a^2 \sin^2 \theta }.(a \cos \theta ) \mathrm d \theta \\ \Rightarrow a^2 \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos^2 \theta \\ =a^2 \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \dfrac{1+\cos 2 \theta }{2} \mathrm d \theta \\ =\dfrac{a^2}{2}[\theta + \dfrac{\sin 2 \theta }{2}]_{\frac{\pi}{4}}^{\frac{\pi}{2}}\\ =\dfrac{a^2}{2}[\dfrac{\pi}{2}+\dfrac{\sin \pi}{2}-\dfrac{\pi}{4}-\dfrac{\sin \frac{\pi}{2}}{2}]\\ `=\dfrac{a^2}{2}[\dfrac{\pi}{4}-\dfrac{1}{2}]\\ =\dfrac{a^2}{4}[\dfrac{\pi}{2}-1]\\ \Rightarrow ABCD =2[\dfrac{a^2}{4}(\dfrac{\pi}{2}-1)]\\ =\dfrac{a^2}{2}(\dfrac{\pi}{2}-1)$$\\$ Therefore, the area of smaller part of the circle,$x^2+y^2=a62$,cut off by the line $x=\dfrac{a}{\sqrt{2}}$,is $\dfrac{a^2}{2}(\dfrac{\pi}{2}-1)$sq. units.

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

Solution :

The line $x = a$ , divides the area bounded by the parabola $x = y ^2$ and $x = 4$ into two equal parts.$\\$ $\therefore $ Area $OADO =$ Area $ABCDA$$\\$

It can be observed that the given area is symmetrical about x-axis.$\\$ Area of $OEDO =\dfrac{1 }{2}$ Area of $OADO$$\\$ Area of $EFCDE =\dfrac{1} {2 }$Area of $ABCDA$$\\$ Therefore, Area $OEDO =$ Area $EFCDE$$\\$ Area $OEDO =\int_0^a y \mathrm dx\\ =\int_0^a \sqrt{x}\mathrm dx \\ \left[\dfrac{\dfrac{x^{\frac{3}{2}}}{3}}{2}\right]_0^a\\ =\dfrac{2}{3}(a)^{\frac{3}{2}} \ \ \ \ \ \ \ \ \ (1)$$\\$ Area of $EFCDE =\int_a^4 y \mathrm dx=\int_a^4 \sqrt{x}\mathrm dx\\ =\left[\dfrac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]\\ =\dfrac{2}{3}[4^{\frac{3}{2}}-a^{\frac{3}{2}}]\\ =\dfrac{2}{3}[8-a^{\frac{3}{2}}] \ \ \ \ \ ....(2)$$\\$ From (1) and (2), we obtain$\\$ $\dfrac{2}{3}(a)^{\frac{3}{2}}=\dfrac{2}{3}[8-(a)^\frac{3}{2}]\\ \Rightarrow 2.(a)^\frac{3}{2}=8\\ \Rightarrow (a)^{\frac{3}{2}}=4\\ \Rightarrow a=(4)^{\frac{2}{3}}$$\\$ Therefore, the value of a is $(4)^{\frac{2}{3}}$$\\$

9   Find the area of the region bounded by the parabola $y = x^ 2$ and $y =| x|$ .

Solution :

The area bounded by the parabola, $x^2=y,$ and the line , $ y=|x|$, can be represented as

The given area is symmetrical about y-axis.$\\$ $\therefore $ Area $ OACO $ = Area $ ODBO$$\\$ The point of intersection of parabola $x^2=y $ and line $ y=x $ is $ A(1,1).$$\\$ Area of $ OACO =$ Area $ \Delta OAM-$ Area $ OMACO$$\\$ $\therefore $ Area of $\Delta OAM=\dfrac{1}{2}*OM*AM =\dfrac{1}{2}*1*1=\dfrac{1}{2}$$\\$ Area of $ OMACO=\int_0^1y \mathrm dx=\int_0^1 x^2 \mathrm dx=[\dfrac{x^3}{3}]_0^1=\dfrac{1}{3}$$\\$ $\Rightarrow $ Area of $OACO=$ Area of $\Delta OAM-$ Area of $OMACO$$\\$ $=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}$$\\$ Therefore, required area $=2[\dfrac{1}{6}]=\dfrac{1}{3} $ sq.units.

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

Solution :

The area bounded by the curve, $x ^2 =4 y$ , and line, $x = 4 y -2$ , is represented by the shaded area $OBAO.$

Let A and B be the points of intersection of the line and parabola.$\\$ Substituting $x=4y-2 $ in $x^2=4y$$\\$ $(4y -2)^2=4y\\ 16y^2-16 y +4=4y\\ 16y^2-20y+4=0\\ 4y^2-5y+1=0\\ (4y-1)(y-1)=0\\ y=\frac{1}{4}, x=-1\\ y=1,x=2$$\\$ Coordinates of point A are $(-1,\dfrac{1}{4})$$\\$ Coordinates of point B are $(2, 1).$$\\$ We draw AL and BM perpendicular to x-axis.$\\$ It can be observed that,$\\$ Area $OBAO =$ Area $OBCO + $Area $OACO \ \ \ ... (1)$$\\$ Area $OMBCO$ = Area under the line $x = 4 y - 2$ between $x = 0$ and $x = 2$$\\$ Area $OMBCO=\int_0^2 \dfrac{x+2}{4}\mathrm dx$$\\$ Area $OMBO $ = Area under the curve $x^ 2 = 4 y$ between $x = 0$ and $x = 2$$\\$ Area $ OMBO =\int_0^2 \dfrac{x^2}{4}\mathrm dx$$\\$ Then, Area $OBCO =$ Area $OMBCO -$ Area $OMBO$$\\$ $=\int_0^2\dfrac{x+2}{4}\mathrm dx-\int_0^2 \dfrac{x^2}{4}\mathrm dx\\ =\dfrac{1}{4}[\dfrac{x^2}{2}+2x]_0^2-\dfrac{1}{4}[\dfrac{x^3}{3}]_0^2\\ =\dfrac{1}{4}[2+4]-\dfrac{1}{4}[\dfrac{8}{3}]\\ =\dfrac{3}{2}-\dfrac{2}{3}=\dfrac{5}{6}$$\\$ Area $OLACO =$ Area under the line $x= 4 y - 2$ between $x=- 1 $ and $x = 0$$\\$ Area $ OLACO =\int_{-1}^{0} \dfrac{x+2}{4}\mathrm dx$$\\$ Area $OLACO $= Area under the curve $x ^2 = 4 y$ between $x =- 1$ and $x =$ Area $OMBO = \int_0^2 \dfrac{x^2}{4} \mathrm dx $$\\$ Area $OACO =$ Area $OLACO-$ Area $OLAO$$\\$ $ \int_{-1}^0\dfrac{x+2}{4} \mathrm dx-\int_{-1}^0\dfrac{x^2}{4}\mathrm d x\\ =\dfrac{1}{4}[\dfrac{x^2}{2}+2x]_{-1}^0-\dfrac{1}{4}[\dfrac{x^3}{3}]_{-1}^0\\ =\dfrac{1}{4}[\dfrac{0}{2}+0-\dfrac{(-1)^2}{2}-2(-1)]-\dfrac{1}{4}[\dfrac{0^3}{0}-\dfrac{(-1)^3}{3}]\\ =-\dfrac{1}{4}[\dfrac{(-1)}{2}+2(-1)]-[-\dfrac{1}{4}(\dfrac{(-1)^3}{3})]\\ =-\dfrac{1}{4}[\dfrac{1}{2}-2]-\dfrac{1}{12}\\ =\dfrac{1}{2}-\dfrac{1}{8}-\dfrac{1}{12}\\ =\dfrac{7}{24}$$\\$ Therefore, required area $=(\dfrac{5}{6}+\dfrac{7}{24})=\dfrac{9}{8}$ sq.units