# 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.

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.

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.$$\\ Area of ABCDA=\int _2^4 x \mathrm{d}y\\ x^2=4y\\ x=2\sqrt{y}\\ \int _2^4 x\mathrm{d}y=\int_2^4 2\sqrt{y}\mathrm{d}y\\ =2\int _2^4 \sqrt{y}\mathrm{d}y\\ =2\left[ \dfrac{\frac{y^\frac{3}{2}}{3}}{2}\right]_2^4\\ =\dfrac{4}{3}[(4)^\frac{3}{2}-(2)^\frac{3}{2}]\\ =\dfrac{4}{3}[8-2\sqrt{2}]\\ =(\dfrac{32-8\sqrt{2}}{3})sq.units 4 Find the area of the region bounded by the ellipse \dfrac{x^2}{16}+\dfrac{y^2}{9}=1 ##### Solution : The given equation of the ellipse,\dfrac{x^2}{16}+\dfrac{y^2}{9}=1can be represented as\\ It can be observed that the ellipse is symmetrical about x-axis and y-axis.\\ \therefore Area bounded by ellipse = 4 * Area of OABO$$\\$ 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 $OAB$

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