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$=4*\dfrac{3\pi}{2}=6\pi sq.units$