Compute the area bounded between $y^2=4 a x$ and $y=m x,(a>0, m>0)$.

solution

$[\sqrt{4a x}-m x] d x=$ elementary area

$\text { Required } \left.A=\int (\sqrt{4 a x}-m x\right) d x$

$m^2 x^2=4 a x$

$\left(m^2 x-4 a\right) x=0$

$x=\frac{4 a}{m^2}, x=0$

Area bounded between $x$-axis,

$y=2\left(x^3-x^2-2 x\right) \text { and } x=-1 \text { to } x=2 \text {.}$

solution:

$y=2 x\left(x^2-x-2\right)$

$y=2 x(x+1)(x-2)$

$-1<x<0 \quad(x+1)>0, x<0$

$(x-2)<0$

$y>0$

$2>x>0 \quad x+1>0 \quad x-2<0$

$y<0,0<x<2$

$y=2(x^3-x^2-2x)$

$A_1=\int_{-1}^0 2\left(x^3-x^2-2 x\right) d x$

$=2\left(\frac{x^4}{4}-\frac{x^3}{3}-2 {\frac{x^2}{2}}\right)_{-1}^0$

$=-2\left(+\frac{1}{4}+\frac{1}{3}-1\right)$

$=-2\left(\frac{7}{12}-1\right)$

$=\frac{-2(-5)}{12}$

$=\frac{5}{6}$

$A_2=\int_0^2 2\left(x^3-x^2-2 x\right) d x$

$=\left.2\left(\frac{x^4}{4}-\frac{x^3}{3}-x^2\right)\right|_0 ^2=2\left(4-\frac{8}{3}-4\right)=-\frac{16}{3}$

Total Area= $A_1 + |A_2|$

$=\frac{5}{6}+\frac{16}{3}=\frac{37}{6}$

Find out area bounded between

$y=2-x^2$ and $y=-x$

$-x=2-x^2$

$x^2-x-2=0$

$x=\frac{1 \pm \sqrt{1+8}}{2}$

$x=2,-1$

$\int_ {-1}^2\left[2-x^2-(-x)\right] d x= \int_{-1}^2[2-x^2+x] d x$

= $2 x-\frac{x^3}{3}+\left.\frac{x^2}{2}\right|_{-1} ^2 = 4-\frac{8}{3}+2-\left(-2+\frac{1}{3}+\frac{1}{2}\right)$

$=8-\frac{8}{3}-\frac{5}{6} =8-\frac{16+5}{6} =8-\frac{21}{6} =\frac{9}{2}$

Area bounded between $y=x \sqrt{4-x^2}$,

$x-axis$ and $x=-2$ and $x=+2$.

solution:

$y=0, \quad x=0$

$y=0, \quad x= \pm 2$

$(0,2), y>0$

$(-2,0), y<0$

$A=\left|A_1\right| +A_2$

$A_2=\int_0^2 x \sqrt{4-x^2} d x$

$\Rightarrow A_2 =\frac{1}{2} \int_0^4 \sqrt{4-t} d t=\frac{1}{2} {\frac{(4-t)^{3 / 2}} {-3/2}}|_0 ^4$

$=-\frac{1}{3}\left[0-(4)^{3 / 2}\right]=+\frac{1}{3} \times 8=\frac{8}{3}$

$A_2=\frac{8}{3}$ $\Rightarrow A_1=-\frac{8}{3}$

$\text { Required } =\left|-\frac{8}{3}\right|+\frac{8}{3}=\frac{16}{3}$

Area bounded between two parabolas

$y=x^2$ & $y=8-x^2$

solution:

[$x^2=8-x^2$

$2 x^2=8$

$x^2=4, x= \pm 2$

$\int_{-a}^a f(x) d x=2 \int_0^a f(x) d x$

$\int_ {-2}^2\left(8-x^2-x^2\right) d x = \int_{-2}^2\left(8-2 x^2\right) d x$

$= 2 \int_ 0^2\left(8-2 x^2\right) d x= \left.2\left(8 x-\frac{2}{3} x^3\right)\right|_0 ^2$

$=2\left(16-\frac{2 \times 8}{3}\right)=32\left(1-\frac{1}{3}\right)=\frac{64}{3}$.

Find out area between $y=(1-\cos x) \sin x$

$x$-axis, $x=0$ and $x=\pi$.

$y={(1-\cos x)}{\sin x}$

$x=0 \quad y=0$

$x=\pi \quad y=0$

$\sin x \geqslant 0 \quad[0, \pi]$

$1-\cos x \geqslant 0 \quad[0, \pi]$

$\int_0^\pi {[(1-\cos x) \sin x d x}$

$\int_0^\pi(\sin x-\sin x \cos x) d x=\int_0^\pi\left(\sin x-\frac{1}{2} \sin 2 x\right) d x$

$=-\cos x+\left.\frac{\cos 2 x}{4}\right|_0 ^\pi$

$=-(-1-1)+\frac{1}{4}(1-1)=2$

Area bounded between $y=\cos ^2 x$ and

$y=1, x=0$ and $x=\pi$

solution:

$A =\int_0^\pi\left[1-\cos ^2 x\right] d x$

$=\int_0^\pi \sin ^2 x d x$

$=\int_0^\pi \frac{1-\cos 2 x}{2} d x$

$A = \frac{x}{2}-\frac{sin2x}{4}|_0 ^ \pi$

=$\frac{\pi}{2} - (0-0)=\frac{\pi}{2}$

$\int_a^b f(x) d x$

$f(x)$ is continuous on $[a, b]$

$a, b$ are finite

1) What if $f(x)$ is discontinuous on $[a, b]$

2) What if interval of integration is not bounded, i.e, $[a, \infty),(-\infty, a]$ OR $(-\infty, \infty)$.

$f(x)$ is discontinuous on $[a, b]$ but $f(x)$ is piecewise continuous.

$\int_{-2}^2[x] d x$

$= \int_{-2}^{-1}(-2) d x+\int_{-1}^0 (-1) d x$ + $\int_0^1 0 d x+\int_1^2 1 d x$

=$(-2)(x)|_ {-2} ^ {-1}$ + (-1) $x|_ {-1} ^ 0$ +0+ $x|_{1}^2$

$=(-2)(-1+2)+(-1)(0-(-1))+2-1$

$=-2-1+1=-2$

Let $f(x)$ be continuous but interval of integration is not bounded:-

1) $\int_0^{\infty}\left(\frac{1}{1+x^2}\right) d x$

2) $\int_{-\infty}^{\infty} \frac{1}{1+x^2} d x$

Function is discontinous but interval is finite $(say [a, b])$.

(discontinous - Infinite Values Some in the interval of $\Rightarrow$ integration).

Functions which are not continuous

also interval of integration is infinite.

For. $\int_{-\infty}^{\infty} \frac{1}{x^2} d x \quad \frac{1}{x^2}$ is not continuous at $x=0$

and interval of integration is also unbounded.

$I =\int_0^{\infty} \frac{1}{1+x^2} d x$

solution:

$=\lim _{a \rightarrow \infty} \int_0^a \frac{1}{1+x^2} d x$

$=\left.\lim _{a \rightarrow \infty} \tan ^{-1} x\right|_0 ^a$

$=\lim _{a \rightarrow \infty}\left[\tan ^{-1} a-\tan ^{-1} 0\right]$

$=\tan ^{-1} \infty=\pi / 2 .$