I= $\int_{-2}^2 \frac{x^3 \sin ^4 x \cos ^2 x}{f(x)} d x$

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

$=0 \text { for odd function }$

$f(-x)=(-x)^3 \sin ^4(-x) \cos ^2(-x)$

$=-x^3 \sin ^4 x \cos ^2 x$

$=-f(x) \text { Integrand is odd. }$

$\text { hence } I=0 \text {. }$

$I=\int_0^1 \frac{\sqrt{x}}{\sqrt{x}+\sqrt{1-x}} d x \cdot-(1) \int_0^a f(x) d x=\int_0^a f(a-x) d x$

$I=\int_0^1 \frac{\sqrt{1-x)}}{\sqrt{1-x}+\sqrt{1-(1-x)}} d x=\int_0^1 \frac{\sqrt{1-x}}{\sqrt{1-x}+\sqrt{x}} d x$

(1) +(2)

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

$=\int_0^1 d x=1$

$I=\frac{1}{2}$

$I=\int_{-\pi / 2}^{\pi / 2} \sin ^2 x d x . \quad \sin ^2 x=\frac{1-\cos 2 x}{2}$

$I=2 \int_0^{\pi / 2} \sin ^2 x d x$

$\sin ^2(-x)=\sin ^2 x$

(1)

$\int_0^a f(x) d x=\int_0^a f(a-x) d x$

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

$I =2 \int_0^{\pi / 2} \cos ^2 x d x-(2)$

(1) + (2)

$2 I=2 \int_0^{\pi / 2}\left(\sin ^2 x+\cos ^2 x\right) d x=2 x \pi$

$2 I=\pi$

$I =\frac{\pi}{2}$

Ex

$I = \int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ (1)

I $=\int_0^{\pi / 2} \log \left(\frac{4+3 \sin \left(\frac{\pi}{2}-x\right)}{4+3 \cos \left(\frac{\pi}{2}-x\right)}\right) d x$

$x=\int_0^\pi f(x-a) d x$

$I =\int_0^{\pi / 2} \log \left(\frac{4+3 \cos x}{4+3 \sin x}\right) d x$ (2)

$2 I =\int_0^{\pi / 2}\left[\log \left(\frac{4+3 \sin x}{4+3 \cos x}\right)+\log \left(\frac{4+3 \cos x}{4+3 \sin x}\right)\right] d x$

$=\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+\cos x} \times \frac{4 x \operatorname{sen} x}{4+\sin x}\right) d x$

$\begin{aligned} & 2 I=\int_0^{\pi / 2} \log 1 d x \ & \Rightarrow 2 I=0 \\ & I=0\end{aligned}$

$\int_0^1 \frac{x}{I} \frac{e^x}{II} d x .$

= $\left.x \cdot e^x\right|_{x=0} ^{x=1}-\int_0^1 1 \cdot e^x d x$

= $1 e^1-0 e^0-\left.e^x\right|_0 ^1$

= $e-0-\left(e-e^0\right)$

= $e-e+1=1 .$

Cases.

Area of elementary

strip =y d x= dA

elementary Area

$A=\int_{x=a}^{x=b} d A=\int_a^b y d x .$

Case 2

elementary area

d A=x d y

A= $\int_{y=c}^{y=d} d A=\int_{y=c}^{y=d} x d y .$

Case 3

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

A<0

Required Area $=|A|$

Case 4

$A_1=\int_a^c f(x) d x>0$

$A_2 =\int_c^b f(x) d x<0$

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

Area of circle

$x^2+y^2=a^2$

Total Area 4A

$=4 \int_{x=0}^{x=a} y d x=4 \int_0^a \sqrt{a^2-x^2} d x$

$x^2+y^2=a^2$

$y= \pm \sqrt{a^2}-\overline{x^2}$

$=4\left[\frac{1}{2} x \sqrt{a^2-x^2}+\frac{1}{2} a^2 \sin ^{-1} \frac{x}{a}\right]_{x=0}^{x=a}$

$=4\left[\frac{1}{2} a^2 \frac{\pi}{2}-0-0\right]=\pi a^2 \text {. }$

Area of circle

=4 $\int_{y=0}^a x d y$

=4 $\int_0^a \sqrt{a^2-y^2} d y$

$x^2+y^2=a^2$

x $= \pm \sqrt{a^2-y^2}$

x $=\sqrt{a^2-y^2}$

$=4\left[\frac{1}{2} y \sqrt{a^2-y^2} +\frac{1}{2} a^2 \sin ^{-1} \frac{y}{a}\right]_{y=0}^a$

$=4\left[\frac{1}{2} a^2 \frac{\pi}{2}-0-0\right]=\pi a^2$

Ex. Area of Ellipse.

Total Area

$=4 A \quad =4 \int_{x=0}^{x=a} y d x$

$=4 \int_0^a \frac{b}{a} \sqrt{a^2-x^2} d x$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad a>b$

y= $\pm b \sqrt{1-\frac{x^2}{a^2}}$

y= $\pm \frac{b}{a} \sqrt{a^2-x^2}$

$=\frac{4 b}{a} \int_0^a \sqrt{a^2-x^2} d x$

$=\frac{4 b}{a}\left[\frac{1}{2} x \sqrt{a^2-x^2}+\frac{1}{2} a^2 \sin^{-1} \frac{x}{a} \right]$

Total Area = $\frac{4 b}{a}\left[\frac{1}{2} x \sqrt{a^2-x^2}+\frac{1}{2} a^2 \sin ^{-1} \frac{x}{a}\right]_{x=0}^a$

$=\frac{4 b}{a} \times\left[\frac{1}{2} a^2 \frac{\pi}{2}-0-0\right]$

$=\frac{4 b}{a} \times \frac{a^2 \pi}{4}$

$=\pi a b$

By using harizontal strip

$x=+a \sqrt{1-\frac{y^2}{b^2}}$

Aea of ellipse

• $=4 \int_{y=0}^b x d y$

$=4 \int_0^b \frac{a}{b} \sqrt{b^2-y^2} d y$

$\frac{x^2}{a^2}=1-\frac{y^2}{b^2}$

$\frac{x}{a}= \pm \sqrt{1-\frac{b^2}{b^2}}$

$=\frac{4 a}{b}\left[\frac{1}{2} y \sqrt{b^2-y^2}+\frac{1}{2} b^2 \sin ^{-1} \frac{y}{b}\right]_0^b$

$=\frac{4 a}{\not b}\left[0+\frac{1}{2} b^2 \frac{\pi}{2}-0-0\right]=\pi a b,$

Area bounded between a line and a curve

Ex.1. Area bounded between $y=1$ & $y=x^2$.

Total Area required

$=2 \times A$

$=2 \int_{y=0}^{y=1} x d y$

$=2 \int_0^1 \sqrt{y} d y=\left.2 \frac{y^{3 / 2}}{3 / 2}\right|_0 ^1=\frac{4}{3}$.

By using vertical stop (elementary rectangles)

Required Area

$=\int_{x=-1}^1 y d x-\int_{y=1}^1 y d x$

$y=x^2$

$y=1, y=x^2 , x= \pm 1$

$=\int_{-1}^1 1 d x-\int_{-1}^1 x^2 d x$

=2-$\frac{x^3}{3} |_1^1=2-\frac{2}{3}=\frac{4}{3}$

Ex. Area bounded between $y=x, y^2=2-x$ and $y=0$ which is in first quadrant.

$x^2=2-x$

$x^2+x-2=0$

$x=\frac{-1 \pm \sqrt{1+0}}{2}=\frac{-1 \pm 3}{2}$

x=-2,+1

$A=A_1+A_2=\int_{x=0)}^1 y d x+\int_{x=1}^2 ydx$

y= $\sqrt{2-x}$

A $=A_1+A_2$

= $\int_0^1 x d x+\int_1^2 \sqrt{2-x} d x$

$=\left.\frac{1}{2} x^2\right|_0 ^1+\left.\frac{(2-x)^{3 / 2}}{-3 / 2}\right|_1 ^2$

$=\frac{1}{2}-0+0-\left(-\frac{2}{3}\right)$

= $\frac{1}{2}+\frac{2}{3}=\frac{7}{6}$