Consider curves $y=\frac{1}{x^2},y=\frac{1}{4\left(x-1\right)}.\text{ Let }\alpha$ be the value of $a\left(a>2\right)$ for which area bounded by curves between $x=2\text{and}x=a\text{ is }1/a\text{ is }{e}^2+1\text{and}\beta \text{ be the of }b\in \left(1,2\right),$ for which the area bounded by curves between x=b and $x=2\text{ is }1-\frac{1}{b},$ then

If y=f(x) is a monotonic function in (a,b), then the area bounded by the ordinates at $x=a,x=b,y=f\left(x\right)\text{and}y=f\left(c\right)\left(\text{where }c\in \left(a,b\right)\right)\text{ is minimum when }c=\frac{a+b}{2}$.
$\text{Proof : }A=\stackrel{c}{\underset{a}{\int }}\left(f\left(c\right)-f\left(x\right)\right)dx+\stackrel{b}{\underset{c}{\int }}\left(f\left(c\right)\right)dx$
$=f\left(c\right)\left(c-a\right)-\stackrel{c}{\underset{a}{\int }}\left(f\left(x\right)\right)dx+\stackrel{b}{\underset{a}{\int }}\left(f\left(x\right)\right)dx-f\left(c\right)\left(b-c\right)$
$\Rightarrow A=\left[2c-\left(a+b\right)\right]f\left(c\right)+\stackrel{b}{\underset{c}{\int }}\left(f\left(x\right)\right)dx-\stackrel{c}{\underset{a}{\int }}\left(f\left(x\right)\right)dx$

Differentiating w.r.t. c, we get
$\frac{dA}{dc}=\left[2c-\left(a+b\right)\right]f^{\prime }\left(c\right)+2f\left(c\right)+0-f\left(c\right)-\left(f\left(c\right)-0\right)$
For maxima and minima , $\frac{dA}{dc}=0$
$\Rightarrow f^{\prime }\left(c\right)\left[2c-\left(a+b\right)\right]=0\left(as{f}^{\prime }\left(c\right)\ne 0\right)$
Hence, $c=\frac{a+b}{2}$
$\text{Also for }c<\frac{a+b}{2},\frac{dA}{dc}<0\text{ and for }c>\frac{a+b}{2},\frac{dA}{dc}>0$
Hence, A is minimum when $c=\frac{a+b}{2}$.
If the area bounded by $f\left(x\right)=\frac{x^3}{3}-x^2+a$ and the straight lines x=0, x=2, and the x-axis is minimum, then the value of a is

$\text{Two curves }{C}_1\equiv {\left[f\left(y\right)\right]}^{2/3}+{\left[f\left(x\right)\right]}^{1/3}=0\text{and}{C}_2\equiv {\left[f\left(y\right)\right]}^{2/3}+{\left[f\left(x\right)\right]}^{2/3}=12,\text{ satisfying the relation }\left(x-y\right)f\left(x+y\right)-\left(x+y\right)f\left(x-y\right)=4xy\left(x^2-y^2\right)$
The area bounded by $C_1\text{and}{C}_2$ is

Consider the two curves $C_1:y=1+\cos x\text{and}{C}_2:y=1+\cos \left(x-\alpha \right)\text{ for }\alpha \in \left(0,\frac{\pi }{2}\right),\text{ where }x\in \left[0,\pi \right].$ Also the area of the figure bounded by the curves $C_1,C_2,\text{and}x=0$ is same as that of the figure bounded by $C_2,y=1,\text{and}x=\pi$.
For the values of $\alpha$, the area bounded by $C_1,C_2,x=0\text{and}x=\pi$ is

Consider the area $S_0,S_1,S_2\dots .$ bounded by the x-axis and half-waves of the curve $y=e^{-x}\sin x,\text{ where }x\ge 0.$
The value of $S_0$ is

$\text{Two curves }{C}_1\equiv {\left[f\left(y\right)\right]}^{2/3}+{\left[f\left(x\right)\right]}^{1/3}=0\text{and}{C}_2\equiv {\left[f\left(y\right)\right]}^{2/3}+{\left[f\left(x\right)\right]}^{2/3}=12,\text{ satisfying the relation }\left(x-y\right)f\left(x+y\right)-\left(x+y\right)f\left(x-y\right)=4xy\left(x^2-y^2\right)$
The area bounded by the curve $C_2\text{and}\left|x\right|+\left|y\right|=\sqrt{12}$ is

Consider the area $S_0,S_1,S_2\dots .$ bounded by the x-axis and half-waves of the curve $y=e^{-x}\sin x,\text{ where }x\ge 0.$
The sequence $S_0,S_1,S_2,\dots ,$ forms a G.P. with common ratio

$\text{Two curves }{C}_1\equiv {\left[f\left(y\right)\right]}^{2/3}+{\left[f\left(x\right)\right]}^{1/3}=0\text{and}{C}_2\equiv {\left[f\left(y\right)\right]}^{2/3}+{\left[f\left(x\right)\right]}^{2/3}=12,\text{ satisfying the relation }\left(x-y\right)f\left(x+y\right)-\left(x+y\right)f\left(x-y\right)=4xy\left(x^2-y^2\right)$
The area bounded by $C_1\text{and}x+y+2=0$ is

Consider the area $S_0,S_1,S_2\dots .$ bounded by the x-axis and half-waves of the curve $y=e^{-x}\sin x,\text{ where }x\ge 0.$
$\stackrel{\infty }{\underset{n=0}{\Sigma }}{S}_n$ is equal to

Consider the two curves $C_1:y=1+\cos x\text{and}{C}_2:y=1+\cos \left(x-\alpha \right)\text{ for }\alpha \in \left(0,\frac{\pi }{2}\right),\text{ where }x\in \left[0,\pi \right].$ Also the area of the figure bounded by the curves $C_1,C_2,\text{and}x=0$ is same as that of the figure bounded by $C_2,y=1,\text{and}x=\pi$.
The value of $\alpha$ is