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

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