If the area bounded by $y=\frac{x^3}{3}-x^2+a$ and the straight and lines x = 0 and x = 2 and the x-axis is minimum,  find the value of a.

Let the area be A.

Then $\begin{array}{l}A={\int }_0^2 \left(\frac{x^3}{3}-x^2+a\right)dx\\ ={\left(\frac{x^4}{12}-\frac{x^3}{3}+ax\right)}_0^2\\ =\left(\frac{4}{3}-\frac{8}{3}+2a\right)\\ =\left(2a-\frac{4}{3}\right)\\ \frac{dA}{da}=2>0\end{array}$

So, the area is minimum

It will be minimum when A = 0.

Thus, $2a-\frac{4}{3}=0$

$\Rightarrow a=\frac{2}{3}$

Hence, the value of $a\text{ is }\frac{2}{3}\text{. }$