If $f\left(x\right)$ is differentiable and ${\int }_0^{t^2} xf\left(x\right)dx=\frac{2}{5}{t}^5,$ then $f\left(\frac{4}{25}\right)$ equals

Solution:

We have,

$\begin{array}{l} {\int }_0^{t^2} xf\left(x\right)dx=\frac{2}{5}{t}^5\\ \Rightarrow \frac{d}{dt}{\int }_0^{t^2} xf\left(x\right)dx=\frac{d}{dt}\left(\frac{2}{5}{t}^5\right)\end{array}$

$\Rightarrow 2t\times t^{2}f\left(t^2\right)=2t^4$                  [Using Leibnitz's rule]

$\Rightarrow f\left(t^2\right)=t\Rightarrow f\left(\frac{4}{25}\right)=\frac{2}{5}$