If $f(x)$ is differentiable and $\int\limits_0^{t^2} x f(x) d x=\frac{2}{5} t^5$, then $f\left(\frac{4}{25}\right)$ equals

2/5

We have,

$\int\limits_0^{t^2} x f(x) d x=\frac{2}{5} t^5$

$\Rightarrow \frac{d}{d t} \int\limits_0^{t^2} x f(x) d x=\frac{d}{d t}\left(\frac{2}{5} t^5\right)$

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

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