Question

$\underset{0}{\overset{t^2}{\int }}xF\left(x\right)dx=\frac{2}{5}{t}^5,thenF\left(\frac{4}{25}\right)$

Answer 1

$\begin{array}{c}{\int }_0^{t^2}xF\left(x\right)dx=\frac{2}{5}{t}^5\\ \frac{d}{dt}\left({\int }_0^{t^2}xF\left(x\right)dx\right)=\frac{2}{5}.5t^4=2t^4\\ =t^{2}F\left(t^2\right).\left(2t\right)=\frac{2}{5}.5t^4=2t^4\\ =F\left(t^2\right)=t\\ \\ when\\ t=\frac{2}{5},F\left(t^2\right)=F\left(\frac{4}{25}\right)\\ =F\left(\frac{4}{25}\right)=\frac{2}{5}\end{array}$