Let f be a continuous function satisfying ${\int }_0^{{\mathrm{t}}^2}\left(\mathrm{f}\left(\mathrm{x}\right)+{\mathrm{x}}^2\right)\mathrm{dx}=\frac{4}{3}{\mathrm{t}}^3,\forall \mathrm{t}>0$ . Then $\mathrm{f}\left(\frac{{\mathrm{\pi }}^2}{4}\right)$ is equal to

Given equation is ${\int }_0^{{\mathrm{t}}^2}\left(\mathrm{f}\left(\mathrm{x}\right)+{\mathrm{x}}^2\right)\mathrm{dx}=\frac{4}{3}{\mathrm{t}}^3,\forall \mathrm{t}>0$

According to Newton Leibnitz theorem we have$\frac{d}{d\mathrm{x}}\left({\int }_{\mathrm{u}\left(\mathrm{x}\right)}^{\mathrm{v}\left(\mathrm{x}\right)}\mathrm{f}\left(\mathrm{t}\right)d\mathrm{t}\right)=\mathrm{f}\left(\mathrm{v}\left(\mathrm{x}\right)\right)\times \mathrm{v}\text{'}\left(\mathrm{x}\right)-\mathrm{f}\left(\mathrm{u}\left(\mathrm{x}\right)\right)\times \mathrm{u}\text{'}\left(\mathrm{x}\right)$

Apply Newtons Leibnitz theorem in the given equation.

$\Rightarrow \left(\mathrm{f}\left({\mathrm{t}}^2\right)+{\mathrm{t}}^4\right)2\mathrm{t}-0=4{\mathrm{t}}^2$

$\Rightarrow \mathrm{f}\left({\mathrm{t}}^2\right)+{\mathrm{t}}^4=2\mathrm{t}$

$\Rightarrow \mathrm{f}\left({\mathrm{x}}^2\right)=-{\mathrm{x}}^4+2\mathrm{x}$

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=-{\mathrm{x}}^2+2\sqrt{\mathrm{x}}$

$\Rightarrow \mathrm{f}\left(\frac{{\mathrm{\pi }}^2}{4}\right)=-\frac{{\mathrm{\pi }}^4}{4^2}+2\times \frac{\mathrm{\pi }}{2}$

$=\frac{-{\mathrm{\pi }}^4}{16}+\mathrm{\pi }$

$=\mathrm{\pi }\left(1-\frac{{\mathrm{\pi }}^3}{16}\right)$

Hence this is the correct option.