Let f be a continuous function satisfying ∫ _ 0 ^ t ^ 2 (f ( x ) +...

Given equation is ${∫}_{0}^{{t}^{2}}\left(f ( x ) + {x}^{2}\right)dx = \dfrac{4}{3}{t}^{3}, \forall t > 0$

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

Apply Newtons Leibnitz theorem in the given equation.

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

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

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

$⇒ f ( x ) = - {x}^{2} + 2\sqrt{x}$

$⇒ f\left(\dfrac{{\pi}^{2}}{4}\right) = - \dfrac{{\pi}^{4}}{{4}^{2}} + 2 \times \dfrac{\pi}{2}$

$= \dfrac{- {\pi}^{4}}{16} + \pi$

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

Hence this is the correct option.