Let f: (0, ∞) → R and F(x) = ∫ _ 0 ^ x tf(t)dt . If F( x ^ 2 ) = x...

Given: 

$F\left(x\right) = \mathop{∫}_{0}^{x}t ⋅ f\left(t\right)dt$

Now, applying Newton Leibnitz Theorem we get,

${F}^{'}\left(x\right) = x ⋅ f\left(x\right)$

 And $F\left({x}^{2}\right) = {x}^{4} + {x}^{5}$

Let ${x}^{2} = t$

$⇒ F\left(t\right) = {t}^{2} + t\dfrac{5}{2}$

$⇒ {F}^{'}\left(t\right) = 2t + \dfrac{5}{2}t\dfrac{3}{2} \left\{\text{as}{F}^{'}\left(x\right) = x ⋅ f\left(x\right)\right\}$

$⇒ t ⋅ f\left(t\right) = 2t + \dfrac{5}{2}t\dfrac{3}{2}$

$⇒ f\left(t\right) = 2 + \dfrac{5}{2}t\dfrac{1}{2}$

$⇒ \mathop{∑}_{r = 1}^{12}f\left({r}^{2}\right) = \mathop{∑}_{r = 1}^{12}2 + \dfrac{5}{2}r$

$⇒ \mathop{∑}_{r = 1}^{12}f\left({r}^{2}\right) = 24 + \dfrac{5}{2}\left[\dfrac{12\left(13\right)}{2}\right]$

$⇒ \mathop{∑}_{r = 1}^{12}f\left({r}^{2}\right) = 219$