Let f : 0 , ∞ → R and F x = ∫ 0 x t f t d t . If F x 2 = x 4 + x 5 , then ∑ r = 1 12 f r 2 is equal to:

Given: 

$F\left(x\right)=\underset{0}{\overset{x}{\int }}t\cdot f\left(t\right)dt$

Now, applying Newton Leibnitz Theorem we get,

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

 And $F\left(x^2\right)=x^4+x^5$

Let $x^2=t$

$\Rightarrow F\left(t\right)=t^2+t^{\frac{5}{2}}$

$\Rightarrow F^{\text{'}}\left(t\right)=2t+\frac{5}{2}{t}^{\frac{3}{2}}$ $\left\{\text{as} F^{\text{'}}\left(x\right)=x\cdot f\left(x\right)\right\}$

$\Rightarrow t\cdot f\left(t\right)=2t+\frac{5}{2}{t}^{\frac{3}{2}}$

$\Rightarrow f\left(t\right)=2+\frac{5}{2}{t}^{\frac{1}{2}}$

$\Rightarrow \sum _{r=1}^{12}f\left(r^2\right)=\sum _{r=1}^{12}2+\frac{5}{2}r$

$\Rightarrow \sum _{r=1}^{12}f\left(r^2\right)=24+\frac{5}{2}\left[\frac{12\left(13\right)}{2}\right]$

$\Rightarrow \sum _{r=1}^{12}f\left(r^2\right)=219$