Let f :(0, ∞) → R be a continuous function such that $F(x)=\int\limits_0^{x^2} t f(t) d t$. If $F\left(x^2\right)=x^4+x^5$, then $\sum\limits_{r=1}^{12} f\left(r^2\right)=$ __________

It is given that f is continuous. Therefore, the integral function f(x) is differentiable. Also,

$F(x)=\int\limits_0^x t(f(t) d t$

$\Rightarrow F(x^2) = \int\limits_0^{x^2} t f(t) d t$

$\Rightarrow x^4+x^5 =\int\limits_0^{x^2} t f(t) d t$

Differentiating with respect to x, we get

$4 x^3+5 x^4=(2 x) x^2 f\left(x^2\right)$

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

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

$\Rightarrow \sum\limits_{r=1}^{12}f\left(r^2\right)=\sum\limits_{r=1}^{12}\left(2+\frac{5}{2} r\right)=24+\frac{5}{2} \times 78=219$