Let f:(0,) R and F(x)= _0^xtf(t)dt.

Question

Let $f:(0,\infty)	o R$ and $F(x)=\int _{0}^{x}tf(t)dt$. If $F({x}^{2})={x}^{4}+{x}^{5}$, then $\sum _{r=1}^{12}f({r}^{2})$ is equal to \_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 219

Accepted Answer

$F(x)=\int _{0}^{x}t\cdot f(t)dt$

$\begin{gathered} {F}^{'}(x)=xf(x) \ F({x}^{2})={x}^{4}+{x}^{5},\, {x}^{2}=t \ F(t)={t}^{2}+{t}^{5/2} \ {F}^{'}(t)=2t+5/2{t}^{3/2} \ t\cdot f(t)=2t+5/2{t}^{3/2} \ f(t)=2+5/2{r}^{1/2} \end{gathered}$

$\begin{gathered} \sum _{r=1}^{12}f({r}^{2})=\sum _{r=1}^{12}2+\frac{5}{2}r \ =24+5/2[\frac{12(13)}{2}] \ =219 \end{gathered}$

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