JEE Main 2024 — 1 February, Shift 2 — Question 23

$F(x)=\int_{0}^{x} t \cdot f(t) d t$

Given $\begin{aligned} & F^{1}(x)=x f(x) & \end{aligned}$

$F\left(x^{2}\right)=x^{4}+x^{5}, \quad \text { let } x^{2}=t$

$F(t)=t^{2}+t^{5 / 2}$

$F^{\prime}(t)=2 t+5 / 2 t^{3 / 2}$

$t \cdot f(t)=2 t+5 / 2 t^{3 / 2}$

$\mathrm{f}(\mathrm{t})=2+5/2 \mathrm{t}^{1 / 2}$

$\sum_{\mathrm{r}=1}^{12} \mathrm{f}\left(\mathrm{r}^{2}\right)=\sum_{\mathrm{r}=1}^{12} 2+\frac{5}{2} \mathrm{r} =24+5 / 2\left[\frac{12(13)}{2}\right] =219$