Question

If ${\int }_0^{t^2}xf\left(x\right)dx=\frac{2}{5}{t}^5,t>0$, then $f\left(\frac{4}{25}\right)$ is?

• A. $\frac{2}{5}$
• B. $\frac{5}{2}$
• C. $-\frac{5}{2}$
• D. None of these

Answer 1

The correct option is A $\frac{2}{5}$

Given, ${\int }_0^{t^2}xf\left(x\right)dx=\frac{2}{5}{t}^5$

Differentiating both side :

$t^{2}f\left(t^2\right)\left(2t\right)=2t^4$

$\Rightarrow f\left(t^2\right)=t$

$\therefore f\left(\frac{4}{25}\right)=\frac{2}{5}$