Question

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

• A. $\frac{2}{5}$
• B. $\frac{5}{2}$
• C. $1$
• D. $\pm \frac{5}{2}$

Answer 1

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

${}^{t^2}{\int }_{0}xf\left(x\right)dx=\frac{2}{5}{t}^5$ (using leneity theorem)

$\left(xt\right).t^{2}f\left(t^2\right)=\frac{2}{5}\times 5t^4$

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

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

$\therefore$ option A is correct.