Question

If $f\left(x\right)$ is differentiable and ${\int }_0^{t^2}xf\left(x\right)dx=\frac{2}{5}{t}^5$ , then $f\left(\frac{4}{25}\right)$ equals :

• A. $\frac{2}{5}$
• B. $-\frac{5}{2}$
• C. $1$
• D. none of these

Answer 1

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

Differentiation of Definite Integral:

$\frac{d}{dt}\left({\int }_a^{b}f\right(x\left)dx\right)=\frac{db}{dt}f\left(b\right)-\frac{da}{dt}f\left(a\right)$

So,

$\frac{d}{dt}\left({\int }_0^{t^2}xf\right(x\left)dx\right)=\frac{d\left(t^2\right)}{dt}{t}^{2}f\left(t^2\right)-0$

$\frac{d}{dt}\left({\int }_0^{t^2}xf\right(x\left)dx\right)$$=2t^{3}f\left(t^2\right)$ .......... (1)

And ,

$\frac{d}{dt}\left(\frac{2}{5}{t}^5\right)=2t^4$ ............(2)

From (1) and (2),

$2t^{3}f\left(t^2\right)=2t^4$

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

Put $t=\frac{2}{5}$

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