Question

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

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

Answer 1

Answer: (A)

$\frac{2}{5}$

${\int }_0^{t^2}xf\left(x\right)dx=\frac{2}{5}{t}^5(Here,t>0)$
Differentiating both sides w.r.t.t
$\Rightarrow t^{2}f\left(t^2\right)\times 2t-0=\frac{2}{5}\times 5t^4\Rightarrow f\left(t^2\right)=tPutt=\frac{2}{5}\Rightarrow f\left(\frac{4}{25}\right)=\frac{2}{5}$