First slide

Evaluation of definite integrals

Question

If, for $t > 0$ the definite integral $\int_{0}^{t^{2}} x f (x) d x = \frac{2}{5} t^{5}$ then $f (\frac{4}{25})$ is equal to

Moderate

A

$\frac{2}{5}$
B

$- \frac{2}{5}$
C

$\frac{2}{\sqrt{5}}$
D

$- \frac{2}{\sqrt{5}}$

Solution

Differentiating both the sides of
$\int_{0}^{t^{2}} x f (x) = \frac{2}{5} t^{5}$
we get
$\begin{array}{l} (2t) (t^{2}) f (t^{2}) = 2 t^{4} \\ \Rightarrow f (t^{2}) = t \\ \Rightarrow f (\frac{4}{25}) = \frac{2}{5} \end{array}$

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