Indefinite Integration 3 Question 6
6. If $f(x)$ is differentiable and $\int _0^{t^{2}} x f(x) d x=\frac{2}{5} t^{5}$, then $f \frac{4}{25}$ equals
(2004, 1M)
(a) $\frac{2}{5}$
(b) $-\frac{5}{2}$
(c) 1
(d) $\frac{5}{2}$
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Answer:
Correct Answer: 6. (a)
Solution:
- Here, $\int _0^{t^{2}} x f(x) d x=\frac{2}{5} t^{5}$
Using Newton Leibnitz’s formula, differentiating both sides, we get
$$ \begin{array}{lll} & t^{2}{f\left(t^{2}\right) } \frac{d}{d t}\left(t^{2}\right)-0 \cdot f(0) \frac{d}{d t}(0)=2 t^{4} \\ \Rightarrow & t^{2} f\left(t^{2}\right) 2 t=2 t^{4} \Rightarrow f\left(t^{2}\right)=t \\ \therefore & f \frac{4}{25}=-\frac{2}{5} & \text { putting } t=\frac{2}{5} \\ \Rightarrow & f \frac{4}{25}=\frac{2}{5} & \end{array} $$
