Let $f$ be a continuous function satisfying $$\int_\limits{0}^{t^{2}}\left(f(x)+x^{2}\right) d x=\frac{4}{3} t^{3}, \forall t > 0$$. Then $f\left(\frac{\pi^{2}}{4}\right)$ is equal to :
Solution
Given that
<br/><br/>$\int\limits_0^{t^2}\left(f(x)+x^2\right) d x=\frac{4}{3} t^3, \forall t>0$
<br/><br/>On differentiating using Leibnitz rule, we get
<br/><br/>$$
\begin{aligned}
& \left(f\left(t^2\right)+t^4\right) \times 2 t=\frac{4}{3} \times 3 t^2 \\\\
& \Rightarrow f\left(t^2\right)+t^4=2 t \\\\
& \Rightarrow f\left(t^2\right)=2 t-t^4
\end{aligned}
$$
<br/><br/>On substituting $\frac{\pi}{2}$ for $t$, we get
<br/><br/>$$
f\left(\frac{\pi^2}{4}\right)=2\left(\frac{\pi}{2}\right)-\frac{\pi^4}{16}=\pi\left(1-\frac{\pi^3}{16}\right)
$$
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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