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 :

Question

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 :

Accepted Answer

Answer: B. \pi\left(1-\frac{\pi^{3}}{16}ight) Given that \int\limits_0^{t^2}\left(f(x)+x^2ight) d x=\frac{4}{3} t^3, \forall t>0 On differentiating using Leibnitz rule, we get \begin{aligned} & \left(f\left(t^2ight)+t^4ight) 	imes 2 t=\frac{4}{3} 	imes 3 t^2 \\ & \Rightarrow f\left(t^2ight)+t^4=2 t \\ & \Rightarrow f\left(t^2ight)=2 t-t^4 \end{aligned} On substituting \frac{\pi}{2} for t, we get f\left(\frac{\pi^2}{4}ight)=2\left(\frac{\pi}{2}ight)-\frac{\pi^4}{16}=\pi\left(1-\frac{\pi^3}{16}ight)

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 :

Answer

Explanation

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