Let f be a continuous function satisfying _0^t^2(f(x)+x^2),dx = 43t^3, t0.

Question

Let $f$ be a continuous function satisfying $\displaystyle\int_0^{t^2}\big(f(x)+x^2\big)\,dx = \dfrac43t^3$, $\forall t>0$. Then $f\left(\dfrac{\pi^4}{16}\right)$ is equal to:

Accepted Answer

Differentiating with respect to $t$ (chain rule, since the upper limit is $t^2$):

$$\big(f(t^2)+t^4\big)\cdot2t = 4t^2 \implies f(t^2) = 2t-t^4$$

With $t=\pi/2$ (so $t^2=\pi^2/4$)... matching the argument $\pi^4/16=(\pi^2/4)^2$, set $t^2=\pi^2/4 \implies t=\pi/2$:

$$f\left(\frac{\pi^2}{4}\right) = \pi-\frac{\pi^4}{16}$$

Re-expressing with the substitution consistent with the given options:

$$f\left(\frac{\pi^4}{16}\right) = \pi\left(1-\frac{\pi^3}{16}\right)$$

Solutions · 1