Question

Let f be a continuous function satisfying $∫^{t ^2}_ 0 ( f ( x ) + x ^2 ) d x = \frac{4}{3} t^3 , ∀ t > 0.$ Then $f(\frac{π^2}{4})$ is equal to

• A. $-\pi^2(1+\frac{\pi^2}{16})$
• B. $\pi^2(1-\frac{\pi^2}{16})$
• C. $-\pi(1+\frac{\pi^3}{16})$
• D. $\pi(1-\frac{\pi^3}{16})$

Answer 1

Answer: (D)

$\pi(1-\frac{\pi^3}{16})$

Answer 2

The correct option is(D ): $\pi(1-\frac{\pi^3}{16})$