Consider the function f(x)=∫0x2sin(t) dtf(x) = \int_{0}^{x^2} \sin(\sqrt{t}) \, dtf(x)=∫0x2sin(t)dt. What is the derivative f′(x)f'(x)f′(x) evaluated at x=πx = \sqrt{\pi}x=π?
0
-2\sqrt{\pi}
2\sqrt{\pi}
1