Question #7c8b2

1 Answer
Feb 7, 2018

pi/2π2

Explanation:

I = int_0 ^(pi/2) sin^(0)(x)dxI=π20sin0(x)dx
You could subst sin^0 (x)sin0(x) with 1, but you must take care when x=0x=0. But, in integration we just need to have a well defined value near to the bounds of integration.
I = lim_(a->0^+) int_a^(pi/2)1dx=lim_(a->0^+) (pi/2-a)=pi/2