Evaluate the integral?

Question

b=3(pi)/2; a=0 ;abs value (sinx) dx

Answer 1

It's not as simple as shaveger made it out to be.
Between 0 and π sin x is positive, so
∫(0..π) | sin x | dx = ∫(0..π) sin x dx = -cos(x)(0..π)
= -cos π + cos(0) = -(-1) + 1 = 2.
But between π and 3π/2 sin x is negative, so
∫ (π, 3π/2) |sin x | dx = -∫(π..3π/2) sin x dx
= cos x (π..3π/2) = cos 3π/2 -cos π = 0 -(-1) = 1.
So the value of your integral is 3.

Answer 2

Since what you have is a definite integral, its simple. Take the antiderivative of the sin function, -cos. evaluate the function as follows: -cos(3pi/2)-(-cos(0))=(0)-(-1)=1 is your answer.