integrate abs(sin x) dx from 0 to 3pi/2
integrate abs(sin x) dx from 0 to 3pi/2
To solve the integral of abs(sin(x)) from 0 to 3π/2, we need to consider the behavior of the absolute value function.
The absolute value function changes the sign of sin(x) based on whether sin(x) is positive or negative.
We know that sin(x) is positive on [0, π] and negative on [π, 3π/2]. Therefore, we split the integral into two parts:
∫[0 to 3π/2] abs(sin(x)) dx = ∫[0 to π] sin(x) dx + ∫[π to 3π/2] -sin(x) dx
Now, compute each part:
1. ∫[0 to π] sin(x) dx = -cos(x) | from 0 to π
= -cos(π) + cos(0) = -(-1) + 1 = 2
2. ∫[π to 3π/2] -sin(x) dx = cos(x) | from π to 3π/2
= cos(3π/2) - cos(π) = 0 - (-1) = 1
So, the total integral is:
2 + 1 = 3
Therefore, the value of the integral is 3.