intergrate with bound pi and o equation sin 2x dx
To integrate the equation ∫sin(2x)dx with bounds from 0 to π, we first apply the integration formula for sin(ax):
∫sin(ax)dx = -1/a * cos(ax) + C,
where C is the constant of integration. In this case, a = 2, so the integral of sin(2x) is:
-1/2 * cos(2x) + C.
Now we can evaluate the definite integral with the bounds of 0 and π:
∫[0,π] sin(2x)dx = [-1/2 * cos(2π) - (-1/2 * cos(0))]
= [-1/2 * cos(2π) + 1/2 * cos(0)]
= [-1/2 * 1 + 1/2 * 1]
= -1/2 + 1/2
= 0.
Therefore, the definite integral of sin(2x) from 0 to π is 0.