intergrate with bound pi and o equation sin 2x dx

Answer 1

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.