intergrate (1+sin(x))dx bound is pi and 0
you just can't keep track of those minus signs, can you?
∫[0, pi] (1 + sin(x)) dx
= ∫[0, pi] 1 dx + ∫[0, pi] sin(x) dx
= [x - cos(x) ] [0, pi]
= (pi - (-1)) - (0 - 1)
= pi + 2
You are correct. I apologize for the mistake in keeping track of the signs. The correct evaluation of the integral ∫[0, pi] (1 + sin(x)) dx is indeed pi + 2. Thank you for pointing out the error.
To integrate (1 + sin(x)) dx from 0 to pi, we can split the integral into two parts and then integrate each part separately.
∫[0, pi] (1 + sin(x)) dx
= ∫[0, pi] 1 dx + ∫[0, pi] sin(x) dx
= ∫[0, pi] 1 dx + [ -cos(x) ] [0, pi]
= [ x ] [0, pi] - [ -cos(x) ] [0, pi]
= pi - (-1)
= pi + 1
Therefore, the integral of (1 + sin(x)) dx from 0 to pi is equal to pi + 1.