Question

Evaluate `int_0^pi(2+ sinx)dx`?

Answer 1

`int_0^pi(2+ sinx)dx = 2pi+2`

Explanation:

You should learn that;

`d/dx sinx = cosx iff int cosx dx = sinx + c`
`d/dx cosx = -sinx iff int sinx = -cosx + c`

So

`int_0^pi(2+ sinx)dx = [2x - cosx]_0^pi`
`\ = (2pi - cospi) - (2*0-cos0)`
`\ = (2pi - (-1)) - (0-1)`
`\ = 2pi+1 +1`
`\ = 2pi+2`