Q: ∫(y)/e^(2y) from 0 to 1.
A: Integration by parts it is!
∫udv = uv – ∫vdu
u = y v = – 1/2 e^(-2y)
du = dy dv = e^(-2y)
So, ∫(y)/e^(2y) = -(1/2) y e^(-2y) – ∫ -(1/2) e^(-2y) dy
With house-keeping gives:
-(1/2) y e^(-2y) + 1/2 ∫ e^(-2y) dy
And integrate the last piece to get:
-(1/2) y e^(-2y) – 1/4 e^(-2y)
Now, from o to 1:
[-(1/2) *1* e^(-2*1) – 1/4 e^(-2*1)] – [-(1/2) *0* e^(-2*0) – 1/4 e^(-2*0)]
With careful simplification gives:
– 3/4 e^(-2) + 1/4