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