I can't figure this one out, can somebody help?
Intergral 0 to 1 (integral 0 to 1 e^(x+y) dy) dx
Use the identity e^(x + y) = e^x * e^y to get ∫[0,1] e^(x + y) dy = e^y ∫[0,1] e^x dx = e^y * e^x (from x = 0 to 1) = e^y * (e - 1). Then
∫[0,1] ∫[0,1] e^(x + y) dy dx = ∫[0,1] e^y * (e - 1) dy = (e - 1) ∫[0,1] e^y dy = (e - 1)(e - 1) = (e - 1)^2.
Comments
Use the identity e^(x + y) = e^x * e^y to get ∫[0,1] e^(x + y) dy = e^y ∫[0,1] e^x dx = e^y * e^x (from x = 0 to 1) = e^y * (e - 1). Then
∫[0,1] ∫[0,1] e^(x + y) dy dx = ∫[0,1] e^y * (e - 1) dy = (e - 1) ∫[0,1] e^y dy = (e - 1)(e - 1) = (e - 1)^2.