Evaluate the integral: ∫ -1 to 1 ( e^(u + 1) ) du
You don't use the power on exponentials, instead $\displaystyle \int e^u \,du = e^u +C $
I made the substitution $\displaystyle v=u+1$. I then differentiated v with respect to u in order to get an integral in terms of dv.
The limits changed because I was integrating with respect to v. There is no reason why I could have left the original limits and then substituted u back in but I find changing the limits easier.
I found the limits by changing the existing limits using the expression for v.
$\displaystyle v(-1) = -1+1 = 0$
$\displaystyle v(1) = 1+1 =2$
Skeeter's method works just as well and is probably simpler for this problem