1. ## Integrating e^(u+1)

Evaluate the integral: ∫ -1 to 1 ( e^(u + 1) ) du

2. Originally Posted by SyNtHeSiS
∫ -1 to 1 ( e^(u + 1) ) du
Let $v = u+1$

Hence $dv = du$

We can also change the limits to 0 and 2 to be in terms of v.

$\int^2_0 e^v\,dv$

which is pretty simple to integrate

3. Originally Posted by SyNtHeSiS
Evaluate the integral: ∫ -1 to 1 ( e^(u + 1) ) du
$\int_{-1}^1 e^{u+1} \, du = e \int_{-1}^1 e^u \, du
$

4. Do it as both e^(i*pi) and skeeter suggest and show that you get the same answer!

5. Originally Posted by e^(i*pi)
Let $v = u+1$

Hence $dv = du$

We can also change the limits to 0 and 2 to be in terms of v.
Why did you differentiate v and u and change the limits to 0 and 2 in terms of v?

What I did was:

∫ -1 to 1 ( e^(u + 1) ) du
= e^(u+2) / u + 2 (basically I added 1 to exponent then divided by new exponent to get anti-derivative). Why cant you do this?

6. Originally Posted by SyNtHeSiS
Why did you differentiate v and u and change the limits to 0 and 2 in terms of v?

What I did was:

∫ -1 to 1 ( e^(u + 1) ) du
= e^(u+2) / u + 2 (basically I added 1 to exponent then divided by new exponent to get anti-derivative). Why cant you do this?
You have been shown how to do it. Now you are strongly advised to review your class notes or textbook on basic integration techniques.

7. Originally Posted by SyNtHeSiS
Why did you differentiate v and u and change the limits to 0 and 2 in terms of v?

What I did was:

∫ -1 to 1 ( e^(u + 1) ) du
= e^(u+2) / u + 2 (basically I added 1 to exponent then divided by new exponent to get anti-derivative). Why cant you do this?
You don't use the power on exponentials, instead $\int e^u \,du = e^u +C$

I made the substitution $v=u+1$. 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.

$v(-1) = -1+1 = 0$
$v(1) = 1+1 =2$

Skeeter's method works just as well and is probably simpler for this problem