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Math Help - Integrating e^(u+1)

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    Integrating e^(u+1)

    Evaluate the integral: ∫ -1 to 1 ( e^(u + 1) ) du
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    Quote Originally Posted by SyNtHeSiS View Post
    ∫ -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
    Last edited by mr fantastic; May 9th 2010 at 02:55 AM. Reason: Small correction.
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    Quote Originally Posted by SyNtHeSiS View Post
    Evaluate the integral: ∫ -1 to 1 ( e^(u + 1) ) du
    \int_{-1}^1 e^{u+1} \, du = e \int_{-1}^1 e^u \, du<br />
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    Do it as both e^(i*pi) and skeeter suggest and show that you get the same answer!
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    Quote Originally Posted by e^(i*pi) View Post
    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?
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  6. #6
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    Quote Originally Posted by SyNtHeSiS View Post
    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.
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    Quote Originally Posted by SyNtHeSiS View Post
    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
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