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Math Help - how do i integrate e^x^1/2

  1. #1
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    how do i integrate e^x^1/2

    SSIA...
    i checked through all theexamples and have no idea how to even begin..i would really appreciate it if someone pointed me in the right direction..thanks
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  2. #2
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    Have you considered using U-Substitution? Look it up, If not.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by twostep08 View Post
    SSIA...
    i checked through all theexamples and have no idea how to even begin..i would really appreciate it if someone pointed me in the right direction..thanks
    How do you integrate e^{\sqrt{x}}? Let z=\sqrt{x} and you'll get a computable integral.
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  4. #4
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    thats what it looks like, but theres no other x's i could use to get the du.

    like u=x^1/2...du=1/2x^(-1/2) dx
    but theres no x's besides the exponent to help me---
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by twostep08 View Post
    thats what it looks like, but theres no other x's i could use to get the du.

    like u=x^1/2...du=1/2x^(-1/2) dx
    but theres no x's besides the exponent to help me---
    z^2=x\implies 2z\:dz=dx\implies \int e^{\sqrt{x}}dz\overbrace{\longmapsto}^{z^2=x}2\int z e^z dz
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  6. #6
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    ok...so i kind of understand, but where did that z inside the integral come from---im getting 2 (integral) e^z dz
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by twostep08 View Post
    ok...so i kind of understand, but where did that z inside the integral come from---im getting 2 (integral) e^z dz
    Because if this was \int e^{\sqrt{x}} I might see where you are coming from but we are integrating \int e^{\sqrt{x}}\color{red}dx so we have to find a replacement for e^{\sqrt{x}} which we do by replacing x with z^2 and we have to replace dx by 2z\:dz
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  8. #8
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    ooh duh. my fault on the bad derivative. im gonna try to work it out from there- well see where i get with integration by parts
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by twostep08 View Post
    ooh duh. my fault on the bad derivative. im gonna try to work it out from there- well see where i get with integration by parts
    P.S.

    Spoiler:

    wolframalpha.com
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  10. #10
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    Drex why you have to bring it to a single term of x could not just leave it as when doing the sub


     dz = -\frac{1}{2\sqrt{x}}dx
    2\sqrt{x} dz = dx

    I never understood when to do things like that
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  11. #11
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    sweet Georgia! that seems like the best site in the world. had i known about that sooner, i wouldve been saved many sleepless nights
    Last edited by mr fantastic; November 18th 2009 at 10:04 PM.
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