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Math Help - Definite Integral By Substitution

  1. #1
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    Definite Integral By Substitution

    hi can you guys help me with this please:

    Evaluate the definite integral, if it exists. \int_{-1}^{0} x^2(x^3+1)^8\! \, \mathrm{d}x.

    My teacher says there's a way that you don't have to bring back the original function, that you can just plug in for u. I got this far:

    u+x^3+1

    \frac{du}{3}=x^2 dx

    \frac{1}{3} \int_{-1}^{0} u^8\! \, \mathrm{d}u.

    where do I go from here to solve using only u?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by RezMan View Post
    hi can you guys help me with this please:

    Evaluate the definite integral, id it exists. \int_{-1}^{0} x^2(x^3+1)^8\! \, \mathrm{d}x.

    My teacher says there's a way that you don't have to bring back the original function, that you can just plug in for u. I got this far:

    u+x^3+1

    \frac{du}{3}=x^2 dx

    \frac{1}{3} \int_{-1}^{0} u^8\! \, \mathrm{d}u.

    where do I go from here to solve using only u?
    ... only a little detail: after the substitution 1+x^{3}=u the integration interval is not any more [-1,0] , it has to be changed...

    Kind regards

    \chi \sigma
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  3. #3
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    I have to plug -1 and 0 into x^3 +1?
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  4. #4
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    Quote Originally Posted by RezMan View Post
    I have to plug -1 and 0 into x^3 +1?
    Correct. What do you get for your new limits?
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  5. #5
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    would it be something like this:

    \frac{1}{3} [\frac{u^9}{9}]^{1}_{0}
    Last edited by RezMan; June 4th 2011 at 06:35 PM.
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  6. #6
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    can I now plug 0 and 1 into u?
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  7. #7
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    Quote Originally Posted by RezMan View Post
    can I now plug 0 and 1 into u?
    Yes

    CB
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  8. #8
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    so would it be
    \frac{1}{27}[1-0]

    1/27?
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  9. #9
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    Quote Originally Posted by RezMan View Post
    so would it be
    \frac{1}{27}[1-0]

    1/27?
    Yes
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