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Math Help - integration

  1. #1
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    integration

    I have the following question that I'm having issues solving

    \int\limits_0^1{(1+\sqrt{x})^8}dx

    I tried using uv substitution with u being {(1+\sqrt{x})^8} and v being just dx. However, it quickly devolved into uv substitution after uv substition that spiralled into a massively negative number, so I must be doing something wrong. My plan was just to uv until I got that exponent down to something I could easily multiply out.

    Any tips, tricks, or something I may have missed?
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  2. #2
    Super Member TheChaz's Avatar
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    Re: integration

    Have you considered using the binomial theorem/expansion? It wouldn't be too ugly ...
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  3. #3
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    Re: integration

    Make a substitution of u=\sqrt{x} and du=\frac{1}{2\sqrt{x}}dx

    Then, make a substitution of s=u+1 and ds=du

    Doing this right will give you the following:

    2\int (s-1)s^{8}ds

    Do the required integration, then substitute back in to get the final answer.
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  4. #4
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    Re: integration

    Quote Originally Posted by cheme View Post
    Make a substitution of u=\sqrt{x} and du=\frac{1}{2\sqrt{x}}dx

    Then, make a substitution of s=u+1 and ds=du

    Doing this right will give you the following:

    2\int (s-1)s^{8}ds

    Do the required integration, then substitute back in to get the final answer.
    Or even just make the substitution \displaystyle u = 1 + \sqrt{x} \implies du = \frac{1}{2\sqrt{x}}\,dx...
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  5. #5
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    Re: integration

    Thanks to you all for being of assistance. @theChaz, I was trying to avoid actually expanding that, though I suppose it's always an option.

    @cheme and Prove It, looks like you two are advocating the same method, with Prove It's being a bit quicker and to the point. I'm afraid I'm not really following, though. I was wondering if you could clarify.

    I understand taking the 1+\sqrt{x} and setting is as u. Then du = \frac{1}{2}x^{-\frac{1}{2}}dx.

    From my understanding, this basically gets us down to \int(u)^8dx, so I need to use the du definition above to replace the dx. But I get stuck here. Can someone show me the rest of the substitution part?

    If I try to work through it, I end up with:
    du = \frac{1}{2}x^{-\frac{1}{2}}dx
    2du = x^{-\frac{1}{2}}dx
    I can't just move the x^{-\frac{1}{2}} to the left side so I get a straight definition for dx.
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: integration

    Let 1+\sqrt{x}=u \Rightarrow \frac{dx}{2\sqrt{x}}=du \Leftrightarrow dx=2\sqrt{x}du \Leftrightarrow dx=2(u-1)du
    Therefore you get:
    2\int u^{8}\cdot(u-1)du=2\int u^9du -2 \int u^8du=...

    Can you finish?

    EDIT:
    Now I notice you're dealing with a definite integral therefore change the integration limits or do the back-substitution at the end.
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  7. #7
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    Re: integration

    Siron, thanks for the assistance. I'm probably missing something blindingly obvious, but how do you get from dx = 2\sqrt{x}du to dx = 2(u-1)du?

    Once you substitute everything back into the integral, I follow the rest of it.

    2\int\limits_1^2{u^8(u-1)du}
    =2\int\limits_1^2{u^9 - u^8du}
    =2(\int\limits_1^2{u^9du} - \int\limits_1^2{u^8du})
    =2([\frac{1}{10}u^10]\limits_1^2 - [\frac{1}{9}u^9]\limits_1^2])
    =2([\frac{1024}{10} - \frac{1}{10}] - [\frac{512}{9} - \frac{1}{9}])
    =\frac{1023}{5} - \frac{1022}{9}
    ...jeez, what a mess. At any rate, assuming I didn't make a mistake somewhere along the line, you end up with a really ugly fraction.
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  8. #8
    MHF Contributor Siron's Avatar
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    Re: integration

    Quote Originally Posted by satis View Post
    Siron, thanks for the assistance. I'm probably missing something blindingly obvious, but how do you get from dx = 2\sqrt{x}du to dx = 2(u-1)du?
    Because the subtitution was 1+\sqrt{x}=u \Leftrightarrow \sqrt{x}=u-1 (1) and you have dx=2\sqrt{x}du (2), substituting (1) in (2) gives dx=2(u-1)du
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  9. #9
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    Re: integration

    Ah, I knew I missed something obvious. Thank you for pointing it out to me.
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