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Math Help - Integral help

  1. #1
    s7b
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    Can someone please explain this integral;

    The integral from 0 to 1 of: 2t(sqrt(1 + 2t^2))dt

    I tried using substitution

    let u=sqrt(1 + 2t^2)

    but I keep getting a different answer than the book...
    Last edited by mr fantastic; April 19th 2009 at 09:58 PM. Reason: Merged posts
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  2. #2
    Flow Master
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    Quote Originally Posted by s7b View Post
    Can someone please explain this integral;

    The integral from 0 to 1 of: 2t(sqrt(1 + 2t^2))dt

    I tried using substitution

    let u=sqrt(1 + 2t^2)

    but I keep getting a different answer than the book...
    Substitute u = 1 + 2 t^2.
    Last edited by mr fantastic; April 20th 2009 at 02:11 AM. Reason: Fixed a typo
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  3. #3
    Newbie Mirado's Avatar
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    solved

    I like fantastic's method, though since I'm new to the posts, I thought I'd get some practice. So, this is the longer version:
    I used 2 substitutions (though you only need one being  u=2+2t^2 ), being  u=t^2  , du=2tdt and p=1+2u dp=2du so, it should look like this:
     \int_0^1 2t\sqrt{1+2t^2} dt
     \int_0^1 \sqrt{1+2u} du           *first substitution*
     (1/2) \int_1^3 \sqrt{p} dp   *second substitution*
     (1/2) \times (2/3) p^{3/2} |_1^3
     (1/3) \times (3^{3/2} - 1)
     \sqrt{3} - (1/3)
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  4. #4
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    oops, my bad, meant to say that the method with 1 sub would be u=1+2t^2, not u=2+2t^2
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  5. #5
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    Quote Originally Posted by s7b View Post
    Can someone please explain this integral;

    The integral from 0 to 1 of: 2t(sqrt(1 + 2t^2))dt

    I tried using substitution

    let u=sqrt(1 + 2t^2)

    but I keep getting a different answer than the book...
    just put t^2=u....you will get answer:sqrt(3)-1/3
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  6. #6
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    Quote Originally Posted by Mathventure View Post
    just put t^2=u....you will get answer:sqrt(3)-1/3
    The substitution suggested in post #2 leads to a simpler calculation.
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