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Math Help - "u sub" problem with no "u"

  1. #1
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    "u sub" problem with no "u"

    I need to know how to integrate 2sqrt(100-x^2) from -10 to 10 by hand. It suggests a u-substitution but no u seems to exist. Please help.

    Bret Norvilitis
    Orchard Park HS
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  2. #2
    Super Member craig's Avatar
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    Quote Originally Posted by bmnorvil View Post
    I need to know how to integrate 2sqrt(100-x^2) from -10 to 10 by hand. It suggests a u-substitution but no u seems to exist. Please help.

    Bret Norvilitis
    Orchard Park HS
    It has been a while since I have done any integration by substitution but I would maybe suggest using u = 100-x^2
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  3. #3
    Super Member Deadstar's Avatar
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    I was never really that sure about substitution but i think this is how it works...

    When it says to do a u-sub it means set x to be equal to some function containing u. This also means that you should change your limits as well.

    In this case, set x = 10\sin(u) and hence dx = 10\cos(u)du then x^2 = 100\sin^2(u).

    So put this into your equation and you get...

    \int 2 \sqrt(100-x^2)dx = \int 2 \sqrt(100-100\sin^2(u))du = \int 2 \sqrt(100(1-\sin^2(u))) 10\cos(u)du.
    = \int 2 \sqrt(100cos^2(u)) 10\cos(u).
    = \int 200cos^2(u)
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  4. #4
    MHF Contributor Amer's Avatar
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    let 10sinu=x
    10cosu(du)=dx
    du=dx/(10cosu)

    cosu=(100-x^2)^1/2/10

    "u sub" problem with no "u"-usub.jpg
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  5. #5
    Super Member Deadstar's Avatar
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    ^^^^^^
    What Amer said.

    Note that Amers integral limits were found by setting 10 and -10 to be equal to 10\sin(u), hence \sin(u) = 1 and -1, so u= \pm \frac{\pi}{2} are the new limits
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