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Math Help - Weird Integral

  1. #1
    Junior Member Evales's Avatar
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    Weird Integral

    A question that I have been given asks me to use a certain substitution to find the answer exactly. The only thing is not only do I think I can manage without it the substitution is the most random thing one I've seen.

    Upper Boundary = 2, Lower Boundary = 0

    Integrate: (4-x)^(1/2).dx

    Use the substitution x = 2sin u to exactly evaluate this definite integral.
    Should I just do it without using that substitution?
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    You just need to solve \int_{0}^{2}{\sqrt{4-x}\,dx}.

    Make the substitution z^2=4-x, and don't forget to get the new integration limits.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Evales View Post
    A question that I have been given asks me to use a certain substitution to find the answer exactly. The only thing is not only do I think I can manage without it the substitution is the most random thing one I've seen.

    Upper Boundary = 2, Lower Boundary = 0

    Integrate: (4-x)^(1/2).dx

    Use the substitution x = 2sin u to exactly evaluate this definite integral.
    Should I just do it without using that substitution?
    the x is not squared. what you wrote does not require a trig sub. a substitution of u = 4 - x will suffice
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  4. #4
    Junior Member Evales's Avatar
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    Quote Originally Posted by Jhevon View Post
    the x is not squared. what you wrote does not require a trig sub. a substitution of u = 4 - x will suffice
    Omg thanks for that. God your sharp yes the x is actually squared.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Evales View Post
    A question that I have been given asks me to use a certain substitution to find the answer exactly. The only thing is not only do I think I can manage without it the substitution is the most random thing one I've seen.

    Upper Boundary = 2, Lower Boundary = 0

    Integrate: (4-x)^(1/2).dx

    Use the substitution x = 2sin u to exactly evaluate this definite integral.
    Should I just do it without using that substitution?
    Quote Originally Posted by Evales View Post
    Omg thanks for that. God your sharp yes the x is actually squared.
    \int_0^2 \sqrt{4-x^2}\,dx

    They tell you to use the substitution x=2\sin u. This implies that dx=2\cos u \ \,du

    Change the limits of integration:

    2=2\sin u \implies u=\arcsin(1)=\frac{\pi}{2}
    0=2\sin u \implies u=\arcsin(0)=0

    Thus, we have:

    \int_{0}^{\frac{\pi}{2}}2\cos u \sqrt{4-4\sin^2u}\,du

    Simplify the integrand:

    2\cos u\sqrt{4-4\sin^2u}=4\cos u \sqrt{1-\sin^2u}=4\cos u\sqrt{\cos^2u}=4\cos^2u

    We now have:

    \int_{0}^{\frac{\pi}{2}}4\cos^2u\,du

    I'll let you take it from here, noting that \cos^2u=\frac{1+\cos (2u)}{2}.

    Hope that this helped!
    Last edited by Chris L T521; May 31st 2008 at 07:26 PM. Reason: typo... :(
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