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Math Help - How do l split this integral ?

  1. #1
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    How do l split this integral ?



    • pg\int_0^2 2\sqrt{4- x^2}(5+x)dx


    Is there a way to split this integral ?


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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by nyasha View Post


    • pg\int_0^2 2\sqrt{4- x^2}(5+x)dx


    Is there a way to split this integral ?


    \int_0^2 \sqrt{4-x^2}(5+x) ~dx=\int_0^2 5 \sqrt{4-x^2} ~dx+\int_0^2 x \sqrt{4-x^2} ~dx
    For the first one, substitute x=2sin(u)
    For the second one, substitute t=sqrt(4-x^2) (or just t=4-x^2)
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    \int_0^2 \sqrt{4-x^2}(5+x) ~dx=\int_0^2 5 \sqrt{4-x^2} ~dx+\int_0^2 x \sqrt{4-x^2} ~dx
    For the first one, substitute x=2sin(u)
    For the second one, substitute t=sqrt(4-x^2) (or just t=4-x^2)
    For the first integral can l substitute x=Rsinθ ?
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  4. #4
    Moo
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    Quote Originally Posted by nyasha View Post
    For the first integral can l substitute x=Rsinθ ?
    that's what I meant... but R is such that Rē=4, so R=2
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  5. #5
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    Quote Originally Posted by Moo View Post
    that's what I meant... but R is such that Rē=4, so R=2

    Will l be correct if l say this integral  \int_0^2 x \sqrt{4-x^2} ~dx is zero so l can try to ignore it ?
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  6. #6
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    That thing does not go to zero.
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  7. #7
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    Quote Originally Posted by nyasha View Post
    Will l be correct if l say this integral  \int_0^2 x \sqrt{4-x^2} ~dx is zero so l can try to ignore it ?
    f(a) = f(b) = 0 does not imply that \int_a^b f(x) \, dx = 0 eg. It should be obvious that \int_0^1 x^2 - x \, dx \neq 0.
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  8. #8
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    Quote Originally Posted by Moo View Post
    Hello,

    \int_0^2 \sqrt{4-x^2}(5+x) ~dx=\int_0^2 5 \sqrt{4-x^2} ~dx+\int_0^2 x \sqrt{4-x^2} ~dx
    For the first one, substitute x=2sin(u)
    For the second one, substitute t=sqrt(4-x^2) (or just t=4-x^2)
    Thanks
    Last edited by nyasha; February 20th 2009 at 09:48 PM.
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