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Math Help - integral problem

  1. #1
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    integral problem

    G(x) = \int_{1}^{x} xt \; dt
    find
    G '(x) =

    using the first fundamental theorem of calc, i got it to be x(x) = 2x.
    the answer is wrong. any ideas?
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  2. #2
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    Quote Originally Posted by viet View Post
    G(x) = \int_{1}^{x} xt \; dt
    find
    G '(x) =

    using the first fundamental theorem of calc, i got it to be x(x) = 2x.
    the answer is wrong. any ideas?
    Note that,
    G(x)=x\int_1^x t dt
    Then, by product rule,
    G'(x)=(x)\left( \int_1^x tdt \right)'+(x)' \left( \int_1^x tdt\right)
    Use fundamental theorem,
    G'(x)=x(x)+G(x)=x^2+G(x)
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  3. #3
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    so answer is  x^2+G(x)
    im not sure how to get  G(x) , isnt it just  x ?
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  4. #4
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    Quote Originally Posted by viet View Post
    so answer is  x^2+G(x)
    im not sure how to get  G(x) , isnt it just  x ?
    No, because you end with after using the product rule,
    G'(x)=x^2+\int_1^x t dt
    But,
    G(x)=\int_1^x tdt
    Thus,
    G'(x)=x^2+G(x)
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    No
    G'(x)=x(x)+G(x)=x^2+G(x)
    Which is a first order ODE for G, which has a general solution of the form:

    G(x)=Ae^{x}+P(x)

    where P(x) is the second degree polynomial for which:

    P'(x)-P(x)=x^2

    which gives:

    P(x)=-x^2-2x-2,

    and so:

    <br />
G(x)=Ae^{x}-x^2-2x-2<br />

    But we also have the initial condition G(1)=0 so:

    A=5/e

    So finaly we have:

    <br />
G(x)=(5/e) e^{x}-x^2-2x-2<br />

    or:

    <br />
\boxed{G(x)=5 e^{x-1}-x^2-2x-2}<br />

    and so:

    <br />
\boxed{G'(x)=5 e^{x-1}-2x-2}<br />

    Now there is probably an error in there somewhere, but you should get the idea.

    RonL
    Last edited by CaptainBlack; November 30th 2006 at 10:14 PM.
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