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Thread: integral problem

  1. #1
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    integral problem

    $\displaystyle G(x) = \int_{1}^{x} xt \; dt$
    find
    $\displaystyle 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
    $\displaystyle G(x) = \int_{1}^{x} xt \; dt$
    find
    $\displaystyle 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,
    $\displaystyle G(x)=x\int_1^x t dt$
    Then, by product rule,
    $\displaystyle G'(x)=(x)\left( \int_1^x tdt \right)'+(x)' \left( \int_1^x tdt\right)$
    Use fundamental theorem,
    $\displaystyle G'(x)=x(x)+G(x)=x^2+G(x)$
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  3. #3
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    so answer is $\displaystyle x^2+G(x) $
    im not sure how to get $\displaystyle G(x) $, isnt it just $\displaystyle x $?
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  4. #4
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    Quote Originally Posted by viet View Post
    so answer is $\displaystyle x^2+G(x) $
    im not sure how to get $\displaystyle G(x) $, isnt it just $\displaystyle x $?
    No, because you end with after using the product rule,
    $\displaystyle G'(x)=x^2+\int_1^x t dt$
    But,
    $\displaystyle G(x)=\int_1^x tdt$
    Thus,
    $\displaystyle G'(x)=x^2+G(x)$
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    No
    $\displaystyle G'(x)=x(x)+G(x)=x^2+G(x)$
    Which is a first order ODE for $\displaystyle G$, which has a general solution of the form:

    $\displaystyle G(x)=Ae^{x}+P(x)$

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

    $\displaystyle P'(x)-P(x)=x^2$

    which gives:

    $\displaystyle P(x)=-x^2-2x-2$,

    and so:

    $\displaystyle
    G(x)=Ae^{x}-x^2-2x-2
    $

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

    $\displaystyle A=5/e$

    So finaly we have:

    $\displaystyle
    G(x)=(5/e) e^{x}-x^2-2x-2
    $

    or:

    $\displaystyle
    \boxed{G(x)=5 e^{x-1}-x^2-2x-2}
    $

    and so:

    $\displaystyle
    \boxed{G'(x)=5 e^{x-1}-2x-2}
    $

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

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