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Thread: Calculating the second derivative of an integral

  1. #1
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    Calculating the second derivative of an integral

    Question:

    $\displaystyle F(x) = \int_0^x[{(2t+3)\int_2^t{f(u)du}]dt$

    Calculate F''(2).

    I believe:

    F'(x) = $\displaystyle [{(2x+3)\int_2^x{f(u)du}]$

    but I'm stuck now.
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  2. #2
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    Re: Calculating the second derivative of an integral

    Now use the product rule! You know both the derivate of (2x + 3), AND the derivative of the integral (by the FTC, which you already used)...

    You're almost there!
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  3. #3
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    Re: Calculating the second derivative of an integral

    This is what I got next.

    (2x+3)(f(x)dx)+2$\displaystyle \int_2^x {f(u)du}$

    and the integral term becomes 0 because it is over the interval 2 to 2, but what happens in the remaining term?
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  4. #4
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    Re: Calculating the second derivative of an integral

    You're close (see P's reply)...
    Just plug in 2 for x (and lose the "dx"...)
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  5. #5
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    Re: Calculating the second derivative of an integral

    I'm down to $\displaystyle {7*f(2)}$. Is this as far as I can go?
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  6. #6
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    Re: Calculating the second derivative of an integral

    No, I don't know what $\displaystyle {f(x)}$. Thanks for your help.
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  7. #7
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    Re: Calculating the second derivative of an integral

    Quote Originally Posted by CountingPenguins View Post
    Question:

    $\displaystyle F(x) = \int_0^x[{(2t+3)\int_2^t{f(u)du}]dt$
    Calculate F''(2).
    I believe:
    F'(x) = $\displaystyle [{(2x+3)\int_2^x{f(u)du}]$
    Correct now:

    $\displaystyle F''=2\int_0^x {f(u)du} + 2xf(x) + 3f(x)$
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