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Math Help - 1st Fundamental Theorem of Calc

  1. #1
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    1st Fundamental Theorem of Calc

    Ok I know the first fundamental theorem of Calc but this question is confusing me because both of the bounds are varibles with exponents. Please help



    Find the derivative of
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    F'(x)=(2x^7-1)^3\cdot 7x^6-(2x^5-1)^3\cdot 5x^4


    Fernando Revilla
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  3. #3
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    You can fix the problem that the lower limit of integration is a variable by breaking it into to integrals:
    \int_{x^5}^{x^7}(2t- 1)^3 dt= \int_a^{x^7}(2t- 1)^3 dt- \int_a^{x^5} (2t- 1)dt
    where "a" is any fixed number- the lower limit won't be relevant to the derivative.

    To fix the problem that the upper limit is not just "x", make a change of variable. If we let u= t^{1/7}, then when [tex]t= x^7[tex], u will be equal to x. t= u^7, of course, and dt= 7t^6 dt so the first integral becomes
    7\int_{a^{1/7}}^x(2u^7- 1)^3t^6 du.

    For the second integral let v= t^{1/5} so that when t= x^5, v= x. t= v^5 and dt= 5v^4dv so the integral becomes
    5\int_{a^{1/5}}^x (2v^5- 1)^3t^4dt

    That is, \int_{x^5}^{x^7}(2x-1)^2dt= 7\int_{a^{1/7}}^x (2u^7- 1)^3u^6du- 5\int_{a^{1/5}}^x (2v^5- 1)v^4 dv
    and now you can easily apply the Fundamental Theorem of Calculus to the two integrals on the right. The derivative with respect to x is, modulo any careless errors,
    7(2x- 1)^3x^6- 5(2x^5-1)x^4

    In fact, that can be generalized to "Leibniz' formula":
    \frac{d}{dx}\left(\int_{\phi(x)}^{\psi(x)} f(x,t)dt= f(x, \psi(x))\frac{d\psi(x)}{dx}- f(x,\phi(x))\frac{d\phi(x)}{dx}+ \int_{\phi(x)}^{\psi(x)}\frac{\partial f(x,t)}{\partial x}dt

    which is, I suspect, how FernandoRevilla got his answer so quickly!
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    ... which is, I suspect, how FernandoRevilla got his answer so quickly!
    I got it in the following way:


    F(x)=\displaystyle\int_{g(x)}^{h(x)}f(t)dt=\displa  ystyle\int_{0}^{h(x)}f(t)dt-\displatytyle\int_{0}^{g(x)}f(t)dt

    Using the Fundamental Theorem of Calculus and Chain's Rule we inmediately obtain:


    F'(x)=f(h(x))h'(x)-f(g(x))g'(x)

    (of course g,h differentiable, etc)


    Fernando Revilla
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