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Thread: fairly simple integral, more interested in how to find it

  1. #1
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    fairly simple integral, more interested in how to find it

    f(x)= int[1,x^3] t^3dt

    f'(x)=
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  2. #2
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    $\displaystyle \int_{4}^{x^{3}} t^3 dt $

    differentiate x using FTC

    $\displaystyle f '(x) = 3x^{11} $
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  3. #3
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    Quote Originally Posted by thedoge View Post
    f(x)= int[1,x^3] t^3dt

    f'(x)=
    If $\displaystyle u=x^3$, then by the FTC:

    $\displaystyle \frac{df}{du}=u^3$,

    and:

    $\displaystyle \frac{df}{dx}=\frac{df}{du}\frac{du}{dx}=u^3 (3 x^2)=3 x^{11}$

    RonL
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