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Math Help - 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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     \int_{4}^{x^{3}} t^3 dt

    differentiate x using FTC

     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 u=x^3, then by the FTC:

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

    and:

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

    RonL
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