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Math Help - Integrals with Functional Limits

  1. #1
    Del
    Del is offline
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    Integrals with Functional Limits

    If then ?

    Please help!
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  2. #2
    Super Member PaulRS's Avatar
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    f(x^{1/4})=\int_{-3}^x{\sqrt[]{t^2+3}dt} for x\geq{0}


    Differentiating (with the chain rule): \frac{x^{-3/4}\cdot{f'(x^{1/4})}}{4}=\sqrt[]{x^2+3}

    Then: f'(x^{1/4})=4\cdot{\sqrt[]{x^2+3}}\cdot{x^{3/4}}

    Let: z=x^{1/4}

    We get f'(z)=4\cdot{\sqrt[]{z^8+3}}\cdot{z^{3}} with z\geq{0}

    In the original function we can see that it has the following property f(x)=f(-x)

    Then it follows that f'(-x)=-f'(x) (imagine the graph of f(x)) which clearly holds in our solution f'(z)=4\cdot{\sqrt[]{z^8+3}}\cdot{z^{3}}

    Thus f'(x)=4\cdot{\sqrt[]{x^8+3}}\cdot{x^{3}}

    I hope it's ok
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by Del View Post
    If then ?
    Here's another attempt:

    By the FTC we have

    f'(x)=\sqrt{(x^4)^2+3}\cdot(x^4)'=4x^3\sqrt{x^8+3}  ,

    which confirms PaulRS answer.

    --

    Of course, if we wish to complicate us the existence, we can integrate \sqrt{t^2+3} then evaluate between t\in[x^4,-3].
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