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Thread: Solving derivative and integral using calculus fundamental theorem

  1. #1
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    Solving derivative and integral using calculus fundamental theorem

    The statement says:using the calculus fundamental theorem find:
    \displaystyle\frac{d}{dx}(\displaystyle\int_{2}^{x  }\displaystyle\frac{t^{3/2}}{\sqrt[ ]{t^2+17}}dt)

    I thought that what I should do is just to apply Barrow, and then I'd have:

    \displaystyle\frac{d}{dx}(\displaystyle\int_{2}^{x  }\displaystyle\frac{t^{3/2}}{\sqrt[ ]{t^2+17}}dt)=-\displaystyle\frac{x^{3/2}}{\sqrt[ ]{x^2+17}}

    Is this right?

    I've tried it on the hard way too, but then:

    t=\sqrt[ ]{17}\sinh u
    dt=\sqrt[ ]{17}\cosh u du

    \displaystyle\frac{d}{dx}(\displaystyle\int_{2}^{x  }\displaystyle\frac{t^{3/2}}{\sqrt[ ]{t^2+17}dt})=\displaystyle\frac{d}{dx}(\sqrt[3]{17}\displaystyle\int_{2}^{x}\sqrt[3]{\sinh^2 u}du)

    I don't know how to solve the last integration.

    Bye there.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ulysses View Post
    The statement says:using the calculus fundamental theorem find:
    \displaystyle\frac{d}{dx}(\displaystyle\int_{2}^{x  }\displaystyle\frac{t^{3/2}}{\sqrt[ ]{t^2+17}}dt)

    I thought that what I should do is just to apply Barrow, and then I'd have:

    \displaystyle\frac{d}{dx}(\displaystyle\int_{2}^{x  }\displaystyle\frac{t^{3/2}}{\sqrt[ ]{t^2+17}}dt)= \displaystyle\frac{x^{3/2}}{\sqrt[ ]{x^2+17}}

    Is this right?
    Yes, this is fine (well, without the negative sign, that is). And this is all you need to do. Forget the hard way, haha

    and what's "Barrow"?
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  3. #3
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    Isaac Barrow - Wikipedia, the free encyclopedia

    Its not needed anyway. I thought of it being x on the "start" of the interval
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  4. #4
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    The fundamental theorem of calculus says, in part,
    \frac{d}{dx}\int_a^x f(t)dt= f(x)

    As jhevon said, the negative sign should not be there.
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