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Derivative and integral using calculus foundamental theorem

  1. Jul 3, 2010 #1
    1. The problem statement, all variables and given/known data
    The statement says:using the calculus fundamental theorem find:
    [tex]\displaystyle\frac{d}{dx}(\displaystyle\int_{2}^{x}\displaystyle\frac{t^{3/2}}{\sqrt[ ]{t^2+17}}dt)[/tex]

    3. The attempt at a solution
    I thought that what I should do is just to apply Barrow, and then I'd have:

    [tex]\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}}[/tex]

    Is this right?

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

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

    [tex]\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)[/tex]

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

    Bye there.
    Last edited: Jul 3, 2010
  2. jcsd
  3. Jul 3, 2010 #2


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    Well, you have at least one typo; do you have two of them? The dt should not be in the denominator. And is the d/dt at the beginning typed correctly or is it supposed to be d/dx?
  4. Jul 3, 2010 #3
    I've corrected the typos. Thanks LCKurtz. Its d/dt.
  5. Jul 3, 2010 #4
    It doesn't make any sense. Are you sure you don't mean d/dx? Otherwise, the answer is zero, assuming x is not a function t.
    \displaystyle\frac{d}{dt}(\displaystyle\int_{2}^{x }\displaystyle\frac{t^{3/2}}{\sqrt[ ]{t^2+17}}dt) = \frac{d}{dt} (F(x) - F(2) ) = 0
  6. Jul 3, 2010 #5
    Sorry, you're right. I'll edit right now.

    How can I find F(x)?
    Last edited: Jul 3, 2010
  7. Jul 3, 2010 #6


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    If you use the fundamental theorem of calculus, you don't need to find F(x). That's the point of the problem.
  8. Jul 3, 2010 #7
  9. Jul 3, 2010 #8


    Staff: Mentor

    You might have this completely figured out by now, but just in case you haven't, that minus sign shouldn't be there on the right side of the equation.
  10. Jul 4, 2010 #9
    Yep, at the beginning I thought of it as [tex]\displaystyle\frac{d}{dx}(\displaystyle\int_{x}^{2}\displaystyle\frac{t^{3/2}}{\sqrt[ ]{t^2+17}}dt)=[/tex]. Now I've realized that it was wrong :P

    Thanks to all of you people.
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