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Derivative of the product of 2 definite integrals

  1. Aug 4, 2012 #1
    1. The problem statement, all variables and given/known data


    Find f'(x) for (integral from 0 to x of cos^5(t)dt)* (integral from x^2 to 1 of e^t^2 dt). No differentiation allowed in the answer
    2. Relevant equations



    3. The attempt at a solution. I used the product rule and integrated then differentiated the first term --> cos^5x* integral of e, etc. but I'm stumped on what to do with the these one term. Obviously I leave it as is for the first part of the product rule. So it's the second half where I'm looking for the differential of the integral of e^t^2
     
  2. jcsd
  3. Aug 4, 2012 #2

    LCKurtz

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    You don't have to integrate then differentiate. Use the fundamental theorem of calculus:$$
    \frac d {dx}\int_a^x f(t)\, dt = f(x)$$
     
  4. Aug 4, 2012 #3
    I apologize for being so dense, but how does that apply to the stated answer for the quiz of cos^5(x) * integral with e + integral with cos^5(t) * (-2x)e^x^4 it's the answer for the last of the four terms in the result that's throwing me. To me, it seems the derivative of the integral of e^t^2 should just be e^x^2.

    Thanks,
     
  5. Aug 4, 2012 #4

    LCKurtz

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    You have given me more information now, so I can see what your problem is. You need to use the chain rule:$$
    \frac d {dx}\int_{x^2}^1 e^{t^2}\ dt = -\frac d {dx}\int_{1}^{x^2} e^{t^2}\ dt$$
    You plug in the ##x^2## for the ##t## in ##e^{t^2}## but you must multiply by the derivative of ##x^2## using the chain rule. Some versions of Leibnitz rule show this.
     
  6. Aug 4, 2012 #5
    [tex] ( \int {cos^t dt})[/tex] *([tex]\int{ e^t^2 dt) } [/tex]. I am trying out the text editor I just found on the site to see if I can more clearly write oute the problem. So I'll post to see if this comes out ok
     
    Last edited: Aug 4, 2012
  7. Aug 4, 2012 #6
    Well my first attempts at using the editor aren't so good. I'm going to go take the laundry out of the dryer and sit and play with your answer for a While to make sure I know what's going on. And the reason you inverted the integral is because it went from a higher number to a lower one, right?
     
  8. Aug 4, 2012 #7

    eumyang

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    You're mixing non-LaTeX symbols with the LaTeX commands. Just use one set of tex tags.
    What you want to type between the tex tags is this:
    \left( \int_{0}^{x} \cos^5 t\ dt \right) \cdot \left( \int_{x^2}^{1} e^{t^2}\ dt \right)
    Which gives you this:
    [tex]\left( \int_{0}^{x} \cos^5 t\ dt \right) \cdot \left( \int_{x^2}^{1} e^{t^2}\ dt \right)
    [/tex]
     
    Last edited: Aug 4, 2012
  9. Aug 4, 2012 #8

    SammyS

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    Dba18,

    Here's how I go about remembering what to do with something like [itex]\displaystyle \frac{d}{dx} \left( \int_{x^2}^{1} e^{t^2}\ dt \right)\ .[/itex]

    Let G(t) be the anti derivative of [itex]\displaystyle e^{t^2}\,, [/itex] so that [itex]\displaystyle G\,'(t)=e^{t^2}\,, [/itex]

    Then [itex]\displaystyle \int_{x^2}^{1} e^{t^2}\ dt =
    G(1)-G(x^2)\ .[/itex]

    Therefore, [itex]\displaystyle \frac{d}{dx} \left( \int_{x^2}^{1} e^{t^2}\ dt \right)=
    \frac{d}{dx}\left(G(1)-G(x^2)\right)=\underline{\ \ \ ?\ \ \ } \ .[/itex]
     
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