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The fundamental theorem of calculus(I think;) )

  1. Nov 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Been doing some old exams lately and found out that something I have problems with is questions of the type ( example):
    Differente the function:

    ∫ (x^2 ),(1), ln(t^2) dt

    Sorry for the bad writing.
    (x^2 ),(1), is the intgral from 1 to X^2

    It should be fairly simple, but my brain somehow clicks when I see this question. Maybe because its integration AND derrivation. Could someone explain the steps they use to solve this exircise.

    And would I get the same answer if I integrated the function ( maybe not the best to do in this case ), solved for the integral and then differented the integral?

    Thank you!:)
     
  2. jcsd
  3. Nov 19, 2012 #2

    HallsofIvy

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    So your problem is "Find the derivative of [itex]\int_1^{x^2} ln(t^2)dt[/itex]".

    You mention the "Fundamental Theorem" and I assume you mean the "Fundamental Theorem of Calculus"! One part of that says that [itex]d/dx \int_a^x f(t)dt= f(x)[/itex]
    Here, the only problem is that the upper limit is [itex]x^2[/itex] rather than x.

    So let [itex]u= x^2[/itex]. The integral becomes [itex]\int_1^u ln(t^2)dt[/itex]. What is the derivative of that with respect to u? And to find the derivative with respect to x, use the chain rule: df/dx= (df/du)(du/dx).
     
  4. Nov 21, 2012 #3
    Im confused. Why does I have to use the chain rule? Cant I just stick X^2 in there and get: ln(X^2)^2 = ln (x)^4 = 4lnx?

    ;-)
     
  5. Nov 21, 2012 #4
    Another way to think about it is that after doing the integration, you will have some expression like [itex]F(x^2)-F(1)[/itex]. Now, when you take the derivative, you will have to use the chain rule (and remember [itex]F(1)[/itex] is a constant) so that
    [itex]\frac{d}{dx}(F(x)-F(1)) = F'(x^2)\frac{d(x^2)}{dx}[/itex]. Now, use the fact that [itex]F'=f[/itex] (which is the FTC you reference).
     
  6. Nov 21, 2012 #5

    Mark44

    Staff: Mentor

    Yes, very much.
    It's hard to know where to start.

    Are you thinking that
    $$ \int ln(t^2) dt = (ln(t^2))^2?$$

    To use the Fundamental Theorem of Calculus, all of the important pieces have to be exactly as in the theorem -- differentiation has to be with respect to the same variable as is present in the upper limit of integration.

    For your problem, it's differentiation with respect to x, but it's x2 as a limit. Reread what HallsOfIvy said you need to do.
     
  7. Nov 21, 2012 #6

    Ray Vickson

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    You can write this as ∫_{1..x^2} ln(t^2) dt, or as ∫{ln(t^2): t=1..x^2}. Anyway, do you know the general formula for
    [tex] \frac{d}{dx} \int_1^{f(x)} F(t) \, dt?[/tex] If you do, just use it. Note that you can find the derivative of the integral, even if you cannot do the integral itself.

    RGV
     
  8. Nov 21, 2012 #7
    Im gonna spend some time working with this, but
    if we had the same exircise but with x instead of X^2, would the answer be ln(x^2)?
     
  9. Nov 21, 2012 #8

    Mark44

    Staff: Mentor

    Yes.
     
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