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Fundamental Theorem of Calculus Problem

  1. Jun 16, 2015 #1
    1. If g(x) = ∫ f(t) dt = xln x, find f(1)
    The ∫ has x^2 on top and 0 on bottom.


    2. g'(x) = f(x) <--FTC1

    3. The attempt at a solution
    g'(x) = f(x) u=x^2
    g'(x) = u*lnu * 2x(derivative of inner function)
    g'(x) = 2x(x^2)ln(x^2)
    f(1) = 2(1)(1^2)ln(1^2)
    f(1) = 0, since ln(1) = 0

    I keep getting 0, and I'm not sure how 0 is not the answer... The answer key says the solution is 1/2. I really don't know what I'm doing wrong :(
     
  2. jcsd
  3. Jun 16, 2015 #2

    mfb

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    That's not what the theorem says. You would need x as upper integration limit to do that.
    You can introduce a new variable to fix that, like your u. Plug it in everywhere in the first equation, then proceed as before.
     
  4. Jun 16, 2015 #3
    Oh, I didn't even have the theorem correct :frown:
    Mmm.. I'm not quite following you...
     
  5. Jun 16, 2015 #4

    mfb

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    u=x2
    You can solve for x and replace all occurences of x in the initial equation. Then it has the right shape to use the theorem because the upper integral limit is the variable itself, not something else.
     
  6. Jun 16, 2015 #5
    I'm going to type out my work, okay? Can you check my work and see if I am doing it correctly?

    u = x^2 --> sqrt(u) = x

    g'(x) = ∫ u*lnu
    with sqrt(u) being the upper limit and 0 being lower

    Am I doing it correctly?
     
  7. Jun 16, 2015 #6

    Mark44

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    In a lot nicer form, this is
    ##g(x) = \int_0^{x^2} f(t)dt = x \ln(x)##
    Since you're taking the derivative with respect to x, it would help if you kept track of what you're given (the equation I wrote above).

    ##g'(x) = d/dx (\int_0^{x^2} f(t)dt )= d/dx(x \ln(x))##
    To take the derivative in the middle expression, you need to use the chain rule.
    The derivative of the expression on the right is straightforward, but you should continue in the pattern above until you get a formula for f.
     
  8. Jun 16, 2015 #7
    I SOLVED IT!
    Thank you so much, mfb and Mark44!

    I didn't know that I had to find the derivative for xlnx!
    Thank you, thank you, thank you!
     
  9. Jun 17, 2015 #8

    mfb

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    That's what I would have done:

    $$\int_0^{u} f(t)dt = \sqrt{u} \ln(\sqrt{u}) = \frac{1}{2} \sqrt{u} \ln(u)$$
    Using the derivative with respect to u on both sides:
    $$f(u) = \frac{d}{du} \left(\frac{1}{2} \sqrt{u} \ln(u)\right)$$
    .. and that is quick to calculate and gives f(1)=1/2.
     
  10. Jun 17, 2015 #9

    Fredrik

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    This calculation makes sense too: $$\frac{d}{dx}\int_0^{x^2} f(t)dt =\frac{d(x^2)}{dx}\frac{d}{d(x^2)}\int_0^{x^2} f(t)dt = 2x f(x^2).$$
     
  11. Jun 17, 2015 #10

    Mark44

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    That's more or less what I said in post #6.
     
  12. Jun 18, 2015 #11

    WWGD

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    In a generalized way, the FTC deals with composite functions, i.e., (EDIT, from Pasmith's post) ##\frac {d}{dx} \int_{h(x)}^{g(x)} f(t)dt= f(g(x))g'(x)-f(h(x))h'(x) ##, by the chain rule.
     
    Last edited: Jun 18, 2015
  13. Jun 18, 2015 #12

    pasmith

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    You have a sign error: The lower limit's derivative should be multiplied by -1, not the upper limit's.
     
  14. Jun 18, 2015 #13

    WWGD

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    Bah, f(x), -f(x), potato, potatoe :) . Just edited, thanks.
     
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