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Homework Help: Fundamental theorem of calculus

  1. Jan 19, 2010 #1
    1. The problem statement, all variables and given/known data

    Does the function F(x)=int(sin(1/t)dt,0,x) (integral of sin(1/t) with lower limit 0 to upper limit x) have a derivative at x=0?

    2. Relevant equations



    3. The attempt at a solution

    I was thinking that F(x) shouldn't have a derivative at x=0 because the integrand isn't even continuous at 0. I tried making this more explicit through using the definition of the derivative along with the convention that F(0)=0 because sin(1/t) is bounded.

    Any suggestions? Is my reasoning correct?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Jan 20, 2010
  2. jcsd
  3. Jan 20, 2010 #2

    Dick

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    Science Advisor
    Homework Helper

    This one's kind of tricky. I had to think about it for a while. F(x) actually IS differentiable at 0. Try to prove this by showing -x^2<=|F(x)|<x^2. Big hint: integrate t*cos(1/t) by parts.
     
  4. Jan 21, 2010 #3
    okay, once I have shown -x^2<=|F(x)|< x^2 then taking the limit as x tends to zero shows that F(x) is zero. My question now is how does this give us any information on the differentiability of F(x)? Is there something that I am missing?

    Thank you for your time
     
  5. Jan 21, 2010 #4
    [tex]|F'(0)| = \lim_{h \rightarrow 0}\left|\frac{F(h) - F(0)}{h}\right| = \lim_{h \rightarrow 0}\frac{|F(h)|}{|h|} \leq \lim_{h \rightarrow 0}\frac{|h|^2}{|h|} = 0.[/tex]

    Hence F'(0) = 0.
     
  6. Jan 21, 2010 #5
    wow, I don't see how I overlooked that haha. I really need to get some sleep. Thank you both for your help!
     
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