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Limit of integral

  1. Jun 22, 2006 #1
    limit as x approaches 0 of [(1/x^3) * integral from 0 to 3 of (sin(t^2))dt]

    is the way to solve this using mean value theorem for integrals?

    [tex]\lim_{x\rightarrow0}\frac{1}{x^3}\int{\sin{t^2}}dt[/tex] the integral is from 0 to 3
    not sure if i did the latex right
     
    Last edited: Jun 23, 2006
  2. jcsd
  3. Jun 22, 2006 #2
    [tex]\lim_{x \rightarrow 0} \frac{1}{x^3} \int_0^3 \sin{t^2} dt[/tex]
     
    Last edited: Jun 23, 2006
  4. Jun 23, 2006 #3

    HallsofIvy

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    If the integral is, in fact, from 0 to 3, then it is just a constant. This is asking for
    [tex]\lim_{x \rightarrow 0} \frac{c}{x^2}[/tex]
    where c is a non-zero constant. Looks pretty easy to me.
     
  5. Jun 23, 2006 #4

    mathman

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    The problem makes more sense (as a problem) if the upper limit of the integral is x, not 3. In that case, the answer is 1/3, assuming that the divisor is x3, not x2.
     
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