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Comparisons of imporper integrals

  1. Feb 15, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the upper bound for

    [tex]\int[/tex] (from 3 to infinity) e^(-x^2)

    [hint: e^(-x^2) < e^(-3x) for x>3]

    3. The attempt at a solution

    I don't understand what they mean by "upper bound"
    Are they asking for the upper bound for the area below e^(-x^2)
    for 3<x<infinity ?

    I knew that the function converges by graphing the function
    y= e^(-x^2) and y=e^(-x) and comparing between those two,
    but I'm not sure what the problem is asking for.
    Pleas help me witht the start!
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 15, 2008 #2
    If for two functions [itex]f,\,g[/itex] you have [itex]f(x)<g(x),\, \forall x>x_0[/itex] what can you say about the integrals [itex]\int_{x_0}^\alpha f(x)\,d\,x,\,\int_{x_0}^\alpha g(x)\,d\,x[/itex]?
  4. Feb 15, 2008 #3
    That [itex]\int_{x_0}^\alpha f(x)\,d\,x,\ < \int_{x_0}^\alpha g(x)\,d\,x[/itex] and that if
    \int_{x_0}^\alpha g(x)\,d\,x[/itex] converges, then [itex]\int_{x_0}^\alpha f(x)\,d\,x,\ converges as well?

    But how do I find upper bound knowing this?
  5. Feb 15, 2008 #4
    I meant "integral of g(x)" for the first one and "integral for f(X)" for the second one. Sorry about the mess
  6. Feb 15, 2008 #5
    Let [itex]f(x)=e^{-x^2},\,g(x)=e^{-3\,x}[/itex]. Can you calculate
    [tex]\int_3^\infty g(x)\,d\,x[/tex]?
  7. Feb 15, 2008 #6
    Yes, I calculaed it and
    lim[b to infinity] (-1/3 e^(-5b)) (1/3)
    so the first part goes to zero when you plug in infinity for b
    therefore it convererges to 1/3

    .. so.. is it right that since f(x) is smaller than g(x) it converges as well
    and therefore the possible upper bound should be 1/3 (or smaller than 1/3) ?
  8. Feb 15, 2008 #7
    I think the result is
    [tex]\int_3^\infty e^{-3\,x}\,d\,x=\frac{1}{3}\,e^{-9}[/tex] :smile:
    And the answer is that one upper bound is [itex]\frac{1}{3}\,e^{-9}[/itex], i.e.

    [tex]\int_3^\infty e^{-x^2}\,d\,x<\frac{1}{3}\,e^{-9}[/tex]
    Last edited: Feb 15, 2008
  9. Feb 15, 2008 #8
    ooops,, I guess I made a calculation mistake somewhere in the middle, I'll figure it out what I did wrong. and THANK YOU for your help!!
  10. Feb 15, 2008 #9
    Your are faster in typing! :smile:
    The result I posted it's the correct one, and one upper bound is [itex]\frac{1}{3}\,e^{-9}[/itex], or every other number bigger than that, not smaller!

    Glad I helped :smile:
  11. Feb 15, 2008 #10
    I noticed that you meant 1/3 e^(-9)
    :) Thanks again!
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