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Evaluate an integral of an integral function

  1. Nov 8, 2007 #1
    1. The problem statement, all variables and given/known data

    What's the best way to evaluate the following integral (the answer is in parenthesis)

    [​IMG]

    3. The attempt at a solution

    Initially I thought I could apply simple chain rule, then I realised it was integration not finding the derivative :yuck: You'll have to excuse me, my integration is a tad rusty...
     
  2. jcsd
  3. Nov 8, 2007 #2
    You'll want to begin with integration by parts. Let [tex] u = e^{\sqrt y} [/tex] and dv = 1.
     
  4. Nov 8, 2007 #3
    Ah! And then apply chain rule to get du/dy

    I see now!
     
  5. Nov 8, 2007 #4
    After doing that, I get the following... (excuse my mspaint)

    [​IMG]

    Where to from here?
     
  6. Nov 9, 2007 #5
    i think you will need to do integration by parts 2 more times (from what i'm seeing)

    so, b4 doing another integration by parts, do a u-substitution for SQRT(Y), which will help get rid of your denominator ... then solve for "y" and you'll end up with something like

    [tex]\int u^{2}e^{u}du[/tex]

    ... ok yes you do have to use integration by parts twice.
     
    Last edited: Nov 9, 2007
  7. Nov 9, 2007 #6
    why not do the following,
    [tex]\int \frac{\sqrt{y}}{\sqrt {y}}e^{\sqrt {y}}dy[/tex]
    everything is less painful from now on. you just take the sub [tex]t={\sqrt {y}[/tex]
    after that as far as i can see u need to apply integration by parts just once.
     
    Last edited: Nov 9, 2007
  8. Nov 9, 2007 #7

    Gib Z

    User Avatar
    Homework Helper

    Well the easiest way would be the differentiate the answer and see if you come up with the integrand.
     
  9. Nov 9, 2007 #8
    well yes if you differentiate the answer you will come up with the integrand.
     
  10. Nov 9, 2007 #9
    shame I don't have the answer in the exam!
     
  11. Nov 9, 2007 #10
    well remember, differentiating the answer is just a means to "prove" that you did integrate properly, it does not solve your porblem though. SO when u differentiate the answer, and come up with the ingegrand, you are sure that your answer is right, if you don't come up with the integrand, than you did some mistakes either when u integrated or when u differentiated.
     
  12. Nov 9, 2007 #11
    If possible, avoid integration by parts. Integration by parts is always a bit messy and can therefore lead to errors. The rule of thumb is that the method of choice should be that method that you can perform in your head.

    In this case, you can substitute y = t^2, which yields:

    [tex]\begin{equation}
    2 \int t\exp(t)dt
    \end{equation}
    [/tex]

    This integral could be evaluated using partial integratuon, but let's not do that. Instead, we first compute this integral:

    [tex]\begin{equation}
    \int \exp(p t)dt = \frac{\exp(p t)}{p}
    \end{equation}
    [/tex]

    And then we differentiate both sides w.r.t. p.:


    [tex]\begin{equation}
    \int t \exp(p t)dt = \left(\frac{t}{p}-\frac{1}{p^{2}}\right)\exp(p t)
    \end{equation}
    [/tex]

    Finally you put p = 1 and [tex]t =\sqrt{y}[/tex].

    Clearly this method is superior to the usual partial integration technique.
     
  13. Nov 9, 2007 #12

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    No you don't. One u-substitution and one integration by parts does the trick.

    Use [itex] u=\surd y[/itex] or [itex] y = u^2[/itex] and thus [itex]dy = 2udu[/itex]. Substituting into the integral

    [tex]\int e^{\surd y}dy = 2\int ue^u du[/tex]

    One application of integrating by parts yields

    [tex]\int ue^u du = (u-1)e^u + C[/tex]

    Converting back to y,

    [tex]\int e^{\surd y}dy = 2(\surd y - 1) e^{\surd y} + C[/tex]
     
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