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Integral calculus problem

  1. Jun 7, 2012 #1
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    Last edited: Jun 7, 2012
  2. jcsd
  3. Jun 7, 2012 #2
    What class is this for, where did the problem come from. A text or written by prof?
     
  4. Jun 7, 2012 #3
    That is just Cal 1 class in college level...I guess it is written by my professor.
    Is that too difficult??lol
     
  5. Jun 7, 2012 #4
    This is plain impossible. The solution can be expressed in terms of the error function if it was negative x squared (which is not elementary):

    $$\displaystyle \frac{\mathrm{erf}(3)\sqrt{\pi}}{2}$$

    But since it is positive, we need to use the Dawson integral function. The answer then becomes

    $$\displaystyle e^9\mathrm{Di}(3)$$

    where Di denotes the Dawson integral.
     
    Last edited: Jun 7, 2012
  6. Jun 7, 2012 #5

    HallsofIvy

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    I suggest you reread the problem and then, possibly, ask your teacher if there has not been a typo. If the function were [itex]xe^{x^2}[/itex], this would be easy- and . But the function [itex]e^{x^2}[/itex] cannot be integrated in terms of elementary functions.
     
  7. Jun 7, 2012 #6
    Your professor is sneaky, you really must make use of all the data:

    [tex] \int f(x) dx=x f(x)-\int f'(x)xdx=xf(x)-\int e^{x^2}(\frac{x^2}{2})'dx=xf(x)-\frac{1}{2} \int (e^{x^2})' dx=xf(x)-\frac{e^{x^2}}{2}+c[/tex]

    Now, since this is a definite integral:

    [tex]\int_0^3 f(x) dx=3f(3)-\frac{e^{3^2}}{2}-0f(0)+\frac{e^{0^2}}{2}=21-\frac{e^{3^2}}{2}+\frac{1}{2}[/tex]
     
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