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A simple question on integration

  1. May 12, 2014 #1
    1. The problem statement, all variables and given/known data

    I want to take an antiderivative of a function with respect to x. But in addition the function includes a term y (x) that is a function of x itself. Do I have to apply the reverse power rule also to y(x) also? The integral can be seen as an indefinite.

    2. Relevant equations

    [itex]A=\int x*y(x)*dx[/itex]

    3. The attempt at a solution

    I think I should just apply the reverse power rule to x. So:

    [itex]A= \frac{x^{2}}{2}*y(x) + C [/itex]
     
  2. jcsd
  3. May 12, 2014 #2

    Ray Vickson

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    You can answer your own question---and that is the best way to learn! Just try it out on some examples. What do you get if you use your formula on the function ##y(x) = x^2##? What happens if you use your formula on ##y(x) = 1/x^2##? Are you getting correct results?
     
    Last edited: May 12, 2014
  4. May 12, 2014 #3
    What you are looking for is something called integration by parts.
     
  5. May 14, 2014 #4
    Okay, then I believe I was mistaken. It seems that I also need to apply the reverse power rule to ##y(x)##.
    So an example:
    ##y(x)=2*x##

    [itex]A=\int^{2}_{1} x*y(x)*dx=\int^{2}_{1} x*2*x*dx[/itex]
    [itex]A=\left|2\frac{x^{3}}{3}\right|^{2}_{1}=4.6667[/itex]

    Is this correct?

    I also looked up the integration by parts and it seems that in the above example it was possible to do without the integration by parts. But can someone give a good example why and when is it necessary to turn to the integration by parts technique?
     
  6. May 14, 2014 #5

    pasmith

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    Try [itex]\displaystyle\int_0^\pi x \sin x\,dx[/itex] or [itex]\displaystyle\int_0^1 x e^{-x}\,dx[/itex].
     
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