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Silly Integration Question

  1. Dec 20, 2008 #1
    Dear all,

    Sorry for the silly integration question. I haven't done calculus in ages and my memory has got really rusty! ='(

    The integral question is as follows:

    [tex] \int\ (X_1-\frac{X_2}{2})^2(\frac{1}{X_2}) dX_1 [/tex]

    Sorry, I could not figure out how to insert the limits on my intergral. It is from 0 to [tex]X_2[/tex].

    My answer is:

    = [tex] \int\ (\frac{X_1^2}{X_2}-X_1 + {\frac{X^2}{4}) dX_1 [/tex]

    Which, after integrating, becomes

    [tex] (\frac{X_2^2}{3}-1/2X_2^2 + X_2^2/8 ] [/tex]

    In the end, I get -X2^2/24, but the answer in the book is X2^2/12. What am I doing wrong? Thanks in advance for your help!
    Last edited: Dec 20, 2008
  2. jcsd
  3. Dec 20, 2008 #2
    So let me rewrite what you have:

    [tex] \int_{0}^{X_2} \left( X_1 - \frac{X_2}{2}\right)^2\frac{1}{X_2} dX_1 [/tex]

    Note that you can take [tex] \frac{1}{X_2} [/tex] outside of the integral since it's a constant w.r.t. [tex]X_1[/tex] (as well as taking out the [tex] \frac{1}{4}[/tex])

    So expanding out the square we get:

    [tex] \frac{1}{4X_2} \int_{0}^{X_2} 4X_{1}^{2} - 4X_{1}X_{2} + X_{2}^{2} dX_{1} = \frac{1}{4X_2} \left( \frac{4X_{1}^{3}}{3} - 2X_{1}^{2}X_{2} + X_{1}X_{2}^{2} \right)_{0}^{X_2} [/tex]

    And finally plugging in the limits we get

    [tex] \frac{1}{4X_2} \left(\frac{4}{3}X_{2}^{3} - 2X_{2}^{3} + X_{2}^{3} \right) = \frac{1}{12} X_{2}^{2} [/tex]
  4. Dec 20, 2008 #3
    Just a side note the notation that you can with limits is

    \int_{a}^{b} f(x) dx or \sum_{i=0}^{n} p(x)
  5. Dec 20, 2008 #4


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    NoMoreExams, that was nicely done but after making the point that X2 was a constant and that you can multiply the square out (or make the substitution v= X1- X2/2) it would have been better to let flying young finish the problem himself.
  6. Dec 20, 2008 #5
    Good point, I will do so in the future.
  7. Dec 21, 2008 #6
    Thank you, NoMoreExams! I finally figured out what I did wrong - a silly integration mistake! Thanks again! You have been an immense help.
  8. Dec 21, 2008 #7
    I know very much to treat X2 as a constant, but if NoMoreExams hasn't written out his steps, I wouldn't have figured out where I have gone wrong in the midst of calculations. I do understand from your stance that simply giving out the answer would do one no good. Perhaps in the future, I will be sure to write out my steps and have the pros here point out what I have done wrong. Thank you!

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