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Homework Help: Definite Integration by Substitution

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data
    [itex]\int^2_1 6x\sqrt{x-1}dx[/itex]

    2. Relevant equations

    3. The attempt at a solution
    Let [itex]u=x-1[/itex].
    Then, [itex]u+1=x[/itex],
    and [itex]du=dx[/itex].

    Continued from problem statement,
    [itex]=6 \int^1_0 (u+1)u^{\frac{1}{2}}du[/itex]
    [itex]=6 \int^1_0 u^{\frac{3}{2}} + u^{\frac{1}{2}}du[/itex]
    [itex]=6(1^{\frac{3}{2}} + 1^{\frac{1}{2}})[/itex]

    My Web submission program (for University) has told me this is wrong.

    What am I doing wrong?

    Thank you for any help!!
    Last edited: Sep 28, 2011
  2. jcsd
  3. Sep 28, 2011 #2


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    Umm, could it be that you forgot to find the antiderivative function...?
  4. Sep 28, 2011 #3
    you have to put x-1 back in for u after you integrate and before you solve it for the definite numbers. Also, why did you change the bounds from 1 to 2, to 0 to 1?
  5. Sep 28, 2011 #4


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    theRukus is doing the definite integral under a u-substitution, so the limits must be changed from x = 1 to x = 2 into u = 0 to u = 1 . That is perfectly legitimate: the problem is that the integration hasn't actually been carried out; tR just put 1 ( and 0 ) into the integrand!
  6. Sep 28, 2011 #5
    interesting! I was never taught that, so I always just used a and b temporarily until I put the original variable back in, and with it the original bounds.

    Yes, integrating an integral is a good idea. (why is it so hard to see the obvious sometimes?)
  7. Sep 28, 2011 #6


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    Working with the transformed definite integral under transformed limits of integration* can save a bit of writing. You are basically still doing the calculations you would have done to evaluate the definite integral in x anyway, but you don't have to take the time writing out the antiderivative in x to set up the evaluation...

    *Common integration error: forgetting to transform those limits... so don't forget.
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