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Evaluating Double Integrals

  1. Oct 17, 2014 #1
    1. The problem statement, all variables and given/known data
    Evaluate the following double integral:
    [itex] V = \int\int \frac{3y}{6x^{5}+1} \,dA[/itex]
    [itex] D = [(x,y) \hspace{1 mm}|\hspace{1 mm} 0<=x<=1 \hspace{5 mm} 0<=y<=x^2] [/itex]

    2. Relevant equations

    3. The attempt at a solution
    [itex] V = \int_{0}^{1} \int_{0}^{x^2} \frac{3y}{6x^{5}+1}\,dy\,dx [/itex]
    [itex] = \frac{1}{2}\int_{0}^{1}\int_{0}^{x^2}\frac{y}{x^{5}+1}\,dy\,dx[/itex]
    [itex] = \frac{1}{2}\int_{0}^{1}\int_{0}^{x^2}\frac{y^2}{2x^{5}+1}\,dx [/itex]
    [itex] = \frac{1}{2}\int_{0}^{1}\frac{x^{4}}{2x^{5}+1}\,dx[/itex]
    [itex] = \frac{1}{4}\int_{0}^{1}\frac{x^{4}}{x^{5}+1}\,dx[/itex]
    To integrate, I made the following substitution:
    [itex]u=x^{5}+1 [/itex]
    Therefore, the new limits for integration are:
    [itex] u=1[/itex]
    [itex] u=2[/itex]
    [itex] = \frac{1}{20}\int_{1}^{2}\frac{du}{u}[/itex]
    [itex] = \frac{1}{20}[lnu] [/itex]
    [itex] = \frac{1}{20}[ln2-ln1] [/itex]
    This is my final (incorrect) answer:
    [itex] \frac{ln2}{20} [/itex]
    Obviously, I am making a mistake somewhere, but I can't see where so I thought a second (or third) set of eyes may help :) Thanks
    Last edited: Oct 17, 2014
  2. jcsd
  3. Oct 17, 2014 #2


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    6x5-1 is not 6(x5-1)
  4. Oct 17, 2014 #3


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    You need to learn how to use parentheses so that your algebra is correct:

    [itex]\frac{y^2}{2x^{5}+1}[/itex] should be [itex]\frac{y^2}{2(x^{5}+1)}[/itex] after you integrate

  5. Oct 17, 2014 #4
    ok, so then the answer will turn out to be:
    [itex] \frac{ln7}{20} [/itex]
    thanks for the help :)
  6. Oct 17, 2014 #5
    Sorry I always forget to show my solution:
    [itex] V = \int_{0}^{1} \frac{3}{6x^{5}+1}\frac{y^2}{2}\,dx[/itex]
    [itex] = \frac{3}{2}\int_{0}^{1}\frac{x^4}{6x^{5}+1}\,dx[/itex]
    [itex] u = 6x^{5}+1 \hspace{5 mm} du = 30x^4[/itex]
    [itex] {when} \hspace{3 mm}x=0, \hspace{3 mm} u = 1 [/itex]
    [itex] {when}\hspace{3 mm} x=1, \hspace{3 mm} u = 7 [/itex]
    [itex] = \frac{1}{20}\int_{1}^{7}\frac{du}{u} [/itex]
    [itex] = \frac{ln(7)}{20} [/itex]
  7. Oct 17, 2014 #6


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    Seems ok to me.
  8. Oct 17, 2014 #7
    thank you
  9. Oct 18, 2014 #8


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    Your final answer is correct but [itex]du= 30x^4 dx[/itex] so [itex](1/30)du= x^4 dx[/itex]

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