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Math Help - another integral

  1. #1
    Super Member Random Variable's Avatar
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    another integral

     \int_{3}^{4} \frac{x^{\frac{1}{7}}}{(7-x)^{\frac{1}{7}} + x^{\frac{1}{7}}} \ dx

    It's extremely simple if you spot the trick (which I didn't).
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Random Variable View Post
     \int_{3}^{4} \frac{x^{\frac{1}{7}}}{(7-x)^{\frac{1}{7}} + x^{\frac{1}{7}}} \ dx

    It's extremely simple if you spot the trick (which I didn't).
    Let I_1=\int_3^4\frac{\left(7-x\right)^{\frac{1}{7}}+x^{\frac{1}{7}}}{\left(7-x\right)^{\frac{1}{7}}+x^{\frac{1}{7}}}\,dx=4-3=1.

    Now,

    I_2=\int_3^4\frac{x^{\frac{1}{7}}}{\left(7-x\right)^{\frac{1}{7}}+x^{\frac{1}{7}}}\,dx

    Let u=7-x\implies -\,du=\,dx.

    So I_2=\int_3^4\frac{x^{\frac{1}{7}}}{\left(7-x\right)^{\frac{1}{7}}+x^{\frac{1}{7}}}\,dx\xright  arrow{u=7-x}{}\int_3^4\frac{\left(7-u\right)^{\frac{1}{7}}}{u^{\frac{1}{7}}+\left(7-u\right)^{\frac{1}{7}}}\,du =\int_3^4\frac{u^{\frac{1}{7}}+\left(7-u\right)^{\frac{1}{7}}}{u^{\frac{1}{7}}+\left(7-u\right)^{\frac{1}{7}}}\,du-\int_3^4\frac{u^{\frac{1}{7}}}{u^{\frac{1}{7}}+\le  ft(7-u\right)^{\frac{1}{7}}}\,du

    Therefore,

    2I_2=I_1=1\implies I_2=\frac{1}{2}
    Last edited by Chris L T521; August 23rd 2009 at 12:01 PM.
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