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j814wong

  • one year ago

Integral Question

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  1. j814wong
    • one year ago
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    \[\int\limits_{0}^{4} \frac{ dx }{ (x-2)^{2/3} }\]

  2. tkhunny
    • one year ago
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    You ARE going to have to cut it up at x = 2 and consider convergence on both sides.

  3. j814wong
    • one year ago
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    Split it and take limit approaching 2 from left and right?

  4. SithsAndGiggles
    • one year ago
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    \[\int_0^4\frac{dx}{(x-2)^{\frac{2}{3}}}=\int_0^2\frac{dx}{(x-2)^{\frac{2}{3}}}+\int_2^4\frac{dx}{(x-2)^{\frac{2}{3}}}\] The integrand is undefined at x = 2, so you'll have to split up the integrals and take limits.

  5. j814wong
    • one year ago
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    How do I take the integral of [1/(x-2)]^(2/3)

  6. tkhunny
    • one year ago
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    Simple exponent. \(\int [1/(x-2)]^{2/3}\;dx = \int [x-2]^{-2/3}\;dx = \dfrac{(x-2)^{1/3}}{1/3} + C\)

  7. j814wong
    • one year ago
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    Gah... How did I miss that. Thanks.

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