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Math Help - Improper Integrals

  1. #1
    Senior Member mollymcf2009's Avatar
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    Improper Integrals

    Can someone explain to me why this is divergent?

    \int\limits^{3}_{-2} \frac{94}{x^4} dx
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  2. #2
    o_O
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    We see that we have a discontinuity at x=0 so we'll split that integral up at that point.

    \int_{-2}^0 \frac{96}{x^4} \ dx \ \ = \ \ \int_{-2}^0 \frac{96}{x^4} \ dx + \int_0^3 \frac{96}{x^4} \ dx

    Now, if either of the two integrals on the right are divergent, then so is our original integral.
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