Can someone explain to me why this is divergent?

$\displaystyle \int\limits^{3}_{-2} \frac{94}{x^4} dx$

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- Feb 17th 2009, 08:19 PMmollymcf2009Improper Integrals
Can someone explain to me why this is divergent?

$\displaystyle \int\limits^{3}_{-2} \frac{94}{x^4} dx$ - Feb 17th 2009, 08:29 PMo_O
We see that we have a discontinuity at $\displaystyle x=0$ so we'll split that integral up at that point.

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

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