# Math Help - Improper Integrals

1. ## Improper Integrals

Can someone explain to me why this is divergent?

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

2. 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.