Sketch the region between -1 and 1 for 1/x^2 and find its area.

So this is integral (-1 to 1) 1/x^2

I can compute this and it comes to -2, but it is not integrable at x=0. So should I put for the area that is is infinite or what would the area be?

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- Oct 5th 2009, 02:54 PMzhupolongjoeSketch the region and find the area
Sketch the region between -1 and 1 for 1/x^2 and find its area.

So this is integral (-1 to 1) 1/x^2

I can compute this and it comes to -2, but it is not integrable at x=0. So should I put for the area that is is infinite or what would the area be? - Oct 5th 2009, 03:11 PMKrizalid
- Oct 5th 2009, 03:30 PMzhupolongjoe
But you also get when you do that, for example in the first integral

-1/x and you can't plug 0 into there...so that brings me back to the question...is it just infinite area? - Oct 5th 2009, 04:24 PMJoshHJ
Try splitting the integral into two and substituting the variable.

So:

$\displaystyle \int_{-1}^0\frac{dx}{x^2}+\int_0^1\frac{dx}{x^2},

$

Substitute the lower limits of integration with variables, and take the limit as those variables approach zero from the left and right respectively.

- Oct 6th 2009, 06:07 AMzhupolongjoe
Still get division by 0

- Oct 9th 2009, 10:59 PMJoshHJ
Not if you take the limit as some variable approaches zero from the right or left. Then you will get infinity.