# Sketch the region and find the area

• Oct 5th 2009, 02:54 PM
zhupolongjoe
Sketch 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 PM
Krizalid
Quote:

Originally Posted by zhupolongjoe
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?

you said it, it's not integrable there because it's not bounded on $\displaystyle [-1,1].$

this splits as $\displaystyle \int_{-1}^0\frac{dx}{x^2}+\int_0^1\frac{dx}{x^2},$ then work this to see what leads to.
• Oct 5th 2009, 03:30 PM
zhupolongjoe
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 PM
JoshHJ
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 AM
zhupolongjoe
Still get division by 0
• Oct 9th 2009, 10:59 PM
JoshHJ
Not if you take the limit as some variable approaches zero from the right or left. Then you will get infinity.