who likes my paint skills? lol

http://i176.photobucket.com/albums/w...ntitled-67.jpg

http://i176.photobucket.com/albums/w...ntitled2-2.jpg

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- Sep 8th 2008, 03:28 PMLegendsn3verdieA negative area?? am i doing this wrong
who likes my paint skills? lol

http://i176.photobucket.com/albums/w...ntitled-67.jpg

http://i176.photobucket.com/albums/w...ntitled2-2.jpg - Sep 8th 2008, 03:50 PMPlato
Have you considered learning to use LaTeX?

- Sep 8th 2008, 03:52 PMmr fantastic
Use symmetry. The enclosed area is twice the area from x = 0 to x = 2:

Enclosed area = $\displaystyle - 2 \int_{0}^{\sqrt{2}} (x^2 - 2) \, dx + 2 \left( 4 - \int_{\sqrt{2}}^{2} (x^2 - 2) \, dx \right)$.

The negative ensures you get a positive number for that section of the area. The 4 is the area of the obvious square. - Sep 8th 2008, 04:01 PMLegendsn3verdie
- Sep 8th 2008, 04:43 PM11rdc11
What part didn't you understand?

- Sep 8th 2008, 04:49 PMLegendsn3verdie
- Sep 8th 2008, 04:58 PMskeeter
- Sep 8th 2008, 05:14 PM11rdc11
Mr. Fantastic just the broke the integral in half because everything below the x axis would be negative. So he found the area below and then added it to area above the x axis from x = 0 to x = 2. That would give you the area in the 1st and 4th quad. Now since this is an even function multiply the area from quad 1 and 4 by 2