# A negative area?? am i doing this wrong

• Sep 8th 2008, 04:28 PM
Legendsn3verdie
A negative area?? am i doing this wrong
• Sep 8th 2008, 04:50 PM
Plato
Have you considered learning to use LaTeX?
• Sep 8th 2008, 04:52 PM
mr fantastic
Quote:

Originally Posted by Legendsn3verdie

Use symmetry. The enclosed area is twice the area from x = 0 to x = 2:

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

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, 05:01 PM
Legendsn3verdie
Quote:

Originally Posted by mr fantastic
Use symmetry. The enclosed area is twice the area from x = 0 to x = 2:

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

The negative ensures you get a positive number for that section of the area. The 4 is the area of the obvious square.

way to advanced for me no idea what you did
• Sep 8th 2008, 05:43 PM
11rdc11
What part didn't you understand?
• Sep 8th 2008, 05:49 PM
Legendsn3verdie
Quote:

Originally Posted by 11rdc11
What part didn't you understand?

how to get the area of the full curve.. did i do it right?
• Sep 8th 2008, 05:58 PM
skeeter
Quote:

Originally Posted by Legendsn3verdie
how to get the area of the full curve.. did i do it right?

no.

$A = \int_{-2}^2 2 - (x^2 - 2) \, dx = 2 \int_0^2 4 - x^2 \, dx = \frac{32}{3}
$
• Sep 8th 2008, 06:14 PM
11rdc11
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