# Thread: A negative area?? am i doing this wrong

1. ## A negative area?? am i doing this wrong

who likes my paint skills? lol

2. Have you considered learning to use LaTeX?

3. Originally Posted by Legendsn3verdie
who likes my paint skills? lol

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.

4. Originally Posted by mr 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.
way to advanced for me no idea what you did

5. What part didn't you understand?

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

how to get the area of the full curve.. did i do it right?

7. Originally Posted by Legendsn3verdie
how to get the area of the full curve.. did i do it right?
no.

$\displaystyle A = \int_{-2}^2 2 - (x^2 - 2) \, dx = 2 \int_0^2 4 - x^2 \, dx = \frac{32}{3}$

8. 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