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.
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