Find the area between the graph of f(x)=8-2x^2 and the x-axis over the interval [-2,2]

Question

anonymous · Answer

Since it gives us the limits of integration, we just need to find $$\int\limits_{-2}^{2} 8-2x ^{2}dx$$

That is simply 8x - (2/3)x^3 evaluated from -2 to 2. 

This is 16 - 16/3 + 16  - 16/3 = 32 - 32/3

Assuming that it does not cross the X-axis somewhere on that interval, which completely breaks everything and means we have to do it differently.

2x^2 = 8
x^2 = 4
It only crosses at the endpoints, how convenient.

Check my math, make sure I did it right. I'm sorta tired.

anonymous · Answer

$$\int\limits_{-2}^2 \left(8-2 x^2\right) \, dx=\frac{64}{3} $$