==============================
Explanation:
f(x) = 1 - x^2
g(x) = integral of f(x)
g(x) = x - (1/3)x^3 + C
---------------
Plug in x = -1 into the g(x) function
g(x) = x - (1/3)x^3 + C
g(-1) = -1 - (1/3)(-1)^3 + C
g(-1) = -1 - (1/3)*(-1) + C
g(-1) = -1 + 1/3 + C
g(-1) = -3/3 + 1/3 + C
g(-1) = -2/3 + C
We'll use this value later in the steps shown in the attached image.
---------------
Repeat for x = 1
g(x) = x - (1/3)x^3 + C
g(1) = 1 - (1/3)(1)^3 + C
g(1) = 1 - (1/3)*(1) + C
g(1) = 1 - 1/3 + C
g(1) = 3/3 - 1/3 + C
g(1) = 2/3 + C
We'll use this value later in the attached image below.
---------------
Repeat for x = 2
g(x) = x - (1/3)x^3 + C
g(2) = 2 - (1/3)(2)^3 + C
g(2) = 2 - (1/3)*(8) + C
g(2) = 2 - 8/3 + C
g(2) = 6/3 - 8/3 + C
g(2) = -2/3 + C
---------------
We now have enough info to finish up the problem. The rest of the steps are shown below in the attached image. The graph is included.
The two regions are set up like that because the parabola dips below the x axis. If you computed the integral of f(x) from x = -1 to x = 2 without doing any sort of region breakup, then you would get an incorrect result of 0. This is because the red region's area (4/3) is exactly equal to that of the blue region's area (4/3), ie the red region cancels out the blue region.
