The two curves intersect when x^2-2x=4-x^2; 2x^2-2x-4=0; x^2-x-2=(x+1)(x-2). f(-1)=3=g(-1) and f(2)=0=g(2). Therefore the intersection points are (-1,3) and (2,0). The limits are x=-1 to 2.
To calculate the area between the graphs we first need to work out the area under g(x) between the limits and the x axis. This curve is like an upturned U parabola that cuts the x axis at -2 and 2. We need the area between -1 and 2. The area is given by integral((4-x^2)dx) for -1<x<2=[4x-x^3/3](2,-1)=8-8/3+4-1/3=12-9/3=9.
We may need to be careful because between x=0 and 2 the curve f(x) dips below the x axis: integral((x^2-2x)dx) for 0<x<2=[x^3/3-x^2](2,0)=(8/3-4)=-4/3. The minus sign places the area below the x axis, so the actual area is 4/3 under the axis.
[x^3/3-x^2](0,-1)=-(-1/3-1)=4/3 is the area to the left of the y axis. This needs to be subtracted from the area under g(x), while the area of f(x) below the x axis has to be added. Area under g(x) is 9 less 4/3 plus 4/3=9.
Now let's see what we get if we subtract the area under f(x) from the area under g(x) over the whole range of x. Area between f(x) and the axis =[x^3/3-x^2](2,-1)=8/3-4-(-1/3-1)=-4/3+4/3=0, so the area above the x axis is the same as the area below and the answer is 9, the area under g(x). There was no need to split the area for f(x).