To find the intersection points of $f$ and $g$, notice that $f(x)-g(x)=$ $x^3-2x^2-(a-b)x=$ $x\left(x^2-2x+(b-a)\right)$ has roots at $0$ and at $1\pm\sqrt{a-b+1}$ for $a-b+1\ge0$. This is three points for $a-b>-1$ (or $ba+1$). Since we are given that there are exactly two intersection points, we therefore conclude that $a-b=-1$, and that $f-g=x(x-1)^2$.
Next, f'(x)-g'(x)=3x^2-4x-(a-b) has roots at $x=\frac{2\pm\sqrt{3(a-b)+4}}{3}$ $=\frac{2\pm1}{3}=\frac13,1$. So the curves enclose a region for $x\in[0,1]$ and are tangent at the left endpoint. Since $b-a=1$, you can get rid of them in the integral because they only occur as a difference:
$ A %=\int_0^1 x(x^2-2x+1)\,dx =\int_0^1 x(x-1)^2\,dx =\int_{-1}^0 (x+1)x^2\,dx %=\int_{-1}^0 (x^3+x^2)\,dx =\left[\frac{x^4}{4}+\frac{x^3}{3}\right]_{-1}^0 %=\frac{0-1^4}{4}+\frac{0+1^3}{3} =\frac13-\frac14 =\frac1{12} $ with a few algebraic manipulations (and one substitution -- see the source for more steps). Of course, you can also do the integral in the more routine way as $ A =\int_0^1 x(x^2-2x+1)\,dx =\int_0^1 (x^3-2x^2+x)\,dx =\left[\frac14x^4-\frac23x^3+\frac12x^2\right]_0^1 =\frac34-\frac23 =\frac1{12} $ and the two together help (as in my case!) to vet errors.