Area and integration question, is this area under the curve? Find the exact area between $x$ and the graph $f(x)=(x-1)(x-2)(x-3)$.
$$f(x) = x^3-6x^2+11x-6$$
I found that this is an odd shaped positive polynomial with a maxima between 1 and 2 and minima between 2 and 3.
I am confused to what the question wants. I naturally want to integrate the expansion of $f(x)$ from 1 to 2 and add it with the absolute value of of $f(x)$ integrate from 2 to 3. 
However I'm worried that the question wants the area under the curve which says not to include the area between 2 and 3 as it is below the $x$ axis.

$$A= \int_{1}^{2}x^3-6x^2+11x-6-\int_{2}^{3}x^3-6x^2+11x-6$$
$$A= \left[\frac{x^4}{4}-{2x^3}+\frac{11x^2}{2}-6x\right]_{1}^{2}-
\left[\frac{x^4}{4}-{2x^3}+\frac{11x^2}{2}-6x\right]_{2}^{3}=\frac{1}{2}$$
