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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.

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    $\begingroup$ If it says the exact area between the $x$-axis and $\dots$ then I would say integrate $(x-1)(x-2)(x-3)$ from $1$ to $2$, add the integral of $-(x-1)(x-2)(x-3)$ from $2$ to $3$. $\endgroup$ Oct 23 '14 at 4:08
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$$A= \int_{1}^{2}x^3-6x^2+11x-6-\int_{2}^{3}x^3-6x^2+11x-6$$

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

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