An area with an absolute value

Idea. There is no antiderivative of $|u|$ that works in one piece. We split the interval where the sign changes.

Sign. $(x-1)(x-2)\ge0$ on $[0,1]$ and $\le0$ on $[1,2]$. Hence
$$\int_0^2|x^2-3x+2|\,dx=\int_0^1(x^2-3x+2)\,dx-\int_1^2(x^2-3x+2)\,dx.$$

Computation. With $F(x)=\frac{x^3}3-\frac{3x^2}2+2x$:
$$F(1)-F(0)=\frac13-\frac32+2=\frac56,\qquad F(2)-F(1)=\left(\frac83-6+4\right)-\frac56=-\frac16.$$
So the integral equals $\frac56+\frac16=1$.

Remark. Without the absolute value, the integral would be $\frac56-\frac16=\frac23$: the signed areas partially cancel. The absolute value counts the total area between the curve and the axis.