Area bounded by the curve $y=(x-1)(x-2)(x-3)$ and x-axis lying between the ordinates $x=0$ and $x=3$ is equal to

$y=(x-1)(x-2)(x-3)$

$= x^3-6x^2+11x-6$

$\therefore$ Required area

$\begin{array}{l}=\left|{\int }_0^1 \left(x^3-6x^2+11x-6\right)dx\right|\\ +{\int }_1 \left(x^3-6x^2+11x-6\right)dx\\ +\left|{\int }_2^3 \left(x^3-6x^2+11x-6\right)dx\right|\end{array}$

$\begin{array}{l}=\left|\frac{1}{4}-2+\frac{11}{2}-6\right|+\left[\frac{15}{4}-14+\frac{33}{2}-6\right]+\left|\frac{65}{4}-38+\frac{55}{2}-6\right|\\ =\frac{9}{4}+\frac{1}{4}+\frac{1}{4}=\frac{11}{4}\end{array}$