The area bounded by the curve y = (x –1) (x – 2) (x – 3) lying between the ordinates x = 0 and x = 3 is

$\frac{11}{4}$ sq. units

Reqd. area $=\int\limits_0^3|y| d x=\int\limits_0^1|y| d x+\int\limits_1^2|y| d x+\int\limits_2^3|y| d x$

$=-\left(\frac{x^4}{4}-2 x^3+\frac{11}{2} x^2-6 x\right)_0^1+\left(\frac{x^4}{4}-2 x^3+\frac{11}{2} x^2-6 x\right)_1^2-\left(\frac{x^4}{4}-2 x^3+\frac{11}{2} x^2-6 x\right)_2^3$

$=\frac{9}{4}+\frac{1}{4}+\frac{1}{4}=\frac{11}{4}$ sq units.