Question

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

• A. $\frac{9}{4}$
• B. $\frac{11}{4}$
• C. $\frac{13}{4}$
• D. $\frac{15}{4}$

Answer 1

The correct option is A $\frac{9}{4}$

from given, we have,

$Area={\int }_0^3(x-1)(x-2)(x-3)dx$

$={\int }_0^3(x^3-6x^2+11x-6)dx$

$={\int }_0^3{x}^{3}dx-{\int }_0^{3}6x^{2}dx+{\int }_0^{3}11xdx-{\int }_0^{3}6dx$

$={\left[\frac{x^4}{4}\right]}_0^3-6{\left[\frac{x^3}{3}\right]}_0^3+11{\left[\frac{x^2}{2}\right]}_0^3-{\left[6x\right]}_0^3$

$=\frac{81}{4}-54+\frac{99}{2}-18$

$=-\frac{9}{4}$

Area $=|-\frac{9}{4}|=\frac{9}{4}$