Question

$f\left(x\right)$ is a cubic polynomial function such that $f\left(x\right)=0$ at $x=1,2,3$ and passing through the point $(0,-6).$ Then the area bounded by the curve $f\left(x\right)$ and the $x-$ axis between $x=0$ and $x=3$ is

Answer 1

$f\left(x\right)=0$ at $x=1,2,3$
$\Rightarrow f\left(x\right)=a(x-1)(x-2)(x-3)$
Since, $f\left(x\right)$ is passing through the point $(0,-6)$
$\Rightarrow f\left(0\right)=-6$
$\Rightarrow a(0-1)(0-2)(0-3)=-6$
$\Rightarrow a=1$
$\Rightarrow f\left(x\right)=(x-1)(x-2)(x-3)$

The graph of the given function for $0\le x\le 3$ is as shown in figure.

Hence, the required area $A=$ shaded area
$=\left|\underset{0}{\overset{1}{\int }}ydx\right|+\left|\underset{1}{\overset{2}{\int }}ydx\right|+\left|\underset{2}{\overset{3}{\int }}ydx\right|...\left(1\right)$
Since, $\int ydx=\int (x-1)(x-2)(x-3)dx$
$=\int (x^3-6x^2+11x-6)dx$
$=\frac{x^4}{4}-2x^3+\frac{11x^2}{2}-6x$
$\therefore$ From (1)
$A=\left|{[\frac{x^4}{4}-2x^3+\frac{11x^2}{2}-6x]}_0^1\right|+\left|{[\frac{x^4}{4}-2x^3+\frac{11x^2}{2}-6x]}_1^2\right|+\left|{[\frac{x^4}{4}-2x^3+\frac{11x^2}{2}-6x]}_2^3\right|$
$=|-\frac{9}{4}|+\left(\frac{1}{4}\right)+\left|\frac{-1}{4}\right|$
$=\frac{11}{4}\text{sq. units}$