Question

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

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

Answer 1

The correct option is B $\frac{11}{4}$sq.unit

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

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

Area $=\left|{\int }_0^{1}ydx\right|+\left|{\int }_1^{2}ydx\right|+\left|{\int }_2^{3}ydx\right|$

$\begin{array}{cc}\int ydx& =\int (x-1)(x-2)(x-3)dx\\ & =\int (x^3-6x^2+11x-6)dx\\ & =\frac{x^4}{4}-\frac{6x^3}{3}+\frac{11x^2}{2}-6x\end{array}$

$\int ydx=\frac{x^4}{4}-2x^3+\frac{11x^2}{2}-6x$

$\begin{array}{cc}\text{Area}=& \left|{[\frac{x^4}{4}-2x^3+\frac{11}{2}{x}^2-6x]}_0^1\right|+\left|{[\frac{x^4}{4}-2x^3+\frac{11x^2}{2}-6x]}_1^2\right|\\ +\left|{[\frac{x^4}{4}-2x^3+\frac{11x^2}{2}-6x]}_2^3\right|& \end{array}$

$\begin{array}{cc}\Rightarrow |\frac{1}{4}-2+\frac{11}{2}-6|+& \left|[4-16+\frac{11}{2}\times 4-12[\frac{1}{4}-2+\frac{11}{2}-6]]\right|+\\ & \mid [{\frac{3}{4}}^4-2\times 27+\frac{11}{2}\times 9-18-[4-16+\frac{11}{2},\times 4-12\left]\right]\end{array}$

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

Option (B) is correct.