Question

The area bounded by the curve $y=(x-1)(x-2)(x-3)$ and x-axis lying between the ordinates $x=0$ and $x=4$ is ________________.

• A. $\frac{7}{4}$
• B. $\frac{9}{4}$
• C. $2$
• D. $\frac{11}{4}$

Answer 1

The correct option is C $2$

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

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

Area ${\int }_0^1(x^3-6x^2+11x-6)dx\mid +{\int }_1^2(x^3-6x^2+16x-6)dx$

$+[{\int }_2^3{x}^3-6x^2+11x-6x)dx\mid +{\int }_3^4(x^3-6x^2+11x-6x)dx$

$\begin{array}{ccc}\text{Area}={[\frac{x^4}{4}-2x^3+\frac{11x^2}{2}-6x]}_0^1=[\frac{1}{4}-2+\frac{11}{2}-6]& =\left[\frac{-9}{4}\right].& =9/4\end{array}$

$\begin{array}{ccc}\text{Area}={[\frac{x^4}{4}-2x^3+\frac{11x^2}{2}-6x]}_1^2& =[\frac{16}{4}-16+22-12]-\left[\frac{9}{4}\right]& \Rightarrow (-2)-\left(\frac{-9}{4}\right)\Rightarrow \frac{9}{4}-2=\frac{1}{4}\end{array}$

Area $={[\frac{84}{4}-2x^3+\frac{11}{2}{x}^2-6x^2]}_2^3=[\frac{81}{4}-54+\frac{99}{2}+18]-[-2]$

$=\frac{81+198-216-72}{4}+2$

$\Rightarrow \frac{-9}{4}+2=\mid \frac{1}{4}\mid =\frac{1}{4}$

Area $={[\frac{x^4}{4}-2x^3+\frac{11x^2}{2}-6x^2]}_3^4$

$=[64-120+88-24]-\left[\frac{-1}{4}\right]=[152-152+\frac{1}{4}]=\frac{1}{4}$

Total Area $=-\frac{9}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}=|-2|=$ 2sq.units