Question

The area(in $\text{sq.units}$) bounded by the curve $y=(x+1)(x+2)(x-1)$ and $x-$axis, lying between the ordinates $x=-2$ and $x=1,$ is:

• A. $\frac{37}{4}$
• B. $\frac{37}{6}$
• C. $\frac{37}{12}$
• D. $\frac{37}{8}$

Answer 1

The correct option is C $\frac{37}{12}$

Given curve is $y=(x+1)(x+2)(x-1)$ which cuts $x-$axis at $x=-1,-2,1.$

$\therefore$ Area bounded by the given curve and the $x-$axis, lying between $x=-2$ and $x=1,$ given by:
$A=A_1+A_2=\underset{-2}{\overset{-1}{\int }}ydx+\left|\underset{-1}{\overset{1}{\int }}ydx\right|$
$\therefore A=\underset{-2}{\overset{-1}{\int }}(x^3+2x^2-x-2)dx+\left|\underset{-1}{\overset{1}{\int }}(x^3+2x^2-x-2)dx\right|$
$={[\frac{x^4}{4}+\frac{2x^3}{3}-\frac{x^2}{2}-2x]}_{-2}^{-1}+\left|{[\frac{x^4}{4}+\frac{2x^3}{3}-\frac{x^2}{2}-2x]}_{-1}^1\right|$
$=[(\frac{1}{4}-\frac{2}{3}-\frac{1}{2}+2)-(4-\frac{16}{3}-2+4)]+|(\frac{1}{4}+\frac{2}{3}-\frac{1}{2}-2)-(\frac{1}{4}-\frac{2}{3}-\frac{1}{2}+2)|$
$=\frac{5}{12}+\frac{8}{3}=\frac{37}{12}$