Question

The area bounded by the curve $y=|x+3|$ between $x=0$ and $x=-6$ is

• A. $\frac{9}{2}$ sq units
• B. 9 sq units
• C. 18 sq units
• D. $\frac{7}{2}$ sq units

Answer 1

Answer: (B)

$y=|x+3|=\{\begin{array}{l}-(x+3)\text{ for }x<-3\\ x+3\text{ for }x\ge -3\end{array}$
When $x<-3$, then $y=-x-3$

x	- 4	- 5	-6
y	1	2	3

When $x\ge -3$

x	- 1	- 2	-3
y	2	1	0

Draw these points on the graph paper and get the required figure

$\therefore$ Required area $=$ Area of region $ABC+$ Area of region $OAD$
$=\underset{-6}{\overset{-3}{\int }}|x+3|dx+\underset{-3}{\overset{0}{\int }}|x+3|dx$
$=\underset{-6}{\overset{-3}{\int }}(-x-3)dx+\underset{-3}{\overset{0}{\int }}(x+3)dx$
$={\left[\frac{-x^2}{2}-3x\right]}_{-6}^{-3}+{\left[\frac{x^2}{2}+3x\right]}_{-3}^0$
$=\left[\left(\frac{-(-3{)}^2}{2}-3\times (-3)\right)-\left(\frac{-(-6{)}^2}{2}-3\times (-6)\right)\right]$
$=\left[\left(\frac{-9}{2}+9\right)-(-18+18)\right]+\left[\frac{9}{2}\right]$
$=\frac{9}{2}+\frac{9}{2}=9$ sq units