Q. The area bounded by the curve $y=|x+3|$ between $x=0$ and $x=-6$ is
Application of Integrals
Solution:
$y=|x+3|=\begin{cases}-(x+3) \text { for } x<-3 \\ x+3 \text { for } x \geq-3\end{cases}$
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 $A B C+$ Area of region $O A D$
$=\int\limits_{-6}^{-3}|x+3| d x+\int\limits_{-3}^0|x+3| d x$
$ =\int\limits_{-6}^{-3}(-x-3) d x+\int\limits_{-3}^0(x+3) d x $
$ =\left[\frac{-x^2}{2}-3 x\right]_{-6}^{-3}+\left[\frac{x^2}{2}+3 x\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
| x | - 4 | - 5 | -6 |
| y | 1 | 2 | 3 |
| x | - 1 | - 2 | -3 |
| y | 2 | 1 | 0 |