Question

Sketch the graph y = | x + 3 |. Evaluate \[\int\limits_{- 6}^0 \left| x + 3 \right| dx\]. What does this integral represent on the graph?

Answer 1

y = | x + 3 | intersect x = 0 and x = −6 at (0, 3) and (−6, 3)
Now, 
\[y = \left| x + 3 \right|\]
\[ = \begin{cases}\left( x + 3 \right)&\text{ For all }x > - 3\\ - \left( x + 3 \right)&\text{ For all }x < - 3\end{cases}\]
Integral represents the area enclosed between x = −6 and x = 0
\[\therefore A = \int_{- 6}^0 \left| y \right| d x\]
\[ = \int_{- 6}^{- 3} \left| y \right| d x + \int_{- 3}^0 \left| y \right| d x\]
\[ = \int_{- 6}^{- 3} - \left( x + 3 \right) d x + \int_{- 3}^0 \left( x + 3 \right) d x\]
\[ = - \left[ \frac{x^2}{2} + 3x \right]_{- 6}^{- 3} + \left[ \frac{x^2}{2} + 3x \right]_{- 3}^0 \]
\[ = - \left[ \frac{9}{2} - 9 - \frac{36}{2} + 18 \right] + \left[ 0 + 0 - \frac{9}{2} + 9 \right]\]
\[ = - \frac{9}{2} + 9 + \frac{36}{2} - 18 - \frac{9}{2} + 9\]
\[ = 9\text{ sq . units }\]