Question

\[\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx\]

 

Answer 1

We know that

\[\left| x + 1 \right| = \begin{cases}x + 1, & \text{if }x + 1 \geq 0 \\ - \left( x + 1 \right), & \text{if }x + 1 < 0\end{cases} = \begin{cases}x + 1, & \text{if }x \geq - 1 \\ - \left( x + 1 \right), & \text{if }x < - 1\end{cases}\]

\[\left| x \right| = \begin{cases}x, & \text{if }x \geq 0 \\ - x, & \text{if }x < 0\end{cases}\]

\[\left| x - 1 \right| = \begin{cases}x - 1, & \text{if }x - 1 \geq 0 \\ - \left( x - 1 \right), & \text{if }x - 1 < 0\end{cases} = \begin{cases}x - 1, & \text{if }x \geq 1 \\ - \left( x - 1 \right), & \text{if }x < 1\end{cases}\]

When

\[- 1 \leq x \leq 0,\]

\[\left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| = x + 1 + \left( - x \right) + \left[ - \left( x - 1 \right) \right] = 2 - x\]

When

\[0 \leq x \leq 1,\]

\[\left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| = x + 1 + x + \left[ - \left( x - 1 \right) \right] = x + 2\]

When

\[1 \leq x \leq 2,\]

\[\left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| = x + 1 + x + x - 1 = 3x\]

\[\therefore \int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx\]
\[ = \int_{- 1}^0 \left( 2 - x \right)dx + \int_0^1 \left( x + 2 \right)dx + \int_1^2 3xdx\]
\[ = \left.\frac{\left( 2 - x \right)^2}{2 \times \left( - 1 \right)}\right|_{- 1}^0 + \left.\frac{\left( x + 2 \right)^2}{2}\right|_0^1 + \left.3 \times \frac{x^2}{2}\right|_1^2 \]
\[ = - \frac{1}{2}\left( 4 - 9 \right) + \frac{1}{2}\left( 9 - 4 \right) + \frac{3}{2}\left( 4 - 1 \right)\]
\[ = \frac{5}{2} + \frac{5}{2} + \frac{9}{2}\]
\[ = \frac{19}{2}\]