Question

Evaluate the following integral:

\[\int\limits_{- 5}^0 f\left( x \right) dx, where\ f\left( x \right) = \left| x \right| + \left| x + 2 \right| + \left| x + 5 \right|\]

 

Answer 1

\[I = \int_{- 5}^0 \left\{ \left| x \right| + \left| x + 2 \right| + \left| x + 5 \right| \right\} dx\]
\[ \Rightarrow I = \int_{- 5}^0 \left| x \right| d x + \int_{- 5}^0 \left| x + 2 \right| d x + \int_{- 5}^0 \left| x + 5 \right| d x\]
\[\text{We know that}, \left| x \right| = \begin{cases} - x &,& - 5 \leq x \leq 0\\x&,& x > 0\end{cases}\]
\[\left| x + 2 \right| = \begin{cases} - \left( x + 2 \right) &,& - 5 \leq x \leq - 2\\x + 2&,& - 2 < x \leq 0\end{cases}\]
\[\left| x + 5 \right| = \begin{cases} - \left( x + 5 \right) &,& - 5 \leq x \leq 0\\x + 5&,& x > - 5\end{cases}\]
\[ \therefore I = - \int_{- 5}^0 x d x - \int_{- 5}^{- 2} \left( x + 2 \right) d x + \int_{- 2}^0 \left( x + 2 \right) d x + \int_{- 5}^0 \left( x + 5 \right) d x\]
\[ \Rightarrow I = - \left[ \frac{x^2}{2} \right]_{- 5}^0 - \left[ \frac{x^2}{2} + 2x \right]_{- 5}^{- 2} + \left[ \frac{x^2}{2} + 2x \right]_{- 2}^0 + \left[ \frac{x^2}{2} + 5x \right]_{- 5}^0 \]
\[ \Rightarrow I = \frac{25}{2} - \left( 2 - 4 - \frac{25}{2} + 10 \right) - 2 + 4 + \left( - \frac{25}{2} + 25 \right)\]
\[ \Rightarrow I = \frac{63}{2}\]