Question

Evaluate the following integral:

\[\int\limits_0^4 \left( \left| x \right| + \left| x - 2 \right| + \left| x - 4 \right| \right) dx\]

Answer 1

\[I = \int_0^4 \left\{ \left| x \right| + \left| x - 2 \right| + \left| x - 4 \right| \right\} dx\]
\[ \Rightarrow I = \int_0^4 \left| x \right| d x + \int_0^4 \left| x - 2 \right| d x + \int_0^4 \left| x - 4 \right| d x\]
\[\text{We know that}, \left| x \right| = \begin{cases} - x &,& - 5 \leq x \leq 0\\x&,& x > 0\end{cases}\]
\[\left| x - 2 \right| = \begin{cases} - \left( x - 2 \right) &,& 0 \leq x \leq 2\\x - 2&,& 2 < x \leq 4\end{cases}\]
\[\left| x - 4 \right| = \begin{cases} - \left( x - 4 \right) &,& 0 \leq x \leq 4\\x - 4&,& x > 4\end{cases}\]
\[ \therefore I = \int_0^4 x d x - \int_0^2 \left( x - 2 \right) d x + \int_2^4 \left( x - 2 \right) d x - \int_0^4 \left( x - 4 \right) d x\]
\[ \Rightarrow I = \left[ \frac{x^2}{2} \right]_0^4 - \left[ \frac{x^2}{2} - 2x \right]_0^2 + \left[ \frac{x^2}{2} - 2x \right]_2^4 - \left[ \frac{x^2}{2} - 4x \right]_0^4 \]
\[ \Rightarrow I = 8 - \left( 2 - 4 \right) + 8 - 8 - 2 + 4 - \left( 8 - 16 \right)\]
\[ \Rightarrow I = 20\]