Question

Evaluate the following integral:

\[\int\limits_1^4 \left\{ \left| x - 1 \right| + \left| x - 2 \right| + \left| x - 4 \right| \right\} dx\]

 

Answer 1

\[I = \int_1^4 \left\{ \left| x - 1 \right| + \left| x - 2 \right| + \left| x - 4 \right| \right\} d x\]
\[ \Rightarrow I = \int_1^4 \left| x - 1 \right| d x + \int_1^4 \left| x - 2 \right| d x + \int_1^4 \left| x - 4 \right| d x\]
\[\text{We know that}, \left| x - 1 \right| = \begin{cases} - \left( x - 1 \right) &,& x \leq 1\\x - 1&,& 1 < x \leq 4\end{cases}\]
\[\left| x - 2 \right| = \begin{cases} - \left( x - 2 \right) &,& 1 \leq x \leq 2\\x - 2&,& 2 < x \leq 4\end{cases}\]
\[\left| x - 4 \right| = \begin{cases} - \left( x - 4 \right) &,& 1 \leq x \leq 4\\x - 4&,& x > 4\end{cases}\]
\[ \therefore I = \int_1^4 \left( x - 1 \right) d x - \int_1^2 \left( x - 2 \right) d x + \int_2^4 \left( x - 2 \right) d x - \int_1^4 \left( x - 4 \right) d x\]
\[ \Rightarrow I = \left[ \frac{x^2}{2} - x \right]_1^4 - \left[ \frac{x^2}{2} - 2x \right]_1^2 + \left[ \frac{x^2}{2} - 2x \right]_2^4 - \left[ \frac{x^2}{2} - 4x \right]_1^4 \]
\[ \Rightarrow I = 8 - 4 - \frac{1}{2} + 1 - \left( 2 - 4 - \frac{1}{2} + 2 \right) + 8 - 8 - 2 + 4 - \left( 8 - 16 - \frac{1}{2} + 4 \right)\]
\[ \Rightarrow I = \frac{23}{2}\]