Question
The value of the integral $$\int_{ - 1}^3 {\left( {\left| x \right| + \left| {x - 1} \right|} \right)dx} $$ is :
A.
$$4$$
B.
$$9$$
C.
$$2$$
D.
$$\frac{9}{2}$$
Answer :
$$9$$
Solution :
We have,
\[\left| x \right| + \left| {x - 1} \right| = \left\{ \begin{array}{l}
- x - \left( {x - 1} \right) = - 2x + 1,\,{\rm{if}}\,x \le 0\\
\,\,\,\,\,x - \left( {x - 1} \right) = 1,\,{\rm{if}}\,0 \le x \le 1\\
\,\,\,\,\,x + x - 1 = 2x - 1,\,{\rm{if}}\,x \ge 1
\end{array} \right.\]
$$\eqalign{
& \therefore \,\int_{ - 1}^3 {\left( {\left| x \right| + \left| {x - 1} \right|} \right)dx} \cr
& = \int_{ - 1}^0 {\left( { - 2x + 1} \right)dx} + \int_0^1 {1\,dx} + \int_1^3 {\left( {2x - 1} \right)dx} \cr
& = \left[ { - {x^2} + x} \right]_{ - 1}^0 + \left[ x \right]_0^1 + \left[ {{x^2} - x} \right]_1^3 \cr
& = 9 \cr} $$