$\int\limits_1^4 {\left[ {\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right|} \right]} dx$

: Let $I = \int\limits_1^4 {\left( {\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right|} \right)} dx$

$\left| {x - 1} \right| = x - 1,$

when $x \ge 1$
$\left| {x - 2} \right| = x - 2,$

when $x \ge 2$
$\left| {x - 2} \right| = - \left( {x - 2} \right),$

when $x \le 2$
$\left| {x - 3} \right| = - \left( {x - 3} \right),$

when $x \le 3$
$\left| {x - 3} \right| = \left( {x - 3} \right),$

when $x \ge 3$
$\Rightarrow$ $I = \int\limits_1^4 {\left( {x - 1} \right)dx} - \int\limits_1^2 {\left( {x - 2} \right)dx} + \int\limits_1^4 {\left( {x - 2} \right)dx} - \int\limits_1^3 {\left( {x - 3} \right)dx} + \int\limits_1^4 {\left( {x - 3} \right)dx}$

$= \left[ {\cfrac{{{x^2}}}{2} - x} \right]_1^4 - \left[ {\cfrac{{{x^2}}}{2} - 2x} \right]_1^2 + \left[ {\cfrac{{{x^2}}}{2} - 2x} \right]_2^4 - \left[ {\cfrac{{{x^2}}}{2} - 3x} \right]_1^3 + \left[ {\cfrac{{{x^2}}}{2} - 3x} \right]_3^4$

$= \left[ {\left( {\cfrac{{16}}{2} - \cfrac{1}{2}} \right) - \left( {4 - 1} \right)} \right] - \left[ {\left( {\cfrac{4}{2} - \cfrac{1}{2}} \right) - \left( {4 - 2} \right)} \right] + \left[ {\left( {\cfrac{{16}}{2} - \cfrac{4}{2}} \right) - \left( {8 - 4} \right)} \right] - \left[ {\left( {\cfrac{9}{2} - \cfrac{1}{2}} \right) - \left( {9 - 3} \right)} \right] + \left[ {\left( {\cfrac{{16}}{2} - \cfrac{9}{2}} \right) - \left( {12 - 9} \right)} \right]$

$= \left[ {\cfrac{{15}}{2} - \cfrac{3}{2} + \cfrac{{12}}{2} - \cfrac{8}{2} + \cfrac{7}{2}} \right] + \left[ { - 3 + 2 - 4 + 6 - 3} \right] = \left[ {\cfrac{{23}}{2}} \right] + \left[ { - 2} \right] = \cfrac{{19}}{2}$