Integrals — Class 12 Maths Solution

Let$I = \int\limits_{ - 5}^5 {\left| {x + 2} \right|} dx$

$\left| {x + 2} \right| = \left\{ \begin{array}{l} - \left( {x + 2} \right),\,\,\,\,\,\,{\rm{if}}\,x < - 2\\x + 2,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,x \ge - 2\end{array} \right.$
$\therefore$ $I = - \int\limits_{ - 5}^{ - 2} {\left( {x + 2} \right)dx} + \int\limits_{ - 2}^5 {\left( {x + 2} \right)dx}$

$= - \left[ {\cfrac{{{{\left( {x + 2} \right)}^2}}}{2}} \right]_{ - 5}^{ - 2} + \left[ {\cfrac{{{{\left( {x + 2} \right)}^2}}}{2}} \right]_{ - 2}^5$

$= - \left[ {\left( {\cfrac{{{{\left( { - 2 + 2} \right)}^2}}}{2} - \cfrac{{{{\left( { - 5 + 2} \right)}^2}}}{2}} \right)} \right] + \left[ {\cfrac{{{{\left( {5 + 2} \right)}^2}}}{2} - \cfrac{{{{\left( { - 2 + 2} \right)}^2}}}{2}} \right]$

$= - \cfrac{1}{2}\left[ {0 - 9} \right] + \cfrac{1}{2}\left[ {49 - 0} \right] = \cfrac{9}{2} + \cfrac{{49}}{2} = \cfrac{{58}}{2} = 29$