\int _3^6\:|12x-10|dx
Solution
"\\int\\limits_3^6 {\\left| {12x - 10} \\right|} \\,dx = 2\\int\\limits_3^6 {\\left| {6x - 5} \\right|} \\,dx"
Then
"\\int {\\left| {6x - 5} \\right|} \\,dx = \\frac{1}{6}\\int {\\left| u \\right|} \\,du\\\\\n = \\frac{1}{6}\\frac{{u\\left| u \\right|}}{2} + C\\\\\n = \\frac{{u\\left| u \\right|}}{{12}} + C\\\\\n = \\frac{{\\left( {6x - 5} \\right)\\left| {6x - 5} \\right|}}{{12}} + C"
Therefore,
"\\begin{array}{l}\n\\int\\limits_3^6 {\\left| {12x - 10} \\right|} \\,dx = \\left[ {2 \\times \\frac{{\\left( {6x - 5} \\right)\\left| {6x - 5} \\right|}}{{12}}} \\right]_3^6\\\\\n = \\frac{1}{6}\\left[ {\\left( {6x - 5} \\right)\\left| {6x - 5} \\right|} \\right]_3^6\\\\\n = \\frac{1}{6}\\left[ {\\left( {6\\left( 6 \\right) - 5} \\right)\\left| {6\\left( 6 \\right) - 5} \\right| - \\left( {6\\left( 3 \\right) - 5} \\right)\\left| {6\\left( 3 \\right) - 5} \\right|} \\right]\\\\\n = \\frac{1}{6}\\left[ {\\left( {31} \\right)\\left| {31} \\right| - \\left( {18} \\right)\\left| {18} \\right|} \\right]\\\\\n = 132\n\\end{array}"
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