QJEEM23Maf360: JEE Main 2023 Mathematics - Class 12 JEEM23Maf360 Solution

Let I = $\int\limits_0^3 {|{x^2} - 3x + 2|dx}$

$= \int\limits_0^3 {|(x - 1)(x - 2)|dx}$

$= \int\limits_0^1 { + ({x^2} - 3x + 2)dx + \int\limits_1^2 { - ({x^2} - 3x + 2)dx + \int\limits_2^3 {({x^2} - 3x + 2)dx} } }$

$= \left[ {{{{x^3}} \over 3} - {{3{x^2}} \over 2} + 2x} \right]_0^1 - \left[ {{{{x^3}} \over 3} - {{3{x^2}} \over 2} + 2x} \right]_1^2 + \left[ {{{{x^3}} \over 3} - {{3{x^2}} \over 2} + 2x} \right]_2^3$

$= \left( {{1 \over 3} - {3 \over 2} + 2} \right) - \left[ {\left( {{8 \over 3} - 6 + 4} \right) - \left( {{1 \over 3} - {3 \over 2} + 2} \right)} \right] + \left[ {\left( {{{27} \over 3} - {{27} \over 2} + 6} \right) - \left( {{8 \over 3} - 6 + 4} \right)} \right]$

$= {5 \over 6} - \left( {{2 \over 3} - {5 \over 6}} \right) + \left( {{3 \over 2} - {2 \over 3}} \right)$

$= {5 \over 6} + {5 \over 6} - {2 \over 3} - {2 \over 3} + {3 \over 2}$

$= {{10} \over 6} - {4 \over 3} + {3 \over 2}$

$= {{10 - 8 + 9} \over 6} = {{11} \over 6}$

$\therefore$ 12I = $12 \times {{11} \over 6}$ = 22