Verified Solution • 2025–26 Session moderate

QJEEM22Ma57db: JEE Main 2022 Mathematics - Class 12 JEEM22Ma57db Solution: , where [t] is the greatest integer function, is equal to :

∫02(∣2x2−3x∣+[x−12])dx\int\limits_{0}^{2}\left(\left|2 x^{2}-3 x\right|+\left[x-\frac{1}{2}\right]\right) \mathrm{d} x , where [t] is the greatest integer function, is equal to :

Multiple Choice Options

  • A 76\frac{7}{6}
  • B 1912\frac{19}{12}
  • C 3112\frac{31}{12}
  • D 32\frac{3}{2}

Step-by-Step Verified Solution

∫02∣2x2−3x∣dx+∫02[x−12]dx\int\limits_0^2 {|2{x^2} - 3x|dx + \int\limits_0^2 {\left[ {x - {1 \over 2}} \right]dx} }

=∫03/2(3x−2x2)dx+∫3/22(2x2−3x)dx+∫01/2−1dx+∫1/23/20 dx+∫3/221dx= \int\limits_0^{3/2} {(3x - 2{x^2})dx + \int\limits_{3/2}^2 {(2{x^2} - 3x)dx + \int\limits_0^{1/2} { - 1dx + \int\limits_{1/2}^{3/2} {0\,dx + \int\limits_{3/2}^2 {1dx} } } } }

=(3x22−2x33)∣03/2+(2x33−3x22)∣3/22−12+12= \left. {\left( {{{3{x^2}} \over 2} - {{2{x^3}} \over 3}} \right)} \right|_0^{3/2} + \left. {\left( {{{2{x^3}} \over 3} - {{3{x^2}} \over 2}} \right)} \right|_{3/2}^2 - {1 \over 2} + {1 \over 2}

=(278−2712)+(163−6−2712+278)= \left( {{{27} \over 8} - {{27} \over {12}}} \right) + \left( {{{16} \over 3} - 6 - {{27} \over {12}} + {{27} \over 8}} \right)

=1912= {{19} \over {12}}

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