\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 :

Question

Accepted Answer

Answer: B. \frac{19}{12} \int\limits_0^2 {|2{x^2} - 3x|dx + \int\limits_0^2 {\left[ {x - {1 \over 2}} \right]dx} } = \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} } } } } = \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} = \left( {{{27} \over 8} - {{27} \over {12}}} \right) + \left( {{{16} \over 3} - 6 - {{27} \over {12}} + {{27} \over 8}} \right) = {{19} \over {12}}

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

Answer

Explanation

More JEE Math questions