JEE Main 202228 Jun, Morning shiftMathematicsMediumIntegral Calculus369× since 2002Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral ∫01[−8x2+6x−1]dx\int\limits_0^1 {[ - 8{x^2} + 6x - 1]dx}0∫1[−8x2+6x−1]dx is equal to :AOption A. −-−1BOption B. −54{{ - 5} \over 4}4−5COption C. 17−138{{\sqrt {17} - 13} \over 8}817−13DOption D. 17−168{{\sqrt {17} - 16} \over 8}817−16CheckSkip · show solution
