JEE Main 202329 Jan, Morning shiftMathematicsHardIntegral Calculus369× since 2002Let [x][x][x] denote the greatest integer ≤x\le x≤x. Consider the function f(x)=max{x2,1+[x]}f(x) = \max \left\{ {{x^2},1 + [x]} \right\}f(x)=max{x2,1+[x]}. Then the value of the integral ∫02f(x)dx\int\limits_0^2 {f(x)dx}0∫2f(x)dx isAOption A. 5+423{{5 + 4\sqrt 2 } \over 3}35+42BOption B. 4+523{{4 + 5\sqrt 2 } \over 3}34+52COption C. 8+423{{8 + 4\sqrt 2 } \over 3}38+42DOption D. 1+523{{1 + 5\sqrt 2 } \over 3}31+52CheckSkip · show solution