Let [x] denote the greatest integer ≤x. Consider the function f(x)=max{x2, 1+[x]}. Then the value of the integral ∫02f(x) dx is [2023]
(1)
Given f(x)=max{x2,1+[x]}
So, ∫02f(x) dx=∫011 dx+∫122 dx+∫22x2 dx
=1+2(2-1)+8-223=5+423