∫12 x3−1dx where [.] denotes the greatest integer function, is equal to

1≤x≤2⇒1≤x3≤8⇒0≤x3−1≤7I=∫12 x3−1dx=∫121/3 x3−1dx+∫21/331/3 x3−1dx+…+∫71/32 x3−1dx now, x∈1,213, then x3∈[1,2) or x3−1=0 and so ontherefore,I=∫121/3 0dx+∫21/331/3 1⋅dx+…∫71/32 6dx =31/3−21/3+241/3−31/3+351/3−41/3+461/3−41/3+561/3−51/3+621/3−71/3=12−71/3+61/3+51/3+41/3+31/3+21/3