Let z = 1 - x, then dz = -dx and x = 1 - z
when x = 0, z = 1 and when x = 1, z = 0
∴ ∫x(1 - x)23 for x ∈ [0,1]
= -∫(1 - z).z23 dz for x ∈ [0,1]
= -∫z23 - z24 dx for x ∈ [0,1]
= - [(z24/25) - (z25/25)] for x ∈ [0,1]
= - [0 - ((1/24) - (1/25))] = 1/600
