sinθ≈θ-θ3/3! for small θ, so if θ=2x3 and 0≤x<1:
sin(2x3)≈2x3-8x9/6=2x3(1-2x6/3).
∛sin(2x3)≈∛(2x3(1-2x6/3))=(x∛2)(1-2x6/3)⅓≈(x∛2)(1-2x6/9).
∫∛sin(2x3)≈∛2∫(x-2x7/9)=∛2[x2/2-x8/36]=(x2∛2/2)(1-x6/18). The upper limit is x<1, so x6<1 and the expression further approximates to x2∛2/2. The lower limit (given) is zero. The upper limit as x→1 of this expression is ∛2/2, making the integral≤∛2/2, QED.