The function $f\left(x\right)\=x^{-1\/3}$ has a vertical asymptote at $x\=0$, the left endpoint of the interval of integration. Hence, the integral is improper and the evaluation must take place as per item 2a in Table 4.5.1.

${\int }_0^1{x}^{-1\/3} \mathit{\ⅆ}x\=\underset{t\to 0\+}{lim}{\int }_t^1{x}^{-1\/3} \mathit{\ⅆ}x$ = $\underset{t\to 0\+}{lim}\left({\left[\frac{x^{2\/3}}{2\/3}\right]}_t^1\right)$ = $\underset{t\to 0\+}{lim}\frac{3}{2}\left(1-t\right)\=\frac{3}{2}$