Answer
$\frac{1}{2}$
Work Step by Step
Setup the integration to find the area between the two functions
$\int_0^1 ((x-1)-\sqrt[3] {x-1})dx + \int_1^2 (\sqrt[3] {x-1} -(x-1))dx$
$ \int_0^1(u-u^{\frac{1}{3}})du + \int_1^2(u^{\frac{1}{3}} -u)du$, use u-substitution, then evaluate the integration
$[\frac{1}{2}u^2 - \frac{3}{4}u^{\frac{4}{3}}]_{-1} ^0 + [\frac{3}{4}u^{\frac{4}{3}} - \frac{1}{2}u^2]_0^1 $
$(0-(\frac{1}{2}-\frac{3}{4}) + (\frac{3}{4}-\frac{1}{2})-0)$
$\frac{1}{2}$
