Answer
$$-\frac{7}{60}$$
Work Step by Step
Using the rules of integrals, we obtain:
$\iint_{R} x-1 d A=\int_{0}^{1} \int_{x^{3}}^{x} -1+x d y d x$
$=\int_{0}^{1}[-y+x y]_{x^{3}}^{x} d x$
$=\int_{0}^{1}-\left[x^{4}-x^{3}\right] d x+\left[x^{2}-x\right]$
$=\int_{0}^{1} x^{2}-x-x^{4}+x^{3} d x$
$=\left[\frac{1}{3} x^{3}-\frac{1}{2} x^{2}-\frac{1}{5} x^{5}+\frac{1}{4} x^{4}\right]_{0}^{1}$
$-\frac{7}{60}=-\frac{1}{2}-\frac{1}{5}+\frac{1}{4}+\frac{1}{3}$
