Answer
$$\frac{9}{4}$$
Work Step by Step
$$\eqalign{
& \int_0^3 {\int_{{x^2}}^{x + 6} {\left( {x - 1} \right)} dydx} \cr
& {\text{Integrate with respect to }}y \cr
& = \int_0^3 {\left[ {\left( {x - 1} \right)y} \right]_{{x^2}}^{x + 6}dx} \cr
& {\text{Evaluating}} \cr
& = \int_0^3 {\left[ {\left( {x - 1} \right)\left( {x + 6} \right) - \left( {x - 1} \right){x^2}} \right]dx} \cr
& = \int_0^3 {\left( {{x^2} + 5x - 6 - {x^3} + {x^2}} \right)dx} \cr
& = \int_0^3 {\left( {2{x^2} + 5x - 6 - {x^3}} \right)dx} \cr
& {\text{Integrate }} \cr
& = \left[ {\frac{2}{3}{x^3} + \frac{5}{2}{x^2} - 6x - \frac{1}{4}{x^4}} \right]_0^3 \cr
& {\text{Evaluating}} \cr
& = \frac{2}{3}{\left( 3 \right)^3} + \frac{5}{2}{\left( 3 \right)^2} - 6\left( 3 \right) - \frac{1}{4}{\left( 3 \right)^4} - 0 \cr
& = \frac{9}{4} \cr} $$
