Evaluate the double integral $\int_{-1}^{0} \int_{1}^{2}(x^2y^2 + xy^3)dydx$ $\int_{-1}^{0} \int_{1}^{2}(x^2y^2 + xy^3)dydx$


\begin{align}
\int_{-1}^{0} \int_{1}^{2}(x^2y^2 + xy^3)dydx = \int_{-1}^{0} \bigg[\frac{x^2}{3}y^3 + \frac{x}{4}y^4\bigg]_{1}^{2}dx = \\ \int_{-1}^{0} \bigg[ \frac{8x^2}{3} + \frac{16x}{4}\bigg] - \bigg[\frac{x^2}{3} - \frac{x}{4} \bigg]dx = \int_{-1}^{0}\bigg( \frac{7x^2}{3} + \frac{17x}{4}\bigg)dx \\= \bigg[\frac{7}{9}x^3 + \frac{17}{8}x^2 \bigg]_{0}^{-1} = 0 - \bigg[ \frac{-7}{9} - \frac{17}{8}\bigg] = \frac{209}{72}
\end{align}

This is my work, but the textbook says the answer is $-\frac{79}{72}$. I double checked my work and everything seems to be checking out for me on my end, wasn't sure if the textbook made a mistake or if there is a super small mistake I am missing here.
Here had been the first out of two sign errors:
$$\bigg[ \frac{8x^2}{3} + \frac{16x}{4}\bigg] - \bigg[\frac{x^2}{3} 
+ \frac{x}{4} \bigg]$$
See the + before the $\frac{x}{4}$. At the end it should be:
$$\bigg[\frac{7}{9}x^3 + \frac{15}{8}x^2 \bigg]_{-1}^{0} = 0 - \bigg[ \frac{-7}{9} + \frac{15}{8}\bigg] = -\frac{79}{72}$$

Note the + instead of the - before the $\frac{15}{8}$.
The textbook is correct.

$$\int_{-1}^{0} \int_{1}^{2} x^2y^2 + xy^3\mathop{dy}\mathop{dx} = \int_{-1}^{0} x^2 \cdot \frac{y^3}{3} + x\cdot \frac{y^{4}}{4}\Big|_{y = 1}^{y = 2} \mathop{dx} $$

$$= \int_{-1}^{0} \left(x^2 \cdot \frac{8}{3} + x \cdot 4\right) - \left(x^2 \cdot \frac{1^{3}}{3} + x \cdot \frac{1}{4}\right) \mathop{dx}$$

$$ = \int_{-1}^{0} \frac{7x^2}{3} + \frac{15x}{4} \mathop{dx}$$

$$ = \left(\frac{7x^{3}}{9} + \frac{15x^{2}}{8}\right)\Big|_{x = -1}^{x = 0}$$

$$ = \left(0\right) - \left(-\frac{7
}{9} + \frac{15}{8}\right) $$

$$= \frac{7}{9} - \frac{15}{8} = \boxed{-\frac{209}{72}}$$