We are asked to evaluate the double integral $\int_{-1}^{4}\int_{1}^{2} (x+y^2) \, dy \, dx$.

1. Problem Description

We are asked to evaluate the double integral $\int_{-1}^{4}\int_{1}^{2} (x+y^2) \, dy \, dx$.

2. Solution Steps

First, we evaluate the inner integral with respect to $y$:

$$\int_{1}^{2} (x+y^2) \, dy = \left[ xy + \frac{y^3}{3} \right]_{1}^{2}$$

$$= (2x + \frac{8}{3}) - (x + \frac{1}{3}) = x + \frac{7}{3}$$

Next, we evaluate the outer integral with respect to $x$:

$$\int_{-1}^{4} (x + \frac{7}{3}) \, dx = \left[ \frac{x^2}{2} + \frac{7}{3}x \right]_{-1}^{4}$$

$$= (\frac{4^2}{2} + \frac{7}{3}(4)) - (\frac{(-1)^2}{2} + \frac{7}{3}(-1))$$

$$= (8 + \frac{28}{3}) - (\frac{1}{2} - \frac{7}{3}) = 8 + \frac{28}{3} - \frac{1}{2} + \frac{7}{3} = 8 + \frac{35}{3} - \frac{1}{2} = 8 + \frac{70}{6} - \frac{3}{6} = 8 + \frac{67}{6} = \frac{48}{6} + \frac{67}{6} = \frac{115}{6}$$

3. Final Answer

$\frac{115}{6}$