\(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x\)
= \([\frac{2x^{2 + 1}}{3} + \frac{x^{1 + 1}}{2}]_{-1} ^{2}\)
= \([\frac{2x^{3}}{3} + \frac{x^{2}}{2}]_{-1} ^{2}\)
= \((\frac{2(2)^{3}}{3} + \frac{2^2}{2}) - (\frac{2(-1)^{3}}{3} + \frac{(-1)^{2}}{2})\)
= \((\frac{16}{3} + 2) - (\frac{-2}{3} + \frac{1}{2})\)
= \(\frac{22}{3} - (-\frac{1}{6})\)
= \(\frac{22}{3} + \frac{1}{6}\)
= \(\frac{15}{2}\)
= \(7\frac{1}{2}\)