To find the integral of \(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x\), we apply the power rule for integration. The integral of \(x^n\) with respect to \(x\) is \(\frac{1}{n+1} x^{n+1} + C\), where \(C\) is the constant of integration. Integrating \(2x^2\) and \(x\) separately, we get:

\(\int (2x^2) \mathrm {d} x = \frac{2}{3} x^3 + C_1\)

\(\int (x) \mathrm {d} x = \frac{1}{2} x^2 + C_2\)

Now, evaluating the definite integral from -1 to 2:

\(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x = \left[ \frac{2}{3} x^3 + \frac{1}{2} x^2 \right]_{-1} ^{2}\)

Substitute the limits of integration:

\(\left[ \frac{2}{3} (2)^3 + \frac{1}{2} (2)^2 \right] - \left[ \frac{2}{3} (-1)^3 + \frac{1}{2} (-1)^2 \right]\)

\(= \left[ \frac{16}{3} + 2 \right] - \left[ -\frac{2}{3} + \frac{1}{2} \right]\)

\(= \frac{22}{3} - \frac{1}{6}\)

\(= \frac{132 - 1}{18}\)

\(= \frac{131}{18}\)

Since \(131\) is not divisible by \(18\), we express it as a mixed fraction:

Therefore, the integral is \(7 \frac{5}{18}\)