Integrate \(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x\)

Explanation

\(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x\)

= \([\frac{2x^{2 + 1}}{3} + \frac{x^{1 + 1}}{2}]_{-1} ^{2}\)

= \([\frac{2x^{3}}{3} + \frac{x^{2}}{2}]_{-1} ^{2}\)

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

= \((\frac{16}{3} + 2) - (\frac{-2}{3} + \frac{1}{2})\)

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

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

= \(\frac{15}{2}\)

= \(7\frac{1}{2}\)

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