Evaluate \(\int_{-1}^{0} (x+1)(x-2) \mathrm{d}x\)

Explanation

Expanding \((x+1)(x-2) = x^{2} - 2x + x - 2 = x^{2} - x - 2\)

\(\int_{-1}^{0} (x^{2} - x - 2) \mathrm{d}x = [\frac{x^{3}}{3} - \frac{x^{2}}{2} - 2x]_{-1}^{0}\)

= \([\frac{0}{3} - \frac{0}{2} - 2\times0 - (\frac{-1^{3}}{3} - \frac{-1^{2}}{2} - 2\times-1)]\)

= \(0 + \frac{1}{3} + \frac{1}{2} - 2 = \frac{-7}{6}\)

Note: This can also be solved using integration by parts. 

\(\int uv \mathrm{d}x = u\int v \mathrm{d}x - \int u'(\int v \mathrm{d}x)\mathrm{d}x\).