Aus Online Mathematik Brückenkurs 2

There is no ready made standard formula for a primitive function to our integrand, but if we expand

\displaystyle \begin{align} & \int\limits_{-1}^{2}{\left( x-2 \right)\left( x+1 \right)\,dx}=\int\limits_{-1}^{2}{\left( x^{2}+x-2x-2 \right)}\,dx \\ & \int\limits_{-1}^{2}{\left( x^{2}-x-2 \right)}\,dx \\ \end{align}

and write the last integral as

\displaystyle \int\limits_{-1}^{2}{\left( x^{2}-x^{1}-2x^{0} \right)}\,dx

we see that the integrand consists of three terms of the type \displaystyle x^{n} and we can directly write down a primitive function:

\displaystyle \begin{align} & \int\limits_{-1}^{2}{\left( x^{2}-x^{1}-2x^{0} \right)}\,dx=\left[ \frac{x^{3}}{3}-\frac{x^{2}}{2}-2\centerdot \frac{x}{1} \right]_{-1}^{2} \\ & =\frac{2^{3}}{3}-\frac{2^{2}}{2}-2\centerdot \frac{2}{1}-\left( \frac{\left( -1 \right)^{3}}{3}-\frac{\left( -1 \right)^{2}}{2}-2\centerdot \frac{\left( -1 \right)}{1} \right) \\ & =\frac{8}{3}-\frac{4}{2}-4-\left( -\frac{1}{3}-\frac{1}{2}+2 \right) \\ & =\frac{16-12-24+2+3-12}{6}=-\frac{27}{6}=-\frac{9}{2} \\ \end{align}