정적분의 계산

$\displaystyle \int _{0} ^{2} {} \left( 3 x ^{2} - 2 x + 3 \right) dx - \int _{2} ^{0} {} \left( 2 x + 1 \right) dx$ $\displaystyle = \int _{0} ^{2} {} \left( 3 x ^{2} - 2 x + 3 \right) dx + \int _{0} ^{2} {} \left( 2 x + 1 \right) dx$ $\displaystyle = \int _{0} ^{2} {} \left\{ \left( 3 x ^{2} - 2 x + 3 \right) + \left( 2 x + 1 \right) \right\} dx$ $\displaystyle = \int _{0} ^{2} {} \left( 3 x ^{2} + 4 \right) dx = \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \begin{matrix} \\ \end{matrix} x ^{3} + 4 x \end{array} \right] _{0} ^{2} = 2 ^{3} + 4 \times 2 = 16$