두 함수 중 큰 값의 정적분

$f \left( x \right) = 2 x ^{3} - 6 x ^{2} + 4 x$, $g \left( x \right) = x ^{2} - 2 x$라 하자. $2 x ^{3} - 6 x ^{2} + 4 x \geq x ^{2} - 2 x$에서 $2 x ^{3} - 7 x ^{2} + 6 x \geq 0$, $x \left( 2 x - 3 \right) \left( x - 2 \right) \geq 0$ $\therefore$ $\displaystyle 0 \leq x \leq \frac{3}{2}$ 또는 $x \geq 2$ 즉 $\displaystyle 0 \leq x \leq \frac{3}{2}$ 또는 $x \geq 2$일 때, $f \left( x \right) \geq g \left( x \right)$이고, 그 외의 구간에서는 $f \left( x \right) < g \left( x \right)$이다. 이때 구하는 값은 $\displaystyle h \left( x \right) = \begin{cases} 2 x ^{3} - 6 x ^{2} + 4 x & & \left( 0 \leq x \leq \frac{3}{2} \right) \\ x ^{2} - 2 x & & \left( \frac{3}{2} < x \leq 2 \right) \end{cases}$ 라 할 때 $\displaystyle \int _{0} ^{2} h \left( x \right) dx$와 같으므로 $\displaystyle \int _{0} ^{2} {} h \left( x \right) dx$ $\displaystyle = \int _{0} ^{\frac{3}{2}} {} \left( 2 x ^{3} - 6 x ^{2} + 4 x \right) dx + \int _{\frac{3}{2}} ^{2} {} \left( x ^{2} - 2 x \right) dx$ $\displaystyle = \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \frac{1}{2} x ^{4} - 2 x ^{3} + 2 x ^{2} \end{array} \right] _{0} ^{\frac{3}{2}} + \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \frac{1}{3} x ^{3} - x ^{2} \end{array} \right] _{\frac{3}{2}} ^{2}$ $\displaystyle = \left\{ \frac{1}{2} \cdot \left( \frac{3}{2} \right) ^{4} - 2 \cdot \left( \frac{3}{2} \right) ^{3} + 2 \cdot \left( \frac{3}{2} \right) ^{2} \right\} +$$\displaystyle \left\{ \left( \frac{8}{3} - 4 \right) - \left( \frac{1}{3} \cdot \left( \frac{3}{2} \right) ^{3} - \left( \frac{3}{2} \right) ^{2} \right) \right\}$ $\displaystyle = \frac{9}{32} - \frac{5}{24}$$\displaystyle = \frac{27 - 20}{96}$ $\displaystyle = \frac{7}{96}$ 따라서 $\displaystyle \int _{0} ^{2} \max \left\{ 2 x ^{3} - 6 x ^{2} + 4 x \right.$, $\left. x ^{2} - 2 x \right\} dx$의 값은 $\displaystyle \frac{7}{96}$이다.