곡선과 직선 넓이 관계

직선 $\displaystyle y = \frac{1}{3} x - \frac{2}{3}$에서 $\displaystyle g \left( x \right) = \frac{1}{3} x - \frac{2}{3}$라 하고, 함수 $y = f \left( x \right)$의 그래프와 직선 $y = g \left( x \right)$의 두 교점의 $x$좌표를 각각 $\alpha$, $\beta$ (단, $\alpha < \beta$)라 하자. 그림에서 세 영역 $A$, $B$, $C$의 넓이를 각각 $S _{1}$, $S _{2}$, $S _{3}$라 하면 $S _{1}$$\displaystyle = \int _{0} ^{\alpha} \left\{ f \left( x \right) - g \left( x \right) \right\} dx$ $S _{2}$$\displaystyle = \int _{\alpha} ^{\beta} \left\{ g \left( x \right) - f \left( x \right) \right\} dx$ $S _{3}$$\displaystyle = \int _{\beta} ^{k} \left\{ f \left( x \right) - g \left( x \right) \right\} dx$ $\left( A \right.$의 넓이$)$$+ \left( C \right.$의 넓이$) = \left( B \right.$의 넓이$)$이므로 $S _{1} + S _{3} = S _{2}$, $S _{1} - S _{2} + S _{3} = 0$ 이때 $S _{1} - S _{2} + S _{3}$ $\displaystyle = \int _{0} ^{\alpha} \left\{ f \left( x \right) - g \left( x \right) \right\} dx$$\displaystyle - \int _{\alpha} ^{\beta} \left\{ g \left( x \right) - f \left( x \right) \right\} dx$$\displaystyle + \int _{\beta} ^{k} \left\{ f \left( x \right) - g \left( x \right) \right\} dx$ $\displaystyle = \int _{0} ^{\alpha} \left\{ f \left( x \right) - g \left( x \right) \right\} dx$$\displaystyle + \int _{\alpha} ^{\beta} \left\{ f \left( x \right) - g \left( x \right) \right\} dx$$\displaystyle + \int _{\beta} ^{k} \left\{ f \left( x \right) - g \left( x \right) \right\} dx$ $\displaystyle = \int _{0} ^{k} \left\{ f \left( x \right) - g \left( x \right) \right\} dx$ 이므로 $\displaystyle \int _{0} ^{k} \left\{ f \left( x \right) - g \left( x \right) \right\} dx$$= 0$ $\cdots$$\cdots$ ㉠ $f \left( x \right) - g \left( x \right)$$\displaystyle = 3 x ^{2} - \frac{22}{3} x + \frac{8}{3}$이고, 이를 ㉠에 대입하면 $\displaystyle \int _{0} ^{k} \left( 3 x ^{2} - \frac{22}{3} x + \frac{8}{3} \right) dx = 0$ $\displaystyle \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} x ^{3} - \frac{11}{3} x ^{2} + \frac{8}{3} x \end{array} \right] _{0} ^{k} = 0$ $\displaystyle k ^{3} - \frac{11}{3} k ^{2} + \frac{8}{3} k = 0$, $k \left( 3 k ^{2} - 11 k + 8 \right) = 0$ $k \left( k - 1 \right) \left( 3 k - 8 \right) = 0$ $k > 2$이므로 $\displaystyle k = \frac{8}{3}$