곡선과 직선 넓이 차

$f \left( x \right) = x ^{3} + 2 x ^{2} + x + 2 = \left( x + 2 \right) \left( x ^{2} + 1 \right)$ 이므로 곡선 $y = f \left( x \right)$와 $x$축이 만나는 점의 좌표는 $\left( - 2 , 0 \right)$ $g \left( x \right) = 4 x + 2$라 하면 $f \left( x \right) = g \left( x \right)$에서 $x ^{3} + 2 x ^{2} + x + 2 = 4 x + 2$ $x \left( x - 1 \right) \left( x + 3 \right) = 0$ $x = - 3$ 또는 $x = 0$ 또는 $x = 1$이므로 점 ${\mathrm{P}}$의 좌표는 $\left( 1 , 6 \right)$, 점 ${\mathrm{Q}}$의 좌표는 $\left( 0 , 2 \right)$ $g \left( x \right) = 0$에서 $4 x + 2 = 0$ $\displaystyle x = - \frac{1}{2}$이므로 점 ${\mathrm{R}}$의 좌표는 $\displaystyle \left( - \frac{1}{2} , 0 \right)$ $\displaystyle A = \int _{- 2} ^{- \frac{1}{2}} f \left( x \right) dx + \int _{- \frac{1}{2}} ^{0} \left( f \left( x \right) - g \left( x \right) \right) dx$ $\displaystyle B = \int _{0} ^{1} \left( g \left( x \right) - f \left( x \right) \right) dx$ $A - B$ $\displaystyle = \int _{- 2} ^{- \frac{1}{2}} f \left( x \right) dx + \int _{- \frac{1}{2}} ^{0} \left( f \left( x \right) - g \left( x \right) \right) dx - \int _{0} ^{1} \left( g \left( x \right) - f \left( x \right) \right) dx$ $\displaystyle = \int _{- 2} ^{- \frac{1}{2}} f \left( x \right) dx + \int _{- \frac{1}{2}} ^{0} \left( f \left( x \right) - g \left( x \right) \right) dx + \int _{0} ^{1} \left( f \left( x \right) - g \left( x \right) \right) dx$ $\displaystyle = \int _{- 2} ^{- \frac{1}{2}} f \left( x \right) dx + \int _{- \frac{1}{2}} ^{1} \left( f \left( x \right) - g \left( x \right) \right) dx$ $\displaystyle = \int _{- 2} ^{1} f \left( x \right) dx - \int _{- \frac{1}{2}} ^{1} g \left( x \right) dx$ $\displaystyle = \int _{- 2} ^{1} \left( x ^{3} + 2 x ^{2} + x + 2 \right) dx - \int _{- \frac{1}{2}} ^{1} \left( 4 x + 2 \right) dx$ $\displaystyle = \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \frac{1}{4} x ^{4} + \frac{2}{3} x ^{3} + \frac{1}{2} x ^{2} + 2 x \end{array} \right] _{- 2} ^{1} - \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} 2 x ^{2} + 2 x \end{array} \right] _{- \frac{1}{2}} ^{1}$ $\displaystyle = \frac{9}{4}$ 따라서 $\displaystyle A - B = \frac{9}{4}$ [참고] 점 ${\mathrm{P}}$에서 $x$축에 내린 수선의 발을 ${\mathrm{H}}$라 하면 $\displaystyle \int _{- \frac{1}{2}} ^{1} g \left( x \right) dx$의 값은 삼각형 ${\mathrm{RHP}}$의 넓이와 같다. $\displaystyle \int _{- \frac{1}{2}} ^{1} g \left( x \right) dx = \frac{1}{2} \times \left\{ 1 - \left( - \frac{1}{2} \right) \right\} \times 6 = \frac{9}{2}$