정삼각형 단면 부피

[출제의도] 정적분을 활용하여 입체도형의 부피 구하는 문제를 해결한다.

$y = e ^{x}$의 역함수는 $y = \ln x$이므로 점 $( 0 , 1 )$은 점 $( 1 , 0 )$으로, 점 $( 0 , e )$는 점 $( e , 0 )$으로 이동한다. 그런데 $x = t$($1 \leq t \leq e$)일 때 정삼각형의 한 변의 길이는 $\ln t$이므로 정삼각형의 넓이를 $S ( t )$라 하면 $S ( t )$$\displaystyle = \frac{\sqrt{3}}{4} \times ( \ln t ) ^{2}$ 따라서 구하는 부피는 $\displaystyle \int _{1} ^{e} S ( t ) dt$$\displaystyle = \frac{\sqrt{3}}{4} \int _{1} ^{e} {( \ln t ) ^{2}} dt$$\displaystyle = \frac{\sqrt{3}}{4} \left\{ \begin{array}{l} \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} t ( \mathrm{\ln} \mathit{t} ) ^{2} \end{array} \right] _{1} ^{e} - \int _{1} ^{e} t \left( 2 \times \frac{1}{t} \times {\ln t} \right) dt \end{array} \right\}$$\displaystyle = \frac{\sqrt{3}}{4} \left\{ \begin{array}{l} \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} t ( \mathrm{\ln} \mathit{t} ) ^{2} \end{array} \right] _{1} ^{e} - 2 \int _{1} ^{e} {\ln t} dt \end{array} \right\}$$\displaystyle = \frac{\sqrt{3}}{4} \left\{ \begin{array}{l} \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} t ( \mathrm{\ln} \mathit{t} ) ^{2} \end{array} \right] _{1} ^{e} - 2 \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} t \mathrm{\ln} \mathit{t} - t \end{array} \right] _{1} ^{e} \end{array} \right\}$ $\displaystyle = \frac{\sqrt{3}}{4} ( e - 2 )$ [다른 풀이] $\displaystyle \int \ln x dx = x \ln x - x + C$(단, $C$는 적분상수)이므로 $\displaystyle \int _{1} ^{e} S ( t ) dt$$\displaystyle = \frac{\sqrt{3}}{4} \int _{1} ^{e} {( \ln t ) ^{2}} dt$ $\displaystyle = \frac{\sqrt{3}}{4} \left\{ \begin{array}{l} \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \mathrm{\ln} \mathit{t} ( t \ln t - t ) \end{array} \right] _{1} ^{e} - \int _{1} ^{e} {( \ln t - 1 )} dt \end{array} \right\}$$\displaystyle = \frac{\sqrt{3}}{4} \left\{ \begin{array}{l} \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \mathrm{\ln} \mathit{t} ( t \ln t - t ) \end{array} \right] _{1} ^{e} - \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} t \mathrm{\ln} \mathit{t} - 2 t \end{array} \right] _{1} ^{e} \end{array} \right\}$$\displaystyle = \frac{\sqrt{3}}{4} ( e - 2 )$ [다른 풀이] 함수 $f \left( x \right) = e ^{x}$의 역함수를 $f ^{- 1} \left( x \right)$라 하고, 입체도형의 부피를 $V$라 하면 $\displaystyle V = \int _{1} ^{e} {\frac{\sqrt{3}}{4} \left\{ f ^{- 1} \left( y \right) \right\} ^{2}} dy$ 이다. $y = f \left( x \right)$에서 $f ^{- 1} \left( y \right) = x$이고 $y = f \left( x \right)$의 양변을 $x$에 대하여 미분하면 $\displaystyle \frac{dy}{dx} = f' \left( x \right) = e ^{x}$이므로 $V$$\displaystyle = \frac{\sqrt{3}}{4} \int _{0} ^{1} {} x ^{2} f' \left( x \right) dx$ $\displaystyle = \frac{\sqrt{3}}{4} \int _{0} ^{1} {} x ^{2} e ^{x} dx$ $\displaystyle = \frac{\sqrt{3}}{4} \left\{ \begin{array}{l} \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} x ^{2} e ^{x} \end{array} \right] _{0} ^{1} - 2 \int _{0} ^{1} {x e ^{x} dx} \end{array} \right\}$ $\displaystyle = \frac{\sqrt{3}}{4} \left\{ \begin{array}{l} \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} x ^{2} e ^{x} \end{array} \right] _{0} ^{1} - 2 \left( \begin{array}{l} \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} xe ^{x} \end{array} \right] _{0} ^{1} - \int _{0} ^{1} {e ^{x}} dx \end{array} \right) \end{array} \right\}$ $\displaystyle = \frac{\sqrt{3}}{4} \left\{ \begin{array}{l} \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} x ^{2} e ^{x} \end{array} \right] _{0} ^{1} - 2 \left( \begin{array}{l} \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} xe ^{x} \end{array} \right] _{0} ^{1} - \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} e ^{x} \end{array} \right] _{0} ^{1} \end{array} \right) \end{array} \right\}$ $\displaystyle = \frac{\sqrt{3}}{4} ( e - 2 )$