역함수 정적분

$\displaystyle \int _{1} ^{e ^{x}} f ^{- 1} \left( g \left( t \right) \right) dt = \left( x - 1 \right) e ^{x} + k$ $\cdots$$\cdots$ ㉠ ㉠의 양변에 $x = 0$을 대입하면 $0 = - 1 + k$, $k = 1$ ㉠의 양변을 $x$에 대하여 미분하면 $f ^{- 1} \left( g \left( e ^{x} \right) \right) \times e ^{x} = xe ^{x}$이고 $e ^{x} > 0$이므로 $f ^{- 1} \left( g \left( e ^{x} \right) \right) = x$ 그러므로 $g \left( e ^{x} \right) = f \left( x \right)$ $g ^{- 1} \left( f \left( x \right) \right) = e ^{x}$ $f ^{- 1} \left( x \right) = u$라 하면 $x = f \left( u \right)$이고 $\displaystyle 1 = f' \left( u \right) \frac{du}{dx}$ $x = f \left( 0 \right)$일 때 $u = 0$, $x = f \left( k \right) = f \left( 1 \right)$일 때 $u = 1$ $\displaystyle \int _{f \left( 0 \right)} ^{f \left( k \right)} \left( f ^{- 1} \left( x \right) + g ^{- 1} \left( x \right) \right) dx$ $\displaystyle = \int _{0} ^{1} \left\{ u + g ^{- 1} \left( f \left( u \right) \right) \right\} f' \left( u \right) du$ $\displaystyle = \int _{0} ^{1} \left( u + e ^{u} \right) f' \left( u \right) du$ $\displaystyle = \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \left( u + e ^{u} \right) f \left( u \right) \end{array} \right] _{0} ^{1} - \int _{0} ^{1} \left( 1 + e ^{u} \right) f \left( u \right) du$ $\displaystyle = \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \left( u + e ^{u} \right) f \left( u \right) \end{array} \right] _{0} ^{1} - \int _{0} ^{1} \left( 1 + e ^{u} \right) \left( u - \frac{1}{e ^{u} + 1} \right) du$ $\displaystyle = \left\{ \left( 1 + e \right) \times \left( 1 - \frac{1}{e + 1} \right) - 1 \times \left( - \frac{1}{2} \right) \right\} - \int _{0} ^{1} \left( u + ue ^{u} - 1 \right) du$ $\displaystyle = \left( e + \frac{1}{2} \right) - \int _{0} ^{1} \left( u - 1 \right) du - \int _{0} ^{1} ue ^{u} du$ $\displaystyle = e + \frac{1}{2} - \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \frac{1}{2} u ^{2} - u \end{array} \right] _{0} ^{1} - \left( \begin{array}{l} \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} ue ^{u} \end{array} \right] _{0} ^{1} - \int _{0} ^{1} e ^{u} du \end{array} \right)$ $\displaystyle = e + \frac{1}{2} - \left( \frac{1}{2} - 1 \right) - \left\{ \left( e - 0 \right) - \left( e - 1 \right) \right\}$ $= e$ 따라서 $\displaystyle \int _{f \left( 0 \right)} ^{f \left( k \right)} \left( f ^{- 1} \left( x \right) + g ^{- 1} \left( x \right) \right) dx = e$