역함수와 정적분 계산

$\displaystyle f \left( x \right) = \int _{0} ^{x} e ^{\cos \pi t} dt$에서 $f \left( 0 \right) = 0$ $f' \left( x \right) = e ^{\cos \pi x}$ 모든 실수 $x$에 대하여 $f' \left( x + 2 \right) = f' \left( x \right)$ $\cdots$$\cdots$ ㉠ $f' \left( - x \right) = f' \left( x \right)$ $\cdots$$\cdots$ ㉡ ㉠에 의하여 $f \left( x + 2 \right) = f \left( x \right) + C$ (단, $C$는 적분상수) $f \left( 2 \right) = f \left( 0 \right) + C = C$이므로 $f \left( x + 2 \right) = f \left( x \right) + f \left( 2 \right)$ $f \left( 2 \right)$$\displaystyle = \int _{0} ^{2} e ^{\cos \pi t} dt$$\displaystyle = \int _{0} ^{2} f' \left( t \right) dt$$\displaystyle = \int _{0} ^{1} f' \left( t \right) dt + \int _{1} ^{2} f' \left( t \right) dt$ $\displaystyle = f \left( 1 \right) + \int _{- 1} ^{0} f' \left( t \right) dt$ ㉡에 의하여 $\displaystyle \int _{- 1} ^{0} f' \left( t \right) dt = \int _{0} ^{1} f' \left( t \right) dt$이므로 $f \left( 2 \right)$$\displaystyle = f \left( 1 \right) + \int _{0} ^{1} f' \left( t \right) dt$$= f \left( 1 \right) + f \left( 1 \right) = 2 f \left( 1 \right)$ $h \left( g \left( t \right) + 2 \right) = 2 t ^{3} + 6 f \left( 1 \right) t ^{2} + 1$ $x = g \left( t \right) + 2$로 치환하면 $\displaystyle 1 = g' \left( t \right) \frac{dt}{dx}$ $g \left( t \right) = x - 2$에서 $t = f \left( x - 2 \right)$ $x = 3$일 때 $t = f \left( 1 \right)$ $x = 7$일 때 $t = f \left( 5 \right)$ 이고 $h' \left( g \left( t \right) + 2 \right) g' \left( t \right) = 6 t ^{2} + 12 f \left( 1 \right) t$이므로 $\displaystyle \int _{3} ^{7} \frac{h' \left( x \right)}{f \left( x \right)} dx$$\displaystyle = \int _{f \left( 1 \right)} ^{f \left( 5 \right)} \frac{h' \left( g \left( t \right) + 2 \right)}{f \left( g \left( t \right) + 2 \right)} g' \left( t \right) dt$ $\displaystyle = \int _{f \left( 1 \right)} ^{f \left( 5 \right)} \frac{6 t ^{2} + 12 f \left( 1 \right) t}{f \left( g \left( t \right) \right) + f \left( 2 \right)} dt$ $\displaystyle = \int _{f \left( 1 \right)} ^{f \left( 5 \right)} \frac{6 t \left\{ t + 2 f \left( 1 \right) \right\}}{t + 2 f \left( 1 \right)} dt$ $\displaystyle = \int _{f \left( 1 \right)} ^{f \left( 5 \right)} 6 t dt$ $\displaystyle = 3 \times \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} t ^{2} \end{array} \right] _{f \left( 1 \right)} ^{f \left( 5 \right)}$ $\displaystyle = 3 \times \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \left\{ f \left( 5 \right) \right\} ^{2} - \left\{ f \left( 1 \right) \right\} ^{2} \end{array} \right]$ $f \left( 5 \right)$$= f \left( 3 \right) + f \left( 2 \right)$$= \left\{ f \left( 1 \right) + f \left( 2 \right) \right\} + f \left( 2 \right) = 5 f \left( 1 \right)$ 이므로 $\displaystyle \int _{3} ^{7} \frac{h' \left( x \right)}{f \left( x \right)} dx$$\displaystyle = 3 \times \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \left\{ 5 f \left( 1 \right) \right\} ^{2} - \left\{ f \left( 1 \right) \right\} ^{2} \end{array} \right]$$= 72 \times \left\{ f \left( 1 \right) \right\} ^{2}$ 따라서 $k = 72$ [다른 풀이] $g \left( f \left( x \right) \right) = x$이므로 $h \left( x + 2 \right) = 2 \left\{ f \left( x \right) \right\} ^{3} + 6 f \left( 1 \right) \left\{ f \left( x \right) \right\} ^{2} + 1$ $h' \left( x + 2 \right) = 6 \left\{ f \left( x \right) \right\} ^{2} f' \left( x \right) + 12 f \left( 1 \right) f \left( x \right) f' \left( x \right)$ $x = t + 2$라 하면 $\displaystyle \int _{3} ^{7} \frac{h' \left( x \right)}{f \left( x \right)} dx$ $\displaystyle = \int _{1} ^{5} \frac{h' \left( t + 2 \right)}{f \left( t + 2 \right)} dt$ $\displaystyle = \int _{1} ^{5} \frac{6 \left\{ f \left( t \right) \right\} ^{2} f' \left( t \right) + 12 f \left( 1 \right) f \left( t \right) f' \left( t \right)}{f \left( t \right) + f \left( 2 \right)} dt$ $\displaystyle = \int _{1} ^{5} \frac{6 f \left( t \right) f' \left( t \right) \left\{ f \left( t \right) + 2 f \left( 1 \right) \right\}}{f \left( t \right) + 2 f \left( 1 \right)} dt$ $\displaystyle = \int _{1} ^{5} 6 f \left( t \right) f' \left( t \right) dt$ $\displaystyle = 3 \times \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \left\{ f \left( 5 \right) \right\} ^{2} - \left\{ f \left( 1 \right) \right\} ^{2} \end{array} \right]$ $f \left( 5 \right)$$= f \left( 3 \right) + f \left( 2 \right)$$= \left\{ f \left( 1 \right) + f \left( 2 \right) \right\} + f \left( 2 \right) = 5 f \left( 1 \right)$ 이므로 $\displaystyle \int _{3} ^{7} \frac{h' \left( x \right)}{f \left( x \right)} dx$$\displaystyle = 3 \times \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \left\{ 5 f \left( 1 \right) \right\} ^{2} - \left\{ f \left( 1 \right) \right\} ^{2} \end{array} \right]$$= 72 \times \left\{ f \left( 1 \right) \right\} ^{2}$ 따라서 $k = 72$