적분 함수방정식

[출제의도] 치환적분법과 부분적분법을 이용하여 정적분에 대한 문제를 해결한다. (나)에서 $x = 0$일 때 $g ( 1 ) = 0$ $\displaystyle g ( x + 1 ) = \mathit{\int} _{0} ^{x} {\left\{ f ( t + 1 ) e ^{t} - f ( t ) e ^{t} + g \left( t \right) \right\}} dt$의 양변을 $x$에 대하여 미분하여 정리하면 $\mathrm{f} ( x + 1 ) - f ( x ) = \left\{ g' \left( x + 1 \right) - g \left( x \right) \right\} e ^{- x}$ 임의의 실수 $t$에 대하여 $\displaystyle \int _{0} ^{t} {\left\{ f \left( x + 1 \right) - f \left( x \right) \right\}} dx$$\displaystyle = \int _{0} ^{t} {\left\{ g' \left( x + 1 \right) - g \left( x \right) \right\} e ^{- x} dx}$ (좌변)$\displaystyle = \int _{0} ^{t} {f ( x + 1 ) dx} - \int _{0} ^{t} {f ( x ) dx}$ $\displaystyle = \int _{1} ^{t + 1} {f ( x ) dx} - \int _{0} ^{t} {f ( x ) dx}$$\displaystyle = \int _{t} ^{t + 1} {f ( x ) dx} - \int _{0} ^{1} {f ( x ) dx}$ …… ㉠ (우변)$\displaystyle = \int _{0} ^{t} {\left\{ g' ( x + 1 ) - g ( x ) \right\} e ^{- x}} dx$ $\displaystyle = \int _{0} ^{t} {g' ( x + 1 ) e ^{- x}} dx - \int _{0} ^{t} {g ( x ) e ^{- x}} dx$ $\displaystyle \int _{0} ^{t} {g' ( x + 1 ) e ^{- x} dx}$에서 $\displaystyle \int _{0} ^{t} {g' ( x + 1 ) e ^{- x} dx}$$\displaystyle = \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} g \left( x + 1 \right) e ^{- x} \end{array} \right] _{0} ^{t} + \int _{0} ^{t} {g ( x + 1 ) e ^{- x} dx}$ (우변)$\displaystyle = \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} g \left( x + 1 \right) e ^{- x} \end{array} \right] _{0} ^{t} + \int _{0} ^{t} {\left\{ g ( x + 1 ) - g \left( x \right) \right\} e ^{- x} dx}$ $\displaystyle = g \left( t + 1 \right) e ^{- t} - g \left( 1 \right) - \int _{0} ^{t} {\pi \left( e + 1 \right) \sin \left( \pi x \right) dx}$ $\displaystyle = g \left( t + 1 \right) e ^{- t} + \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \left( e + 1 \right) \cos \left( \pi x \right) \end{array} \right] _{0} ^{t}$ $= g \left( t + 1 \right) e ^{- t} + \left( e + 1 \right) \cos \left( \pi t \right) - \left( e + 1 \right)$ …… ㉡ ㉠, ㉡에서 $\displaystyle \int _{t} ^{t + 1} {f \left( x \right) dx}$$\displaystyle = \int _{0} ^{1} {f \left( x \right)} dx + g \left( t + 1 \right) e ^{- t} + \left( e + 1 \right) \cos \left( \pi t \right) - \left( e + 1 \right)$ $g \left( x + 1 \right) = g \left( x \right) - \pi \left( e + 1 \right) \sin \left( \pi x \right) e ^{x}$에서 $g \left( 0 \right) = g \left( 1 \right) = g \left( 2 \right) = \cdots = g \left( 9 \right) = 0$ $\displaystyle \int _{1} ^{10} {f \left( x \right)} dx$$\displaystyle = \sum\limits _{n = 1} ^{9} \int _{n} ^{n + 1} {f \left( x \right)} dx$ $\displaystyle = \sum\limits _{n = 1} ^{9} \left\{ \int _{0} ^{1} {f \left( x \right)} dx + g \left( n + 1 \right) e ^{- n} + \left( e + 1 \right) \cos \left( \pi n \right) - \left( e + 1 \right) \right\}$ $\displaystyle = 9 \int _{0} ^{1} {f \left( x \right)} dx + 0 + \left( e + 1 \right) \sum\limits _{n = 1} ^{9} \left\{ \cos \left( \pi n \right) - 1 \right\}$ $\displaystyle = 9 \left( \frac{10}{9} e + 4 \right) + \left( e + 1 \right) \times \left( - 10 \right)$$= 26$