적분 조건 함수 계산

[출제의도] 이해능력-적분법 $\displaystyle \int _{0} ^{1} {f ( x + t ) dt}$에서 $x + t = s$로 놓으면 $\displaystyle \frac{dt}{ds} = 1$이고 $t = 0$일 때, $s = x$, $t = 1$일 때 $s = x + 1$이므로 $\displaystyle \int _{0} ^{1} {f ( x + t ) dt} = \int _{x} ^{x + 1} {f ( s )} ds$ $\displaystyle f ( x ) = e ^{x} + \int _{0} ^{1} {f ( x + t ) dt = e ^{x} + \int _{x} ^{x + 1} {f ( s )} ds}$ 에서 양변을 $x$에 대하여 미분하면 $f' ( x ) = e ^{x} + f ( x + 1 ) - f ( x )$ 즉, $f ( x + 1 ) = f ( x ) + f' ( x ) - e ^{x} \cdots \text{㉠}$ 따라서 $\displaystyle \int _{1} ^{2} {xf ( x ) dx} - \int _{0} ^{1} {xf ( x )} dx$ $\displaystyle = \int _{0} ^{1} {( x + 1 ) f ( x + 1 ) dx - \int _{0} ^{1} {xf ( x ) dx}}$ $\displaystyle = \int _{0} ^{1} {( x + 1 ) \left\{ f ( x ) + f' ( x ) - e ^{x} \right\} dx} - \int _{0} ^{1} {xf ( x ) dx ( \because \text{㉠} )}$ $\displaystyle = \int _{0} ^{1} {( x + 1 ) f ( x ) dx + \int _{0} ^{1} {( x + 1 ) f' ( x ) dx} - \int _{0} ^{1} {( x + 1 ) e ^{x} dx}}$ $\displaystyle - \int _{0} ^{1} {xf ( x ) dx}$ $\displaystyle = \int _{0} ^{1} {xf ( x ) dx} + \int _{0} ^{1} {f ( x )} dx + \int _{0} ^{1} {( x + 1 ) f' ( x ) dx}$ $\displaystyle - \int _{0} ^{1} {( x + 1 ) e ^{x}} dx - \int _{0} ^{1} {xf ( x ) dx}$ $\displaystyle = \int _{0} ^{1} {f ( x )} dx + \left[ ( x + 1 ) f ( x ) \right] _{0} ^{1} - \int _{0} ^{1} {f ( x ) dx} - \left[ xe ^{x} \right] _{0} ^{1}$ $= 2 f ( 1 ) - f ( 0 ) - e$ $= 2 ( 2 e + 3 ) - 1 - e = 3 e + 5$