적분방정식 보기

[출제의도] 정적분의 적분법을 이용하여 정적분의 값을 추론한다. ㄱ. (가)의 $F ( x ) = f ( x ) - x$를 (나)의 $F ( x )$에 대입하면 $\displaystyle \int _{0} ^{1} F ( x ) dx$$\displaystyle = \int _{0} ^{1} \left\{ f ( x ) - x \right\} dx$ $\displaystyle = \int _{0} ^{1} f ( x ) dx - \int _{0} ^{1} x dx$ $\displaystyle = \int _{0} ^{1} f ( x ) dx - \frac{1}{2}$ $\displaystyle = F ( 1 ) - \frac{1}{2}$ 이므로 $\displaystyle F ( 1 ) = e - \frac{5}{2} + \frac{1}{2}$$= e - 2$ (거짓) ㄴ. (가)의 $F ( x ) = f ( x ) - x$를 $\displaystyle \int _{0} ^{1} xF ( x ) dx$의 $F ( x )$에 대입하면 $\displaystyle \int _{0} ^{1} xF ( x ) dx$$\displaystyle = \int _{0} ^{1} x \left\{ f ( x ) - x \right\} dx$ $\displaystyle = \int _{0} ^{1} \left\{ xf ( x ) - x ^{2} \right\} dx$ $\displaystyle = \int _{0} ^{1} xf ( x ) dx - \int _{0} ^{1} x ^{2} dx$$\displaystyle = \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} xF ( x ) \end{array} \right] _{0} ^{1} - \int _{0} ^{1} F ( x ) dx - \frac{1}{3}$ $\displaystyle = F ( 1 ) - \left( e - \frac{5}{2} \right) - \frac{1}{3}$ $\displaystyle = e - 2 - e + \frac{5}{2} - \frac{1}{3}$$\displaystyle = \frac{1}{6}$ (참) ㄷ. $\displaystyle F \left( 0 \right) = \int _{0} ^{0} {f \left( t \right)} dt$$= 0$이고, $\displaystyle F \left( x \right) = \int _{0} ^{x} {f \left( t \right)} dt$의 양변을 $x$에 대하여 미분하면 $F' \left( x \right) = f \left( x \right)$ (가)에 의해서 $\displaystyle \int _{0} ^{1} \left\{ F ( x ) \right\} ^{2} dx$$\displaystyle = \int _{0} ^{1} F ( x ) \left\{ f ( x ) - x \right\} dx$ $\displaystyle = \int _{0} ^{1} F ( x ) f ( x ) dx - \int _{0} ^{1} x F ( x ) dx$$\displaystyle = \int _{0} ^{1} F ( x ) F' ( x ) dx - \int _{0} ^{1} x F ( x ) dx$ $=$$\displaystyle \left[ \frac{1}{2} \left\{ F ( x ) \right\} ^{2} \right] _{0} ^{1} - \frac{1}{6}$ $\displaystyle = \frac{1}{2} \left\{ F ( 1 ) \right\} ^{2} - \frac{1}{2} \left\{ F ( 0 ) \right\} ^{2} - \frac{1}{6}$ $\displaystyle = \frac{1}{2} ( e - 2 ) ^{2} - \frac{1}{6}$ $\displaystyle = \frac{1}{2} e ^{2} - 2 e + \frac{11}{6}$ (참) 따라서 옳은 것은 ㄴ, ㄷ이다. [다른 풀이] $\displaystyle F \left( x \right) = \int _{0} ^{x} {f \left( t \right)} dt$의 양변을 $x$에 대하여 미분하면 $F' \left( x \right) = f \left( x \right)$ 조건 (가)의 $F \left( x \right) = f \left( x \right) - x$의 양변을 $x$에 대하여 두 번 미분하면 $f' ( x ) = f'' ( x )$ $\displaystyle \frac{f'' ( x )}{f' ( x )} = 1$이므로 양변을 $x$에 대하여 적분하면 $\ln | f' ( x ) | = x + C _{1}$(단, $C _{1}$는 적분상수) $f' ( x ) = C _{2} e ^{x}$(단, $C _{2}$는 상수)이므로 양변을 $x$에 대하여 적분하면 $f ( x ) = C _{2} e ^{x} + C _{3}$(단, $C _{3}$은 적분상수) $0 = F ( 0 ) = f ( 0 ) - 0$, $f ( 0 ) = 0$ $f ( x ) = f' ( x ) - 1$이므로 $f' ( 0 ) = 1$ $f' ( x ) = C _{2} e ^{x}$, $f ( x ) = C _{2} e ^{x} + C _{3}$에 $x = 0$을 대입하면 $f' ( 0 ) = C _{2} = 1$, $f ( 0 ) = C _{2} + C _{3} = 0$, $C _{3} = \mathit{-} 1$ $f \left( x \right) = e ^{x} - 1$ 이를 조건 (가)에 대입하면 $F \left( x \right) = e ^{x} - x - 1$ $\displaystyle \int _{0} ^{1} {F \left( x \right)} dx$$\displaystyle = \int _{0} ^{1} {\left( e ^{x} - x - 1 \right)} dx$ $\displaystyle = \left[ e ^{x} - \frac{1}{2} x ^{2} - x \right] _{0} ^{1}$ $\displaystyle = \left( e - 1 \right) - \frac{1}{2} - 1$ $\displaystyle = e - \frac{5}{2}$ 이므로 함수 $F \left( x \right) = e ^{x} - x - 1$은 조건 (나)를 만족시킨다. ㄱ. $F \left( 1 \right) = e ^{1} - 1 - 1 = e - 2$ (거짓) ㄴ. $\displaystyle \int _{0} ^{1} {x F \left( x \right)} dx$$\displaystyle = \int _{0} ^{1} {x \left( e ^{x} - x - 1 \right)} dx$ $\displaystyle = \int _{0} ^{1} {x e ^{x}} dx - \int _{0} ^{1} {x ^{2}} dx - \int _{0} ^{1} {x} dx$ $\displaystyle = \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} x e ^{x} \end{array} \right] _{0} ^{1} - \int _{0} ^{1} {e ^{x}} dx - \left[ \frac{1}{3} x ^{3} \right] _{0} ^{1} - \left[ \frac{1}{2} x ^{2} \right] _{0} ^{1}$ $\displaystyle = \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} x e ^{x} \end{array} \right] _{0} ^{1} - \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} e ^{x} \end{array} \right] _{0} ^{1} - \left[ \frac{1}{3} x ^{3} \right] _{0} ^{1} - \left[ \frac{1}{2} x ^{2} \right] _{0} ^{1}$ $\displaystyle = e - \left( e - 1 \right) - \frac{1}{3} - \frac{1}{2}$$\displaystyle = \frac{1}{6}$ (참) ㄷ. $\displaystyle \int _{0} ^{1} {\left\{ F \left( x \right) \right\} ^{2}} dx$$\displaystyle = \int _{0} ^{1} {\left( e ^{x} - x - 1 \right) ^{2}} dx$ $\displaystyle = \int _{0} ^{1} {\left\{ e ^{2 x} + x ^{2} + 1 - 2 x e ^{x} + 2 x - 2 e ^{x} \right\}} dx$$\displaystyle = \int _{0} ^{1} {e ^{2 x}} dx + \int _{0} ^{1} {x ^{2}} dx + \int _{0} ^{1} {1} dx$ $\displaystyle - 2 \int _{0} ^{1} {x e ^{x}} dx + \int _{0} ^{1} {2 x} dx - 2 \int _{0} ^{1} {e ^{x}} dx$$\displaystyle = \left[ \frac{e ^{2 x}}{2} \right] _{0} ^{1} + \left[ \frac{1}{3} x ^{3} \right] _{0} ^{1} + \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} x \end{array} \right] _{0} ^{1}$$\displaystyle - 2 \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} x e ^{x} \end{array} \right] _{0} ^{1} + 2 \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} e ^{x} \end{array} \right] _{0} ^{1} + \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} x ^{2} \end{array} \right] _{0} ^{1} - 2 \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} e ^{x} \end{array} \right] _{0} ^{1}$$\displaystyle = \frac{1}{2} \left( e ^{2} - 1 \right) + \frac{1}{3} + 1 - 2 \times 1 + 1 - 2 \left( e - 1 \right)$$\displaystyle = \frac{1}{2} e ^{2} - 2 e + \frac{11}{6}$ (참) 따라서 옳은 것은 ㄴ, ㄷ이다.