치환적분과 부분적분

ㄱ. (참) $\displaystyle \int _{0} ^{1} {} \left\{ f' ( x ) g ( 1 - x ) - g' ( x ) f ( 1 - x ) \right\} dx = k \cdots \cdots \text{㉠}$에서 $1 - x = t$로 놓으면 $- dx = dt$이고, $x = 0$이면 $t = 1$, $x = 1$이면 $t = 0$이므로 ㉠에서 $\displaystyle - \int _{1} ^{0} \left\{ f' ( 1 - t ) g ( t ) - g' ( 1 - t ) f ( t ) \right\} dt = k$ $\displaystyle \int _{1} ^{0} \left\{ g' ( 1 - t ) f ( t ) - f' ( 1 - t ) g ( t ) \right\} dt = k$ $\displaystyle \int _{0} ^{1} \left\{ f ( x ) g' ( 1 - x ) - g ( x ) f' ( 1 - x ) \right\} dx$$= - k$ ㄴ. (참) $\displaystyle \int _{0} ^{1} \left\{ f' ( x ) g ( 1 - x ) - g' ( x ) f ( 1 - x ) \right\} dx$ $\displaystyle = \left[ f ( x ) g ( 1 - x ) \right] _{0} ^{1} + \int _{0} ^{1} f ( x ) g' ( 1 - x ) dx$ $\displaystyle - {\left\{ \left[ g ( x ) f ( 1 - x ) \right] _{0} ^{1} + \int _{0} ^{1} g ( x ) f' ( 1 - x ) dx \right\}}$ $= 2 \left\{ f ( 1 ) g ( 0 ) - f ( 0 ) g ( 1 ) \right\}$ $\displaystyle + \int _{0} ^{1} \left\{ f ( x ) g' ( 1 - x ) - g ( x ) f' ( 1 - x ) \right\} dx$ $= 2 \left\{ f ( 1 ) g ( 0 ) - f ( 0 ) g ( 1 ) \right\} - k$ $\therefore 2 k = 2 \left\{ f ( 1 ) g ( 0 ) - f ( 0 ) g ( 1 ) \right\} \cdots \cdots \text{㉡}$ 이때, $f ( 0 ) = f ( 1 )$, $g ( 0 ) = g ( 1 )$이므로 $k = 0$ ㄷ. (참) $g ( 0 ) = \sin 0 = 0$, $g ( 1 ) = \sin \pi = 0$이므로 ㉡에서 $2 k = 0$ ∴ $k = 0$ 따라서, ㄱ, ㄴ, ㄷ 모두 옳다.