적분 방정식

[출제의도] 부분적분법을 이용하여 문제해결하기 $\displaystyle g ( x ) = \int _{0} ^{\ln x ^{2}} {} f ( e ^{t} ) dt = \int _{1} ^{x ^{2}} {} \frac{f ( s )}{s} ds$ $\displaystyle g' ( x ) = \frac{2 f ( x ^{2} )}{x}$ $\displaystyle \int _{1} ^{e} {} g ( x ) dx = \left[ x g ( x ) \right] _{1} ^{e} - \int _{1} ^{e} {} x g' ( x ) dx$ $\displaystyle = e \times g ( e ) - g ( 1 ) - \int _{1} ^{e} {} 2 f ( x ^{2} ) dx$ $\displaystyle = e \int _{1} ^{e ^{2}} {} \frac{f ( x )}{x} dx - 2 \left\{ 2 e \times f ( e ) + 4 \right\}$ $\displaystyle = 2 e \int _{1} ^{e} {} \frac{f ( t ^{2} )}{t} dt - 4 e \times f ( e ) - 8$ 조건 (가)의 양변을 $x$에 대하여 미분하면 $f ( x ^{2} ) = 2 f ( x ) + 2 x f' ( x )$이므로 $\displaystyle \int _{1} ^{e} {} g ( x ) dx$$\displaystyle = 2 e \int _{1} ^{e} {} \frac{f ( t ^{2} )}{t} dt - 4 e \times f ( e ) - 8$ $\displaystyle = 2 e \int _{1} ^{e} {} \frac{2 f ( t ) + 2 t f' ( t )}{t} dt - 4 e \times f ( e ) - 8$ $\displaystyle = 4 e \int _{1} ^{e} {} \frac{f ( t )}{t} dt + 4 e \int _{1} ^{e} {} f' ( t ) dt - 4 e \times f ( e ) - 8$ $\displaystyle = 4 e \left( 1 + \frac{1}{e ^{2}} - \frac{3}{e ^{4}} \right) + 4 e \left[ f ( t ) \right] _{1} ^{e} - 4 e \times f ( e ) - 8$ $\displaystyle = 4 e + \frac{4}{e} - \frac{12}{e ^{3}} + 4 e \times f ( e ) - 4 e \times f ( 1 ) - 4 e \times f ( e ) - 8$ $\displaystyle = 12 e + \frac{4}{e} - \frac{12}{e ^{3}} - 8$ $k _{1} = 12 , k _{2} = 4 , k _{3} = - 12 , k _{4} = - 8$ $\therefore \left| k _{1} \right| + \left| k _{2} \right| + \left| k _{3} \right| + \left| k _{4} \right| = 36$