정적분으로 정의된 함수의 계수

$f ( x ) = ( x - 1 ) ^{4} ( x + 1 )$ 에서 $f' ( x ) = 4 ( x - 1 ) ^{3} ( x + 1 ) + ( x - 1 ) ^{4} = ( x - 1 ) ^{3} ( 5 x + 3 )$ $f'' ( x ) = 3 ( x - 1 ) ^{2} ( 5 x + 3 ) + 5 ( x - 1 ) ^{3}$$= ( x - 1 ) ^{2} ( 20 x + 4 )$ 이차함수 $g ( x ) = ax ^{2} + bx + c$ 라하고 $\displaystyle f ( x ) = g ( x ) + \int _{0} ^{x} ( x - t ) ^{2} h ( t ) dt$ $\displaystyle = g ( x ) + x ^{2} \int _{0} ^{x} h ( t ) dt - 2 x \int _{0} ^{x} t h ( t ) dt + \int _{0} ^{x} t ^{2} h ( t ) dt$ $x = 0$을 대입하면, $f ( 0 ) = g ( 0 )$, $\therefore$ $c = 1$ 이제 양변을 미분하면 $( x - 1 ) ^{3} ( 5 x + 3 )$ $= 2 ax + b$$\displaystyle + 2 x \int _{0} ^{x} h ( t ) dt$$+ x ^{2} h ( x )$$\displaystyle - 2 \int _{0} ^{x} t h ( t ) dt$$- 2 x ^{2} h ( x )$$+ x ^{2} h ( x )$ $\displaystyle = 2 ax + b + 2 x \int _{0} ^{x} h ( t ) dt - 2 \int _{0} ^{x} t h ( t ) dt$ $x = 0$을 대입하면, $- 3 = b$ 또 양변을 미분하면 $( x - 1 ) ^{2} ( 20 x + 4 )$ $\displaystyle = 2 a + 2 \int _{0} ^{x} {} h ( t ) dt + 2 xh ( x ) - 2 xh ( x )$ $\displaystyle = 2 a + 2 \int _{0} ^{x} h ( t ) dt$ $x = 0$을 대입하면, $4 = 2 a$, $\therefore$ $a = 2$ 또 양변을 미분하면 $12 ( x - 1 ) ( 5 x - 1 ) =$$2 h ( x )$ 그러므로 $g ( x ) = 2 x ^{2} - 3 x + 1$, $h ( x ) = 6 ( x - 1 ) ( 5 x - 1 )$ $\therefore$ $g ( 2 ) + h ( 2 ) = 3 + 54 = 57$