자연계열 3번

[문제 3-1]

$\displaystyle f ( x ) = x^{2} - \int_{0}^{x} ( x - t ) f ^{\prime} ( t ) d t = x^{2} - x \int_{0}^{x} f ^{\prime} ( t ) d t + \int_{0}^{x} t f ^{\prime} ( t ) d t\cdots \cdots$(1)

에서 양변을 $\displaystyle x$에 대하여 미분하면

$\displaystyle f ^{\prime} ( x ) = 2 x - \int_{0}^{x} f ^{\prime} ( t ) d t - x f ^{\prime} ( x ) + x f ^{\prime} ( x ) = 2 x - \int_{0}^{x} f ^{\prime} ( t ) d t$

$\displaystyle = 2 x - \Biggl[ f ( t ) \Biggr]_{0}^{x} = 2 x - f ( x ) + f ( 0 ) = 2 x - f ( x )\cdots \cdots$(2) ( ( 1 ) 에$\displaystyle x = 0$을 대입하면 ,$\displaystyle f ( 0 ) = 0 )$

$\displaystyle \left( e^{x} f ( x ) \right) ^{\prime} = e^{x} f ( x ) + e^{x} f ^{\prime} ( x ) = e^{x} f ( x ) + e^{x} 2 x - e^{x} f ( x ) = 2 x e^{x}\cdots \cdots$(3)

$\displaystyle e^{x} f ( x ) = \int 2 x e^{x} d x = 2 x e^{x} - \int 2 e^{x} d x = 2 x e^{x} - 2 e^{x} + C\cdots \cdots$(4)

$\displaystyle f ( x ) = 2 x - 2 + C e^{- x}$, $\displaystyle f ( 0 ) = 0$이므로 $\displaystyle C = 2$

$\displaystyle f ( x ) = 2 x - 2 + 2 e^{- x}\cdots \cdots$(5)

[문제 3-2]

$\displaystyle \int_{- 1}^{1} 2 x f ( x ) d x = \int_{- 1}^{1} \left( 4 x e^{- x} + 4 x^{2} - 4 x \right) d x = 4 \int_{- 1}^{1} \left( x e^{- x} + x^{2} - x \right) d x\cdots \cdots$(6)

$\displaystyle = 4 \Biggl[ - x e^{- x} - e^{- x} + \frac{1}{3} x^{3} - \frac{1}{2} x^{2} \Biggr]_{- 1}^{1}\cdots \cdots$(7)

$\displaystyle = 4 \left[ \left( - e^{- 1} - e^{- 1} + \frac{1}{3} - \frac{1}{2} \right) - \left( e - e - \frac{1}{3} - \frac{1}{2} \right) \right]\cdots \cdots$(8)

$\displaystyle = 4 \left( \frac{2}{3} - \frac{2}{e} \right) = \frac{8}{3} - \frac{8}{e}\cdots \cdots$(9)