자연계열 3번

[문제 $\displaystyle 3$]

(1) $\displaystyle g ( f ( t ) ) = \begin{cases}e^{- t} - \frac{1}{e^{4}} & ( t \le 4 ) \\ 0 & ( t > 4 )\end{cases}$

(2) $\displaystyle x \le 2$일 때

$\displaystyle h ( x ) = \int_{x}^{x + 2} g \left( e^{- t} - \frac{1}{e^{4}} \right) d t = \int_{x}^{x + 2} \left( e^{- t} - \frac{1}{e^{4}} \right) d t = \Biggl[ - e^{- t} - \frac{t}{e^{4}} \Biggr]_{x}^{x + 2} = - e^{- x - 2} + e^{- x} - \frac{2}{e^{4}}$

(3) $\displaystyle 2 < x \le 4$일 때 $\displaystyle g \left( e^{- t} - \frac{1}{e^{4}} \right)$가 $\displaystyle t > 4$일 때 $\displaystyle 0$이므로

$\displaystyle h ( x ) = \int_{x}^{x + 2} g \left( e^{- t} - \frac{1}{e^{4}} \right) d t = \int_{x}^{4} \left( e^{- t} - \frac{1}{e^{4}} \right) d t = \Biggl[ - e^{- t} - \frac{t}{e^{4}} \Biggr]_{x}^{4} = e^{- x} + \frac{x - 5}{e^{4}}$

(4) $\displaystyle x > 4$일 때 $\displaystyle g \left( e^{- t} - \frac{1}{e^{4}} \right)$가 모든 적분 구간에서 $\displaystyle 0$이므로 $\displaystyle h ( x ) = 0$

(5) 따라서 $\displaystyle h ( x ) = \begin{cases}- e^{- x - 2} + e^{- x} - \frac{2}{e^{4}} & ( x \le 2 ) \\ e^{- x} + \frac{x - 5}{e^{4}} & ( 2 < x \le 4 ) \\ 0 & ( x > 4 )\end{cases}$

(6) $\displaystyle h ( 1 ) = - \frac{1}{e^{3}} + \frac{1}{e} - \frac{2}{e^{4}}$

(7) $\displaystyle h ( 3 ) = \frac{1}{e^{3}} - \frac{2}{e^{4}}$

(8) $\displaystyle h ( 5 ) = 0$

(9) 따라서 $\displaystyle h ( 1 ) - h ( 3 ) - h ( 5 ) = \frac{1}{e} - \frac{2}{e^{3}}$