절댓값 적분 함수

$f \left( x \right) = \left( x - 2 \right) e ^{x}$에서 $f' \left( x \right) = e ^{x} + \left( x - 2 \right) e ^{x} = \left( x - 1 \right) e ^{x}$ 이므로 $\displaystyle g \left( x \right) = \int _{0} ^{x} {} \left| f' \left( t \right) \right| dt = \int _{0} ^{x} {} \left| t - 1 \right| e ^{t} dt$ 이때 $u \left( t \right) = t - 1$, $v' \left( t \right) = e ^{t}$으로 놓으면 $u' \left( t \right) = 1$, $v \left( t \right) = e ^{t}$ (ⅰ) $0 \leq x < 1$일 때 $g \left( x \right)$$\displaystyle = \int _{0} ^{x} {} \left( 1 - t \right) e ^{t} dt$ $\displaystyle = - \int _{0} ^{x} {} \left( t - 1 \right) e ^{t} dt$ $\displaystyle = - \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \left( t - 1 \right) e ^{t} \end{array} \right] _{0} ^{x} + \int _{0} ^{x} {} e ^{t} dt$ $\displaystyle = - \left\{ \left( x - 1 \right) e ^{x} - \left( - 1 \right) \right\} + \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} e ^{t} \end{array} \right] _{0} ^{x}$ $= \left( 1 - x \right) e ^{x} - 1 + e ^{x} - 1$ $= \left( 2 - x \right) e ^{x} - 2$ (ⅱ) $x \geq 1$일 때 $g \left( x \right)$$\displaystyle = \int _{0} ^{1} {} \left( 1 - t \right) e ^{t} dt + \int _{1} ^{x} {} \left( t - 1 \right) e ^{t} dt$ $\displaystyle = - \int _{0} ^{1} {} \left( t - 1 \right) e ^{t} dt + \int _{1} ^{x} {} \left( t - 1 \right) e ^{t} dt$ $\displaystyle = - \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \left( t - 1 \right) e ^{t} \end{array} \right] _{0} ^{1} + \int _{0} ^{1} {} e ^{t} dt$ $\displaystyle + \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \left( t - 1 \right) e ^{t} \end{array} \right] _{1} ^{x} - \int _{1} ^{x} {} e ^{t} dt$ $\displaystyle = - \left\{ 0 - \left( - 1 \right) \right\} + \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} e ^{t} \end{array} \right] _{0} ^{1} + \left\{ \left( x - 1 \right) e ^{x} - 0 \right\} - \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} e ^{t} \end{array} \right] _{1} ^{x}$ $= - 1 + e - 1 + \left( x - 1 \right) e ^{x} - e ^{x} + e$ $= \left( x - 2 \right) e ^{x} + 2 e - 2$ (ⅰ), (ⅱ)에서 $\displaystyle g \left( x \right) = {\begin{cases} \left( 2 - x \right) e ^{x} - 2 & & \left( 0 \leq x < 1 \right) \\ \left( x - 2 \right) e ^{x} + 2 e - 2 & & \left( x \geq 1 \right) \end{cases}}$ 이므로 $\displaystyle \int _{0} ^{3} {} g \left( x \right) dx$$\displaystyle = \int _{0} ^{1} {} g \left( x \right) dx + \int _{1} ^{3} {} g \left( x \right) dx$ $\displaystyle = \int _{0} ^{1} {} \left\{ \left( 2 - x \right) e ^{x} - 2 \right\} dx$ $\displaystyle + \int _{1} ^{3} {} \left\{ \left( x - 2 \right) e ^{x} + 2 e - 2 \right\} dx$ 이때 $p \left( x \right) = x - 2$, $q' \left( x \right) = e ^{x}$으로 놓으면 $p' \left( x \right) = 1$, $q \left( x \right) = e ^{x}$ 이므로 $\displaystyle \int _{0} ^{1} {} \left\{ \left( 2 - x \right) e ^{x} - 2 \right\} dx$ $\displaystyle = \int _{0} ^{1} {} \left( 2 - x \right) e ^{x} dx + \int _{0} ^{1} {} \left( - 2 \right) dx$ $\displaystyle = - \int _{0} ^{1} {} \left( x - 2 \right) e ^{x} dx - \int _{0} ^{1} {} 2 dx$ $\displaystyle = - \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \left( x - 2 \right) e ^{x} \end{array} \right] _{0} ^{1} + \int _{0} ^{1} {} e ^{x} dx - \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} 2 x \end{array} \right] _{0} ^{1}$ $\displaystyle = - \left\{ \left( - e \right) - \left( - 2 \right) \right\} + \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} e ^{x} \end{array} \right] _{0} ^{1} - \left( 2 - 0 \right)$ $= e - 2 + \left( e - 1 \right) - 2$ $= 2 e - 5$ 이고 $\displaystyle \int _{1} ^{3} {} \left\{ \left( x - 2 \right) e ^{x} + 2 e - 2 \right\} dx$ $\displaystyle = \int _{1} ^{3} {} \left( x - 2 \right) e ^{x} dx + \int _{1} ^{3} {} \left( 2 e - 2 \right) dx$ $\displaystyle = \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \left( x - 2 \right) e ^{x} \end{array} \right] _{1} ^{3} - \int _{1} ^{3} {} e ^{x} dx + \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \left( 2 e - 2 \right) x \end{array} \right] _{1} ^{3}$ $\displaystyle = \left\{ e ^{3} - \left( - e \right) \right\} - \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} e ^{x} \end{array} \right] _{1} ^{3} + \left\{ 6 e - 6 - \left( 2 e - 2 \right) \right\}$ $= e ^{3} + e - \left( e ^{3} - e \right) + 4 e - 4$ $= 6 e - 4$ 따라서 $\displaystyle \int _{0} ^{3} {} g \left( x \right) dx$$= \left( 2 e - 5 \right) + \left( 6 e - 4 \right)$$= 8 e - 9$ 이므로 $a = 8$, $b = - 9$ $\therefore$ $a - b = 8 - \left( - 9 \right) = 17$