절댓값 적분 함수

$- 1 \leq x \leq 0$일 때, $\left| f ( x ) \sin x \right|$$= f ( x ) \sin x$ $0 \leq x \leq 1$일 때. $\left| f ( x ) \sin x \right| =$$- f ( x ) \sin x$ $\therefore$ $\displaystyle \int _{- 1} ^{0} {\left| f ( x ) \sin x \right| dx = 2}$, $\displaystyle \int _{0} ^{1} {\left| f ( x ) \sin x \right| dx = 3}$ $g ( x )$에 $x = 0$을 대입하면 $\displaystyle g ( 0 ) = \int _{- 1} ^{0} {\left| f ( x ) \sin x \right|} dx = 2$ $\therefore$ $g ( 0 ) = 2$ $g ( x )$에 $x = 1$을 대입하면 $g ( 1 )$$\displaystyle = \int _{- 1} ^{1} {\left| f ( t ) \sin t \right| dt}$ $\displaystyle = \int _{- 1} ^{0} {\left| f ( x ) \sin x \right|} dx + \int _{0} ^{1} {\left| f ( x ) \sin x \right|} dx$$= 2 + 3$ $\therefore$ $g ( 1 ) = 2 + 3 = 5$ $g ( x )$에 $x = - 1$을 대입하면 $\displaystyle g ( - 1 ) = \int _{- 1} ^{- 1} {\left| f ( x ) \sin x \right|} dx = 0$ $\therefore$ $g ( - 1 ) = 0$ 이때 $g' ( x ) = \left| f ( x ) \sin x \right|$가 $- 1 \leq x \leq 0$일 때, $\left| f ( x ) \sin x \right|$$= f ( x ) \sin x$ $0 \leq x \leq 1$일 때. $\left| f ( x ) \sin x \right| =$$- f ( x ) \sin x$이므로 $\displaystyle \int _{- 1} ^{1} {f ( - x ) g ( - x ) \sin xdx}$에서 $- x = t$로 치환하면 $dx = - dt$, $- 1 \leq t \leq 1$ $\therefore$ $\displaystyle \int _{- 1} ^{1} {f ( - x ) g ( - x ) \sin x dx}$ $\displaystyle = \int _{1} ^{- 1} {f ( t ) g ( t ) \sin ( - t ) ( - 1 ) dt}$ $\displaystyle = - \int _{- 1} ^{1} {f ( t ) g ( t ) \sin t dt}$ $\displaystyle = - \int _{- 1} ^{0} {g' ( t ) g ( t )} dt + \int _{0} ^{1} {g' ( t ) g ( t )} dt$ $\displaystyle = - \left[ \frac{1}{2} \left\{ g ( t ) \right\} ^{2} \right] _{- 1} ^{0} + \left[ \frac{1}{2} \left\{ g ( t ) \right\} ^{2} \right] _{0} ^{1}$ $\displaystyle = \frac{1}{2} \left[ \left\{ g ( 1 ) \right\} ^{2} - \left\{ g ( 0 ) \right\} ^{2} \right] - \frac{1}{2} \left[ \left\{ g ( 0 ) \right\} ^{2} - \left\{ g ( - 1 ) \right\} ^{2} \right]$ $\displaystyle = \frac{1}{2} ( 5 ^{2} - 2 ^{2} ) - \frac{1}{2} ( 2 ^{2} - 0 ^{2} )$ $\displaystyle = \frac{17}{2}$