도함수 절댓값 적분

[출제의도] 정적분으로 표시된 함수의 미분을 이용하여 문제해결하기 $f' ( x ) = 3 ( x + 1 ) ( x - 1 ) = 3 x ^{2} - 3$ $f ( x ) = x ^{3} - 3 x + C$ (단, $C$는 적분상수)이고, (가), (나)에 의하여 $\alpha < - 1 < \beta < 1 < \gamma$ $\displaystyle g ( \gamma ) - g ( \alpha ) = \int _{- 1} ^{\gamma} {\left| f' ( x ) \right|} dx - \int _{- 1} ^{\alpha} {\left| f' ( x ) \right| dx}$ $\displaystyle = \int _{- 1} ^{\gamma} {\left| f' ( x ) \right|} dx + \int _{\alpha} ^{- 1} {\left| f' ( x ) \right| dx}$ $\displaystyle = \int _{\alpha} ^{\gamma} {\left| f' ( x ) \right|} dx$ $\displaystyle = \int _{\alpha} ^{- 1} {f' ( x )} dx - \int _{- 1} ^{1} {f' ( x )} dx + \int _{1} ^{\gamma} {f' ( x )} dx$ $= \left[ f ( x ) \right] _{\alpha} ^{- 1} - \left[ f ( x ) \right] _{- 1} ^{1} + \left[ f ( x ) \right] _{1} ^{\gamma}$ $= f ( - 1 ) - f ( \alpha ) - f ( 1 ) + f ( - 1 ) + f ( \gamma ) - f ( 1 )$ $= 2 \left\{ f ( - 1 ) - f ( 1 ) \right\}$ $= 8$