치환적분 함숫값 차

$\displaystyle f' ( x ) = \frac{3 x - 4}{\sqrt{x - 1}}$이므로 $\displaystyle f ( x ) = \int {\frac{3 x - 4}{\sqrt{x - 1}}} dx$에서 $x - 1 = t$로 치환하면 $\displaystyle 1 = \frac{dt}{dx}$이므로 $\displaystyle \int {\frac{3 x - 4}{\sqrt{x - 1}}} dx = \int {\frac{3 t - 1}{\sqrt{t}}} dt$$\displaystyle = \int {\left( 3 t ^{\frac{1}{2}} - t ^{- \frac{1}{2}} \right)} dt$ $\displaystyle = 2 t ^{\frac{3}{2}} - 2 t ^{\frac{1}{2}} + C$ ($C$는 적분상수) $\displaystyle = 2 ( x - 1 ) ^{\frac{3}{2}} - 2 ( x - 1 ) ^{\frac{1}{2}} + C$ $f ( 5 ) = 12 + C$, $f ( 2 ) = C$ 따라서 $f ( 5 ) - f ( 2 ) = 12$