치환 정적분

$y = e ^{x}$로 고치면, $\displaystyle g ( y ) = {\begin{cases} f ( \ln y ) & \left( 1 \leq y \leq e \right) \\ g \left( \frac{y}{e} \right) + 5 & ( e \leq y \leq e ^{2} ) \end{cases}}$ $\displaystyle \int _{1} ^{e ^{2}} {g ( y ) dy} = \int _{1} ^{e} {f ( \ln y )} dy + \int _{e} ^{e ^{2}} {\left\{ g \left( \frac{y}{e} \right) + 5 \right\} dy} = 6 e ^{2} + 4$ $\displaystyle \frac{y}{e} = t , dy = e dt$라 두면, $\displaystyle \int _{e} ^{e ^{2}} {\left\{ g \left( \frac{y}{e} \right) + 5 \right\} dy} = e \int _{1} ^{e} {g ( t )} dt + 5 e ^{2} - 5 e$ $\displaystyle \int _{1} ^{e} {g ( t ) dt = \int _{1} ^{e} {f ( \ln t ) dt}}$이므로 $\displaystyle \int _{1} ^{e ^{2}} {g ( y ) dy} = ( 1 + e ) \int _{1} ^{e} {f ( \ln y ) dy + 5 e ^{2} - 5 e = 6 e ^{2} + 4}$ ∴$\displaystyle \int _{1} ^{e} {f ( \ln y ) dy} = e + 4$ ∴$a = 1 , b = 4$ ∴$a ^{2} + b ^{2} = 17$