역함수 정적분

$g \left( 0 \right) = 0$, $g \left( 1 \right) = 1$이므로 $g ^{- 1} \left( 0 \right) = 0$, $g ^{- 1} \left( 1 \right) = 1$이고 $\displaystyle \left( g ^{- 1} \right)' \left( x \right) = \frac{1}{g' \left( g ^{- 1} \left( x \right) \right)}$이므로 $\displaystyle \int _{0} ^{1} {g ^{- 1} \left( x \right) dx}$$\displaystyle = \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} x \times g ^{- 1} \left( x \right) \end{array} \right] _{0} ^{1} - \int _{0} ^{1} {x \times \left( g ^{- 1} \right)' \left( x \right) dx}$ $\displaystyle = 1 - \int _{0} ^{1} {\frac{x}{g' \left( g ^{- 1} \left( x \right) \right)} dx}$ $x = g \left( t \right)$라 하면 $\displaystyle \frac{dx}{dt} = g' \left( t \right)$이고, $x = 0$일 때 $t = 0$, $x = 1$일 때 $t = 1$이므로 $\displaystyle \int _{0} ^{1} {\frac{x}{g' \left( g ^{- 1} \left( x \right) \right)} dx}$$\displaystyle = \int _{0} ^{1} {\frac{g \left( t \right)}{g' \left( g ^{- 1} \left( g \left( t \right) \right) \right)} \times g' \left( t \right) dt}$$\displaystyle = \int _{0} ^{1} {g \left( t \right) dt}$ $\displaystyle \int _{0} ^{1} {g ^{- 1} \left( x \right) dx} = 1 - \int _{0} ^{1} {g \left( x \right) dx}$ $\displaystyle \int _{0} ^{1} {g ^{- 1} \left( x \right) dx} = 2 \int _{0} ^{1} {f' \left( 2 x \right) \sin \pi x dx} + \frac{1}{4}$에서 $\displaystyle 1 - \int _{0} ^{1} {g \left( x \right) dx} = 2 \int _{0} ^{1} {\left\{ g \left( x \right) - x \right\} dx} + \frac{1}{4}$ $\displaystyle \int _{0} ^{1} {g \left( x \right) dx} = \frac{1}{4} + \frac{2}{3} \int _{0} ^{1} {x dx} = \frac{7}{12}$ $\displaystyle \int _{0} ^{1} {\left\{ f' \left( 2 x \right) \sin \pi x + x \right\} dx} = \frac{7}{12}$ $\displaystyle \int _{0} ^{1} {x dx} = \frac{1}{2}$이므로 $\displaystyle \int _{0} ^{1} {f' \left( 2 x \right) \sin \pi x dx} = \frac{1}{12}$ $\displaystyle \int _{0} ^{1} {f' \left( 2 x \right) \sin \pi x dx}$ $\displaystyle = \left[ \frac{1}{2} f \left( 2 x \right) \sin \pi x \right] _{0} ^{1} - \frac{1}{2} \int _{0} ^{1} {f \left( 2 x \right) \times \pi \cos \pi x dx}$ $\displaystyle = - \frac{\pi}{2} \int _{0} ^{1} {f \left( 2 x \right) \cos \pi x dx}$ $\displaystyle \int _{0} ^{1} {f \left( 2 x \right) \cos \pi x dx} = - \frac{1}{6 \pi}$ $2 x = s$라 하면 $\displaystyle \frac{ds}{dx} = 2$이고 $x = 0$일 때 $s = 0$, $x = 1$일 때 $s = 2$이므로 $\displaystyle \int _{0} ^{1} {f \left( 2 x \right) \cos \pi x dx}$$\displaystyle = \frac{1}{2} \int _{0} ^{2} {f \left( s \right) \cos \frac{\pi}{2} s ds}$ $\displaystyle \therefore \int _{0} ^{2} {f \left( x \right) \cos \frac{\pi}{2} x dx} = - \frac{1}{3 \pi}$