자연계열 I 4번

(1) $\displaystyle t=x-S_n$으로 치환하면 $\displaystyle \int_0^{a_n}f(x-S_n)\,dx=\int_{-S_n}^{-S_n+a_n}f(t)\,dt$이고, $\displaystyle -S_n+a_n=-S_{n-1}$이므로$\displaystyle \int_{-S_n}^{-S_n+a_n}f(t)\,dt=\int_{-S_n}^{-S_{n-1}}f(t)\,dt$$\displaystyle a_n$은 모두 0보다 크거나 같으므로 $\displaystyle -S_n\le0$이다. 조건 (i)에 의하여$\displaystyle \int_{-S_n}^{-S_{n-1}}f(t)\,dt=\int_{-S_n}^{-S_{n-1}}g(t)\,dt$따라서$\displaystyle \begin{gathered}\int_{-S_n}^{-S_{n-1}}g(x)\,dx=\int_0^{a_n}f(x-S_n)\,dx=1,\\\int_{-S_{n-1}}^{-S_n}g(x)\,dx=-1\end{gathered}$

(2) 정적분의 성질에 의하여$\displaystyle \int_0^{2023}f(x)\,dx=\sum_{n=1}^{2023}\int_{n-1}^n f(x)\,dx$$\displaystyle n$이 1보다 크거나 같은 정수일 때$\displaystyle \int_{n-1}^n f(x)\,dx=\int_{n-1}^n g(-S_n+(n-1-x)a_{n+1})\,dx$$\displaystyle t=-S_n+(n-1-x)a_{n+1}$라고 두면$\displaystyle \begin{aligned}\int_{n-1}^n f(x)\,dx&=\int_{n-1}^n g(-S_n+(n-1-x)a_{n+1})\,dx\\&=\int_{-S_n}^{-S_n-a_{n+1}}\left(-\frac{g(t)}{a_{n+1}}\right)\,dt\\&=\int_{-S_n-a_{n+1}}^{-S_n}\frac{g(t)}{a_{n+1}}\,dt\end{aligned}$그런데 $\displaystyle -S_n-a_{n+1}=-S_{n+1}$이고 $\displaystyle n\ge2$일 때 문항 (1)의 결과에 의하여 $\displaystyle \int_{-S_n}^{-S_{n-1}}g(t)\,dt=1$이므로, $\displaystyle n\ge1$일 때 $\displaystyle \int_{-S_{n+1}}^{-S_n}g(t)\,dt=1$이다.그러므로$\displaystyle \int_{-S_n-a_{n+1}}^{-S_n}\frac{g(t)}{a_{n+1}}\,dt=\frac1{a_{n+1}}\int_{-S_{n+1}}^{-S_n}g(t)\,dt=\frac1{a_{n+1}}$따라서 $\displaystyle \int_{n-1}^n f(x)\,dx=\frac1{a_{n+1}}=\frac1{n(n+1)}=\frac1n-\frac1{n+1}$이고,$\displaystyle \begin{aligned}\int_0^{2023}f(x)\,dx&=\sum_{n=1}^{2023}\int_{n-1}^n f(x)\,dx=\sum_{n=1}^{2023}\left(\frac1n-\frac1{n+1}\right)\\&=1-\frac1{2024}=\frac{2023}{2024}\end{aligned}$