구분구적법 보기

ㄱ. $\displaystyle \lim\limits _{n \rightarrow \infty} {\sum\limits _{k = 1} ^{n}} S _{k}$ $\displaystyle = \lim\limits _{n \rightarrow \infty} {\sum\limits _{k = 1} ^{n}} \left\{ \frac{1}{2 n} \times f \left( \frac{k}{2 n} \right) \right\}$ $\displaystyle = \lim\limits _{n \rightarrow \infty} {\sum\limits _{k = 1} ^{n}} \left\{ \frac{1}{2 n} \times \left( \frac{k}{2 n} \right) ^{2} \right\}$ $\displaystyle = \lim\limits _{n \rightarrow \infty} {\sum\limits _{k = 1} ^{n}} \left\{ \frac{1}{2} \bullet \frac{1}{n} \times \left( \frac{1}{2} \bullet \frac{k}{n} \right) ^{2} \right\}$ $\displaystyle = \int _{0} ^{\frac{1}{2}} {x ^{2} dx}$ (참) ㄴ. $\displaystyle \sum\limits _{k = 1} ^{n} \left( S _{2 k} - S _{2 k - 1} \right) = \sum\limits _{k = 1} ^{n} \left\{ \frac{1}{2 n} f \left( \frac{2 k}{2 n} \right) - \frac{1}{2 n} f \left( \frac{2 k - 1}{2 n} \right) \right\}$ $\displaystyle = \frac{1}{2 n} \sum\limits _{k = 1} ^{n} \left\{ \left( \frac{2 k}{2 n} \right) ^{2} - \left( \frac{2 k - 1}{2 n} \right) ^{2} \right\}$ $\displaystyle = \frac{1}{8 n ^{3}} \sum\limits _{k = 1} ^{n} \left( 4 k - 1 \right)$ $\displaystyle = \frac{2 n ^{2} + n}{8 n ^{3}}$ $\displaystyle \therefore \lim\limits _{n \rightarrow \infty} {} \sum\limits _{k = 1} ^{n} \left( S _{2 k} - S _{2 k - 1} \right)$$\displaystyle = \lim\limits _{n \rightarrow \infty} {} \frac{2 n ^{2} + n}{8 n ^{3}} = 0$ (참) ㄷ. $\displaystyle \lim\limits _{n \rightarrow \infty} {\sum\limits _{k = 1} ^{n}} S _{2 k}$ $\displaystyle = \lim\limits _{n \rightarrow \infty} {\sum\limits _{k = 1} ^{n}} \left\{ \frac{1}{2 n} \times f \left( \frac{2 k}{2 n} \right) \right\}$$\displaystyle = \lim\limits _{n \rightarrow \infty} {\sum\limits _{k = 1} ^{n}} \left\{ \frac{1}{2 n} \times \left( \frac{k}{n} \right) ^{2} \right\}$ $\displaystyle = \frac{1}{2} \lim\limits _{n \rightarrow \infty} {\sum\limits _{k = 1} ^{n}} \left\{ \left( \frac{k}{n} \right) ^{2} \frac{1}{n} \right\}$ $\displaystyle = \frac{1}{2} \int _{0} ^{\frac{1}{2}} {x ^{2} dx}$ (참) 따라서 보기 중 옳은 것은 ㄱ, ㄴ, ㄷ이다.