주기함수 정적분 극한

[출제의도] 수열의 극한값 추론하기 $\displaystyle \int _{0} ^{2} {f \left( x \right) dx} = \int _{0} ^{1} {2 \left( 1 - x \right) dx} + \int _{1} ^{2} {2 \left( x - 1 \right) dx} = 2$ $\displaystyle \int _{0} ^{2} {f \left( x \right) dx} = \int _{2} ^{4} {f \left( x \right) dx} = \cdots = \int _{2 \left( k - 1 \right)} ^{2 k} {f \left( x \right) dx} = 2$ 이므로 $\displaystyle \int _{0} ^{2 k} {f \left( x \right) dx} = 2 k$($k$는 자연수) $\displaystyle g \left( k \right) = \int _{2 k} ^{4 k} {f \left( x \right) dx}$ $\displaystyle = \int _{0} ^{4 k} {f \left( x \right) dx} - \int _{0} ^{2 k} {f \left( x \right) dx} = 4 k - 2 k = 2 k$ $\therefore$ $\displaystyle \lim\limits _{n \rightarrow \infty} {\frac{1}{n ^{2}} \sum\limits _{k = 1} ^{n} g \left( k \right) =} \lim\limits _{n \rightarrow \infty} {\frac{1}{n ^{2}} \sum\limits _{k = 1} ^{n} 2 k}$ $\displaystyle = \lim\limits _{n \rightarrow \infty} {\frac{n \left( n + 1 \right)}{n ^{2}}} = 1$