절댓값 함수의 거듭제곱의 정적분의 합

$\displaystyle f ( x ) = \begin{cases} 1 - | x | & ( | x | \leq 1 ) \\ 0 & ( | x | > 1 ) \end{cases}$이므로 $f ( x ) = f ( - x )$가 성립한다. ∴ $\displaystyle \int _{-n} ^{n} \frac{\left\{ f ( x ) \right\} ^{n}}{n} dx$$\displaystyle = 2 \int _{-n} ^{0} \frac{\left\{ f ( x ) \right\} ^{n}}{n} dx$ $\displaystyle = 2 \int _{-1} ^{0} \frac{\left\{ f ( x ) \right\} ^{n}}{n} dx$ $\displaystyle = 2 \int _{-1} ^{0} \frac{( x + 1 ) ^{n}}{n} dx$ $\displaystyle = 2 \int _{0} ^{1} \frac{x ^{n}}{n} dx$ $\displaystyle = 2 \left[ \frac{x ^{n+1}}{n ( n + 1 )} \right] _{0} ^{1}$ $\displaystyle = \frac{2}{n ( n + 1 )}$ $\displaystyle = 2 \left( \frac{1}{n} - \frac{1}{n + 1} \right)$ ∴ $\displaystyle \sum\limits _{n = 1} ^{10} \int _{- n} ^{n} \frac{\left\{ f ( x ) \right\} ^{n}}{n} dx$ $\displaystyle = \sum\limits _{n=1} ^{10} 2 \left( \frac{1}{n} - \frac{1}{n + 1} \right)$ $\displaystyle = 2 \left\{ \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \cdots + \left( \frac{1}{10} - \frac{1}{11} \right) \right\}$ $\displaystyle = \frac{20}{11}$ [다른 풀이] $y = x ^{n}$와 $y = ( 1 - x ) ^{n}$은 직선 $\displaystyle x = \frac{1}{2}$에 대하여 대칭이므로 $\displaystyle \int _{0} ^{1} x ^{n} dx = \int _{0} ^{1} ( 1 - x ) ^{n} dx$가 성립한다. 따라서 앞의 풀이에서 $\displaystyle \int _{-n} ^{n} \frac{\left\{ f ( x ) \right\} ^{n}}{n} dx$$\displaystyle = 2 \int _{0} ^{n} \frac{\left\{ f ( x ) \right\} ^{n}}{n} dx$ $\displaystyle = 2 \int _{0} ^{1} \frac{\left\{ f ( x ) \right\} ^{n}}{n} dx$ $\displaystyle = 2 \int _{0} ^{1} \frac{( 1 - x ) ^{n}}{n} dx$ $\displaystyle = 2 \int _{0} ^{1} \frac{x ^{n}}{n} dx$ ⋮ 로 식을 변형하여 문제를 해결할 수 있다.