구간별 함수 적분 극한

[출제의도] 주어진 조건을 만족시키는 함수에 대하여 수열 $\left\{ a _{n} \right\}$을 정하고 극한값을 이용하여 미지수의 값을 구할 수 있는가? 직선 $y = x$와 $x$축 및 직선 $x = n$으로 둘러싸인 삼각형의 넓이가 $\displaystyle \frac{n ^{2}}{2}$이다. $\displaystyle 2 \lim\limits _{n \rightarrow \infty} {\left( a _{n} - \frac{n ^{2}}{2} \right)} = \frac{241}{768}$이므로 $F \left( x \right)$$= h \left( x \right) - x$$\displaystyle = {\begin{cases} g \left( x \right) - x & \left( 0 \leq x < 5 , x \geq k \right) \\ x - g \left( x \right) & \left( 5 \leq x < k \right) \end{cases}}$ 로 놓으면 $\displaystyle 2 \lim\limits _{n \rightarrow \infty} {} \int _{0} ^{n} {F \left( x \right) dx} = \frac{241}{768}$이다. $n \leq x < n + 1$일 때 $g \left( x \right) - x$$\displaystyle = \frac{1}{2 ^{n}} \left\{ f \left( x - n \right) - \left( x - n \right) \right\}$ $\displaystyle = \frac{1}{2 ^{n}} \left\{ \frac{3}{2} \left( x - n \right) - \frac{1}{2} \left( x - n \right) ^{2} - \left( x - n \right) \right\}$ $\displaystyle = \frac{1}{2 ^{n}} \left\{ \frac{1}{2} \left( x - n \right) - \frac{1}{2} \left( x - n \right) ^{2} \right\}$ $\displaystyle = \frac{1}{2 ^{n + 1}} \left( x - n \right) \left( 1 + n - x \right)$ $\displaystyle = - \frac{1}{2 ^{n + 1}} \left( x - n \right) \left( x - \left( n + 1 \right) \right)$ 이므로 $\displaystyle \int _{n} ^{n + 1} {\left\{ g \left( x \right) - x \right\} dx}$ $\displaystyle = \int _{n} ^{n + 1} {\left\{ - \frac{1}{2 ^{n + 1}} \left( x - n \right) \left( x - \left( n + 1 \right) \right) \right\} dx}$ $\displaystyle = \int _{0} ^{1} {\left\{ - \frac{1}{2 ^{n + 1}} x \left( x - 1 \right) \right\} dx}$ $\displaystyle = - \frac{1}{2 ^{n + 1}} \int _{0} ^{1} {x \left( x - 1 \right) dx}$ $\displaystyle = \frac{1}{2 ^{n + 1}} \times \frac{1}{6}$ $\cdots \cdots$ ㉠ 한편, $\displaystyle \int _{0} ^{n} {F \left( x \right) dx}$ $\displaystyle = \int _{0} ^{1} {\left( x - g \left( x \right) \right) dx} + \cdots + \int _{4} ^{5} {\left( x - g \left( x \right) \right) dx}$ $\displaystyle + \int _{5} ^{6} {\left( x - g \left( x \right) \right) dx} + \cdots + \int _{k - 1} ^{k} {\left( x - g \left( x \right) \right) dx}$ $\displaystyle + \int _{k} ^{k + 1} {\left( x - g \left( x \right) \right) dx} + \cdots + \int _{n - 1} ^{n} {\left( x - g \left( x \right) \right) dx}$ 이때, ㉠은 $n = 0$일 때도 성립하므로 ㉠에서 $\displaystyle 2 \int _{0} ^{n} {F \left( x \right) dx}$ $\displaystyle = \frac{1}{3} \times \left\{ \frac{1}{2} + \left( \frac{1}{2} \right) ^{2} + \cdots + \left( \frac{1}{2} \right) ^{5} \right\}$$\displaystyle - \frac{1}{3} \times \left\{ \left( \frac{1}{2} \right) ^{6} + \cdots + \left( \frac{1}{2} \right) ^{k} \right\}$ $\displaystyle + \frac{1}{3} \times \left\{ \left( \frac{1}{2} \right) ^{k + 1} + \cdots + \left( \frac{1}{2} \right) ^{n} \right\}$ $\displaystyle = \frac{1}{3} \times \left\{ \frac{1}{2} + \left( \frac{1}{2} \right) ^{2} + \cdots + \left( \frac{1}{2} \right) ^{n} \right\}$$\displaystyle - \frac{2}{3} \times \left\{ \left( \frac{1}{2} \right) ^{6} + \cdots + \left( \frac{1}{2} \right) ^{k} \right\}$ $\displaystyle = \frac{1}{3} \times \frac{\frac{1}{2} \times \left\{ 1 - \left( \frac{1}{2} \right) ^{n} \right\}}{1 - \frac{1}{2}}$$\displaystyle - \frac{2}{3} \times \frac{\left( \frac{1}{2} \right) ^{6} \times \left\{ 1 - \left( \frac{1}{2} \right) ^{k - 5} \right\}}{1 - \frac{1}{2}}$ $\displaystyle = \frac{1}{3} \times \left\{ 1 - \left( \frac{1}{2} \right) ^{n} \right\} - \frac{2}{3} \times \left( \frac{1}{2} \right) ^{5} \times \left\{ 1 - \left( \frac{1}{2} \right) ^{k - 5} \right\}$ $\displaystyle = \frac{1}{3} \times \left\{ 1 - \left( \frac{1}{2} \right) ^{n} \right\} - \frac{1}{48} \times \left\{ 1 - \left( \frac{1}{2} \right) ^{k - 5} \right\}$ 따라서 $\displaystyle \lim\limits _{n \rightarrow \infty} {\left( \frac{1}{2} \right) ^{n}} = 0$이므로 $\displaystyle 2 \lim\limits _{n \rightarrow \infty} {} \int _{0} ^{n} {F \left( x \right) dx}$$\displaystyle = \frac{1}{3} - \frac{1}{48} \times \left\{ 1 - \left( \frac{1}{2} \right) ^{k - 5} \right\}$ $\displaystyle = \frac{15}{48} + \frac{1}{48} \times \left( \frac{1}{2} \right) ^{k - 5}$ 즉, $\displaystyle \frac{15}{48} + \frac{1}{48} \times \left( \frac{1}{2} \right) ^{k - 5} = \frac{241}{768}$에서 $\displaystyle \left( \frac{1}{2} \right) ^{k - 5} = \frac{1}{16}$이므로 $k - 5 = 4$에서 $k = 9$