+40점 · k \displaystyle k k ∫ 0 k π ∣ sin x 2 n ∣ d x \displaystyle \int_{0}^{k\pi}\left|\sin\frac{x}{2n}\right|dx ∫ 0 k π sin 2 n x d x 먼저 자연수 k = 1 , 2 , ⋯ , 3 n \displaystyle k=1,\,2,\,\cdots,\,3n k = 1 , 2 , ⋯ , 3 n ∫ 0 k π ∣ sin x 2 n ∣ d x \displaystyle \int_{0}^{k\pi}\left|\sin\frac{x}{2n}\right|dx ∫ 0 k π sin 2 n x d x 을 구하자.
(i) k = 1 , 2 , ⋯ , 2 n \displaystyle k=1,\,2,\,\cdots,\,2n k = 1 , 2 , ⋯ , 2 n 0 ≤ x ≤ 2 n π \displaystyle 0\le x\le2n\pi 0 ≤ x ≤ 2 nπ sin x 2 n ≥ 0 \displaystyle \sin\dfrac{x}{2n}\ge0 sin 2 n x ≥ 0 ∫ 0 k π ∣ sin x 2 n ∣ d x = ∫ 0 k π sin x 2 n d x = [ − 2 n cos x 2 n ] 0 k π = 2 n ( 1 − cos k π 2 n ) \displaystyle \int_{0}^{k\pi}\left|\sin\frac{x}{2n}\right|dx=\int_{0}^{k\pi}\sin\frac{x}{2n}\,dx=\Biggl[-2n\cos\frac{x}{2n}\Biggr]_{0}^{k\pi}=2n\left(1-\cos\frac{k\pi}{2n}\right) ∫ 0 k π sin 2 n x d x = ∫ 0 k π sin 2 n x d x = [ − 2 n cos 2 n x ] 0 k π = 2 n ( 1 − cos 2 n k π ) 이다.
(ii) k = 2 n + 1 , 2 n + 2 , ⋯ , 3 n \displaystyle k=2n+1,\,2n+2,\,\cdots,\,3n k = 2 n + 1 , 2 n + 2 , ⋯ , 3 n 0 ≤ x < 2 n π \displaystyle 0\le x<2n\pi 0 ≤ x < 2 nπ sin x 2 n ≥ 0 \displaystyle \sin\dfrac{x}{2n}\ge0 sin 2 n x ≥ 0 2 n π ≤ x ≤ 3 n π \displaystyle 2n\pi\le x\le3n\pi 2 nπ ≤ x ≤ 3 nπ sin x 2 n ≤ 0 \displaystyle \sin\dfrac{x}{2n}\le0 sin 2 n x ≤ 0 ∫ 0 k π ∣ sin x 2 n ∣ d x = ∫ 0 2 n π sin x 2 n d x + ∫ 2 n π k π ( − sin x 2 n ) d x = [ − 2 n cos x 2 n ] 0 2 n π + [ 2 n cos x 2 n ] 2 n π k π = 2 n ( 3 + cos k π 2 n ) \displaystyle \int_{0}^{k\pi}\left|\sin\frac{x}{2n}\right|dx=\int_{0}^{2n\pi}\sin\frac{x}{2n}\,dx+\int_{2n\pi}^{k\pi}\left(-\sin\frac{x}{2n}\right)dx=\Biggl[-2n\cos\frac{x}{2n}\Biggr]_{0}^{2n\pi}+\Biggl[2n\cos\frac{x}{2n}\Biggr]_{2n\pi}^{k\pi}=2n\left(3+\cos\frac{k\pi}{2n}\right) ∫ 0 k π sin 2 n x d x = ∫ 0 2 nπ sin 2 n x d x + ∫ 2 nπ k π ( − sin 2 n x ) d x = [ − 2 n cos 2 n x ] 0 2 nπ + [ 2 n cos 2 n x ] 2 nπ k π = 2 n ( 3 + cos 2 n k π ) 이다.
+35점 · T n \displaystyle T_{n} T n 위 (i)과 (ii)에 의해 T n = ∑ k = 1 n 1 n 2 { ∫ 0 k π ∣ sin x 2 n ∣ d x + ∫ 0 ( n + k ) π ∣ sin x 2 n ∣ d x + ∫ 0 ( 2 n + k ) π ∣ sin x 2 n ∣ d x } = ∑ k = 1 n 1 n [ 2 ( 1 − cos k π 2 n ) + 2 { 1 − cos ( n + k ) π 2 n } + 2 { 3 + cos ( 2 n + k ) π 2 n } ] = 10 + 1 n ∑ k = 1 n ( − 4 cos k π 2 n + 2 sin k π 2 n ) \displaystyle \begin{aligned} T_{n} &=\sum_{k=1}^{n}\frac{1}{n^{2}}\left\{\int_{0}^{k\pi}\left|\sin\frac{x}{2n}\right|dx+\int_{0}^{(n+k)\pi}\left|\sin\frac{x}{2n}\right|dx+\int_{0}^{(2n+k)\pi}\left|\sin\frac{x}{2n}\right|dx\right\} \\ &=\sum_{k=1}^{n}\frac{1}{n}\left[2\left(1-\cos\frac{k\pi}{2n}\right)+2\left\{1-\cos\frac{(n+k)\pi}{2n}\right\}+2\left\{3+\cos\frac{(2n+k)\pi}{2n}\right\}\right] \\ &=10+\frac{1}{n}\sum_{k=1}^{n}\left(-4\cos\frac{k\pi}{2n}+2\sin\frac{k\pi}{2n}\right) \end{aligned} T n = k = 1 ∑ n n 2 1 { ∫ 0 k π sin 2 n x d x + ∫ 0 ( n + k ) π sin 2 n x d x + ∫ 0 ( 2 n + k ) π sin 2 n x d x } = k = 1 ∑ n n 1 [ 2 ( 1 − cos 2 n k π ) + 2 { 1 − cos 2 n ( n + k ) π } + 2 { 3 + cos 2 n ( 2 n + k ) π } ] = 10 + n 1 k = 1 ∑ n ( − 4 cos 2 n k π + 2 sin 2 n k π ) 이다.
+40점 · 정적분과 급수의 합 사이의 관계로부터 lim n → ∞ T n \displaystyle \lim_{n\to\infty}T_{n} n → ∞ lim T n 따라서 정적분과 급수의 합 사이의 관계로부터 lim n → ∞ T n = 10 + lim n → ∞ 1 n ∑ k = 1 n ( − 4 cos k π 2 n + 2 sin k π 2 n ) = 10 + ∫ 0 1 ( − 4 cos π 2 x + 2 sin π 2 x ) d x = 10 + [ − 8 π sin π 2 x − 4 π cos π 2 x ] 0 1 = 10 − 4 π \displaystyle \begin{aligned} \lim_{n\to\infty}T_{n} &=10+\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\left(-4\cos\frac{k\pi}{2n}+2\sin\frac{k\pi}{2n}\right) \\ &=10+\int_{0}^{1}\left(-4\cos\frac{\pi}{2}x+2\sin\frac{\pi}{2}x\right)dx \\ &=10+\Biggl[-\frac{8}{\pi}\sin\frac{\pi}{2}x-\frac{4}{\pi}\cos\frac{\pi}{2}x\Biggr]_{0}^{1}=10-\frac{4}{\pi} \end{aligned} n → ∞ lim T n = 10 + n → ∞ lim n 1 k = 1 ∑ n ( − 4 cos 2 n k π + 2 sin 2 n k π ) = 10 + ∫ 0 1 ( − 4 cos 2 π x + 2 sin 2 π x ) d x = 10 + [ − π 8 sin 2 π x − π 4 cos 2 π x ] 0 1 = 10 − π 4 이다.
수리논술, 이게 다다.