Bài 112327

Question

Tính các giới hạn sau :
a) $ I = \mathop {\lim }\limits_{ } \left ( \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}\right )$
b) $ J = \mathop {\lim }\limits_{ } \frac{1^6+2^6+...+n^6}{n^7}$
c) $ K = \mathop {\lim }\limits_{ } \frac{1}{2n+1}\left ( \sin \frac{\pi}{n}+\sin \frac{2\pi}{n}+...+ \sin \frac{n-1}{n}\pi\right ).$

score 0 · Answer 1

a) Xét $f(x) = \frac{1}{x+1}, x \in [0;1]$
   * Phân hoạch đoạn $[0;1]$ :
            $ x_0 = 0 < x_1 = \frac{1}{n} < x_2=\frac{2}{n}<...<x_i = \frac{i}{n}<...< x_n = 1$
   * Chọn $ \xi _i = x_i =\frac{i}{n}, i = \overline{1,n}$
   * Tổng Riemann của $f$ :
                $\sum\limits_{i = 1}^{n}f(\xi _i)(x_i - x_{x-1}) = \sum\limits_{i = 1}^{n} \frac{1}{n+1} = \mathop {\lim }\limits_{n \to +\infty} \sum\limits_{i = 1}^{n}f(\xi _i)(x_i - x_{i-1} )$
                         $=\int\limits_{0}^{1} f(x)dx = \int\limits_{0}^{1}\frac{dx}{x+1} = \ln 2$
b) Xét $f(x) = x^6, x \in [0;1]$
    * Phân hoạch đoạn $[0;1]$ :
            $ x_0 = 0 < x_1 = \frac{1}{n} < x_2=\frac{2}{n}<...<x_i = \frac{i}{n}<...< x_n = 1$
   * Chọn $ \xi _i = x_i =\frac{i}{n}, i = \overline{1,n}$
   * Tổng Riemann của $f$ :
$\sum\limits_{i = 1}^{n}f(\xi _i)(x_i - x_{x-1}) = \sum\limits_{i = 1}^{n}\left ( \frac{i}{n} \right )^6.\frac{i}{n} =   \sum\limits_{i = 1}^{n}\frac{i^6}{n^7}$
$ \Rightarrow J = \mathop {\lim }\limits_{n \to +\infty} \sum\limits_{i = 1}^{n}\frac{i^6}{n^7} = \mathop {\lim }\limits_{n \to +\infty} \sum\limits_{i = 1}^{n}f(\xi _i)(x_i -x_{i-1} = \int\limits_{0}^{1}f(x)dx = \int\limits_{0}^{1}x^6dx$
     $ = \frac{x^7}{7} \left| \begin{array}{l}
1\\
0
\end{array} \right. = \frac{1}{7}.$
c) Xét $f(x) = \sin x, x \in [0;\pi ]$
    * Phân hoạch đoạn $[0;1]$ :
         $ x_0 <x_1 = \frac{\pi }{n} <x_2 = \frac{2\pi}{n}<...< x_i = \frac{i\pi}{n}<...< x_n = \pi $
    * Chọn $\xi _i = x_i = \frac{i\pi}{n}, i = \overline{1,n}$
    * Tổng Riemann của $f$ :
$\sum\limits_{i = 1}^{n}f(\xi_i)(x_i - x_{i-1}) = \sum\limits_{i = 1}^{n}\frac{\pi}{n}\sin \frac{i\pi}{n} = \frac{\pi}{n}\sum\limits_{i = 1}^{n-1}\sin \frac{i\pi}{n}$
$\Rightarrow \mathop {\lim }\limits_{n \to +\infty} \frac{\pi}{n}\sum\limits_{i = 1}^{n-2}\sin \frac{i\pi}{n} = \mathop {\lim }\limits_{n \to +\infty} \sum\limits_{i = 1}^{n}f(\xi _i)(x_i-x_{i-1}) = \int\limits_{0}^{\pi }f(x)dx = \int\limits_{0}^{\pi }\sin xdx$
                $ = - \cos \left| \begin{array}{l}
\pi \\
2
\end{array} \right. = 2$
$\Rightarrow K = \frac{1}{\pi }\mathop {\lim }\limits_{n \to +\infty} \left [ \frac{n}{2n+1}\left ( \frac{\pi}{n}\sum\limits_{i = 1}^{n-1}\sin \frac{i\pi}{n}   \right )\right ] = \frac{1}{\pi }\mathop {\lim }\limits_{n \to +\infty} \frac{n}{2n+1}. \mathop {\lim }\limits_{n \to +\infty} \frac{1}{n}\sum\limits_{i = 1}^{n-1}\sin \frac{i\pi}{n}$
                $ = \frac{1}{2\pi }.2 = \frac{1}{\pi}.$

Bài 112327

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Bài 112327

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