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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