Bài 112324

Question

Cho $f(x) = \frac{1}{4}x^2 - \frac{1}{2}\ln x - \frac{1}{4}, (x >0).$ Tính $ \mathop {\lim }\limits_{ }\left [ \frac{1}{n}\sum\limits_{i = 1}^{n}f \left ( \frac{i}{n} \right )\right ].$

score 0 · Answer 1

Ta có : $ f'(x) = \frac{x}{2} - \frac{1}{2x} = \frac{x^2-1}{2x} <0 \forall x \in (0;1)$
$ \Rightarrow f$ giảm trên $(0;1)$
Khi đó : với $ x \in \left [ \frac{i}{n};\frac{i+1}{n}\right ], i \in [1;2;...;n-1], 2 \leq n \in Z$, ta có :
         $ f \left ( \frac{i+1}{n} \right ) < f(x) < f \left (\frac{i}{n} \right )$
$\Rightarrow \int\limits_{\frac{i}{n} }^{\frac{i+1}{n} } f \left ( \frac{i+1}{n} \right )dx <   \int\limits_{\frac{i}{n} }^{\frac{i+1}{n} }f(x)dx < \int\limits_{\frac{i}{n} }^{\frac{i+1}{n} }f \left ( \frac{i}{n} \right )dx.$
$\Rightarrow \frac{1}{n}f \left ( \frac{i+1}{n} \right ) < \int\limits_{\frac{i}{n} }^{\frac{i+1}{n} }f(x)dx < \frac{1}{n}f \left ( \frac{i}{n} \right )$
$\Rightarrow \frac{1}{n}\sum\limits_{i = 1}^{n - 1} \int\limits_{\frac{i}{n} }^{\frac{i+1}{n} }f(x)dx < \frac{1}{n} \sum\limits_{i = 1}^{n - 1}f \left ( \frac{1}{n} \right )$
$\Rightarrow \frac{1}{n}\sum\limits_{i = 1}^{n}f \left ( \frac{i}{n} \right )-\frac{1}{n}f \left ( \frac{i}{n} \right ) \leq \int\limits_{\frac{1}{n} }^{1}f(x)dx \leq \frac{1}{n} \sum\limits_{i = 1}^{n}f \left ( \frac{1}{n} \right )-\frac{1}{n}f(1)$
$\Rightarrow \int\limits_{\frac{1}{n} }^{1}f(x)dx + \frac{1}{n}f(1) \leq \frac{1}{n}\sum\limits_{i = 1}^{n}f \left ( \frac{1}{2n} \right ) \leq \int\limits_{\frac{1}{n}}^{1}f(x)dx + \frac{1}{n}f \left ( \frac{1}{n} \right )$               (1)
Hơn nữa :
* $ f(1) = 0$
* $\frac{1}{n}f \left ( \frac{1}{n} \right )   = \frac{1}{n} \left [ \frac{1}{4n^2}-\frac{1}{2} \ln \frac{1}{n}-\frac{1}{4} \right ] = \frac{1}{4n^3} + \frac{1}{2n}\ln n - \frac{1}{4n}$
$ \int\limits_{\frac{1}{n} }^{1}f(x)dx = \frac{x^3}{12} - \frac{1}{2}(x\ln x - x) - \frac{1}{4} x\left| \begin{array}{l}
1\\
\frac{1}{n}
\end{array} \right. = \frac{1}{3}- \frac{1}{12n^3}+\frac{\ln n}{2n}-\frac{1}{4n}$
$\Rightarrow \mathop {\lim }\limits_{n \to +\infty }\left [\int\limits_{\frac{1}{n} }^{1} f(x)dx + \frac{1}{n}f(1)\right ] = \mathop {\lim }\limits_{n \to +\infty} \left [ \int\limits_{\frac{1}{n} }^{1}f(x)dx + \frac{1}{n}f \left ( \frac{1}{n} \right )\right ] = \frac{1}{3}$           (2)
Từ (1) và (2) $\Rightarrow \mathop {\lim }\limits_{n \to +\infty}\sum\limits_{i = 1}^{n}f \left ( \frac{i}{n}\right ) = \frac{1}{3}.$

Bài 112324

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Bài 112324

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