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本帖最后由 永远 于 2022-4-29 11:18 编辑
利用定积分的定义,计算\(\displaystyle\int_0^1 {{e^x}} dx\)的值。
解:令\(f(x) = {e^x}\)。
(1)分割
在区间\([0,1]\)上等间隔地插入\(n - 1\) 个分点,把区间\([0,1]\)等分成\(n\) 个小区间\([\frac{{i - 1}}{n},\frac{i}{n}]\;(i = 1,2, \cdots ,n)\),每个小区间的长度为\(\Delta x = \frac{i}{n} - \frac{{i - 1}}{n} = \frac{1}{n}\)。
(2)近似代替、作和
取\(\xi = \frac{i}{n}\;(i = 1,2, \cdots ,n)\),则
\(\begin{align}
\int_0^1 {{e^x}} dx \approx {S_n}& = \sum\limits_{i = 1}^n {f(\frac{i}{n})} \bullet \Delta x \\
&= \sum\limits_{i = 1}^n {{e^{\frac{i}{n}}}} \bullet \frac{1}{n} \\
&= \frac{1}{n}\sum\limits_{i = 1}^n {{e^{\frac{i}{n}}}} \\
&= \frac{{(1 - e){e^{\frac{1}{n}}}}}{{n(1 - {e^{\frac{1}{n}}})}} \\
\end{align}\)
(3) 取极限
\(\begin{align}
\int_0^1 {{e^x}} dx = \mathop {\lim }\limits_{n \to \infty } {S_n} &= \mathop {\lim }\limits_{n \to \infty } \frac{{(1 - e){e^{\frac{1}{n}}}}}{{n(1 - {e^{\frac{1}{n}}})}} \\
&= (1 - e)\mathop {\lim }\limits_{n \to \infty } \frac{{{e^{\frac{1}{n}}}}}{{n(1 - {e^{\frac{1}{n}}})}} \\
&= (1 - e)\mathop {\lim }\limits_{\lambda \to 0} \frac{{{e^\lambda }}}{{\frac{1}{\lambda }(1 - {e^\lambda })}} \\
&= (1 - e)\mathop {\lim }\limits_{\lambda \to 0} \frac{{\lambda {e^\lambda }}}{{1 - {e^\lambda }}} \\
&= (1 - e)\mathop {\lim }\limits_{\lambda \to 0} \frac{{(\lambda + 1){e^\lambda }}}{{ - {e^\lambda }}} \\
&= (1 - e) \times [ - \mathop {\lim }\limits_{\lambda \to 0} (\lambda + 1)] \\
& = e - 1 \\
\end{align} \) |
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