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# 复习广义积分

## 第一套题：用无界端点求反常积分

$\begin{array}{rl}{\int }_{1}^{b}\frac{1}{{x}^{2}}\phantom{\rule{0.167em}{0ex}}dx& ={\int }_{1}^{b}{x}^{-2}\phantom{\rule{0.167em}{0ex}}dx\\ \\ & ={\left[\frac{{x}^{-1}}{-1}\right]}_{1}^{b}\\ \\ & ={\left[-\frac{1}{x}\right]}_{1}^{b}\\ \\ & =-\frac{1}{b}-\left(-\frac{1}{1}\right)\\ \\ & =1-\frac{1}{b}\end{array}$

$\begin{array}{rl}\underset{b\to \mathrm{\infty }}{lim}{\int }_{1}^{b}\frac{1}{{x}^{2}}\phantom{\rule{0.167em}{0ex}}dx& =\underset{b\to \mathrm{\infty }}{lim}\left(1-\frac{1}{b}\right)\\ \\ & =1-0\\ \\ & =1\end{array}$

${\int }_{1}^{\mathrm{\infty }}\frac{1}{{x}^{3}}\phantom{\rule{0.167em}{0ex}}dx=?$

## 第二套题：用无界函数求反常积分

$\begin{array}{rl}{\int }_{a}^{1}\frac{1}{\sqrt{x}}\phantom{\rule{0.167em}{0ex}}dx& ={\int }_{a}^{1}{x}^{{}^{-\frac{1}{2}}}\phantom{\rule{0.167em}{0ex}}dx\\ \\ & ={\left[\frac{{x}^{{}^{\frac{1}{2}}}}{\frac{1}{2}}\right]}_{a}^{1}\\ \\ & =\left[2\sqrt{x}{\right]}_{a}^{1}\\ \\ & =2\sqrt{1}-2\sqrt{a}\\ \\ & =2-2\sqrt{a}\end{array}$

$\begin{array}{rl}\underset{a\to 0}{lim}{\int }_{a}^{1}\frac{1}{\sqrt{x}}\phantom{\rule{0.167em}{0ex}}dx& =\underset{a\to 0}{lim}\left(2-2\sqrt{a}\right)\\ \\ & =2-2\cdot 0\\ \\ & =2\end{array}$

${\int }_{0}^{8}\frac{1}{\sqrt[3]{\phantom{A}x}}\phantom{\rule{0.167em}{0ex}}dx=?$