搜集到的数学分析例题（不断更新）

设方程 $\sin x-x\cos x=0$ 在 $(0,+\infty )$ 中的第 $n$ 个解为 ${{x}_{n}}$ ，证明：

$n\pi +\frac{\pi }{2}-\frac{1}{n\pi }<{{x}_{n}}<n\pi +\frac{\pi }{2}$

证明：设 $f(x)=\sin x-x\cos x$ ，则 $f‘(x) = x\sin x\left\{\begin{array}{ll} > 0, & \hbox{$x \in {I_{2n}}$;} \\ < 0, & \hbox{$x \in {I_{2n + 1}}$.} \end{array} \right.$

其中 ${{I}_{n}}=(n\pi ,(n+1)\pi )$ ，又

$f(0)=0,f(2n\pi )=-2n\pi <0(n\ge 1),f((2n+1)\pi )=(2n+1)\pi >0$

于是 $f(x)=0$ 在 $(0,+\infty )$ 中的第 $n$ 个解 ${{x}_{n}}\in {{I}_{n}}$ ，再注意到

$f(2n\pi +\frac{\pi }{2}-\frac{1}{2n\pi })=\cos \frac{1}{2n\pi }-(2n\pi +\frac{\pi }{2}-\frac{1}{2n\pi })\sin \frac{1}{2n\pi }$

$<(2n\pi +\frac{\pi }{2}-\frac{1}{2n\pi })\cos \frac{1}{2n\pi }\cdot \left( \frac{1}{(2n\pi +\frac{\pi }{2}-\frac{1}{2n\pi })}-\frac{1}{2n\pi } \right)$

$<0$

于是

${{x}_{2n}}\in (2n\pi +\frac{\pi }{2}-\frac{1}{2n\pi },2n\pi +\frac{\pi }{2})$

同理，由

$f(2n\pi +\frac{\pi }{2}-\frac{1}{(2n+1)\pi })=-\cos \frac{1}{(2n+1)\pi }-\left[ (2n+1)\pi +\frac{\pi }{2}-\frac{1}{(2n+1)\pi } \right]\sin \frac{1}{(2n+1)\pi }$

于是

${{x}_{2n+1}}\in \left( (2n+1)\pi +\frac{\pi }{2}-\frac{1}{(2n+1)\pi },(2n+1)\pi +\frac{\pi }{2} \right)$

求 $\int_{\Gamma }{{{y}^{2}}}ds$ ，其中 $\Gamma$ 是由 $\left\{\begin{array}{ll} {x^2} + {y^2} + {z^2} = {a^2} \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} x + z = a \end{array} \right.$ 决定

解：由于 $\Gamma :$ $\left\{\begin{array}{ll} {\left( {x - \frac{a}{2}} \right)^2} + {y^2} + {\left( {z - \frac{a}{2}} \right)^2} = \frac{{{a^2}}}{2} \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} (x - \frac{a}{2}) + (z - \frac{a}{2}) = 0 \end{array} \right.$

作变换 $\left\{\begin{array}{ll} u = x - \frac{a}{2} \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} v = y\\ w = z - \frac{a}{2} \end{array} \right.$

则令 $l:$ $\left\{\begin{array}{ll} {u^2} + {v^2} + {w^2} = \frac{{{a^2}}}{2} \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} u + w = 0 \end{array} \right.$

$\int_{\Gamma }{{{y}^{2}}}ds=\int_{l}{{{v}^{2}}}ds=\int_{0}^{2\pi }{\frac{{{a}^{2}}}{2}}{{\sin }^{2}}\theta \sqrt{{{\left( \frac{du}{d\theta } \right)}^{2}}+{{\left( \frac{dv}{d\theta } \right)}^{2}}+{{\left( \frac{dw}{d\theta } \right)}^{2}}}d\theta$

$=\int_{0}^{2\pi }{\frac{{{a}^{2}}}{2}{{\sin }^{2}}\theta \cdot \frac{a}{\sqrt{2}}}d\theta$

$=\frac{{{a}^{3}}\pi }{2\sqrt{2}}$

其中 $\left\{\begin{array}{ll} u = \frac{a}{2}\cos \theta \\ v = \frac{a}{{\sqrt 2 }}\sin \theta \\ w = - \frac{a}{2}\cos \theta \end{array} \right.$ $0 \le \theta \le 2\pi$

搜集到的数学分析例题（不断更新）,布布扣,bubuko.com

搜集到的数学分析例题（不断更新）

原文：http://www.cnblogs.com/Colgatetoothpaste/p/3674872.html