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}}

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

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

   \[=(2n\pi +\frac{\pi }{2}-\frac{1}{2n\pi })\cos \frac{1}{2n\pi }\cdot \left( \frac{1}{(2n\pi +\frac{\pi }{2}-\frac{1}{2n\pi })}-\tan \frac{1}{2n\pi } \right)\]

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

$<0$

\[f(2n\pi +\frac{\pi }{2})=1>0\]

于是

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

同理，由

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

\[=-\left[ (2n+1)\pi +\frac{\pi }{2}-\frac{1}{(2n+1)\pi } \right]\cos \frac{1}{(2n+1)\pi }\cdot \left( \frac{1}{(2n+1)\pi +\frac{\pi }{2}-\frac{1}{(2n+1)\pi }}-\tan \frac{1}{(2n+1)\pi } \right)\]$>0$

\[f\left( (2n+1)\pi +\frac{\pi }{2} \right)=-1<0\]

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 $

转载于:https://www.cnblogs.com/Colgatetoothpaste/p/3674872.html