搜集到的数学分析例题（不断更新）

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

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

证明：设 $f(x)=sin x-xcos x$，则$f'(x) = xsin xleft{egin{array}{ll}
> 0, & hbox{$x in {I_{2n}}$;} \
< 0, & hbox{$x in {I_{2n + 1}}$.}
end{array}
ight.$

其中${{I}_{n}}=(npi ,(n+1)pi )$，又

$f(0)=0,f(2npi )=-2npi <0(nge 1),f((2n+1)pi )=(2n+1)pi >0$

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

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

   [=(2npi +frac{pi }{2}-frac{1}{2npi })cos frac{1}{2npi }cdot left( frac{1}{(2npi +frac{pi }{2}-frac{1}{2npi })}- an frac{1}{2npi } ight)]

$<(2npi +frac{pi }{2}-frac{1}{2npi })cos frac{1}{2npi }cdot left( frac{1}{(2npi +frac{pi }{2}-frac{1}{2npi })}-frac{1}{2npi } ight)$

$<0$

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

于是

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

同理，由

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

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

[fleft( (2n+1)pi +frac{pi }{2} ight)=-1<0]

于是

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

求$int_{Gamma }{{{y}^{2}}}ds$，其中$Gamma $是由$left{egin{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}
ight.$决定

解：由于$Gamma :$$
left{egin{array}{ll}
{left( {x - frac{a}{2}} ight)^2} + {y^2} + {left( {z - frac{a}{2}} ight)^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}
ight.$

作变换$
left{egin{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}
ight.$

则令$l:$$
left{egin{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}
ight.$

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

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

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

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