武汉大学2011年数学分析试题解答

武汉⼤学2011年数学分析试题解答

1：计算题

（1） 解：原极限\ext{=}\set{n\o \\infty }{\\mathop{\\lim }}\\,\\frac{\\sqrt[n]{n}}{n}\\cdot {{n}^{1-\\alpha }}={{e}^{-1}}\set{n\o \\infty }

{\\mathop{\\lim }}\\,{{n}^{1-\\alpha }}=\\left\\{\\begin{array} +\\infty, & \\hbox{$0<\\alpha<1$;} \\\\ e^{-1}, & \\hbox{$\\alpha=1$;} \\\\ 0, & \\hbox{$\\alpha>1$.}\\end{array} \\right.

(解释⼀下：\set{n\o \\infty }{\\mathop{\\lim }}\\,\\frac{\\sqrt[n]{n}}{n}={{e}^{\set{n\o \\infty }{\\mathop{\\lim }}\\,\\frac{1}{n}\\sum\\limits_{i=1}^{n}{In\\frac{i}{n}}}}={{e}^{\\int\\limits_{0}^{1}{Inxdx}}}={{e}^{-1}} （来源于数学分析上册第⼆章课后习题）（2） 解：考虑等价⽆穷⼩：\set{x\o 0}{\\mathop{\\lim }}\\,\\frac{1-\\cos x}{\\frac{1}{2}{{x}^{2}}}=1 ，则

1-\\cos \\sqrt{\an x-\\sin x}=2{{\\sin }^{2}}\\frac{\\sqrt{\an x-\\sin x}}{2}\\sim \\frac{1}{2}(\an x-\\sin x)=\\frac{\\sin x}{2\\cos x}(1-\\cos x)\\sim \\frac{1}{4}{{x}^{3}}

另⼀⽅⾯：\\sqrt[3]{1+{{x}^{3}}}-\\sqrt[3]{1-{{x}^{3}}}=\\frac{2{{x}^{3}}}{{{(\\sqrt[3]{1+{{x}^{3}}})}^{2}}+\\sqrt[3]{(1+{{x}^{3}})(1-{{x}^{3}})}+{{(\\sqrt[3]{1-{{x}^{3}}})}^{2}}}\\sim \\frac{2{{x}^{3}}}{3}从⽽原式 =\\frac{3}{8}（3） 法⼀：

解：原式=\\int{\\frac{1+\\cos x}{\\sqrt{1+\\cos x}}dx=}\\int{\\frac{2{{\\cos }^{2}}\\frac{x}{2}}{\\sqrt{1+\\cos x}}dx=}\\int{\\frac{2{{\\cos }^{2}}\\frac{x}{2}(1-{{\\sin}^{2}}\\frac{x}{2})}{{{\\cos }^{2}}\\frac{x}{2}\\sqrt{1+\\cos x}}dx}

=\\int{\\frac{1+\\cos x-\\sin x\\sin \\frac{x}{2}\\cos \\frac{x}{2}}{{{\\cos }^{2}}\\frac{x}{2}\\sqrt{1+\\cos x}}dx=}\\int{\\sqrt{1+\\cos x}{{\\sec }^{2}}\\frac{x}{2}+\\frac{-\\sin x}{\\sqrt{1+\\cos x}}\an \\frac{x}{2}dx}

=2\\int{d(\\sqrt{1+\\cos x}\an \\frac{x}{2})=2}\\sqrt{1+\\cos x}\an \\frac{x}{2}+C法⼆：由于\\int{\\sqrt{\ext{1+}\\cos x}dx}=\\sqrt{2}\\int{\\left| \\cos \\frac{x}{2} \\right|}dx考虑到\\int{\\left| x \\right|}dx=\\frac{{{x}^{2}}}{2}sgn x+C

于是\\int{\\sqrt{\ext{1+}\\cos x}dx}=2\\sqrt{2}\\sin \\frac{x}{2}sgn (\\cos \\frac{x}{2})+C（C为常数）法三：由于\\int{\\sqrt{\ext{1+}\\cos x}dx}=\\sqrt{2}\\int{\\left| \\cos \\frac{x}{2} \\right|}dx\\overset{t=\\frac{x}{2}}{\\mathop{=}}\\,2\\sqrt{2}\\int{\\left| \\cos t \\right|}dt⽽

\\int {\\left| {\\cos t} \\right|} dt=\\left\\{\\begin{array}{ll} \\sin t + {c_k}, & \\hbox{$- \\frac{\\pi }{2} + 2k\\pi \\le t \\le \\frac{\\pi }{2} + 2k\\pi$;} \\\\ - \\sin t + {d_k}, &\\hbox{$\\frac{\\pi }{2} + 2k\\pi \\le t \\le \\frac{{3\\pi }}{2} + 2k\\pi$.} \\end{array} \\right.+C为连续函数，其中C为常数于是

\\left\\{\\begin{array}{ll} {c_k} + 1 = - 1 + {d_k} \\\\ {c_{k + 1}} - 1 = 1 + {d_k} \\end{array} \\right.，令{{c}_{0}}=0

则{{c}_{k}}=4k,{{d}_{k}}=4k+2于是

\\int {\\left| {\\cos t} \\right|} dt=\\left\\{\\begin{array}{ll} \\sin t + 4k, & \\hbox{$ - \\frac{\\pi }{2} + 2k\\pi \\le t \\le \\frac{\\pi }{2} + 2k\\pi$;} \\\\ - \\sin t + 4k + 2, &\\hbox{$\\frac{\\pi }{2} + 2k\\pi \\le t \\le \\frac{{3\\pi }}{2} + 2k\\pi$.} \\end{array} \\right.+C即

\\int {\\sqrt {{\\rm{1 + }}\\cos x} dx}=\\left\\{\\begin{array}{ll} 2\\sqrt 2 (\\sin \\frac{x}{2} + 4k), & \\hbox{$ - \\pi + 4k\\pi \\le x \\le \\pi + 4k\\pi$;} \\\\ 2\\sqrt 2 ( - \\sin\\frac{x}{2} + 4k + 2), & \\hbox{$\\pi + 4k\\pi \\le x \\le 3\\pi + 4k\\pi$.} \\end{array} \\right.+C，其中C为常数（4） 解：F(x,y)=x\\int_{\\frac{y}{x}}^{xy}{zf(z)dz-y\\int_{\\frac{y}{x}}^{xy}{f(z)dz}}

则F_{x}^{'}=\\int_{\\frac{y}{x}}^{xy}{zf(z)dz+x[xyf(xy)\\cdot y-\\frac{y}{x}f(\\frac{y}{x})\\cdot \\frac{-y}{{{x}^{2}}}]-y[f(xy)\\cdot y-f(\\frac{y}{x})\\cdot \\frac{-y}{{{x}^{2}}}]}

=\\int_{\\frac{y}{x}}^{xy}{zf(z)dz+({{x}^{2}}-1){{y}^{2}}f(xy)}

F_{xx}^{''}=xyf(xy)\\cdot y-\\frac{y}{x}f(\\frac{y}{x})\\cdot \\frac{-y}{{{x}^{2}}}+2x{{y}^{2}}f(xy)+({{x}^{2}}-1){{y}^{2}}f'(xy)\\cdot y

=3x{{y}^{2}}f(xy)+\\frac{{{y}^{2}}}{{{x}^{3}}}f(\\frac{y}{x})+({{x}^{2}}-1){{y}^{3}}f'(xy)

（5） 解：原式=\\int\\limits_{0}^{1}{dy\\int\\limits_{-1}^{{{y}^{2}}}{({{y}^{2}}-x)dx+}}\\int\\limits_{0}^{1}{dy\\int\\limits_{{{y}^{2}}}^{1}{(-{{y}^{2}}+x)dx}}=\\frac{6}{5}

2：说明，原版的试卷中的题⽬可能有点问题，原版试题如下：

已知f(x),g(x)在[a,b]上连续，在(a,b)上可微，且g'(x)在(a,b)上⽆零点，证明：\\exists \\xi \\in (a,b),st \\frac{f'(\\xi )}{g'(\\xi )}=\\frac{f(b)-g(\\xi )}{g(\\xi )-g(a)}

如果有思路的话，欢迎补充！

证明:作辅助函数F\\left( x \\right)=f\\left( x \\right)g\\left( x \\right)-g\\left( b \\right)f\\left( x \\right)-f\\left( a \\right)g\\left( x \\right)虽然F\\left( x \\right) 在[a,b]上连续，在(a,b)上可微, F\\left( a \\right)=F\\left( b \\right)=-f\\left( a \\right)g\\left( b \\right)由罗尔中值定理,存在\\xi \\in \\left( a,b \\right)使得{F}'\\left( \\xi \\right)=0

即{f}'\\left( \\xi \\right)g\\left( \\xi \\right)+f\\left( \\xi \\right){g}'\\left( \\xi \\right)-g\\left( b \\right){f}'\\left( \\xi \\right)-f\\left( a \\right){g}'\\left( \\xi \\right)=0整理\\left[ f\\left( a \\right)-f\\left( \\xi \\right) \\right]{g}'\\left( \\xi \\right)-{f}'\\left( \\xi \\right)\\left[ g\\left( \\xi \\right)-g\\left( b \\right) \\right]=0即\\frac{f\\left( a \\right)-f\\left( \\xi \\right)}{g\\left( \\xi \\right)-g\\left( b \\right)}=\\frac{{f}'\\left( \\xi \\right)}{{g}'\\left( \\xi \\right)} ,证得3：（⽅法⼀）

证明：\\left| \\frac{{{a}_{1}}{{b}_{n}}+{{a}_{2}}{{b}_{n-1}}+\\cdots +{{a}_{n}}{{b}_{1}}}{{{a}_{1}}+{{a}_{2}}+\\cdots +{{a}_{n}}}-b \\right|=\\left|\\frac{{{a}_{1}}({{b}_{n}}-b)+{{a}_{2}}({{b}_{n-1}}-b)+\\cdots +{{a}_{n}}({{b}_{1}}-b)}{{{a}_{1}}+{{a}_{2}}+\\cdots +{{a}_{n}}} \\right|

\\le \\frac{{{a}_{1}}\\left| {{b}_{n}}-b \\right|+\\cdots +{{a}_{n-N}}\\left| {{b}_{N+1}}-b \\right|}{{{a}_{1}}+\\cdots +{{a}_{n}}}+\\frac{{{a}_{n-N+1}}\\left|{{b}_{N}}-b \\right|+\\cdots +{{a}_{n}}\\left| {{b}_{1}}-b \\right|}{{{a}_{1}}+\\cdots +{{a}_{n}}}

\\le \set{N+1\\le k\\le n}{\\mathop \\max }\\,\\left| {{b}_{k}}-b \\right|+\set{N+1\\le k\\le n}{\\mathop \\max }\\,\\left| {{b}_{k}}-b \\right|\\cdot\\frac{N}{n-N+1}=I_{1}^{n}+I_{2}^{n}

从⽽对\\forall \\varepsilon >0,先取定N使得I_{1}^{n}<\\frac{\\varepsilon }{2}，后让n充分⼤即有I_{2}^{n}<\\frac{\\varepsilon }{2}，于是有结论成⽴。（⽅法⼆）

证明：设{{t}_{nk}}=\\frac{{{a}_{n-k+1}}}{{{a}_{1}}+{{a}_{2}}+\\cdots +{{a}_{n}}},k=1,2,\\cdots ,n,n=1,2,\\cdots则{{t}_{nk}}>0且\\sum\\limits_{k=1}^{n}{{{t}_{nk}}}=1

再由{{a}_{n}}\\ge 0,{{a}_{n}}\\le {{a}_{n-1}}\\Rightarrow {{a}_{1}}+{{a}_{2}}+\\cdots \ext{+}{{a}_{n}}\\ge {{a}_{1}}+{{a}_{2}}+\\cdots \ext{+}{{a}_{n-k+1}}\\ge (n-k+1){{a}_{n-k+1}}

于是0\\le {{t}_{nk}}\\le \\frac{{{a}_{n-k+1}}}{(n-k+1){{a}_{n-k+1}}}=\\frac{1}{(n-k+1)}\o 0(n\o +\\infty )由迫敛性知：\set{n\o +\\infty }{\\mathop{\\lim }}\\,{{t}_{nk}}=0再由\set{n\o +\\infty }{\\mathop{\\lim }}\\,{{b}_{n}}=b，可知\\exists M>0,s.t对任意的n\\in {{N}^{*}},\\left| {{a}_{n}}-a \\right|同时，对任意的\\varepsilon >0,\\exists {{N}_{1}}\\in {{N}^{*}}，当n>{{N}_{1}}时，有\\left| {{a}_{n}}-a \\right|<\\frac{\\varepsilon }{2}固定{{N}_{1}}，由\set{n\o +\\infty }{\\mathop{\\lim }}\\,{{t}_{nk}}=0可知：

\\exists {{N}_{2}}\\in {{N}^{*}}，当n>{{N}_{2}}时，有\\left| {{a}_{n}}-a \\right|<\\frac{\\varepsilon }{2{{N}_{1}}M},k=1,2,\\cdots ,{{N}_{1}}令N=\\max \\{{{N}_{1}},{{N}_{2}}\\}，当n>N时，有

\\left| \\frac{{{a}_{1}}{{b}_{n}}+{{a}_{2}}{{b}_{n-1}}+\\cdots +{{a}_{n}}{{b}_{1}}}{{{a}_{1}}+{{a}_{2}}+\\cdots +{{a}_{n}}}-b \\right|=\\left|

\\sum\\limits_{k=1}^{n}{{{t}_{nk}}{{b}_{k}}-b} \\right|=\\left| \\sum\\limits_{k=1}^{n}{{{t}_{nk}}{{b}_{k}}-\\sum\\limits_{k=1}^{n}{{{t}_{nk}}}b} \\right|\\le \\sum\\limits_{k=1}^{n}{{{t}_{nk}}}\\left| {{b}_{k}}-b \\right|即\set{n\o +\\infty }{\\mathop{\\lim }}\\,\\frac{{{a}_{1}}{{b}_{n}}+{{a}_{2}}{{b}_{n-1}}+\\cdots +{{a}_{n}}{{b}_{1}}}{{{a}_{1}}+{{a}_{2}}+\\cdots +{{a}_{n}}}=b

（⽅法三，书上并⽆介绍此公式，不建议使⽤）：直接利⽤Stolz进⾏计算即可

4：证明：{{a}_{k}}=\\frac{1}{2\\pi }\\int\\limits_{-\\pi }^{\\pi }{f(x)\\cos kxdx=}\\frac{1}{2k\\pi }\\int\\limits_{-\\pi }^{\\pi }{f(x)d(sinkx)=}\\frac{-1}{2k\\pi}\\int\\limits_{-\\pi }^{\\pi }{f'(x)sinkxdx}

=\\frac{-1}{2{{k}^{2}}\\pi }\\int\\limits_{-k\\pi }^{k\\pi }{f'(\\frac{t}{k})sintdt} 则

{{a}_{2n}}=\\frac{-1}{8{{n}^{2}}\\pi }\\int\\limits_{-2n\\pi }^{2n\\pi }{f'(\\frac{t}{2n})\\operatorname{sint}dt=}\\frac{-1}{8{{n}^{2}}\\pi }\\sum\\limits_{m=-n}^{n-1}{[\\int\\limits_{-2m\\pi }^{(2m+1)\\pi }{f'(\\frac{t}{2n})\\operatorname{sint}dt+\\int\\limits_{(2m+1)\\pi }^{2(m+1)\\pi }{f'(\\frac{t}{2n})\\operatorname{sint}dt]=}}}

\\ge \\frac{-1}{8{{n}^{2}}\\pi }\\sum\\limits_{m=-n}^{n-1}{[\\int\\limits_{2m\\pi }^{(2m+1)\\pi }{f'(\\frac{t}{2n})\\sin tdt+}}\\int\\limits_{(2m+1)\\pi }^{2(m+1)\\pi }{\\operatorname{sintdt}]} ⽽

\\sin t\\left\\{\\begin{array}{ll} > 0, & \\hbox{$ 2m\\pi < x < (2m + 1)\\pi$;} \\\\ < 0, & \\hbox{$(2m + 1)\\pi < x < 2(m + 1)\\pi$.} \\end{array} \\right. 且f为凸函数

从⽽原式\\ge \\frac{-1}{8{{n}^{2}}\\pi }\\sum\\limits_{m=-n}^{n-1}{f'(\\frac{2m+1}{2n}\\pi )\\int\\limits_{2m\\pi }^{2(m+1)\\pi }{\\sin tdt=0}}另⼀⽅⾯：

{{a}_{2n+1}}=\\frac{-1}{2{{(2n+1)}^{2}}\\pi }\\int\\limits_{-(2n+1)\\pi }^{(2n+1)\\pi }{f'(\\frac{t}{2n+1})\\sin tdt}

=\\frac{-1}{2{{(2n+1)}^{2}}\\pi }\\sum\\limits_{m=-n}^{n}{[\\int\\limits_{(2m-1)\\pi }^{2m\\pi }{f'(\\frac{t}{2n+1})\\sin tdt+\\int\\limits_{2m\\pi }^{(2m+1)\\pi }{f'(\\frac{t}{2n+1})\\sin tdt}}}

\\le \\frac{-1}{2{{(2n+1)}^{2}}\\pi }\\sum\\limits_{m=-n}^{n}{[\\int\\limits_{(2m+1)\\pi }^{2m\\pi }{f'(\\frac{2m}{2n+1})\\sin tdt}}+\\int\\limits_{2m\\pi}^{(2m+1)\\pi }{f'(\\frac{2m}{2n+1})\\sin tdt}]

==\\frac{-1}{2{{(2n+1)}^{2}}\\pi }\\sum\\limits_{m=-n}^{n}{f'(\\frac{2m}{2n+1}\\pi )}\\int\\limits_{(2m-1)\\pi }^{(2m+1)\\pi }{\\sin tdt=0}5：证明：

（1） 设{{f}_{k}}(x)=\\sum\\limits_{n=1}^{k}{{{u}_{n}}(x),}因为

\\left| f(x)-f(y) \\right|\\le \\left| f(x)-{{f}_{K}}(x) \\right|+\\left| {{f}_{K}}(x)-{{f}_{K}}(y) \\right|+\\left| {{f}_{K}}(y)-f(y) \\right| \\le 2\\cdot \\sup \\left| \\left| {{f}_{K}}(z)-f(z) \\right| \\right|+\\left| {{f}_{K}}(x)-{{f}_{K}}(y) \\right|\\equiv I_{1}^{K}+I_{2}^{K}

从⽽对\\forall \\varepsilon >0,可先选定K使I_{1}^{K}<\\frac{\\varepsilon }{2}，后取\\delta >0 使\\left| x-y \\right|<\\delta \\Rightarrow I_{2}^{K}<\\frac{\\varepsilon }{2}

（2） 若条件改为逐点连续，则结论不成⽴。为此，仅需构造出{{f}_{k}}(x)\o f(x)其中每个{{f}_{k}}(x)⼀致收敛，但f(x)不⼀致收敛，甚⾄不连续的例⼦很多，例如：

{f_k}(x)\\left\\{\\begin{array}{ll} - 1, & \\hbox{$x \\le \\frac{{ - 1}}{k}$;} \\\\ kx, & \\hbox{$\\frac{{ - 1}}{k} < x < \\frac{1}{k}$;} \\\\ 1, & \\hbox{$x \\ge \\frac{1}{k}$.}\\end{array} \\right.

，恒有{{f}_{k}}(x)\o sgn (x)6：证明：

（1）\\int\\limits_{c}^{d}{f(x,y)dy}⼀致收敛

\\Leftrightarrow \\exists F(x),\\forall \\varepsilon >0,\\exists \\delta >0,s.t.\\left\\{\\begin{array}{ll} x \\in [a,b] \\\\ \\delta ' \\in (0,\\delta ) \\end{array} \\right.其有Cauchy准则：

\\forall \\varepsilon >0,\\exists \\delta >0,s.t.

\\left\\{\\begin{array}{ll} x \\in [a,b] \\\\ \\delta ',\\delta '' \\in (0,\\delta ) \\end{array} \\right.\\Rightarrow \\left| \\int\\limits_{c+\\delta '}^{c+\\delta ''}{f(x,y)dy} \\right|<\\varepsilon

（3） 仅需注意到\\left| \\int\\limits_{c+\\delta '}^{c+\\delta }{f(x,y)g(x,y)dy} \\right|\\le \set{[a,b]\imes [c,d]}{\\mathop{\\max }}\\,\\left| g(x,y)\\right|\\cdot \\int\\limits_{c+\\delta '}^{c+\\delta ''}{\\left| f(x,y) \\right|}dy 后利⽤Cauchy即可得证

7：（1）解：g=f(\\frac{x}{z})\\mp zf(\\frac{z}{x})+f(\\frac{y}{z})\\mp zf(\\frac{z}{y})

{{g}_{x}}=\\frac{1}{z}f''(\\frac{x}{z})\\mp \\frac{{{z}^{2}}}{{{x}^{2}}}f''(\\frac{z}{x})

{{g}_{xx}}=\\frac{1}{{{z}^{2}}}f''(\\frac{x}{z})\\mp \\frac{2{{z}^{2}}}{{{x}^{3}}}f'(\\frac{z}{x})\\mp \\frac{{{z}^{3}}}{{{x}^{4}}}f''(\\frac{z}{x}) {{g}_{yy}}=\\frac{1}{{{z}^{2}}}f''(\\frac{y}{z})\\mp \\frac{{{z}^{2}}}{{{y}^{3}}}f'(\\frac{z}{y})\\mp \\frac{{{z}^{3}}}{{{y}^{4}}}f''(\\frac{z}{y}) {{g}_{z}}=\\frac{-1}{{{z}^{2}}}f'(\\frac{x}{z})\\mp f(\\frac{z}{x})\\mp \\frac{z}{x}f'(\\frac{z}{x}) =\\frac{-y}{{{z}^{2}}}f'(\\frac{y}{z})\\mp f(\\frac{z}{y})\\mp \\frac{z}{y}f'(\\frac{z}{y})

{{g}_{zz}}=\\frac{2x}{{{z}^{3}}}f'(\\frac{x}{z})+\\frac{{{x}^{2}}}{{{z}^{4}}}f''(\\frac{x}{z})\\mp \\frac{1}{x}f'(\\frac{z}{x})\\mp \\frac{1}{x}f'(\\frac{z}{x})\\mp\\frac{z}{{{x}^{2}}}f''(\\frac{z}{x})

+\\frac{2y}{{{z}^{3}}}f'(\\frac{y}{z})+\\frac{{{y}^{2}}}{{{z}^{4}}}f''(\\frac{y}{z})\\mp \\frac{1}{y}f'(\\frac{z}{y})\\mp \\frac{1}{y}f'(\\frac{z}{y})\\mp \\frac{z}{{{y}^{2}}}f''(\\frac{z}{y}) 于是

{{x}^{2}}{{g}_{xx}}+{{y}^{2}}{{g}_{yy}}-{{z}^{2}}{{g}_{zz}}=-\\frac{2x}{z}f(\\frac{x}{z})-\\frac{2y}{z}f(\\frac{y}{z})（2）设0<{{a}_{1}}<{{a}_{2}},0<{{b}_{1}}<{{b}_{2}},1<{{c}_{1}}<{{c}_{2}}.则 原式=-2\\iiint_{\\Omega }{[\\frac{x}{z}f'(\\frac{x}{z})+\\frac{y}{z}f'(\\frac{y}{z})dxdydz}

=-2\\int\\limits_{{{a}_{1}}}^{{{a}_{2}}}{du\\int\\limits_{{{b}_{1}}}^{{{b}_{2}}}{dv\\int\\limits_{{{c}_{1}}}^{{{c}_{2}}}{[\\frac{1}{u}f'(\\frac{1}{u})+\\frac{1}{v}f'(\\frac{1}{v})]\\cdot \\frac{2}{3}{{u}^{-2}}{{v}^{-2}}dw}}} （其中令u=\\frac{z}{x},v=\\frac{z}{y},w={{z}^{3}} ）

\\frac{-2({{c}_{2}}-{{c}_{1}})}{3}\\int\\limits_{{{a}_{1}}}^{{{a}_{2}}}{du\\int\\limits_{{{b}_{1}}}^{{{b}_{2}}}{[\\frac{1}{{{u}^{2}}v}f'(\\frac{1}{u})+\\frac{1}{u{{v}^{2}}}f'(\\frac{1}{v})}}]dv

=\\frac{-2({{c}_{2}}-{{c}_{1}})}{3}\\int\\limits_{{{a}_{1}}}^{{{a}_{2}}}{\\{\\frac{1}{{{u}^{2}}}f'(\\frac{1}{u})In\\frac{{{b}_{2}}}{{{b}_{1}}}-\\frac{1}{u}[f(\\frac{1}{{{b}_{2}}})-f(\\frac{1}{{{b}_{1}}})]In\\frac{{{a}_{2}}}{{{a}_{1}}}\\}du}

=\\frac{-2({{c}_{2}}-{{c}_{1}})}{3}\\left\\{ -In\\frac{{{b}_{2}}}{{{b}_{1}}}[ \\right.f(\\frac{1}{{{a}_{2}}})-f(\\frac{1}{{{a}_{1}}})]-[f(\\frac{1}{{{b}_{2}}})-f(\\frac{1}{{{b}_{1}}})]In\\frac{{{a}_{2}}}{{{a}_{1}}}\\}

=\\frac{2({{c}_{2}}-{{c}_{1}})}{3}\\left\\{ In\\frac{{{b}_{2}}}{{{b}_{1}}}[ \\right.f(\\frac{1}{{{a}_{2}}})+f(\\frac{1}{{{a}_{1}}})]-[f(\\frac{1}{{{b}_{2}}})-f(\\frac{1}{{{b}_{1}}})]In\\frac{{{a}_{2}}}{{{a}_{1}}}\\}

8：证明：\\iint_{D}{(x\\frac{\\partial u}{\\partial x}}+y\\frac{\\partial u}{\\partial y})dxdy=\\int\\limits_{0}^{1}{dr}\\int_{{{x}^{2}}+{{y}^{2}}={{r}^{2}}}{(}x{{u}_{x}}+y{{u}_{y}})dS

=\\int\\limits_{0}^{1}{r}\\int_{{{x}^{2}}+{{y}^{2}}={{r}^{2}}}{\\frac{\\partial u}{\\partial n}}dS=\\int\\limits_{0}^{1}{r}\\iint_{{{x}^{2}}+{{y}^{2}}\\le{{r}^{2}}}{\\nabla udxdy}

=\\int\\limits_{0}^{1}{r}\\iint_{{{x}^{2}}+{{y}^{2}}\\le {{r}^{2}}}{\\cos (\\pi ({{x}^{2}}+{{y}^{2}}))dxdy} =\\int\\limits_{0}^{1}{r}\\iint_{{{x}^{2}}+{{y}^{2}}\\le {{r}^{2}}}{\\cos (\\pi {{s}^{2}})\\cdot 2\\pi sdS} =\\int\\limits_{0}^{1}{r\\sin (\\pi r)dr=\\frac{1}{\\pi }}

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