1. (1) 求极限 \dps{I=\lim_{x\to 0^+} \cfrac{ 1-\cos x}{\int_0^x \cfrac{\ln(1+xy)}{y}\rd y}}<script type="math/tex">\dps{I=\lim_{x\to 0^+} \cfrac{ 1-\cos x}{\int_0^x \cfrac{\ln(1+xy)}{y}\rd y}}</script>. (2) 计算含参变量广义积分 \dps{F(x)=\int_0^\infty \cfrac{\sin (xy)}{ye^y}\rd y}<script type="math/tex">\dps{F(x)=\int_0^\infty \cfrac{\sin (xy)}{ye^y}\rd y}</script>.

解答: (1) \bex I=\lim_{x\to 0^+}\cfrac{\sin x}{\cfrac{\ln(1+x^2)}{x}} =\lim_{x\to 0^+}\cfrac{x^2}{\ln(1+x^2)}=1. \eex<script type="math/tex; mode=display">\bex I=\lim_{x\to 0^+}\cfrac{\sin x}{\cfrac{\ln(1+x^2)}{x}} =\lim_{x\to 0^+}\cfrac{x^2}{\ln(1+x^2)}=1. \eex</script> (2) \bex F(x)=\int_0^\infty e^{-y}\int_0^x \cos ty\rd t\rd y =\int_0^x \rd t\int_0^\infty e^{-y}\cos ty\rd y =\int_0^x \cfrac{1}{1+t^2}\rd t =\arctan x. \eex<script type="math/tex; mode=display">\bex F(x)=\int_0^\infty e^{-y}\int_0^x \cos ty\rd t\rd y =\int_0^x \rd t\int_0^\infty e^{-y}\cos ty\rd y =\int_0^x \cfrac{1}{1+t^2}\rd t =\arctan x. \eex</script>

2. 设 R(x,y)<script type="math/tex">R(x,y)</script> 是曲线 C: \cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1<script type="math/tex">C: \cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1</script> 上点 (x,y)<script type="math/tex">(x,y)</script> 的曲率半径 (a,b>0<script type="math/tex">a,b>0</script>). 计算第一型曲线积分 \dps{\int_C R(x,y)\rd s}<script type="math/tex">\dps{\int_C R(x,y)\rd s}</script>.

解答: 曲线 \bex C:\quad x=a\cos t,\quad y=b\sin t\quad (0\leq t\leq 2\pi). \eex<script type="math/tex; mode=display">\bex C:\quad x=a\cos t,\quad y=b\sin t\quad (0\leq t\leq 2\pi). \eex</script> 而 \beex \bea \int_C R(x,y)\rd s&=\int_0^{2\pi}\cfrac{[x‘^2+y‘^2]^{3/2}}{|x‘y‘‘-x‘‘y‘|}\cdot (x‘^2+y‘^2)^{1/2}\rd t\\ &=\int_0^{2\pi} \cfrac{(a^2\sin^2t+b^2\cos^2t)^2}{ab}\rd t\\ &=\cfrac{3a^4+2a^2b^2+3b^4}{4ab}\pi. \eea \eeex<script type="math/tex; mode=display">\beex \bea \int_C R(x,y)\rd s&=\int_0^{2\pi}\cfrac{[x‘^2+y‘^2]^{3/2}}{|x‘y‘‘-x‘‘y‘|}\cdot (x‘^2+y‘^2)^{1/2}\rd t\\ &=\int_0^{2\pi} \cfrac{(a^2\sin^2t+b^2\cos^2t)^2}{ab}\rd t\\ &=\cfrac{3a^4+2a^2b^2+3b^4}{4ab}\pi. \eea \eeex</script>

3. 设 \Omega<script type="math/tex">\Omega</script> 是椭球体 \cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}+\cfrac{z^2}{c^2}\leq 1<script type="math/tex">\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}+\cfrac{z^2}{c^2}\leq 1</script> 在第一卦限中的部分 (a,b,c>0)<script type="math/tex">a,b,c>0)</script>. 计算三重积分 \bex I=\iiint_\Omega \cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}+\cfrac{z^2}{c^2} \rd x\rd y\rd z. \eex<script type="math/tex; mode=display">\bex I=\iiint_\Omega \cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}+\cfrac{z^2}{c^2} \rd x\rd y\rd z. \eex</script>

解答: \beex \bea I&=abc\iiint_{u^2+v^2+w^2\leq 1\atop u,v,w>0} (u^2+v^2+v^2)\rd u\rd v\rd w\\ &=\cfrac{abc}{8}\iiint_{u^2+v^2+w^2\leq 1} (u^2+v^2+v^2)\rd u\rd v\rd w\\ &=\cfrac{abc}{8}\int_0^1 r^2\cdot 4\pi r^2\rd r\\ &=\cfrac{\pi abc}{10}. \eea \eeex<script type="math/tex; mode=display">\beex \bea I&=abc\iiint_{u^2+v^2+w^2\leq 1\atop u,v,w>0} (u^2+v^2+v^2)\rd u\rd v\rd w\\ &=\cfrac{abc}{8}\iiint_{u^2+v^2+w^2\leq 1} (u^2+v^2+v^2)\rd u\rd v\rd w\\ &=\cfrac{abc}{8}\int_0^1 r^2\cdot 4\pi r^2\rd r\\ &=\cfrac{\pi abc}{10}. \eea \eeex</script>

4. 求幂级数 \dps{\vsm{n}(-1)^{n-1}n^2x^n}<script type="math/tex">\dps{\vsm{n}(-1)^{n-1}n^2x^n}</script> 的收敛域及其和函数.

解答: \beex \bea \vsm{n}(-1)^{n-1}n^2x^n &=x\sum_{n=1}^\infty n^2(-x)^{n-1}\\ &=x\sez{(-x)\sum_{n=1}^\infty n(n-1)(-x)^{n-2}+\vsm{n} n(-x)^{n-1}}\\ &=x\sez{ (-x)\sex{\sum_{n=0}^\infty t^n}‘‘|_{t=-x} \sex{\sum_{n=0}^\infty t^n}‘|_{t=-x} }\\ &=x\sez{(-x)\cfrac{2}{(1+x)^3}+\cfrac{1}{(1+x)^2}}\\ &=\cfrac{x(1-x)}{(1+x)^3},\quad |x|<1. \eea \eeex<script type="math/tex; mode=display">\beex \bea \vsm{n}(-1)^{n-1}n^2x^n &=x\sum_{n=1}^\infty n^2(-x)^{n-1}\\ &=x\sez{(-x)\sum_{n=1}^\infty n(n-1)(-x)^{n-2}+\vsm{n} n(-x)^{n-1}}\\ &=x\sez{ (-x)\sex{\sum_{n=0}^\infty t^n}‘‘|_{t=-x} \sex{\sum_{n=0}^\infty t^n}‘|_{t=-x} }\\ &=x\sez{(-x)\cfrac{2}{(1+x)^3}+\cfrac{1}{(1+x)^2}}\\ &=\cfrac{x(1-x)}{(1+x)^3},\quad |x|<1. \eea \eeex</script>

5. 在椭球面 \cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}+\cfrac{z^2}{c^2}=1<script type="math/tex">\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}+\cfrac{z^2}{c^2}=1</script> 上确定一点 (x_0,y_0,z_0)<script type="math/tex">(x_0,y_0,z_0)</script> (其中 x_0,y_0,z_0>0<script type="math/tex">x_0,y_0,z_0>0</script>) 使得过该点的切平面与三个坐标面所围成的四面体的体积最小. 并求该最小体积.

解答: 由 Lagrange 乘数法即知当 \bex x_0=\cfrac{a}{\sqrt{3}},\quad y_0=\cfrac{b}{\sqrt{3}},\quad z_0=\cfrac{c}{\sqrt{3}} \eex<script type="math/tex; mode=display">\bex x_0=\cfrac{a}{\sqrt{3}},\quad y_0=\cfrac{b}{\sqrt{3}},\quad z_0=\cfrac{c}{\sqrt{3}} \eex</script> 时体积最小, 为 \cfrac{\sqrt{3}abc}{2}<script type="math/tex">\cfrac{\sqrt{3}abc}{2}</script>.

6. 设 f<script type="math/tex">f</script> 在 [a,\infty)<script type="math/tex">[a,\infty)</script> 上连续, \dps{\lim_{x\to \infty}f(x)=A}<script type="math/tex">\dps{\lim_{x\to \infty}f(x)=A}</script>. 证明: f<script type="math/tex">f</script> 在 [a,\infty)<script type="math/tex">[a,\infty)</script> 上有界.

证明: 由 \dps{\lim_{x\to \infty}f(x)=A}<script type="math/tex">\dps{\lim_{x\to \infty}f(x)=A}</script> 知 \bex \exists\ X>a,\st x>X\ra |f(x)-A|\leq\cfrac{|A|+1}{2}\ra |f(x)|\leq \cfrac{3|A|+1}{2}. \eex<script type="math/tex; mode=display">\bex \exists\ X>a,\st x>X\ra |f(x)-A|\leq\cfrac{|A|+1}{2}\ra |f(x)|\leq \cfrac{3|A|+1}{2}. \eex</script> 又 f<script type="math/tex">f</script> 在 [a,X]<script type="math/tex">[a,X]</script> 上有界, 而 \bex \exists\ M>\cfrac{3|A|+1}{2},\st |f(x)|\leq M. \eex<script type="math/tex; mode=display">\bex \exists\ M>\cfrac{3|A|+1}{2},\st |f(x)|\leq M. \eex</script> 故在 [a,\infty)<script type="math/tex">[a,\infty)</script> 上总有 |f(x)|\leq M<script type="math/tex">|f(x)|\leq M</script>.

7. 设函数 f<script type="math/tex">f</script>, g<script type="math/tex">g</script> 在 [a,b]<script type="math/tex">[a,b]</script> 上可微且 g\neq 0<script type="math/tex">g\neq 0</script>. 证明: 存在 \xi\in (a,b)<script type="math/tex">\xi\in (a,b)</script>, 使得 \bex \cfrac{f(a)-f(\xi)}{g(\xi)-g(b)}=\cfrac{f‘(\xi)}{g‘(\xi)}. \eex<script type="math/tex; mode=display">\bex \cfrac{f(a)-f(\xi)}{g(\xi)-g(b)}=\cfrac{f‘(\xi)}{g‘(\xi)}. \eex</script>

证明: 设 \bex F(x)=f(a)g(x)-f(x)g(x)+g(b)f(x), \eex<script type="math/tex; mode=display">\bex F(x)=f(a)g(x)-f(x)g(x)+g(b)f(x), \eex</script> 则 F(a)=f(a)g(b)=F(b)<script type="math/tex">F(a)=f(a)g(b)=F(b)</script>. 由 Rolle 定理即知存在 \xi\in (a,b)<script type="math/tex">\xi\in (a,b)</script>, 使得 \beex \bea 0&=F‘(\xi)\\ &=f(a)g‘(\xi) -f‘(\xi)g(\xi)-f(\xi)g‘(\xi)+g(b)f‘(\xi)\\ &=[f(a)-f(\xi)]g‘(\xi) -[g(\xi)-g(b)]f‘(\xi). \eea \eeex<script type="math/tex; mode=display">\beex \bea 0&=F‘(\xi)\\ &=f(a)g‘(\xi) -f‘(\xi)g(\xi)-f(\xi)g‘(\xi)+g(b)f‘(\xi)\\ &=[f(a)-f(\xi)]g‘(\xi) -[g(\xi)-g(b)]f‘(\xi). \eea \eeex</script>

8. 设函数列 \sed{u_n(x)}<script type="math/tex">\sed{u_n(x)}</script> 在 [a,b]<script type="math/tex">[a,b]</script> 上一致收敛于 u(x)<script type="math/tex">u(x)</script>, 并且对 n=1,2,\cdots,<script type="math/tex">n=1,2,\cdots,</script> u_n(x)<script type="math/tex">u_n(x)</script> 在区间 [a,b]<script type="math/tex">[a,b]</script> 上至少有一个零点, 证明 u(x)<script type="math/tex">u(x)</script> 在区间 [a,b]<script type="math/tex">[a,b]</script> 上也至少有一个零点.

证明: 用反证法. 若 u<script type="math/tex">u</script> 没有零点, 则 \bex m\equiv \min |u|>0. \eex<script type="math/tex; mode=display">\bex m\equiv \min |u|>0. \eex</script> 由一致收敛即知当 n\gg 1<script type="math/tex">n\gg 1</script> 时, \bex \min |u_n|>\cfrac{m}{2}>0. \eex<script type="math/tex; mode=display">\bex \min |u_n|>\cfrac{m}{2}>0. \eex</script> 这与 u_n<script type="math/tex">u_n</script> 有零点矛盾. 故有结论.

 9. 设函数 f(x)<script type="math/tex">f(x)</script> 在区间 [0,1]<script type="math/tex">[0,1]</script> 上二阶连续可微, 且在区间 (0,1)<script type="math/tex">(0,1)</script> 内存在极值, 证明: \bex \max_{x\in [0,1]}|f‘(x)|\leq \int_0^1 |f‘(x)|\rd x. \eex<script type="math/tex; mode=display">\bex \max_{x\in [0,1]}|f‘(x)|\leq \int_0^1 |f‘(x)|\rd x. \eex</script>

证明: 设 \xi\in (a,b)<script type="math/tex">\xi\in (a,b)</script> 为 f<script type="math/tex">f</script> 的极值点, 则 f‘(\xi)=0<script type="math/tex">f‘(\xi)=0</script>. 于是 \bex |f‘(x)|=|f‘(x)-f‘(\xi)| =\sev{\int_\xi^x f‘‘(t)\rd t} \leq \int_0^1 |f‘‘(t)|\rd t,\quad \forall\ x\in [0,1]. \eex<script type="math/tex; mode=display">\bex |f‘(x)|=|f‘(x)-f‘(\xi)| =\sev{\int_\xi^x f‘‘(t)\rd t} \leq \int_0^1 |f‘‘(t)|\rd t,\quad \forall\ x\in [0,1]. \eex</script>

10. 证明: 对每个自然数 n<script type="math/tex">n</script>, 方程 x^{n+1}+x^n+\cdots+x=1<script type="math/tex">x^{n+1}+x^n+\cdots+x=1</script> 在开区间 (0,1)<script type="math/tex">(0,1)</script> 内存在唯一实根 x_n<script type="math/tex">x_n</script>, 并证明 \sed{x_n}<script type="math/tex">\sed{x_n}</script> 收敛.

证明: 设 f(x)=x^{n+1}+x^n+\cdots+x-1<script type="math/tex">f(x)=x^{n+1}+x^n+\cdots+x-1</script>, 则 f(0)=-1<script type="math/tex">f(0)=-1</script>, f(1)=n\geq 1<script type="math/tex">f(1)=n\geq 1</script>. 由连续函数的介值定理及 f‘(x)=(n+1)x^n+\cdots+1>0<script type="math/tex">f‘(x)=(n+1)x^n+\cdots+1>0</script>, x\in (0,1)<script type="math/tex">x\in (0,1)</script> 知 f(x)=0<script type="math/tex">f(x)=0</script> 在 (0,1)<script type="math/tex">(0,1)</script> 内有且仅有一个实根 x_n<script type="math/tex">x_n</script>. 又 \bex 1=x_n^{n+1}+\cdots+x_n=x_{n+1}^{n+2}+\cdots+x_{n+1}>x_{n+1}^{n+1}+\cdots+x_{n+1}, \eex<script type="math/tex; mode=display">\bex 1=x_n^{n+1}+\cdots+x_n=x_{n+1}^{n+2}+\cdots+x_{n+1}>x_{n+1}^{n+1}+\cdots+x_{n+1}, \eex</script> 我们有 x_n>x_{n+1}<script type="math/tex">x_n>x_{n+1}</script> (否则, ``\leq<script type="math/tex">\leq</script>‘‘). 于是 \sed{x_n}<script type="math/tex">\sed{x_n}</script> 单调递减有下界, 是收敛的.