1. 计算下列定积分: (1) $\dps{\int_{-\pi}^\pi \frac{x\sin x \arctan e^x}{1+\cos^2x}\rd x}$; (2) $\dps{\int_{\frac{1}{2}}^2 \sex{1+x-\frac{1}{x}}e^{x+\frac{1}{x}}\rd x}$.
解答: $$\beex \bea &\quad\int_{-\pi}^\pi \frac{x\sin x \arctan e^x}{1+\cos^2x}\rd x =\int_{-\pi}^\pi\frac{-t\sin (-t)\arctan e^{-t}}{1+\cos^2(-t)}\rd (-t)\quad(x=-t)\\ &=\int_{-\pi}^\pi \frac{x\sin x\arctan e^{-x}}{1+\cos^2x}\rd x =\frac{1}{2}\int_{-\pi}^\pi \frac{x\sin x}{1+\cos^2x}[\arctan e^x+\arctan e^{-x}]\rd x\\ &=\frac{\pi}{4}\int_{-\pi}^\pi\frac{x\sin x}{1+\cos^2x}\rd x \quad\sex{y(t)=\arctan t+\arctan t^{-1}\ra y'(t)=0\ra y(t)=y(1)=\frac{\pi}{2}}\\ &=\frac{\pi}{4} \sez{\int_{-\pi}^0\frac{x\sin x}{1+\cos^2x}\rd x +\int_0^\pi\frac{x\sin x}{1+\cos^2x}\rd x}\\ &=\frac{\pi}{4} \sez{ \int_0^\pi\frac{(t-\pi)\sin(t-\pi)}{1+\cos^2(t-\pi)}\rd t +\int_0^\pi\frac{x\sin x}{1+\cos^2x}\rd x }\quad\sex{x+\pi=t}\\ &=\frac{\pi^2}{4}\int_0^\pi \frac{\sin x}{1+\cos^2x}\rd x =\frac{\pi^2}{4}\arctan(\cos x)|_0^\pi =\frac{\pi^3}{8}. \eea \eeex$$ (2) $$\beex \bea \int_{\frac{1}{2}}^2 \sex{1+x-\frac{1}{x}}e^{x+\frac{1}{x}}\rd x &=\int_\frac{1}{2}^2 \sez{\sex{1+x}\sex{1-\frac{1}{x^2}}+\frac{1}{x^2}}e^{x+\frac{1}{x}}\rd x\\ &=\int_{\frac{1}{2}}^2 (1+x)\rd e^{x+\frac{1}{x}} +\int_{\frac{1}{2}}^2 \frac{1}{x^2}e^{x+\frac{1}{x}}\rd x\\ &=\frac{3}{2}e^{\frac{5}{2}} -\int_{\frac{1}{2}}^2 e^{x+\frac{1}{x}}\rd x +\int_{\frac{1}{2}}^2 \frac{1}{x^2}e^{x+\frac{1}{x}}\rd x\\ &=\frac{3}{2}e^{\frac{5}{2}} -\int_{\frac{1}{2}}^2 \rd e^{x+\frac{1}{x}}\\ &=\frac{3}{2}e^{\frac{5}{2}}. \eea \eeex$$
2. 设当 $x>-1$ 时, 可微函数 $f(x)$ 满足条件 $$\bex f'(x)+f(x)-\cfrac{1}{x+1}\int_0^x f(t)\rd t=0, \eex$$ 且 $f(0)=1$. 试证: 当 $x\geq 0$ 时, $$\bex e^{-x}\leq f(x)\leq 1. \eex$$
证明: $$\beex \bea &\quad f'(x)+f(x)-\cfrac{1}{x+1}\int_0^x f(t)\rd t=0\\ &\ra f'(x)+f(0)+\int_0^x f'(s)\rd s -\cfrac{1}{x+1}\int_0^x \sez{f(0)+\int_0^t f'(s)\rd s}\rd t=0\\ &\ra f'(x)+1+\int_0^x f'(s)\rd s -\cfrac{x}{x+1}-\cfrac{1}{x+1}\int_0^x (x-s)f'(s)\rd s=0\\ &\ra f'(x)+\cfrac{1}{x+1}+\cfrac{1}{x+1}\int_0^x (s+1)f'(s)\rd s=0\\ &\ra (x+1)f'(x)+1+\int_0^x(s+1)f'(s)\rd s=0\\ &\ra F'(x)+1+F(x)=0\quad \sex{F(x)=\int_0^x (s+1)f'(s)\rd s}\\ &\ra [e^xF(x)]'=-e^x\\ &\ra e^xF(x)=1-e^x\\ &\ra F(x)=e^{-x}-1\\ &\ra (x+1)f'(x)=-e^{-x}\\ &\ra f'(x)=-\cfrac{e^{-x}}{x+1}\\ &\ra -e^{-x}\leq f'(x)\leq 0\\ &\ra e^{-x}\leq f(x)=f(0)+\int_0^xf'(t)\rd t\leq 1. \eea \eeex$$
3. 设 $f:[0,1]\to [-a,b]$ 连续, 且 $\dps{\int_0^1 f^2(x)\rd x=ab}$. 试证: $$\bex 0\leq \frac{1}{b-a}\int_0^1 f(x)\rd x\leq \frac{1}{4}\sex{\frac{a+b}{a-b}}^2. \eex$$
证明: 对 $\forall\ 0\leq c\leq b$, $$\bex -a-c\leq f(x)-c\leq b-c. \eex$$ 而 $$\bex 0\leq |f(x)-c|\leq \max\sed{a+c,b-c}. \eex$$ 取 $c=\cfrac{b-a}{2}$, 则 $$\beex \bea 0&\leq \sev{f(x)-\cfrac{b-a}{2}}\leq \cfrac{a+b}{2},\\ 0&\leq \int_0^1 \sev{f(x)-\cfrac{b-a}{2}}^2\rd x\leq \cfrac{(a+b)^2}{4},\\ 0&\leq ab-(b-a)\int_0^1 f(x)\rd x+\cfrac{(b-a)^2}{4}\leq \cfrac{(a+b)^2}{4},\\ 0&=\cfrac{ab+\cfrac{(b-a)^2}{4}-\cfrac{(a+b)^2}{4}}{b-a}\leq \int_0^1 f(x)\rd x \leq \cfrac{ab+\cfrac{(b-a)^2}{4}}{b-a}=\cfrac{(a+b)^2}{4(b-a)}. \eea \eeex$$
4. 求极限: $$\bex \vlm{n}\sez{\sqrt{n}\sex{\sqrt{n+1}-\sqrt{n}}+\cfrac{1}{2}}^{\cfrac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}}. \eex$$
解答: $$\beex \bea \mbox{原极限} &=\vlm{n}\sez{\cfrac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}+\cfrac{1}{2}} ^{(\sqrt{n+1}+\sqrt{n})^2}\\ &=\exp\sez{\vlm{n} (\sqrt{n+1}+\sqrt{n})^2\ln \sex{\cfrac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}+\cfrac{1}{2}} }\\ &=\exp\sez{\vlm{n} \cfrac{\ln \sex{\cfrac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}+\cfrac{1}{2}}} {(\sqrt{n+1}-\sqrt{n})^2} }\\ &=\exp\sez{\vlm{n} \cfrac{\cfrac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}-\cfrac{1}{2}} {(\sqrt{n+1}-\sqrt{n})^2} }\quad\sex{\ln(1+x)\sim x\ (x\to 0)}\\ &=\exp\sez{\vlm{n} \cfrac{ \cfrac{\sqrt{n}-\sqrt{n+1}}{2(\sqrt{n+1}+\sqrt{n})} }{ (\sqrt{n+1}-\sqrt{n})^2 } }\\ &=e^{-\frac{1}{2}}. \eea \eeex$$
5. 设 $a_1,\cdots,a_n$ 为非负实数. 试证： $$\bex \sev{\sum_{k=1}^n a_k\sin kx}\leq \sev{\sin x},\quad \forall\ x\in\bbR \eex$$ 的充分必要条件是 $$\bex \sum_{k=1}^n ka_k\leq 1. \eex$$
证明: $\ra$: $$\beex \bea 1&\geq \lim_{x\to 0}\sev{\cfrac{\sum_{k=1}^n a_k\sin kx}{\sin x}}\\ &=\sev{\sum_{k=1}^n a_k\lim_{x\to 0} \cfrac{\sin kx}{\sin x}}\\ &=\sev{\sum_{k=1}^n ka_k}. \eea \eeex$$ $\la$: 先用数学归纳法证明 $$\bex |\sin kx|\leq k|\sin x|,\quad \forall\ x\in\bbR. \eex$$ 当 $k=1$ 时, 结论显然成立. 假设当 $k=n$ 时结论成立, 则 $$\beex \bea |\sin (n+1)x|&=|\sin nx\cos x+\cos nx\sin x|\\ &\leq n|\sin x|+|\sin x|\\ &=(n+1)|\sin x|. \eea \eeex$$ 往证充分性: $$\beex \bea \sev{\sum_{k=1}^n a_k\sin kx} &\leq \sum_{k=1}^n a_k|\sin kx|\\ &\leq \sum_{k=1}^n ka_k|\sin x|\\ &\leq |\sin x|. \eea \eeex$$
6. 设 $f(x)$ 在 $[-1,1]$ 上二阶连续可微, 试证: 存在 $\xi\in (-1,1)$ 使得 $$\bex \int_{-1}^1 xf(x)\rd x=\cfrac{2}{3}f''(\xi)+\cfrac{1}{3}\xi f''(\xi). \eex$$
证明: 仅须证明 $$\bex \int_{-1}^1 xf(x)\rd x=(xf(x))''|_{x=\xi}. \eex$$ 为此, 记 $g(x)=xf(x)$, 则 $g(0)=0$, $g'(0)=f(0)$. 于是 $$\beex \bea g(x)&=\int_0^x g'(t)\rd t\\ &=\int_0^x \sez{f(0)+\int_0^t g''(s)\rd s}\rd t\\ &=f(0)x+\int_0^x \int_0^t g''(s)\rd s\rd t. \eea \eeex$$ 积分而有 $$\beex \bea \int_{-1}^1 g(x)\rd x &=\int_{-1}^1 \int_0^x \int_0^t g''(s)\rd s\rd t\rd x\\ &=\iiint_\Omega g''(s)\rd s\rd t\rd x\\ &\quad\sex{\Omega=\sed{(x,t,s);\ {{0\leq x\leq 1,\ 0\leq t\leq x,\ 0\leq s\leq t}\atop{-1\leq x\leq 0,\ -x\leq t\leq 0,\ -t\leq s\leq 0}}}}\\ &=g''(\xi)|\Omega|\\ &=g''(\xi)\cdot 2\int_0^1 \rd x\int_0^x \rd t\int_0^t\rd s\\ &=\frac{1}{3}g''(\xi). \eea \eeex$$
7. 已知 $f(x)$ 在 $[0,1]$ 上三阶可导, 且 $$\bex f(0)=-1,\quad f(1)=0,\quad f'(0)=0. \eex$$ 试证: $$\bex \forall\ x\in (0,1),\ \exists\ \xi\in (0,1),\st f(x)=-1+x^2+\cfrac{x^2(x-1)}{3!}f'''(\xi). \eex$$
证明: 令 $$\bex F(t)=f(t)+1-t^2-\cfrac{f(x)+1-x^2}{x^2(x-1)}t^2(t-1), \eex$$ 则 $F(0)=F(x)=F(1)=0$. 由 Rolle 定理, $$\bex \exists\ 0<\eta<x<\zeta<1,\st F'(\eta)=F'(\zeta)=0. \eex$$ 又 $F'(0)=0$, 对条件 $$\bex F'(0)=F'(\eta)=F'(\zeta)=0 \eex$$ 应用 Rolle 定理两次后即可发现 $$\bex \exists\ \xi\in (0,1),\st F'''(\xi)=0. \eex$$
8. 设 $f(x)$ 在 $[0,1]$ 上连续, 试证: $$\bex \sev{\int_0^1 \cfrac{f(x)}{t^2+x^2}\rd x}^2\leq \cfrac{\pi}{2t}\int_0^1 \cfrac{f^2(x)}{t^2+x^2}\rd x,\quad t>0. \eex$$
证明: $$\beex \bea LHS&=\sez{\int_0^1 \cfrac{f(x)}{\sqrt{t^2+x^2}}\cdot \cfrac{1}{\sqrt{t^2+x^2}}\rd x}^2\\ &\leq \int_0^1 \cfrac{f^2(x)}{t^2+x^2}\rd x\cdot \int_0^1 \cfrac{1}{t^2+x^2}\rd x\\ &\leq \int_0^1 \cfrac{f^2(x)}{t^2+x^2}\rd x\cdot \cfrac{1}{t} \int_0^\infty \cfrac{1}{1+\sex{\cfrac{x}{t}}^2}\rd \cfrac{x}{t}\\ &\leq RHS. \eea \eeex$$
9. 设 $f(x)$ 在 $[0,1]$ 上连续, 在 $(0,1)$ 内可导, 且 $f(0)=f(1)=0$. 试证: 对任意正数 $a,b$, 均存在不同的两点 $\xi,\eta\in (0,1)$, 使得 $$\bex \cfrac{a}{f'(\xi)}+\cfrac{b}{f'(\eta)}=a+b. \eex$$
证明: 由介值定理, $$\bex \exists\ \zeta\in(0,1),\st f(\zeta)=\cfrac{a}{a+b}. \eex$$ 又由 Lagrange 中值定理, $$\beex \bea \exists\ \xi\in(0,\zeta),\st &f'(\xi)=\cfrac{f(\zeta)-f(0)}{\zeta-0}=\cfrac{a}{(a+b)\zeta},\\ \exists\ \eta\in (\zeta,1),\st &f'(\eta)=\cfrac{f(1)-f(\zeta)}{1-\zeta}=\cfrac{b}{(a+b)(1-\zeta)}. \eea \eeex$$ 于是 $$\bex \cfrac{a}{f'(\xi)}+\cfrac{b}{f'(\zeta)} =(a+b)\zeta+(a+b)(1-\zeta)=a+b. \eex$$
