1. 计算下列定积分: (1) $dps{int_{-pi}^pi frac{xsin x arctan e^x}{1+cos^2x} d x}$; (2) $dps{int_{frac{1}{2}}^2 sex{1+x-frac{1}{x}}e^{x+frac{1}{x}} d x}$.
解答: $$eex ea &quadint_{-pi}^pi frac{xsin x arctan e^x}{1+cos^2x} d x =int_{-pi}^pifrac{-tsin (-t)arctan e^{-t}}{1+cos^2(-t)} d (-t)quad(x=-t)\ &=int_{-pi}^pi frac{xsin xarctan e^{-x}}{1+cos^2x} d x =frac{1}{2}int_{-pi}^pi frac{xsin x}{1+cos^2x}[arctan e^x+arctan e^{-x}] d x\ &=frac{pi}{4}int_{-pi}^pifrac{xsin x}{1+cos^2x} d x quadsex{y(t)=arctan t+arctan t^{-1} a y'(t)=0 a y(t)=y(1)=frac{pi}{2}}\ &=frac{pi}{4} sez{int_{-pi}^0frac{xsin x}{1+cos^2x} d x +int_0^pifrac{xsin x}{1+cos^2x} d x}\ &=frac{pi}{4} sez{ int_0^pifrac{(t-pi)sin(t-pi)}{1+cos^2(t-pi)} d t +int_0^pifrac{xsin x}{1+cos^2x} d x }quadsex{x+pi=t}\ &=frac{pi^2}{4}int_0^pi frac{sin x}{1+cos^2x} d x =frac{pi^2}{4}arctan(cos x)|_0^pi =frac{pi^3}{8}. eea eeex$$ (2) $$eex ea int_{frac{1}{2}}^2 sex{1+x-frac{1}{x}}e^{x+frac{1}{x}} d x &=int_frac{1}{2}^2 sez{sex{1+x}sex{1-frac{1}{x^2}}+frac{1}{x^2}}e^{x+frac{1}{x}} d x\ &=int_{frac{1}{2}}^2 (1+x) d e^{x+frac{1}{x}} +int_{frac{1}{2}}^2 frac{1}{x^2}e^{x+frac{1}{x}} d x\ &=frac{3}{2}e^{frac{5}{2}} -int_{frac{1}{2}}^2 e^{x+frac{1}{x}} d x +int_{frac{1}{2}}^2 frac{1}{x^2}e^{x+frac{1}{x}} d x\ &=frac{3}{2}e^{frac{5}{2}} -int_{frac{1}{2}}^2 d e^{x+frac{1}{x}}\ &=frac{3}{2}e^{frac{5}{2}}. eea eeex$$
2. 设当 $x>-1$ 时, 可微函数 $f(x)$ 满足条件 $$ex f'(x)+f(x)-cfrac{1}{x+1}int_0^x f(t) d t=0, eex$$ 且 $f(0)=1$. 试证: 当 $xgeq 0$ 时, $$ex e^{-x}leq f(x)leq 1. eex$$
证明: $$eex ea &quad f'(x)+f(x)-cfrac{1}{x+1}int_0^x f(t) d t=0\ & a f'(x)+f(0)+int_0^x f'(s) d s -cfrac{1}{x+1}int_0^x sez{f(0)+int_0^t f'(s) d s} d t=0\ & a f'(x)+1+int_0^x f'(s) d s -cfrac{x}{x+1}-cfrac{1}{x+1}int_0^x (x-s)f'(s) d s=0\ & a f'(x)+cfrac{1}{x+1}+cfrac{1}{x+1}int_0^x (s+1)f'(s) d s=0\ & a (x+1)f'(x)+1+int_0^x(s+1)f'(s) d s=0\ & a F'(x)+1+F(x)=0quad sex{F(x)=int_0^x (s+1)f'(s) d s}\ & a [e^xF(x)]'=-e^x\ & a e^xF(x)=1-e^x\ & a F(x)=e^{-x}-1\ & a (x+1)f'(x)=-e^{-x}\ & a f'(x)=-cfrac{e^{-x}}{x+1}\ & a -e^{-x}leq f'(x)leq 0\ & a e^{-x}leq f(x)=f(0)+int_0^xf'(t) d tleq 1. eea eeex$$
3. 设 $f:[0,1] o [-a,b]$ 连续, 且 $dps{int_0^1 f^2(x) d x=ab}$. 试证: $$ex 0leq frac{1}{b-a}int_0^1 f(x) d xleq frac{1}{4}sex{frac{a+b}{a-b}}^2. eex$$
证明: 对 $forall 0leq cleq b$, $$ex -a-cleq f(x)-cleq b-c. eex$$ 而 $$ex 0leq |f(x)-c|leq maxsed{a+c,b-c}. eex$$ 取 $c=cfrac{b-a}{2}$, 则 $$eex ea 0&leq sev{f(x)-cfrac{b-a}{2}}leq cfrac{a+b}{2},\ 0&leq int_0^1 sev{f(x)-cfrac{b-a}{2}}^2 d xleq cfrac{(a+b)^2}{4},\ 0&leq ab-(b-a)int_0^1 f(x) d x+cfrac{(b-a)^2}{4}leq cfrac{(a+b)^2}{4},\ 0&=cfrac{ab+cfrac{(b-a)^2}{4}-cfrac{(a+b)^2}{4}}{b-a}leq int_0^1 f(x) d x leq cfrac{ab+cfrac{(b-a)^2}{4}}{b-a}=cfrac{(a+b)^2}{4(b-a)}. eea eeex$$
4. 求极限: $$ex vlm{n}sez{sqrt{n}sex{sqrt{n+1}-sqrt{n}}+cfrac{1}{2}}^{cfrac{sqrt{n+1}+sqrt{n}}{sqrt{n+1}-sqrt{n}}}. eex$$
解答: $$eex ea mbox{原极限} &=vlm{n}sez{cfrac{sqrt{n}}{sqrt{n+1}+sqrt{n}}+cfrac{1}{2}} ^{(sqrt{n+1}+sqrt{n})^2}\ &=expsez{vlm{n} (sqrt{n+1}+sqrt{n})^2ln sex{cfrac{sqrt{n}}{sqrt{n+1}+sqrt{n}}+cfrac{1}{2}} }\ &=expsez{vlm{n} cfrac{ln sex{cfrac{sqrt{n}}{sqrt{n+1}+sqrt{n}}+cfrac{1}{2}}} {(sqrt{n+1}-sqrt{n})^2} }\ &=expsez{vlm{n} cfrac{cfrac{sqrt{n}}{sqrt{n+1}+sqrt{n}}-cfrac{1}{2}} {(sqrt{n+1}-sqrt{n})^2} }quadsex{ln(1+x)sim x (x o 0)}\ &=expsez{vlm{n} cfrac{ cfrac{sqrt{n}-sqrt{n+1}}{2(sqrt{n+1}+sqrt{n})} }{ (sqrt{n+1}-sqrt{n})^2 } }\ &=e^{-frac{1}{2}}. eea eeex$$
5. 设 $a_1,cdots,a_n$ 为非负实数. 试证： $$ex sev{sum_{k=1}^n a_ksin kx}leq sev{sin x},quad forall xinbR eex$$ 的充分必要条件是 $$ex sum_{k=1}^n ka_kleq 1. eex$$
证明: $ a$: $$eex ea 1&geq lim_{x o 0}sev{cfrac{sum_{k=1}^n a_ksin kx}{sin x}}\ &=sev{sum_{k=1}^n a_klim_{x o 0} cfrac{sin kx}{sin x}}\ &=sev{sum_{k=1}^n ka_k}. eea eeex$$ $la$: 先用数学归纳法证明 $$ex |sin kx|leq k|sin x|,quad forall xinbR. eex$$ 当 $k=1$ 时, 结论显然成立. 假设当 $k=n$ 时结论成立, 则 $$eex ea |sin (n+1)x|&=|sin nxcos x+cos nxsin x|\ &leq n|sin x|+|sin x|\ &=(n+1)|sin x|. eea eeex$$ 往证充分性: $$eex ea sev{sum_{k=1}^n a_ksin kx} &leq sum_{k=1}^n a_k|sin kx|\ &leq sum_{k=1}^n ka_k|sin x|\ &leq |sin x|. eea eeex$$
6. 设 $f(x)$ 在 $[-1,1]$ 上二阶连续可微, 试证: 存在 $xiin (-1,1)$ 使得 $$ex int_{-1}^1 xf(x) d x=cfrac{2}{3}f''(xi)+cfrac{1}{3}xi f''(xi). eex$$
证明: 仅须证明 $$ex int_{-1}^1 xf(x) d x=(xf(x))''|_{x=xi}. eex$$ 为此, 记 $g(x)=xf(x)$, 则 $g(0)=0$, $g'(0)=f(0)$. 于是 $$eex ea g(x)&=int_0^x g'(t) d t\ &=int_0^x sez{f(0)+int_0^t g''(s) d s} d t\ &=f(0)x+int_0^x int_0^t g''(s) d s d t. eea eeex$$ 积分而有 $$eex ea int_{-1}^1 g(x) d x &=int_{-1}^1 int_0^x int_0^t g''(s) d s d t d x\ &=iiint_Omega g''(s) d s d t d x\ &quadsex{Omega=sed{(x,t,s); {{0leq xleq 1, 0leq tleq x, 0leq sleq t}atop{-1leq xleq 0, -xleq tleq 0, -tleq sleq 0}}}}\ &=g''(xi)|Omega|\ &=g''(xi)cdot 2int_0^1 d xint_0^x d tint_0^t d s\ &=frac{1}{3}g''(xi). eea eeex$$
7. 已知 $f(x)$ 在 $[0,1]$ 上三阶可导, 且 $$ex f(0)=-1,quad f(1)=0,quad f'(0)=0. eex$$ 试证: $$ex forall xin (0,1), exists xiin (0,1),st f(x)=-1+x^2+cfrac{x^2(x-1)}{3!}f'''(xi). eex$$
证明: 令 $$ex F(t)=f(t)+1-t^2-cfrac{f(x)+1-x^2}{x^2(x-1)}t^2(t-1), eex$$ 则 $F(0)=F(x)=F(1)=0$. 由 Rolle 定理, $$ex exists 0<eta<x<zeta<1,st F'(eta)=F'(zeta)=0. eex$$ 又 $F'(0)=0$, 对条件 $$ex F'(0)=F'(eta)=F'(zeta)=0 eex$$ 应用 Rolle 定理两次后即可发现 $$ex exists xiin (0,1),st F'''(xi)=0. eex$$
8. 设 $f(x)$ 在 $[0,1]$ 上连续, 试证: $$ex sev{int_0^1 cfrac{f(x)}{t^2+x^2} d x}^2leq cfrac{pi}{2t}int_0^1 cfrac{f^2(x)}{t^2+x^2} d x,quad t>0. eex$$
证明: $$eex ea LHS&=sez{int_0^1 cfrac{f(x)}{sqrt{t^2+x^2}}cdot cfrac{1}{sqrt{t^2+x^2}} d x}^2\ &leq int_0^1 cfrac{f^2(x)}{t^2+x^2} d xcdot int_0^1 cfrac{1}{t^2+x^2} d x\ &leq int_0^1 cfrac{f^2(x)}{t^2+x^2} d xcdot cfrac{1}{t} int_0^infty cfrac{1}{1+sex{cfrac{x}{t}}^2} d cfrac{x}{t}\ &leq RHS. eea eeex$$
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 $xi,etain (0,1)$, 使得 $$ex cfrac{a}{f'(xi)}+cfrac{b}{f'(eta)}=a+b. eex$$
证明: 由介值定理, $$ex exists zetain(0,1),st f(zeta)=cfrac{a}{a+b}. eex$$ 又由 Lagrange 中值定理, $$eex ea exists xiin(0,zeta),st &f'(xi)=cfrac{f(zeta)-f(0)}{zeta-0}=cfrac{a}{(a+b)zeta},\ exists etain (zeta,1),st &f'(eta)=cfrac{f(1)-f(zeta)}{1-zeta}=cfrac{b}{(a+b)(1-zeta)}. eea eeex$$ 于是 $$ex cfrac{a}{f'(xi)}+cfrac{b}{f'(zeta)} =(a+b)zeta+(a+b)(1-zeta)=a+b. eex$$