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第11章 一元函数积分学的应用(二)——积分等式与积分不等式

基础部分

  1. a>0a>0则在[0,a][0,a]上方程0x4a2t2dt+ax14a2t2dt=0\int_{0}^{x}\sqrt{4a^{2}-t^{2}}dt+\int_{a}^{x}\frac{1}{\sqrt{4a^{2}-t^{2}}}dt=0实根个数为() A. 0 B. 1 C. 2 D. 3
  2. 设函数f(x)f(x)具有二阶导数,f(x)>0f'(x)>0f(x)<0f''(x)<0,记I1=ππf(x)sinxdxI_1=\int_{-\pi}^{\pi}f(x)\sin xdxI2=ππf(x)cosxdxI_2=\int_{-\pi}^{\pi}f(x)\cos xdx,则() A. I1>0,I2<0I_1>0,I_2<0 B. I1<0,I2>0I_1<0,I_2>0 C. I1<0,I2<0I_1<0,I_2<0 D. I1>0,I2>0I_1>0,I_2>0
  3. f(x)f(x)为连续函数,T>0T>0,证明:f(x)f(x)TT为周期的充分必要条件是任给常数aaaa+Tf(x)dx\int_{a}^{a+T}f(x)dx为常数
  4. f(x)f(x)连续,证明:0x[0tf(u)du]dt=0xf(t)(xt)dt\int_{0}^{x}[\int_{0}^{t}f(u)du]dt=\int_{0}^{x}f(t)(x-t)dt
  5. f(x),g(x)f(x),g(x)连续,x[a,b]x\in[a,b],证明至少存在一点ξ(a,b)\xi\in(a,b),使f(ξ)aξg(t)dt=g(ξ)ξbf(t)dtf(\xi)\int_{a}^{\xi}g(t)dt=g(\xi)\int_{\xi}^{b}f(t)dt
  6. 证明:0+dx1+x4=0+x21+x4dx=24π\int_{0}^{+\infty}\frac{dx}{1+x^4}=\int_{0}^{+\infty}\frac{x^2}{1+x^4}dx=\frac{\sqrt{2}}{4}\pi
  7. 已知函数f(x),g(x)f(x),g(x)可导,且f(x)>0f'(x)>0g(x)<0g'(x)<0,则() A. 10f(x)g(x)dx>01f(x)g(x)dx\int_{-1}^{0}f(x)g(x)dx>\int_{0}^{1}f(x)g(x)dx B. 10f(x)g(x)dx>01f(x)g(x)dx\int_{-1}^{0}|f(x)g(x)|dx>\int_{0}^{1}|f(x)g(x)|dx C. 10f[g(x)]dx>01f[g(x)]dx\int_{-1}^{0}f[g(x)]dx>\int_{0}^{1}f[g(x)]dx D. 10f[f(x)]dx>01g[g(x)]dx\int_{-1}^{0}f[f(x)]dx>\int_{0}^{1}g[g(x)]dx
  8. f(x),g(x)f(x),g(x)[0,1][0,1]上连续,则使得01f(x)dx01g(x)dx01f(x)g(x)dx\int_{0}^{1}f(x)dx\int_{0}^{1}g(x)dx\geq\int_{0}^{1}f(x)g(x)dx成立的条件是() A. f(x),g(x)f(x),g(x)均为增函数 B. f(x),g(x)f(x),g(x)均为减函数 C. f(x)f(x)为减函数,g(x)g(x)为偶函数 D. f(x)f(x)为奇函数,g(x)g(x)为增函数
  9. f(x)f'(x)[a,b][a,b]上连续,f(a)=f(b)=0f(a)=f(b)=0,证明:当x(a,b)x\in(a,b)时,f(x)12abf(x)dx|f(x)|\leq\frac{1}{2}\int_{a}^{b}|f'(x)|dx
  10. f(x)f'(x)[0,1][0,1]上连续,且f(0)=0f(0)=0,证明:01f(x)dxM2|\int_{0}^{1}f(x)dx|\leq\frac{M}{2},其中M=max0x1f(x)M=\max\limits_{0\leq x\leq1}|f'(x)|
  11. f(x)=xx+1sinu2duf(x)=\int_{x}^{x+1}\sin u^2du,证明:当x>0x>0时,有f(x)1x|f(x)|\leq\frac{1}{x}

强化部分

  1. f(x)f(x)[0,1][0,1]上可导,当0x10\leq x\leq1时,f(x)+f2(x)0f'(x)+f^2(x)\geq0f(0)>0f(0)>0,则() A. 01f(x)dxlnf(1)f(0)\int_{0}^{1}f(x)dx\leq\ln\frac{f(1)}{f(0)} B. 01f(x)dxlnf(0)f(1)\int_{0}^{1}f(x)dx\geq\ln\frac{f(0)}{f(1)} C. 01f(x)dxlnf(1)\int_{0}^{1}f(x)dx\leq\ln f(1) D. 01f(x)dxlnf(0)\int_{0}^{1}f(x)dx\geq\ln f(0)
  2. 设函数f(x)f(x)[0,1][0,1]上的连续函数,利用分部积分法证明:01[x2xf(t)dt]dx=01(xx2)f(x)dx\int_{0}^{1}[\int_{x^2}^{\sqrt{x}}f(t)dt]dx=\int_{0}^{1}(\sqrt{x}-x^2)f(x)dx
  3. f(x)f(x)[0,1][0,1]上的可导函数,f(0)=f(1)=1f(0)=f(1)=1max0x1f(x)=1\max\limits_{0\leq x\leq1}|f'(x)|=1,则() A. 14<01f(x)dx<12\frac{1}{4}<\int_{0}^{1}f(x)dx<\frac{1}{2} B. 12<01f(x)dx<34\frac{1}{2}<\int_{0}^{1}f(x)dx<\frac{3}{4} C. 34<01f(x)dx<54\frac{3}{4}<\int_{0}^{1}f(x)dx<\frac{5}{4} D. 54<01f(x)dx<74\frac{5}{4}<\int_{0}^{1}f(x)dx<\frac{7}{4}
  4. 证明:01(xxsinttdt)dx=1sin1\int_{0}^{1}(\int_{x}^{\sqrt{x}}\frac{\sin t}{t}dt)dx=1-\sin1
  5. 设函数f(x),g(x)f(x),g(x)在区间[a,b][a,b]上连续,对任意的x[a,b]x\in[a,b],满足axf(t)dtaxg(t)dt\int_{a}^{x}f(t)dt\leq\int_{a}^{x}g(t)dt,证明:abxf(x)dxabxg(x)dx\int_{a}^{b}xf(x)dx\leq\int_{a}^{b}xg(x)dx
  6. f(x)f(x)[a,b][a,b]上可导,且f(x)f'(x)连续,f(a)=0f(a)=0,证明:abf2(x)dx(ba)22ab[f(x)]2dx\int_{a}^{b}f^2(x)dx\leq\frac{(b-a)^2}{2}\int_{a}^{b}[f'(x)]^2dx
  7. f(x)f(x)[0,1][0,1]上单调增加的连续函数,则() A. 001ex2dtf(x)dx01f(x)ex2dx\int_{0}^{\int_{0}^{1}e^{-x^2}dt}f(x)dx\geq\int_{0}^{1}f(x)e^{-x^2}dx B. 001ex2dtf(x)dx01f(x)ex2dx\int_{0}^{\int_{0}^{1}e^{-x^2}dt}f(x)dx\leq\int_{0}^{1}f(x)e^{-x^2}dx C. 001ex2dtf(x)dx01f(x)dx\int_{0}^{\int_{0}^{1}e^{-x^2}dt}f(x)dx\geq\int_{0}^{1}f(x)dx D. 001ex2dtf(x)dx01f(x)dx\int_{0}^{\int_{0}^{1}e^{-x^2}dt}f(x)dx\leq\int_{0}^{1}f(x)dx
  8. f(x)f(x)[0,32π][0,\frac{3}{2}\pi]上连续,在(0,32π)(0,\frac{3}{2}\pi)内是函数sinx\sin x的一个原函数,f(0)=0f(0)=0 (1)证明f(32π)>0f(\frac{3}{2}\pi)>0;(2)求方程1xsinttdt=lnx2\int_{1}^{x}\frac{\sin t}{t}dt=\ln x^2的实根个数
  9. f(x)f(x)[0,1][0,1]上具有二阶导数,f(0)=f(1)=0f(0)=f(1)=0f(x)<0f''(x)<00f(x)10\leq f(x)\leq1,记曲线y=f(x)y=f(x)[0,1][0,1]上的长度为aa,证明:(1)存在ξ(0,1)\xi\in(0,1),使得对任意x(0,ξ)x\in(0,\xi),有f(x)>0f'(x)>0;(2)a<3a<3