张宇1000题 第11章 一元函数积分学的应用(二)——积分等式与积分不等式 On this page
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
设a > 0 a>0 a > 0 则在[ 0 , a ] [0,a] [ 0 , a ] 上方程∫ 0 x 4 a 2 − t 2 d t + ∫ a x 1 4 a 2 − t 2 d t = 0 \int_{0}^{x}\sqrt{4a^{2}-t^{2}}dt+\int_{a}^{x}\frac{1}{\sqrt{4a^{2}-t^{2}}}dt=0 ∫ 0 x 4 a 2 − t 2 d t + ∫ a x 4 a 2 − t 2 1 d t = 0 实根个数为()
A. 0 B. 1 C. 2 D. 3
设函数f ( x ) f(x) f ( x ) 具有二阶导数,f ′ ( x ) > 0 f'(x)>0 f ′ ( x ) > 0 ,f ′ ′ ( x ) < 0 f''(x)<0 f ′′ ( x ) < 0 ,记I 1 = ∫ − π π f ( x ) sin x d x I_1=\int_{-\pi}^{\pi}f(x)\sin xdx I 1 = ∫ − π π f ( x ) sin x d x ,I 2 = ∫ − π π f ( x ) cos x d x I_2=\int_{-\pi}^{\pi}f(x)\cos xdx I 2 = ∫ − π π f ( x ) cos x d x ,则()
A. I 1 > 0 , I 2 < 0 I_1>0,I_2<0 I 1 > 0 , I 2 < 0 B. I 1 < 0 , I 2 > 0 I_1<0,I_2>0 I 1 < 0 , I 2 > 0 C. I 1 < 0 , I 2 < 0 I_1<0,I_2<0 I 1 < 0 , I 2 < 0 D. I 1 > 0 , I 2 > 0 I_1>0,I_2>0 I 1 > 0 , I 2 > 0
设f ( x ) f(x) f ( x ) 为连续函数,T > 0 T>0 T > 0 ,证明:f ( x ) f(x) f ( x ) 以T T T 为周期的充分必要条件是任给常数a a a ,∫ a a + T f ( x ) d x \int_{a}^{a+T}f(x)dx ∫ a a + T f ( x ) d x 为常数
设f ( x ) f(x) f ( x ) 连续,证明:∫ 0 x [ ∫ 0 t f ( u ) d u ] d t = ∫ 0 x f ( t ) ( x − t ) d t \int_{0}^{x}[\int_{0}^{t}f(u)du]dt=\int_{0}^{x}f(t)(x-t)dt ∫ 0 x [ ∫ 0 t f ( u ) d u ] d t = ∫ 0 x f ( t ) ( x − t ) d t
设f ( x ) , g ( x ) f(x),g(x) f ( x ) , g ( x ) 连续,x ∈ [ a , b ] x\in[a,b] x ∈ [ a , b ] ,证明至少存在一点ξ ∈ ( a , b ) \xi\in(a,b) ξ ∈ ( a , b ) ,使f ( ξ ) ∫ a ξ g ( t ) d t = g ( ξ ) ∫ ξ b f ( t ) d t f(\xi)\int_{a}^{\xi}g(t)dt=g(\xi)\int_{\xi}^{b}f(t)dt f ( ξ ) ∫ a ξ g ( t ) d t = g ( ξ ) ∫ ξ b f ( t ) d t
证明:∫ 0 + ∞ d x 1 + x 4 = ∫ 0 + ∞ x 2 1 + x 4 d x = 2 4 π \int_{0}^{+\infty}\frac{dx}{1+x^4}=\int_{0}^{+\infty}\frac{x^2}{1+x^4}dx=\frac{\sqrt{2}}{4}\pi ∫ 0 + ∞ 1 + x 4 d x = ∫ 0 + ∞ 1 + x 4 x 2 d x = 4 2 π
已知函数f ( x ) , g ( x ) f(x),g(x) f ( x ) , g ( x ) 可导,且f ′ ( x ) > 0 f'(x)>0 f ′ ( x ) > 0 ,g ′ ( x ) < 0 g'(x)<0 g ′ ( x ) < 0 ,则()
A. ∫ − 1 0 f ( x ) g ( x ) d x > ∫ 0 1 f ( x ) g ( x ) d x \int_{-1}^{0}f(x)g(x)dx>\int_{0}^{1}f(x)g(x)dx ∫ − 1 0 f ( x ) g ( x ) d x > ∫ 0 1 f ( x ) g ( x ) d x
B. ∫ − 1 0 ∣ f ( x ) g ( x ) ∣ d x > ∫ 0 1 ∣ f ( x ) g ( x ) ∣ d x \int_{-1}^{0}|f(x)g(x)|dx>\int_{0}^{1}|f(x)g(x)|dx ∫ − 1 0 ∣ f ( x ) g ( x ) ∣ d x > ∫ 0 1 ∣ f ( x ) g ( x ) ∣ d x
C. ∫ − 1 0 f [ g ( x ) ] d x > ∫ 0 1 f [ g ( x ) ] d x \int_{-1}^{0}f[g(x)]dx>\int_{0}^{1}f[g(x)]dx ∫ − 1 0 f [ g ( x )] d x > ∫ 0 1 f [ g ( x )] d x
D. ∫ − 1 0 f [ f ( x ) ] d x > ∫ 0 1 g [ g ( x ) ] d x \int_{-1}^{0}f[f(x)]dx>\int_{0}^{1}g[g(x)]dx ∫ − 1 0 f [ f ( x )] d x > ∫ 0 1 g [ g ( x )] d x
设f ( x ) , g ( x ) f(x),g(x) f ( x ) , g ( x ) 在[ 0 , 1 ] [0,1] [ 0 , 1 ] 上连续,则使得∫ 0 1 f ( x ) d x ∫ 0 1 g ( x ) d x ≥ ∫ 0 1 f ( x ) g ( x ) d x \int_{0}^{1}f(x)dx\int_{0}^{1}g(x)dx\geq\int_{0}^{1}f(x)g(x)dx ∫ 0 1 f ( x ) d x ∫ 0 1 g ( x ) d x ≥ ∫ 0 1 f ( x ) g ( x ) d x 成立的条件是()
A. f ( x ) , g ( x ) f(x),g(x) f ( x ) , g ( x ) 均为增函数 B. f ( x ) , g ( x ) f(x),g(x) f ( x ) , g ( x ) 均为减函数
C. f ( x ) f(x) f ( x ) 为减函数,g ( x ) g(x) g ( x ) 为偶函数 D. f ( x ) f(x) f ( x ) 为奇函数,g ( x ) g(x) g ( x ) 为增函数
设f ′ ( x ) f'(x) f ′ ( x ) 在[ a , b ] [a,b] [ a , b ] 上连续,f ( a ) = f ( b ) = 0 f(a)=f(b)=0 f ( a ) = f ( b ) = 0 ,证明:当x ∈ ( a , b ) x\in(a,b) x ∈ ( a , b ) 时,∣ f ( x ) ∣ ≤ 1 2 ∫ a b ∣ f ′ ( x ) ∣ d x |f(x)|\leq\frac{1}{2}\int_{a}^{b}|f'(x)|dx ∣ f ( x ) ∣ ≤ 2 1 ∫ a b ∣ f ′ ( x ) ∣ d x
设f ′ ( x ) f'(x) f ′ ( x ) 在[ 0 , 1 ] [0,1] [ 0 , 1 ] 上连续,且f ( 0 ) = 0 f(0)=0 f ( 0 ) = 0 ,证明:∣ ∫ 0 1 f ( x ) d x ∣ ≤ M 2 |\int_{0}^{1}f(x)dx|\leq\frac{M}{2} ∣ ∫ 0 1 f ( x ) d x ∣ ≤ 2 M ,其中M = max 0 ≤ x ≤ 1 ∣ f ′ ( x ) ∣ M=\max\limits_{0\leq x\leq1}|f'(x)| M = 0 ≤ x ≤ 1 max ∣ f ′ ( x ) ∣
设f ( x ) = ∫ x x + 1 sin u 2 d u f(x)=\int_{x}^{x+1}\sin u^2du f ( x ) = ∫ x x + 1 sin u 2 d u ,证明:当x > 0 x>0 x > 0 时,有∣ f ( x ) ∣ ≤ 1 x |f(x)|\leq\frac{1}{x} ∣ f ( x ) ∣ ≤ x 1
强化部分
设f ( x ) f(x) f ( x ) 在[ 0 , 1 ] [0,1] [ 0 , 1 ] 上可导,当0 ≤ x ≤ 1 0\leq x\leq1 0 ≤ x ≤ 1 时,f ′ ( x ) + f 2 ( x ) ≥ 0 f'(x)+f^2(x)\geq0 f ′ ( x ) + f 2 ( x ) ≥ 0 ,f ( 0 ) > 0 f(0)>0 f ( 0 ) > 0 ,则()
A. ∫ 0 1 f ( x ) d x ≤ ln f ( 1 ) f ( 0 ) \int_{0}^{1}f(x)dx\leq\ln\frac{f(1)}{f(0)} ∫ 0 1 f ( x ) d x ≤ ln f ( 0 ) f ( 1 ) B. ∫ 0 1 f ( x ) d x ≥ ln f ( 0 ) f ( 1 ) \int_{0}^{1}f(x)dx\geq\ln\frac{f(0)}{f(1)} ∫ 0 1 f ( x ) d x ≥ ln f ( 1 ) f ( 0 )
C. ∫ 0 1 f ( x ) d x ≤ ln f ( 1 ) \int_{0}^{1}f(x)dx\leq\ln f(1) ∫ 0 1 f ( x ) d x ≤ ln f ( 1 ) D. ∫ 0 1 f ( x ) d x ≥ ln f ( 0 ) \int_{0}^{1}f(x)dx\geq\ln f(0) ∫ 0 1 f ( x ) d x ≥ ln f ( 0 )
设函数f ( x ) f(x) f ( x ) 是[ 0 , 1 ] [0,1] [ 0 , 1 ] 上的连续函数,利用分部积分法证明:∫ 0 1 [ ∫ x 2 x f ( t ) d t ] d x = ∫ 0 1 ( x − x 2 ) f ( x ) d x \int_{0}^{1}[\int_{x^2}^{\sqrt{x}}f(t)dt]dx=\int_{0}^{1}(\sqrt{x}-x^2)f(x)dx ∫ 0 1 [ ∫ x 2 x f ( t ) d t ] d x = ∫ 0 1 ( x − x 2 ) f ( x ) d x
设f ( x ) f(x) f ( x ) 是[ 0 , 1 ] [0,1] [ 0 , 1 ] 上的可导函数,f ( 0 ) = f ( 1 ) = 1 f(0)=f(1)=1 f ( 0 ) = f ( 1 ) = 1 ,max 0 ≤ x ≤ 1 ∣ f ′ ( x ) ∣ = 1 \max\limits_{0\leq x\leq1}|f'(x)|=1 0 ≤ x ≤ 1 max ∣ f ′ ( x ) ∣ = 1 ,则()
A. 1 4 < ∫ 0 1 f ( x ) d x < 1 2 \frac{1}{4}<\int_{0}^{1}f(x)dx<\frac{1}{2} 4 1 < ∫ 0 1 f ( x ) d x < 2 1 B. 1 2 < ∫ 0 1 f ( x ) d x < 3 4 \frac{1}{2}<\int_{0}^{1}f(x)dx<\frac{3}{4} 2 1 < ∫ 0 1 f ( x ) d x < 4 3
C. 3 4 < ∫ 0 1 f ( x ) d x < 5 4 \frac{3}{4}<\int_{0}^{1}f(x)dx<\frac{5}{4} 4 3 < ∫ 0 1 f ( x ) d x < 4 5 D. 5 4 < ∫ 0 1 f ( x ) d x < 7 4 \frac{5}{4}<\int_{0}^{1}f(x)dx<\frac{7}{4} 4 5 < ∫ 0 1 f ( x ) d x < 4 7
证明:∫ 0 1 ( ∫ x x sin t t d t ) d x = 1 − sin 1 \int_{0}^{1}(\int_{x}^{\sqrt{x}}\frac{\sin t}{t}dt)dx=1-\sin1 ∫ 0 1 ( ∫ x x t s i n t d t ) d x = 1 − sin 1
设函数f ( x ) , g ( x ) f(x),g(x) f ( x ) , g ( x ) 在区间[ a , b ] [a,b] [ a , b ] 上连续,对任意的x ∈ [ a , b ] x\in[a,b] x ∈ [ a , b ] ,满足∫ a x f ( t ) d t ≤ ∫ a x g ( t ) d t \int_{a}^{x}f(t)dt\leq\int_{a}^{x}g(t)dt ∫ a x f ( t ) d t ≤ ∫ a x g ( t ) d t ,证明:∫ a b x f ( x ) d x ≤ ∫ a b x g ( x ) d x \int_{a}^{b}xf(x)dx\leq\int_{a}^{b}xg(x)dx ∫ a b x f ( x ) d x ≤ ∫ a b xg ( x ) d x
设f ( x ) f(x) f ( x ) 在[ a , b ] [a,b] [ a , b ] 上可导,且f ′ ( x ) f'(x) f ′ ( x ) 连续,f ( a ) = 0 f(a)=0 f ( a ) = 0 ,证明:∫ a b f 2 ( x ) d x ≤ ( b − a ) 2 2 ∫ a b [ f ′ ( x ) ] 2 d x \int_{a}^{b}f^2(x)dx\leq\frac{(b-a)^2}{2}\int_{a}^{b}[f'(x)]^2dx ∫ a b f 2 ( x ) d x ≤ 2 ( b − a ) 2 ∫ a b [ f ′ ( x ) ] 2 d x
设f ( x ) f(x) f ( x ) 是[ 0 , 1 ] [0,1] [ 0 , 1 ] 上单调增加的连续函数,则()
A. ∫ 0 ∫ 0 1 e − x 2 d t f ( x ) d x ≥ ∫ 0 1 f ( x ) e − x 2 d x \int_{0}^{\int_{0}^{1}e^{-x^2}dt}f(x)dx\geq\int_{0}^{1}f(x)e^{-x^2}dx ∫ 0 ∫ 0 1 e − x 2 d t f ( x ) d x ≥ ∫ 0 1 f ( x ) e − x 2 d x
B. ∫ 0 ∫ 0 1 e − x 2 d t f ( x ) d x ≤ ∫ 0 1 f ( x ) e − x 2 d x \int_{0}^{\int_{0}^{1}e^{-x^2}dt}f(x)dx\leq\int_{0}^{1}f(x)e^{-x^2}dx ∫ 0 ∫ 0 1 e − x 2 d t f ( x ) d x ≤ ∫ 0 1 f ( x ) e − x 2 d x
C. ∫ 0 ∫ 0 1 e − x 2 d t f ( x ) d x ≥ ∫ 0 1 f ( x ) d x \int_{0}^{\int_{0}^{1}e^{-x^2}dt}f(x)dx\geq\int_{0}^{1}f(x)dx ∫ 0 ∫ 0 1 e − x 2 d t f ( x ) d x ≥ ∫ 0 1 f ( x ) d x
D. ∫ 0 ∫ 0 1 e − x 2 d t f ( x ) d x ≤ ∫ 0 1 f ( x ) d x \int_{0}^{\int_{0}^{1}e^{-x^2}dt}f(x)dx\leq\int_{0}^{1}f(x)dx ∫ 0 ∫ 0 1 e − x 2 d t f ( x ) d x ≤ ∫ 0 1 f ( x ) d x
设f ( x ) f(x) f ( x ) 在[ 0 , 3 2 π ] [0,\frac{3}{2}\pi] [ 0 , 2 3 π ] 上连续,在( 0 , 3 2 π ) (0,\frac{3}{2}\pi) ( 0 , 2 3 π ) 内是函数sin x \sin x sin x 的一个原函数,f ( 0 ) = 0 f(0)=0 f ( 0 ) = 0
(1)证明f ( 3 2 π ) > 0 f(\frac{3}{2}\pi)>0 f ( 2 3 π ) > 0 ;(2)求方程∫ 1 x sin t t d t = ln x 2 \int_{1}^{x}\frac{\sin t}{t}dt=\ln x^2 ∫ 1 x t s i n t d t = ln x 2 的实根个数
设f ( x ) f(x) f ( x ) 在[ 0 , 1 ] [0,1] [ 0 , 1 ] 上具有二阶导数,f ( 0 ) = f ( 1 ) = 0 f(0)=f(1)=0 f ( 0 ) = f ( 1 ) = 0 ,f ′ ′ ( x ) < 0 f''(x)<0 f ′′ ( x ) < 0 ,0 ≤ f ( x ) ≤ 1 0\leq f(x)\leq1 0 ≤ f ( x ) ≤ 1 ,记曲线y = f ( x ) y=f(x) y = f ( x ) 在[ 0 , 1 ] [0,1] [ 0 , 1 ] 上的长度为a a a ,证明:(1)存在ξ ∈ ( 0 , 1 ) \xi\in(0,1) ξ ∈ ( 0 , 1 ) ,使得对任意x ∈ ( 0 , ξ ) x\in(0,\xi) x ∈ ( 0 , ξ ) ,有f ′ ( x ) > 0 f'(x)>0 f ′ ( x ) > 0 ;(2)a < 3 a<3 a < 3