[题目]设 (x,y)=(int )_(0)^1f(t)|xy-t|dt, 其中f(t)在[0,1]上-|||-连续, leqslant xleqslant 1 ,0≤y≤1, 求 dfrac ({partial )^2u}(partial {x)^2}cdot dfrac ({partial )^2u}(partial {y)^2},

25.f(x)在[0,1]上连续,在(0,1)内可导,且 (int )_(0)^1f(t)dt=0, 证明:存在 xi in (0,1) 使得 (xi )=(i

设 f 在[0,1]上是单调增正值函数，令=dfrac ({int )_(0)^1tf(t)dt}({{int )_(0)^1}f(t)dt}，证明：=dfra

4、设 (int )_(0)^1f(x)dx=a, = (x,y)|0leqslant xleqslant 1,0leqslant yleqslant 1 则

【题目】12、设函数f(x)在 [0,1] 上连续,且 f(x)0F(x)=∫_0^xf(t)dt+∫_1^x1/(f(t))dt, x∈[0,1]证明:方程F

2.假设f(x)在[0,1 ]上导数连续, f(1)=0 .|f(x)|leqslant 1 ,证明: |(int )_(0)^1f(x)ds|leqslant

int_(0)^1f^2(x)dxleqslantint_(0)^1xdxcdotint_(0)^1f^prime(}^2(t)dt=(1)/(2)int_{0

设f(x)连续,且 (x)=x+2(int )_(0)^1f(t)dt, 则 f(x)= __

[题目]设f(x)是连续函数,且 (x)=x+2(int )_(0)^1f(t)dt,-|||-则 f(x)= __

[例18]设f(x )连续,试求下列函数的导数.-|||-(1)f(t)dt;-|||-(2) (int )_(0)^x(x-t)f(t)dt ;-|||-(3

2.设函数f(t)在[0,1]上连续，且f(x)<1，证明方程 2x-int_(0)^xf(t)dt=1在(0,1)内有且仅有一实根.2.设函数f(t)在[0,