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：=dfrac ({int )_(0)^1tf(t)dt}({{int )_(0)^1}f(t)dt}.

证明 :(int )_(x)^1dfrac (dt)(1+{t)^2}=(int )_(1)^dfrac (1{x)}dfrac (dt)(1+{t)^2}(x

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

[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

3.证明: (int )_(x)^1dfrac (dt)(1+{t)^2}=(int )_(1)^dfrac (1{x)}dfrac (dt)(1+{t)^2}

3.证明: (int )_(x)^1dfrac (dt)(1+{t)^2}=(int )_(1)^dfrac (1{x)}dfrac (dt)(1+{t)^2}

(1) lim _(xarrow 0)dfrac ({({int )_(0)^x(e)^(t^2)dt)}^2}({int )_(0)^xt(e)^2(t^2)

6.证明 (x)=(int )_(1)^xsqrt (1+{t)^3}dt 在 [ -1,+infty ) 上是单调增加函数,并求 ((f)^-1)(0).

6.证明 (x)=(int )_(1)^xsqrt (1+{t)^3}dt 在 [ -1,+infty ) 上是单调增加函数,并求 ((f)^-1)(0).

[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,