int_(0)^1f^2(x)dxleqslantint_(0)^1xdxcdotint_(0)^1f^prime(}^2(t)dt=(1)/(2)int_{0)^1f^prime{}^2(t)dt.注 本题若再加一个条件f(1)=0，便可证明int_(0)^1f^2(x)dxleqslant(1)/(4)int_(0)^1f^prime(}^2(x)dx及int_{0)^1f^2(x)dxleqslant(1)/(8)int_(0)^1f^prime{}^2(x)dx.

(4)已知 (x)=(int )_(1)^x(e)^-(t^2)dt, 则 (int )_(0)^1f(x)dx=

[2019数学二](13)已知函数f(x)=xint_(1)^x(sin t^2)/(t)dt,则int_(0)^1f(x)dx=____[2019数学二](1

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

【例4】已知函数f(x)在[-1,2]上连续，且int_(-1)^0f(x)dx=2，int_(0)^1f(2x)dx=1，则int_(-1)^2f(x)dx=

2.(2020山东高数Ⅲ)已知函数f(x)在[-1,2]上连续，且int_(-1)^0f(x)dx=2,int_(0)^1f(2x)dx=1,则int_(-1)

设 f ( x ) 是连续奇函数且(int )_(0)^1f(x)dx=-2 则 (int )_(0)^1f(x)dx=-2设f(x)是连续奇函数且则

设(x)=dfrac (1)(1+{x)^2}+sqrt (1-{x)^2}(int )_(0)^1f(x)dx, 则 (int )_(0)^1f(x)dx=设

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

(B) (int )_(-1)^1f(x)dxlt 0.-|||-(C) (int )_(-1)^0f(x)dxgt (int )_(0)^1f(x)dx. (