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

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

【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=

设函数f(x)在 (-infty ,+infty ) 上连续,且 (x)=(x)^2-x(int )_(0)^1f(x)dx, 则f(x)为 (-|||-

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)连续,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. (

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

8.设f(x)满足等式 (x)-f(x)=sqrt (2x-{x)^2}, 且 (1)=4, 则 (int )_(0)^1f(x)dx= __

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

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