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

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

(4)若 (int )_(0)^1f(x)dx=agt 0, 则 (int )_(0)^1dfrac (1)(sqrt {x)}f(sqrt (x))dx=()

已知(x)=dfrac (xcos x)(1+{x)^4}+(x)^2-(int )_(-1)^1f(x)dx,则(x)=dfrac (xcos x)(1+{x

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=

(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. (

设(x)=arcsin ((x-1))^2 (0)=0, 求 (int )_(0)^1f(x)dx.

【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=

求-|||-(1) (int )_(-1)^1f(x)dx;-|||-(2)f(x)dx;-|||-(3) (int )_(3)^-1g(x)dx;-|||-(