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

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

(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 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

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

设 f(2)=4 , (int )_(0)^2f(x)dx=1, 则 (int )_(0)^2xf(x)dx=

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

[例9]设f(x)在[0,1]上连续, (0)=0, (int )_(0)^1f(x)dx=0.-|||-求证:存在 xi in (0,1), 使 (int )

设f(x)为连续函数，则(int )_(0)^1f(dfrac (x)(2))dx等于( )．(int )_(0)^1f(dfrac (x)(2))dx设f(x

求int xsqrt (x-1)dx.求.