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

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=()

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 (-|||-

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=

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

2.设 (x)=2, 且 f(0)=0 ,则 int f(x)dx= __-|||-

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

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