数值求积公式(int )_(-1)^1f(x)dxapprox dfrac (2)(3)[ f(-dfrac (1)(sqrt {2)})+f(0)+f(dfrac (1)(sqrt {2)})] 的代数精度为（）

求积公式(int )_(0)^2f(x)dxapprox dfrac (1)(3)f(0)+dfrac (4)(3)f(1)+dfrac (1)(3)f(2)的

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

_(1)-(F)_(2)-|||-C. dfrac (2)(3)(F)_(1)+dfrac (1)(3)(F)_(2)-|||-D. dfrac (2)(3)(

设f(x)连续，xgt 0，且(int )_(1)^(x^2)f(t)dt=(x)^2(1+x)，则f(2)=(,,,,,)A、4B、2sqrt (2)+12C

16、设 (int )_(0)^xf(t)dt=dfrac (1)(2)f(x)-dfrac (1)(2), 其中f(x)为连续函数,则 f(x)=()-|||

11.设 (x)=dfrac (x)(sqrt {1+{x)^2}} 求 [ f(x)] , f[ f(x)] .

2.假设f(x)在[0,1 ]上导数连续, f(1)=0 .|f(x)|leqslant 1 ,证明: |(int )_(0)^1f(x)ds|leqslant

1.设 (x)=dfrac (1)(1-{x)^2}, 求 (-x),f[ f(x)] ,f[ dfrac (1)(f(x))]

(3)设 (x)=dfrac (x)(sqrt {{x)^2+1}}, 则 f(f(x))= __ (2012计算机