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

(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

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=

[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 )

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

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

24.证明: (int )_(0)^1dx(int )_(0)^xf(x-y)dy=(int )_(0)^1f(x)(1-x)dx-|||-"f(x-y)dy=

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