3.已知函数f(x )= {x)^2,xgt 1f(x+1)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

[题目]已知函数 (x)=(x)^2+3x+1, 则 f(x+1)=-|||-__

函数 f(x)= ) 3x-1 xgeqslant 1 (x)^2 xlt 1f(x)=()

(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.

3.已知 (2x)=x(e)^x, 则 (int )_(-1)^1|f(x)|dx= __

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