求-|||-(1) (int )_(-1)^1f(x)dx;-|||-(2)f(x)dx;-|||-(3) (int )_(3)^-1g(x)dx;-|||-(4) (int )_(-1)^3dfrac (1)(5)[ 4f(x)+3g(x)] dx.

设int_(-1)^13f(x)dx=18，int_(-1)^3f(x)dx=4，int_(-1)^3g(x)dx=3。则int_(-1)^3(1)/(5)[4

若f(x)的一个原函数是F(x)，则int f(3x-1)dx=()int f(3x-1)dx=()int f(3x-1)dx=()int f(3x-1)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

【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(int )_(-1)^13f(x)dx=9 , (int )_(-1)^13f(x)dx=9 , (int )_(-1)^13f(x)dx=9， 则 (in

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

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

(3)若 int f(x)dx=(x)^2+C, 则 int xf(1-(x)^2)dx= __ ;

int ((x)^3+1-dfrac (1)(x))dx=（ ） A int ((x)^3+1-dfrac (1)(x))dx= B int ((