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

[题目]设连续函数f(x)满足 (x)=(x)^2-(int )_(0)^2f(x)dx ,则-|||-(int )_(0)^2f(x)dx=

2、设f`(x)为连续函数,则 int (sin )^2xf(cos x)dx-int cos xf(cos x)dx=

3.已知 (int )_(0)^2f(x)dx=3 ,则 (int )_(0)^2[ f(x)+6] dx= __ 一

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

【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(x)为连续函数，则(int )_(0)^1f(dfrac (x)(2))dx等于( )．(int )_(0)^1f(dfrac (x)(2))dx设f(x

已知 f(0)=1 ,f(2)=3 ,f(2)=5 ,则 (int )_(0)^2xf(x)dx= ()-|||-(A)12; (B)8; (C)7; (D)6

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

[2023年真题]设连续函数f(x)满足： f(x+2)-f(x)=x,int_(0)^2f(x)dx=0,则 int_(1)^3f(x)dx=[2023年真题

已知 (x)cdot (int )_(0)^2f(x)dx=8, 且 (0)=0, (x)geqslant 0, 则 f(x)= __