【题目】-|||-设f(x)在 (-infty ,+infty ) 内连续,则 (int )_(2)^3f(x)dx+(int )_(3)^2f(t)dt+(int )_(1)^2dx= () .-|||-

[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=

[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

[题目]-|||-若 int f(x)dx=F(x)+C, 则 int f(2x-3)dx= __ .

[例5] 设函数f(x,y)连续,则 (int )_(1)^2dx(int )_(x)^2f(x,y)dy+(int )_(1)^2dy(int )_(y)^4

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

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

19.设f(x)连续,且 (int )_(0)^xtf(2x-t)dt=dfrac (1)(2)arctan (x)^2 (1)=1, 则 (int )_(1)

设函数f(x)在 (-infty ,+infty ) 上连续,且 (x)=(x)^2-x(int )_(0)^1f(x)dx, 则f(x)为 (-|||-

设函数 f(x) 连续，则 (d)/(dx) int_(0)^x t f(x^2-t^2)dt = ( )A. $xf\left(x^{2}\right)$.B

【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=