设f(x)连续,则 dfrac (d)(dx)(int )_(0)^xtf((x)^2-(t)^2)dt= ()-|||-A、xf(x^2)-|||-B、 -xf((x)^2)-|||-C、2xf(x^2)-|||-D、 -2xf((x)^2)

设f(x)为连续函数，则(d)/(dx)int_(0)^2xtf(x^2+t^2)dt=（） (A.)2xf(5x^2) (B.)2xf(5x^2)-

设(x)=dfrac ({x)^2}(x-a)(int )_(a)^xf(t)dt, 其中f(x)设(x)=dfrac ({x)^2}(x-a)(int )_(

(8)设f(x)连续,则 dfrac (d)(dx)(int )_(0)^xtf((x)^2-(t)^2)dt= __

[题目]设函数f(x)连续,则 dfrac (d)(dx)(int )_(0)^xtf((t)^2-(x)^2)dt= __-|||-_.

16、设 (int )_(0)^xf(t)dt=dfrac (1)(2)f(x)-dfrac (1)(2), 其中f(x)为连续函数,则 f(x)=()-|||

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

(2)若连续函数f(x)满足关系式 (x)=(int )_(0)^2xf(dfrac (t)(2))dt+ln 2, 则 f(x)为 ()-|||-(A)e^x

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

10 若 f(x)= int_(0)^2xf((t)/(2))dt+4, 则 int_(0)^pi f(x) sin xdx= ___.10 若 $f(x)=

分) 设 (x)=(int )_({x)^2}dfrac (t)(sqrt {1+{t)^3}}dt 求 =(int )_(0)^1xF(x)dx