设f(x)连续 则d∫(0,2x)xf(t)dt/dx=?

设f(x)连续 则d∫(0,2x)xf(t)dt/dx=? 设函数f(x)连续,则d∫xf(x^2)dx=? 设函数f(x)连续,则积分区间(0->x),d/dx{∫tf(x^2-t^2)dt} = ()A.2xf(x^2)设函数f(x)连续,则积分区间（0->x）,d/dx{∫tf(x^2-t^2)dt} = （）A.2xf(x^2)B.-2xf(x^2)C.xf(x^2)D.-xf(x^2) 设f''(x)在[0,1]上连续,f'(1)=0,且f(1)-f(2)=2,则∫(0,1)xf''(x)dx= 设f''(x)在[0,1]连续,且f(0)=1,f(2)=3,f'(2)=5,求∫[0,1]xf''(2x)dx 设f(x)连续,若f(x)满足∫（0,1）xf(t)dt=f(x)+xe^x,求f(x). 设当x>0时,函数f(x)连续且满足f(x)=x+∫(1,x)1/xf(t)dt,求f(x) 设f(x)具有二阶连续导数,求∫xf''(x)dx 高数不定积分选择：设函数f(x)连续,且∫xf(x)dx=x^2*e^x +C,则∫f(x)dx=( )A.(x+1)e^x +C B.(x-1)e^x +C C.(x+2)e^x +C D.(2-x)e^x +C 设f(x)连续,证明（积分区间为0到π）∫xf(sinx)dx=(π/2)∫f(sinx)dx 设f(x)连续,证明（积分区间为0到2π）∫xf(cosx)dx=π∫f(sinx)dx 一道数学题：设f(x)连续,满足f(x)=x+2∫0xf(t)dt(从0到x积分),求f(x).答案是1/2（e^2x-1),这是怎么做出来的, 设f(x)在[a,b]上连续,f(a)=f(b)=0,定积分f^2(x)从b到a等于1,则定积分xf(x)f'(x)=-1/2. 设f(x)满足∫[0,x]t^2f(tx)dt=xf(x)-1,求f(x) 设函数f(x)在R上的导函数为f'(x),且2f(x)+xf'(x)>x^2,则下列不等式在R内恒成立的是A.xf'(x)>0B.xf'(x)=0 设f(x)在[a,b]连续且f′(x)>0,证明∫(a,b) xf(x)dx≥(a+b)/2 ∫(a,b)f(x)dx 设L为正向圆周:(x-a)^2+(y-a)^2=R^2,函数f(x)连续且恒f(x)>0,证明：∫(L)xf(y)dy-y/f(x)dx>=2πR^2 设f(x)在[a,b]上连续,f(a)=f(b)=0,定积分f^2(x)从b到a等于1,则定积分xf(x)f'(x)等于多少A.1/2 B.1 C.0 D.-1/2