如果f(x)在[0,1]上连续,证明：∫[0->1][∫[0->x]f(t)dt]dx=∫[0->1](1-x)f(x)

如果f(x)在[0,1]上连续,证明：∫[0->1][∫[0->x]f(t)dt]dx=∫[0->1](1-x)f(x) 设函数f(x)在区间[0,1]上连续,证明∫[∫f(t)dt]dx=∫(1-x)f(x)dx 函数f(x)>0在[a,b]上连续,令F(x)=∫(0到x)f(t)dt+∫(0到x)1/f(t)dt,证明方程F(x) 设f(x)在区间【0,1】上有连续导数,证明x∈【0,1】,有|f(x)|≤∫(|f(t)|+|f′(t)|)dt f(x)连续且f(x)=x+(x^2)∫ (0,1)f(t)dt,求f(x) f(x)在[a,b]上连续,在(a,b) 内可导,且 f '(x)≤0,F(x)=1/(x-a)∫(x-a)f(t)dt 证明：设f(x)在(-∞,+∞)连续,则函数F(x)=∫(0,1)f(x+t)dt可导,并求F'(x) 设f(x)在(-无穷,+无穷)内连续,证明(d/dx)∫(0~x)(x-t)f'(t)dt=f(x)-f(a) f(x)在(0.1)上连续且单调增,证明∫[0,1]f(x)dx 设f(x)在区间[0,1]上连续,且满足f(x)=x²∫(0,1)f(t)dt+3,求∫(0,1)f(x)dx f(x)在[a,b]上连续可导,f'(x)≤0 若F(x)=1/x-a,定积分∫f(t)dt[a,x] 证明在(a,b) 那个关于定积分的题目的答案看不懂啊 设函数f(x)在区间[0,1]上连续,证明∫[∫f(t)dt]dx=∫(1-x)f(