设f(x)=1,且f(0)=0,则∫f(x)dx=

一道高数题,设函数f(x)在[0,+∞)上连续,且f(x)=x(e^-x)+(e^x)∫(0,1) f(x)dx,则f( ,设f(x)是连续函数,且f(x)=x+2∫[1,0]f(t)dt ,则∫[1,0]f(x)dx=? 设f'(x)=cosx/(1+sinx^2),且f(0)=0,则∫f'(x)/(1+f(x)^2)dx= 设f(x)为连续函数,且满足f(x)=3x^2-x∫（1,0）f(x)dx求f(x) 设f(x)为连续函数,且满足f(x)=1+[(1-x^2)^1/2]*∫﹙0→1﹚f(x)dx,求f(x) 设函数f(x)在(-∞,+∞)上连续,且f(x)=e^x+1/e∫(0,1)f(x)dx,求f(x) 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)为连续函数,且满足设f(x)=x+∫(0,1)xf(x)dx,求f(x) 设f（x）=e^x,则∫(0,1)f＇(x)f＇＇(x)dx=? 设2f(x)cos x=d/dx [f(x)]²,f(0)=1,则f(x)= 设f(x)在[0,1]上连续,且单调不增,证明∫(α,0)f(x)dx>=α∫(1,0)f(x)dx （0 设f(x)是连续函数,且满足∫[0,x]f(x-t)dt=e^(-2x)-1,求定积分∫[0,1]f(x)dx