f(x)在〔0,1〕内连续,证明:∫{0,1}∫{x,1}∫{x,y}f(x)f(y)f(z)dxdydz={[∫{0,
f(x)在〔0,1〕内连续,证明:∫{0,1}∫{x,1}∫{x,y}f(x)f(y)f(z)dxdydz={[∫{0,
设函数f(x)在[0,1]上连续,证明:∫(0->1)dx∫(0->1)dy∫(x->y)f(x)f(y)f(z)dz=
设函数f(x)在(-∞,+∞)内有定义,f(0)不等于0,f(xy)=f(x)f(y),证明:f(x)=1
f(x)在[0,1]上连续,证明:∫[0,1]f(x)dx∫[x,1]f(y)dy=1/2(∫[0,1]f(x)dx)的
设f(x)在【0,1】上连续且∫(0,1)f(x)dx=A,证明∫(0,1)dx∫(x,1)f(x)f(y)dy=A∧2
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)
f(x)连续且f(x)=x+(x^2)∫ (0,1)f(t)dt,求f(x)
f(x)在(0.1)上连续且单调增,证明∫[0,1]f(x)dx
设f(x)在[0,1]上连续,证明在(0,1)内至少存在一点ξ,使∫f(x)dx=(1-ξ)f(ξ)
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)
【高数】定积分 设f(x)连续,f(0)=1,则曲线y=∫(上限x,下限0) f(x)dx 在(0
f(x)具有二阶连续导数,f(0)=1,f'(0)=-1,且[xy(x+y)-f(x)y]dx+[f'(x)+x^2y]