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)的平方
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设原函数为F(x),
∫[x,1]f(y)dy=F(1)-F(x)∫[0,1]f(x)dx∫[x,1]f(y)dy=∫[0,1]f(x)(F(1)-F(x))dx=F(1)∫[0,1]f(x)dx - ∫[0,1]F(x)d(F(x))=F(1)(F(1)-F(0)) - 1/2 [(F(1))^2 - F(0)^2]=1/2(F(1)^2 - 2F(1)F(0) + F(0)^2] =1/2(∫[0,1]f(x)dx)^2不懂可追问