題目沒有問題
∫{0,1}xⁿ*f(x)dx=∫{0,1-1/√n}xⁿ*f(x)dx+∫{1-1/√n,1}xⁿ*f(x)dx
由于f(x)在[0,1]上連續(xù),xⁿ在[0,1]上不變號,且在[0,1]上可積
對f(x)在[0,1-1/√n]上運用積分第一中值定理,存在一點ξ₁∈[0,1-1/√n],使得
∫{0,1-1/√n}xⁿ*f(x)dx=f(ξ₁)*∫{0,1-1/√n}xⁿdx
=f(ξ₁)*[x^(n+1)/(n+1)]| {0,1-1/√n}
=f(ξ₁)*(1-1/√n)^(n+1)/(n+1)
對f(x)在[1-1/√n,1]上運用積分第一中值定理,存在一點ξ₂∈[1-1/√n,1],使得
∫{1-1/√n,1}xⁿ*f(x)dx=f(ξ₂)*∫{1-1/√n,1}xⁿdx
=f(ξ₂)*[1/(n+1)-(1-1/√n)^(n+1)/(n+1)]
=f(ξ₂)/(n+1)[1-(1-1/√n)^(n+1)]
故lim{n→∞}n*∫{0,1}xⁿ*f(x)dx
=lim{n→∞}n*f(ξ₁)*(1-1/√n)^(n+1)/(n+1)+lim{n→∞}n* f(ξ₂)/(n+1)[1-(1-1/√n)^(n+1)]
由于lim{n→∞}(1-1/√n)^(n+1)
=lim{n→∞}(1-1/√n)^n
=lim{x→0+}(1-x)^(1/x²)
=lim{x→0+}e^[1/x²*ln(1-x)]
=e^{lim{x→0+}[1/x²*(-x)]} a→0,ln(1+a)~a
=0
故lim{n→∞}n*f(ξ₁)*(1-1/√n)^(n+1)/(n+1)
=lim{n→∞}f(ξ₁)*lim{n→∞}[n/(n+1)]*lim{n→∞}(1-1/√n)^(n+1)
=lim{n→∞}f(ξ₁)*1*0
=0
注：∵f(x)在[0,1]上連續(xù),∴f(x)有界,∴l(xiāng)im{n→∞}f(ξ₁)為有限值
∵當(dāng)n→∞時,1-1/√n→1,∴ξ₂→1
故lim{n→∞} n* f(ξ₂)/(n+1)[1-(1-1/√n)^(n+1)]
=lim{n→∞}f(ξ₂)*lim{n→∞}[n/(n+1)]*lim{n→∞}[1-(1-1/√n)^(n+1)]
=lim{ξ₂→1}f(ξ₂)*1*(1-0)
=f(1) 由連續(xù)性
因此,lim{n→∞}n*∫{0,1}xⁿ*f(x)dx=f(1),證畢
本題若直接根據(jù)積分中值定理,得到存在一點ξ∈(0,1),使得
∫{0,1}xⁿ*f(x)dx=ξⁿ*f(ξ),這里0