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549

Part of 2010 Today's Calculation Of Integral

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Today's calculation of Integral 549

Source: 2010 Tokyo University of Science entrance exam/Biological Sciences

2/5/2010
Let f(x) f(x) be a function defined on [0, 1] [0,\ 1]. For n=1, 2, 3, ⋯ n=1,\ 2,\ 3,\ \cdots, a polynomial Pn(x) P_n(x) is defined by Pn(x)=∑k=0nnCkf(kn)xk(1−x)n−k P_n(x)=\sum_{k=0}^n {}_nC{}_k f\left(\frac{k}{n}\right)x^k(1-x)^{n-k}. Prove that lim⁡n→∞∫01Pn(x)dx=∫01f(x)dx \lim_{n\to\infty} \int_0^1 P_n(x)dx=\int_0^1 f(x)dx.
calculusintegrationfunctionalgebrapolynomiallimitcalculus computations