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Fbyj nxx cxg. Z b a f(x)dx Solution Let ">0 Since f n!funiformly, N2N such that jf n(x) f(x)j. Proof Put P(x) = C 0x C 1 2 x2 C n 1 n xn C n n 1 xn1 P n is a polynomial, so it is di erentiable everywhere on R It is clear that P(0) = 0, and P(1) = 0 by the hypothesis By the Mean Value Theorem, there is c2(0;1) such that. ʐ^ N b N ƁA e N ́u N X } X C x g v ̗l q ܂ B 19 18.

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