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Unformatted text preview: ANSWERS TO SELECTED EXERCISES: HW2 Exercise 6.2.12 Let f : [0 , 1] R be a continuous function. Prove that lim n Z 1 f ( x n ) dx = f (0) . Proof : First approach. Let I n = Z 1 f ( x n ) dx . It is natural to apply the mean value theorem for the continuous function f ( x n ) on [0 , 1]; we get a sequence { c n } such that Z 1 f ( x n ) dx = f ( c n n ) , for all n N . It is desirable that lim n c n n = 0, so that lim n f ( c n n ) = 0, by continuity of f . However, this cannot be deduced from 0 < c n < 1, so this approach fails. If instead of 0 < c n < 1 we had 0 < c n 1 for some very small (but fixed!) value of , then a simple application of the squeeze theorem would yield lim n c n n = 0. What if we force these inequalities to be true? We can make this happen if we consider I n, = Z 1 f ( x n ) dx instead of I n = Z 1 f ( x n ) dx . It is natural then to work as follows: Second approach. We split the integral into two parts: Z 1 f ( x n ) dx = Z 1 f ( x n ) dx + Z 1 1 f ( x n ) dx....
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 Winter '08
 hitrik

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