1873_solutions

# F b d e e n n u p n f d kd a b en a

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: bc bc a. Suppose that f is a continuous function on an interval a, b , that g is nonnegative and integrable on a, b . Prove that, if m and M are, respectively, the minimum and maximum values of f, then mÞ g b a and deduce that there exists a number c Þ a fg MÞ g b b a a, b such that Þa f g  f b Þ a g. b c This fact is sometimes called the mean value theorem for integrals. Since m fx M for every x a, b we have Þ a mg Þ a fg Þ a Mg b 315 b b which gives us mÞ g Þ a fg MÞ g Þ a fg b M. b a b a and so b m Þa g b The existence of the number c follows at once from the Bolzano intermediate value theorem. b. Given that f is continuous on an interval a, b , prove that there exits a number c Þa f  f c a, b such that b 6. b a. a. Given that f is a nonnegative continuous function on an interval a, b where a  b and that Þ f  0, a prove that f is the constant zero function. To obtain a contradiction, suppose that c a, b and that f c  0. Using the fact that f is continuous at c, choose   0 such that the inequality fc fx  2 holds for every number x a, b c , c   . The set a, b c , c   is an interval with positive length that we shall write as c, d . In fact, c is the larger of the two numbers a and c  and d is the smaller of the two numbers b and c  . Since b d d c c Þa f Þc f Þc f 2  f 2 d c  0 which gives us our desired contradiction. b b. Given that f is a continuous function on an interval a, b where a  b and that Þ f  0, for every a x a, b , prove that f is the constant zero function. We know that if x Fx  Þa f x for each x then F x  f x for each x. Since F is the constant function zero we conclude that f x  0 for each x. 7. In this exercise we consider another proof of the “u decreasing” form of the change of variable theorem. a. Given that f is an integrable function on an interval a, b and that g t  f t whenever b give a direct proof that g is integrable on the interval b, a and that t a, Þ b g t dt  Þ a f x dx. a b Suppose first that f is a step function on a, b that steps within the partition P  x 0 , x 1 , , x n taking the constant value  j on each interval x j 1 , x j . We define Q  x n , x n 1 , , x 1 , x 0 and observe that Q is a partition of the interval b, a . Now we define g t  f t for each t a, b and we observe that g is a step function that takes the value  j on each interval x j , x j 1 . We see at once that Þa f  b n n j xj xj 1 j1  xj j1 1 xj j  Þ bg a We can now handle the general case. Using the fact that f is integrable on the interval a, b , choose a pair of sequences of step functions that squeezes f on the interval b, a . In other words, 316 sn f Sn for each n and Þa Sn b lim nÝ s n  0. For each n we define s n t  s n t and S n t  S n t and we observe that sn g Sn and, by the case we have already considered we deduce that Þ b Sn a sn  Þa Sn b as n Ý. It follows that the function g is integrable on the interval lim Þ b g  n Ý Þ b sn a a 0 sn nÝ lim b, a a...
View Full Document

Ask a homework question - tutors are online