1873_solutions

F is a nonnegative integrable function on an interval

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Unformatted text preview: unction 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 Þ a fg b a and deduce that there exists a number c MÞ g b b a a, b such that Þa f g  f Þ a g. b 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 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 279 t a, give a direct proof that g is integrable on the interval b, a and that 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  n b n j xj xj  1 xj j1 xj j  1 j1 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, 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 as n Þa Sn b sn  Ý. It follows that the function g is integrable on the interval lim Þ b g  n Ý Þ b sn a 1. 0 sn a nÝ lim b, a and that Þa sn  Þa f . b b a. Suppose that u is a decreasing diffe...
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