1873_solutions

We choose a partition p 2 of a b such that if we

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Unformatted text preview: a, b . Since the function log is uniformly continuous on the interval 1, Ý , the integrability of g follows at once from the composition theorem. 4. Suppose that f is integrable with respect to an increasing function  on an interval a, b and that  f x  for every x a, b . Show how the junior version of the composition theorem for integrability can be used to show that if h is any continuous function on the interval ,  then the function h f is integrable with respect to  on a, b . The result follows at once from the fact that any continuous function on the interval ,  must be uniformly continuous. Exercises on the Change of Variable Theorem 1. a. Given that f is a continuous function on the interval 2 Þ0 Solution: We define u x 1, 1 , prove that f sin x cos xdx  0.  sin x for every number x 2 Þ0 f sin x cos xdx   2 Þ0 0, 2 and observe that f u x u x dx u 2 Þu 0 f Þ0 f  0 0 b. Given that f is a continuous function on the interval 0, 1 , prove that /2 Þ0 Solution: We define u x /2 Þ0  f sin x dx   Þ /2 f sin x dx. x for every number x /2 f sin x dx  Þ0  Þ0  Þu 0  /2 f sin   2 and observe that x u x dx f sin u x u /2 0, u x dx f sin t dt  /2 Þ f sin t dt  Þ /2 f sin t dt Of course, it makes no difference whether we write t or x in the integral Þ c. Given that   0, prove that /2 Þ0 Solution: We define u x sin  xdx  2  Þ  2x for all x 314 /2 sin  x cos  xdx. 0 0, /2 and observe that  /2 f sin t dt. 2 Þ /2 0 sin  x cos  xdx  /2 Þ0 /2 Þ0 2  sin  x cos  xdx  /2 1 2 Þ0 1 2 Þu 0 /2 Þ0 sin  2x 2dx  1 2 u /2 2 sin x cos x  dx sin  u x u x dx  Þ 0 sin  tdt sin  tdt  1 2 /2   1 Þ sin  tdt  Þ sin  tdt 2 0 /2 and from part b we deduce that the latter expression is equal to 1 2 /2 Þ0 sin  tdt  Þ /2 sin  tdt /2 Þ0  0 sin  tdt. 2. Given that u is a differentiable function on an interval a, b and that its derivative u is integrable on a, b and given that u a  u b and that f is integrable on the range of u, prove that Þa f u t b u t dt  0. From the change of variable theorem we see that Þa f u t b Þu a ub u t dt  Þu a ua f x dx  f x dx  0. 3. Given that f is integrable on an interval a, b and that c is any number, prove that b c Þ a f t dt  Þ ac f t b c. We observe that u t  1 for every t. Now For every number t we define u t  t b c Þ a c f t c dt  c dt b c Þ a c f u t u t dt  Þu a ub f x dx  Þ a f t dt. b I have changed the name of the dummy variable back to t to match the expression in the exercise. 4. Given that a, b and c are real numbers, that ac  bc and that f is a continuous function on the interval ac, bc , prove that Þ ac f t dt  c Þ a f ct dt. bc b Hint: Look at the definition u t  ct for each t. We assume that c  0. We define u t  ct for every number t. We see that u t  c for each t and we deduce from the change of variable theorem that c Þ f ct dt  b a 5. Þa f u t b u t dt  Þ ac f x dx  Þ ac f t dt...
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