1873_solutions

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

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: 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 ac 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...
View Full Document

Ask a homework question - tutors are online