1873_solutions

# X is an integer multiple of 2 1 n cos nx n the series

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Unformatted text preview: lim w /2 g. Þ /2 x cos x sin x x2 0 sin x dx  lim 2 w /2 cos x  2. 2 cos w dx This integral is /2 Þ 0 w lim w h. Þ Ý 1 x cos x sin x dx  lim sin /2 w 0 /2 x2 sin w w 2  1 e x sin xdx This integral is lim Þ 1 e x sin xdx  w Ý w lim wÝ 2. 1 e w cos w 2 cos 1  sin 1  2e 2e 1e 2 w sin w 1 e 1 cos 1 2 1 e 1 sin 1 2 a. Prove that the integral Þ0 1 converges when p  1 and diverges when p As long as p 1 we have 1 dx xp 1. Þ0 1 dx  lim Þ 1 1 dx  lim 1 p xp 1p w0 w0 wx if p  1 and is Ý if p  1. We see also that 1 This limit is 1 1p Þ0 1 1 dx  lim log 1 x w0 351 1 1 log w  Ý. p w1 p . b. Prove that the integral Þ1 converges when p  1 and diverges when p As long as p 1 we have Ý 1 dx xp 1. Ý Þ1 This limit is 1 p1 1 dx  lim Þ w 1 dx  lim 1 w1 w Ý 1 xp wÝ 1 xp p if p  1 and is Ý if p  1. We see also that Ý Þ1 1 dx  lim log w x wÝ 1 p 1 . p log 1  Ý. 3. Prove that the integral Þ2 Ý 1 x log x 1. dx p is convergent when p  1 and divergent when p As long as p 1 we have Ý w 1 1 lim Þ 2 x log x p dx  w Ý Þ 2 x log x p dx 1 log w 1 1p when p  1 and is Ý when p  1. wÝ lim This limit is p11 log 2 We see also that 1p Ý Þ2 1 dx  lim log log w wÝ x log x 1 p 1 p log 2 1p log log 2  Ý. 4. Interpret the integral Þ0 2 1 dx x 1 1/3 as the sum of two improper integrals and evaluate it. 1 2 1 1 dx and Þ dx and can easily be seen to be 0. The two integrals are Þ 1/3 1/3 0 1 x1 x1 5. Prove that if f is bounded on an interval a, b and is improper Riemann integrable on a, b then f is Riemann integrable on a, b and Þa b f Þ a f. b  0. Using the fact that f To show that f satisfies the second criterion for integrability, suppose that is integrable on the interval a, b 2 choose a partition P  x 0 , x 1 , , x n of the interval a, b 2 such that if E w P, f x 2 2 then m E  2 . Now we extend the partition P to make a partition Q of the interval a, b by adding the point b. Thus Q  x 0 , x 1 , , x n , b and since x x a, b a, b w Q, f x 2 the measure of the latter set is less than . Now to show that 352 2 EÞ b 2 ,b lim Þa f  Þa f  Þw f w wb we need only note that if a b w  b then Þa f Þa f b w b and that the latter expression approaches 0 as w sup|f | b w b. 6. In the discussion of improper integrals that appears above we used the words ... Somewhat less precisely, we also say that the integral Þ b a f x dx is convergent. ... Why is the statement that the integral Þ f x dx is convergent less precise than the statement that f is a improper Riemann integrable on a, b ? Hint: In our study of infinite series we made a careful distinction Ý between the symbols a n and n1 a n . b The symbol Þ b f x dx stand for the limit a Þ a f x dx w lim wb which is a number. When we say that Þ b a f x dx is convergent we do not really mean what we are saying. Literally, the statement that Þ f x dx is convergent is the statement that the value of the a limit b Þ a f x dx w lim wb is convergent. Thus, if the l...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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