Math 132, Fall 2009 Exam 2: Solutions

Transcription

1 Math 3, Fall 009 Exam : Solutios () a) ( poits) Determie for which positive real umbers p, is the followig improper itegral coverget, ad for which it is diverget. Evaluate the itegral for each value of p, for which it coverges, ad express its value i terms of p. Justify your aswer ad show all your algebraic steps. Hit: Do ot forget to cosider the case p = as well. (l(x)) p x dx e Let u = l(x) so du = /x dx. The e (l(x)) p x N dx e l(n) N (l(x)) p x dx u p du u p du Case : p =. e (l(x)) p x N dx u du [l u ]N (l(n) l()) = So, for p =, the itegral is diverget.

2 Case : p. Now so e (l(x)) p x N dx = p u du p u p N p ( N p ) { 0 if p > N p = if p < p ( N p ) { if p > = p if p < The the itegral is diverget for p < ad coverget for p >. We have e { (l(x)) p x dx = if p > p diverget if p b) (0 poits) Determie whether the followig improper itegral coverges or diverges. Evaluate it, showig all your algebraic steps, if it is coverget. Otherwise, explai why it is diverget. 8 dx (x 8) (/3) N dx (x 8) /3 N 8 0 dx (x 8) /3 N 8 3(x 8)/3 N 0 N 8 ( 3(N 8) /3 3( 8) /3) = 6

3 () (4) For each of the followig sequeces (ot series) determie whether the sequece coverges or diverges. If it coverges, fid the it, showig all your algebraic steps. Otherwise, explai why it diverges. + 3 a) a = 3,. a The the sequece coverges to /3. = (3 ) ( + 3) b) a = si()e,. Sice si() ad e 0, Now, Also, e si()e e e xe x x x x e x = 0 x e x by L Hospital s rule e = e = 0.

4 4 By the Squeeze Theorem, si()e = 0. Therefore, the sequece coverges to 0. (3) Cosider the series = 6. (a) (9 poits) Use the itegral test to show that the series is coverget. Show that all the hypothesis of the test are satisfied. Show all your algebraic steps. (Credit will ot be give for a aswer usig aother test). Let f(x) = so that f() = for each iteger. x 6 6. f(x) = /x 6 is cotiuous for x.. f(x) = /x 6 0 for x. 3. f (x) = 6x 7 < 0 for x. So, f is decreasig for x. N dx x6 x dx 6 N x (N ) = Sice the itegral = is also coverget. 6 dx is coverget, by the itegral test, the series x6

5 (b) (9 poits) Let s be the sum of the series i part 3a. Fid the miimal umber of terms of the series, for which we kow that s s , by the error estimate of the itegral test. Justify your aswer, showig all your algebraic steps. From part a, the fuctio f(x) = /x 6 is cotiuous, positive ad decreasig for x. Sice f() = / 6, by the error estimate for the itegral test, Now So, s s x 6 dx N dx x6 x dx 6 N x (N ) = Sice we wat s s , s s s s or / 7. Therefore, we eed at least = 8 terms to coclude s s from the error estimate for the itegral test.

6 6 (4) (4 poits) Cosider the series = 3 + (a) ( poits) Use the compariso test to show that the series coverges. For each, Ad The series = ( ) is a geometric series with r = / < = =, so it is coverget. Therefore, by the compariso test, the series is also coverget. 3 + = (b) (9 poits) The sum s 0 of the first 0 terms of the series, rouded to five decimal digits, is.7. You do ot eed to verify this. Show that s.7 is less tha 0.0. Justify your aswer! Solutio. For a coverget geometric series the partial sums are t = ar, r <, = t = a( r ) r ad the sum is t = a r.

7 7 Let a = / ad r = / so that t = = = = = / / = ( ) ad t = (/)( (/) ) (/) = ( (/) ) Sice 3+ for all, we have s s t t for all. The for = 0, s s 0 t t 0 = ( (/) 0 ) = / 0 < 0.0

8 8 Solutio. Let t =, t be the partial sums ad f(x) = / x. =. f(x) = / x is cotiuous for all x.. f(x) = / x > 0 for all x. 3. f(x) = ( x ) so f (x) = (l ) x < 0. The f is decreasig for all x. Therefore, by the error estimate for the itegral test, t t Sice 3+ for all, we also have for all. So for = 0, x dx s s t t s s 0 t t 0 dx x 0 N 0 N 0 = ( 0 ) l < 0.0 x dx ( x ) dx N l x 0 ( N 0) l

9 () (3 poits) Determie whether the followig series coverge absolutely, coverge coditioally, or diverge. Name each test you use ad idicate why all the coditios eeded for it to apply actually hold. 9 (a) = Let a = ad b =. The a, b > 0 ad a b ( ) (0 3 + ) 3 ( ) / 0 4/ + / 3 ) =. a Sice = is positive ad fiite, by the it compariso test b either both series are coverget or both series are diverget. The series is a p-series with p = > so it is coverget. Sice =, the series is also coverget. Thus, the series is coverget. Sice the terms are positive, the = series is absolutely coverget.

10 0 (b) ( ) 4 ( )! = Let a = ( ) ( 4! ). The ( ) a + a ( ) 4 + (+)! ( ) ( ) 4! ( ) ( ) 4 +! ( + )! 4 = 0 <. + 4 Thus, the series is absolutely coverget by the ratio test. (c) = ( ) ( ) Let a = ( ) ( ) The

11 a ( = ( = ( = = (/3) = 4/9 <. ( ) + ( ) ( ) ( ) ) ( + )/ (3 + 7)/ + / 3 + 7/ ) ) (d) Thus, the series is absolutely coverget by the root test. = ( ) + ( ) + 4 This is a alteratig series = ( ) b with b = b / + 4/ = 0.

12 . Let f(x) = x+ x +4 so that f() = b. f (x) = (x + 4) x(x + ) (x + 4) = x 4x + 4 (x + 4) x 4x 4x 4 for x. So, x 4x for x. The f is decreasig for x which implies b + b for. Hece, by the alteratig series test, the series is coverget. Let a = ( ) ( + +4 = ) so that a = + +4 > 0. Let c = a c / + 4/ = > 0. The a Sice = is positive ad fiite, by the it compariso test b either both series are coverget or both series are diverget. The series is a p-series with p = so it is diverget. Thus, the series + is also diverget. + 4 Therefore, the series ( ) + ( ) is coditioally coverget. + 4 =