MTH 201 Homework 2 Solution, Fall 2016

!"#$%&'()*+,( ( ( ( ( ( ( ( ( ( ( ( ( ( ( 17. ; lim lim 2 tan u u x du ' ' ' 0 0 0 b b 0 dx 2 du 2 du (1 x) x dx 2 x u 1 u 1 � � � �" È È – — È c d œ œ Ä œ œ b b Ä _ Ä _ lim 2 tan b 2 tan 0 2 2(0) œ � œ � œ b Ä _ a b ˆ ‰ �" �" # 1 1 25. x ln x dx lim ln x ln 1 lim ln b lim 0 ' 0 1 1 b œ � œ � � � œ � � � b b b Ä ! Ä ! Ä ! ’ “ Š ‹ ˆ ‰ x x b b ln b 4 4 4 4 # # # " " " Š 2 ‹ b lim lim 0 œ � � œ � � œ � � œ � " " " " � 4 4 4 4 4 b b b Ä ! Ä ! Š ‹ Š ‹ b 4 b Š ‹ 29. lim sec s sec 2 lim sec b 0 ' 1 2 b ds s s 1 3 3 È � �" �" �" œ œ � œ � œ b 1 b 1 Ä Ä c d 1 1 39. x e dx; y e dy lim e lim e e ' ' ' 0 1 ln 2 ln 2 1 ln 2 1 x y y b 1 ln 2 b 1 ln 2 �# " � � ‘ c d c d c d x y y e dy œ Ä œ œ � œ � � � y b b Ä _ Ä _ 0 e e , so the integral converges. œ � 1 ln 2 œ 1 ln 2 43. and lim ln lim ln 0 , which ' ' ' ' 0 0 1 0 2 1 2 1 b 0 dx dx dx dx 1 x b 1 x 1 x 1 x 1 x 1 x 1 b � � � � # � # � " � " " � œ � œ œ � œ _ b 1 b 1 Ä Ä � ‘ � ‘ ¸ ¸ ¸ ¸ diverges diverges as well. Ê ' 0 2 dx 1 x � 49. ; lim lim lim 1 and lim 2 v , ' ' 2 2 b dv dv v 1 v 1 1 0 v 1 v È È È È Š ‹ Š ‹ È É � � � " " � v v v b Ä _ Ä _ Ä _ Ä _ 1 v 1 v œ œ œ œ œ œ _ v � ‘ È which diverges diverges by the Limit Comparison Test. Ê ' 2 dv v 1 È � 57. ; lim 1 and lim 4t lim 2 2 converges ' ' ' 4 4 4 b 4 2 dt t 2 dt 4 2 dt t 1 t 1 t t b � � �"Î# � t b b Ä _ Ä _ Ä _ œ œ � œ � œ Ê � ‘ Š ‹ È converges by the Limit Comparison Test. Ê ' 4 2 dt t 1 � 59. dx; 0 for x 1 and diverges diverges by the Direct Comparison Test. ' ' ' 1 1 1 e e dx e dx x x x x x x x x � � � Ê " 61. ; lim lim lim 1; e dx ' ' ' 1 1 1 x 2 dx dx e x e x e 1 1 0 È È È È Š ‹ Š ‹ È É x x x � � " " � � x x x Ä _ Ä _ Ä _ e x e x e x x x x œ œ œ œ œ e lim 2e 2e 2e e dx converges converges œ � � � œ Ê Ê b b lim Ä _ œ Ä _ c d a b x 2 b 2 1 2 x 2 b 1 1 2 dx e e x È È ' ' x � by the Limit Comparison Test. MATH 201 Homework II Solutions - October 20, 2011MATH ( ( ( -./#$%#"(01".#%2"2,( ( ( ( ( ( 65. (a) ; t ln x lim t lim (ln 2) ' ' 1 0 2 ln 2 1 p 1 p ln 2 b dx dt b x(ln x) t p 1 p 1 1 p p p 1 p c d ’ “ œ Ä œ œ � b b Ä ! Ä ! " " � � � � the integral converges for p 1 and diverges for p 1 Ê � (b) ; t ln x and this integral is essentially the same as in Exercise 65(a): it converges ' ' 2 ln 2 dx dt x(ln x) t p p c d œ Ä for p 1 and diverges for p 1 � Ÿ 59. x e dx lim x e 2xe 2e lim b e 2be 2e ( 2) 0 2 2 ' 0 x x x x b b b b 0 # # # œ � � � œ � � � � � œ � œ b b Ä _ Ä _ c d a b 61. 2 lim tan lim tan tan (0) ' ' ' dx dx dx 2 2x 2 2b 4x 9 4x 9 3 3 3 3 3 x � � # # # " " " " � �" �" �" œ œ œ œ � 0 0 b 0 9 4 b b Ä _ Ä _ � ‘ � ‘ ˆ ‰ ˆ ‰ 0 œ � œ " # # ˆ ‰ 2 3 6 † 1 1 65. dz dz dz lim 0 lim ' ' ' 1 1 e e e b 1 e ln z ln z ln z 1 z z z 2 2 2 (ln z) (ln z) (ln b) œ � œ � œ � � � œ _ ’ “ ’ “ Š ‹ ’ “ b b Ä _ Ä _ # # " diverges Ê 67. 2 converges converges ' ' ' ' 2 dx 2 dx 4 dx 2 dx e e e e e e e x x x x x x x � � � œ � Ê 0 0 1. Is it true that the improper integral � ∞ a f(x) converges, whenever f(x) → 0. Prove or give a counterexample. This is NOT true. For example consider � ∞ 1 1 x dx. We know from the p-integral test that this integral diverges (case p = 1). Yet, limx→∞ 1/x = 0. 2. Evaluate the following integrals: Both integrals are improper of Type I. In both cases, we can explicitly compute the answer. • � ∞ e 1 x(ln x+3)2 dx We need to compute lim b→∞ � b e 1 x(ln x + 3)2 dx. We first evaluate the integral by substitution by letting u = ln x, du dx = 1/x, to get (up to a constant) � du (u + 3)2 = −(u + 3)−2+1 = −(ln x + 3)−1 We now compute the limit: lim b→∞ � − (ln x + 3)−1�b e = 0 + 1/4 = 1/4 • � ∞ 0 √ tan−1(x) 1+x2 dx We need to compute lim b→∞ � b 0 � tan−1(x) 1 + x2 dx. We evaluate this integral by substitution. Let u = tan−1 x, then du = 1 1+x2dx, obtaining (up to a constant): � � tan−1(x) 1 + x2 dx = � u1/2du = (2/3)u3/2 = (2/3)(tan−1 x)2/3. We can now take the limit lim b→∞ � (2/3)(tan−1 x)2/3�b 0 = (2/3 lim b→∞[(tan−1 b)3/2−(tan−1 0)3/2] = (2/3) � (π 2 )3/2−0 � 1 3. Determine whether the following integrals diverge or converge: • � ∞ 1 x (x3+2)2dx We first observe that this integral is roughly like ∼ � x x6 dx = � 1 x5 dx. This suggests to us that we can use the limit comparison test (LCT) with g(x) = 1 x5 (remember that f(x) = x (x3+2)2). lim x→∞ f(x) g(x) = lim x→∞ x6 (x3 + 2)2 = 1, using l’Hopital’s rule. So the corresponding integrals behave the same. But we know by the p-integral test that � ∞ 1 1 x5 dx converges. So the original integral converges. • � −3 −∞ 1 x3 dx We first observe that since the integrand is an odd function, we can write � −3 −∞ 1 x3 dx = − � ∞ 3 1 x3 dx. We know that � ∞ 3 1 x3 dx converges being a p− integral with p = 3. So the original integral converges. • � ∞ −∞ x ex+e−x dx Since the integrand is odd, it is enough to analyze � ∞ 0 x ex+e−x dx. We can do LCT with g(x) = x ex, and conclude that the integrals behave a like. So let us now look at � ∞ 0 x ex dx. One way would be to do DCT with � ∞ 0 e−x/2 dx and conclude it is convergent so � ∞ 0 x ex+e−x dx converges to a finite limit L > 0. This means that � 0 −∞ x ex+e−x dx converges to −L, which means that the original integral goes to 0. • � ∞ 2 1 √x ln x dx We first note that for all x > 0, ln x < √x. Using this fact, we see that � ∞ 2 1 √x ln x dx ≥ � ∞ 2 1 √x√x dx = � ∞ 2 1 x dx. So the original integral diverges by the direct comparison test and the fact that � ∞ 2 1 x dx is a divergent p− integral (p = 1). • � ∞ 1 x−ex x2 dx. We first observe that since the numerator is dominated by the exponential, we 2 suspect this integral to diverge. We can prove so by using the limit comparison test and comparing the integral with � ex x2. But before doing that, we need to insure that the integrand is positive (to be able to use the limit comparison test). We first rewrite � ∞ 1 x − ex x2 dx = − � ∞ 1 ex − x x2 dx. and work with � ∞ 1 ex − x x2 dx. Now let g(x) = ex x2 and f(x) = ex−x x2 . We compute lim x→∞ f(x) g(x) = lim x→∞ ex − x ex = 1. So the integrals of f and g behave alike. Then we see that � ∞ 1 ex x2 dx diverges because the integrand tends to infinity as x → ∞. So we conclude that the original integral diverges. • This is an integral of mixed type, so we split it as � 1 0 1 x2 dx + � ∞ 1 1 x2 dx. The second is a convergent p-integral. As for the first one, we integrate it explicitly and evaluate the limit lim b→0 � 1 b x−2 dx = lim b→0 −x−1��1 b = ∞. We conclude that the original integral diverges. 3