Download Solved Problems on the Calculus II - Assignment | MATH 230 and more Assignments Calculus in PDF only on Docsity!
Homework solutions, Stewart, 5th edition
Section 7. #3. (a) π/ 3 (b) −π/ 4 #16. Graphs of y = tan x (−π/ 2 < x < π/2) and y = arctan(x) are symmetric about the line y = x.
#26. f ′(x) = ln(arctan x) + (^) (1 + x (^2) ) arctanx x.
#29. y′^ = −^2 e
2 x √ 1 − e^4 x^.
#30. y′^ = √ 1 − −x x 2 + arccos x + √ 1 x− x 2 = arccos x.
#45. (^) xlim→∞ ex^ = ∞, so limx→∞ arctan(ex) = π/2.
#50. Done in class. #66. u = arctan(x) gives (arctan x)^2 /2 + C.
#72. V = π
∫ (^2) 0
x^2 + 4 dx^ =^
π 2 arctan(x/2)
∣∣ ∣∣^2 0 = π^2 /8.
Section 7. #6. Cancel x + 2 in the fraction, or use l’Hˆopital to get −1. #9. Done in class, −∞. #15. (^) xlim→∞ 1 /x = 0.
#26. lim x→ 0 cos 3 xx^2 − 1 = lim x→ 0 −^ sin 6 x x= − 1 /6 (I made a mistake here at first).
#28. (^) xlim→∞ 2 /x = 0, so lim x→∞^ 2 lnx x= 0.
#41. Write as lim x→∞^ x
3 ex^2 and apply l’Hˆopital’s rule twice to get lim^ x→∞
4 xex^2 = 0.
#48. Write as lim x→ 1 x (^ x− −^1 1) ln−^ ln xx , then use l’Hˆopital’s rule twice (simplify as you go!) to get
xlim→ (^1) 2 + ln^1 x = 1/2.
#60. Done in class; lim x→ 0 5 ln(cos 3 x x)= 0, so the answer is e^0 = 1.
#88. For any p > 0 we have an indeterminate form ∞/∞, so l’Hˆopital’s rule applies. Since
xlim→∞ xpx^1 p−^1 = limx→∞ px^1 p^ is 0 for any positive^ p, the original limit was 0.
Section 8. #3. Letting v = 15 sin(5x) I get 15 x sin(5x) + 251 cos(5x) + C.
#8. Parts twice, each time expressing the trig function as a derivative( v′: first write cos mx = m^1 sin^ mx)′^ , then sin^ mx^ =^ (^ −^ m^1 cos^ mx)′^. You should get something like 1 mx
(^2) sin mx + 2 m^2 x^ cos^ mx^ −^
m^3 sin^ mx^ +^ C.
#14. Parts three times, each time writing et^ = (et)′; should be t^3 et^ − 3 t^2 et^ + 6tet^ − 6 et^ + C. #25. Use v′^ = (x)′^ and then substitution u = 1 − x^2 to get x arccos x −
1 − x^2. Evaluates to π/ 6 − √ 3 /2 + 1. #26. Write 5x^ = (5x/ ln 5)′. I’m getting 5/ ln 5 − 4 /(ln 5)^2. #36. u = x^2 to get 12 ∫^ u^2 eu^ du, then do just like #14. #52.∫ 5 5 ln x = x ln x when x = 1 and x = 5, and in that region 5 ln x ≥ x ln x, so we need 1 5 ln^ x^ −^ x^ ln^ x dx, which - if I’m right - is 5(x^ ln^ x^ −^ x)^ −^12 (x^2 ln^ x^ −^12 x^2 ) evaluated between 1 and 5. Simplified that works out to be 12.5 ln 5 − 14 ' 6 .118.
#55. 2 π ∫^01 x cos(πx/2) dx = 4
( x sin(πx/2) + (^) π^2 cos(πx/2)
)∣∣ ∣^10. I hope that gives the same answer as the book.
Section 8. #2. Separate one copy of the cosine and use u = sin x to get ∫^ u^6 (1 − u^2 ) dx
#9. Parts or double-angle formula twice: sin^4 (3t) =
( (^1) 2 (1^ −^ cos(6t)
) 2 = 14 − 12 cos(6t) + 18 (1 + cos(12t), which is easy to integrate. #24. Rewrite tan^4 (x) = tan^2 (x)(sec^2 (x) − 1) = tan^2 (x) sec^2 (x) − tan^2 (x) = tan^2 (x) sec^2 (x) − sec^2 (x) + 1. The first part is easy to integrate with u = tan(x), the second just gives tan(x), and the last gives x. #29. Separate sec(x) tan(x) and do u = sec(x). Then you’ll get ∫^ tan^2 (x) du = ∫^ u^2 − 1 du. #36. To get substitution going, write ∫^ cot^2 (x) cot(x) dx = ∫^ (cot^2 (x) − 1) cot(x) + cot(x) dx = − ∫^ (cot(x))′^ cot(x) dx + ∫^ cot(x) dx. The first piece is −^12 cot^2 (x) by substitution u = cot(x), the second, by another substitution v = sin x, is ln | sin x|. To evaluate use cot(π/2) = 0, cot(π/4) = 1, sin(π/2) = 1, sin(π/4) =
#40. Done in class; start with parts writing the integral as − ∫^ (cot x)′^ csc(x) dx.
#60. V = ∫^0 π/ 4 π tan^4 (x) dx and use #24.
Section 8. #4. Let x = 4 sin(u). This gives 64 ∫^ sin^3 (u) du, which is done using z = cos(u). This one works out to be −64(cos(u) − 13 cos^3 (u)). Since u = arcsin(x/4), cos(u) =
16 − x^2 /4 (draw a triangle representing u). Evaluating from 0 to 2
3 I get 13^13. #7. Use x = 5 sin(u) to get 251 ∫^ csc^2 (u) du = − 251 cot(u). #8. x = a sec(u) gives dx = a sec(u) tan(u)du, so that the integral becomes (^) a^13 ∫^ sin^2 (u) cos(u) du. Use√ z = sin(u), then the fact that u = arcsec(x/a), so the picture shows that sin(u) = x^2 − a^2 /x.
#66. u
u^3 − u^2 =^
u^3 − u^2 + u^2 + 1 u^3 − u^2 = 1 +^
u^2 + 1 u^3 − u^2 , and^
u^2 + 1 u^3 − u^2 =^
u^2 + 1 u^2 (u − 1) =^ −^
u −^
u^2 +^
u − 1.
#74. u = ex^ will give
u^2 − 1 du^ =
u − 1 −^
u + 1 du.
Section 9. #2. Done in class.
#4. 1 + (f ′(x))^2 = x
4 4 +
2 +^
4 x^4 =
( (^) x 2 2 +^
2 x^2
) 2 , so find
∫ (^1) 1 / 2
x^2 2 +^
2 x^2 dx.
#12.
∫ √ 3 1
x^2 + 1 x dx. Use^ x^ = tan^ u^ first to get the integral of sec
(^3) (u)/ tan(u). If you multiply
the numerator and denominator by tan(u), you’ll be able to separate z′^ = sec(u) tan(u) and use substitution z = sec(u). This gives z^2 /(z^2 − 1) = 1 + 1/(z + 1)(z − 1), which is done by partial fractions. There might be a simpler way, but I don’t see it.
#20. y = (^) ab
a^2 − x^2 for the upper half of the ellipse. So L =
∫ (^) a −a
√√ √√ 1 + b
(^2) x 2 a^2 (a^2 − x^2 ) dx.
Section 8. #2. (b) discontinuity/asymptote at x = 1/2; (c) infinite interval; (d) asymptote at x = 1.
#8. lim t→∞ − (^2) t (^21) + 4 +^14 = 1/4.
#28. (^) tlim→ 0 + − 2 /
t = ∞.
#30. (^) tlim→ 9 −^32 (t − 9)^2 /^3 − 6 = 6.
#50. Since e−x^ > 0, we have 2 +^ e
−x x >^
x, and we know that
∫ (^) ∞ 1
integral in question is divergent as well.^ x dx^ is divergent, so the
#54. For all x ≥ 0, e−x^ ≤ 1, so on the interval [0, 1] we have e
−x √x ≤ √^1 x. Since
∫ (^1) 0 x
− 1 / (^2) dx is
convergent, the integral in the problem is also convergent.
Section 12. #20. an = √n (^) 1+1^1 /√n → ∞.
#30. an = (−1)nn, divergent. #31. an → 0 by the squeeze law. #41. divergent (the even-numbered terms approach 1, the others approach −1). #56. Several approaches: a) you can compute an+1 − an = 17/(3n + 7)(3n + 4) and observe that it’s always positive; b) find the derivative of f (x) = (2x − 3)/(3x + 4) and verify that f ′(x) > 0, so f is increasing, so an is increasing as well; c) write an = (^23)
( 1 − (^6) n^17 +
) , which makes it obvious that as n grows, an also grows. The last way is the best, because it gives you the answer to the second question: an is always less than 2/3.
Section 12. #2. limn→∞ Sn = 5, where Sn = a 1 + a 2 +... + an. #5. limn→∞ an doesn’t exist (it isn’t entirely trivial to prove!) #8. Telescoping series, just like in example 6; convergent. #11. Looks like ∑∞ n=0 2 n/ 3 n−^1 , geometric series with ratio 2/3, although it’s hard to be sure just by looking at the first four terms... #18. Again, geometric series with r = 1/
2, convergent. #20. Geometric series with r = e/3, convergent. #30. an → ln(1/2) 6 = 0, divergent. #36. 73/ 99 #43. | 4 x| < 1, i.e. − 1 / 4 < x < 1 /4. #49. an = Sn − Sn− 1 = 2/(n^2 + n), and limn→∞ Sn = 1.
Section 12.3 (incl. a few that weren’t assigned) #2. ∑^6 n=2 an ≤ ∫^16 f (x) ≤ ∑^5 n=1 an. #4. ∫^1 ∞ x−^1 /^4 dx = limt→∞ 43 (t^3 /^4 − 1) = ∞, divergent (∑^ an is a p-series with p = 1/ 4 < 1). #5. lim t → ∞^13 (ln(3t + 1) − ln 4), divergent. #7. Compute the derivative to make sure f (x) = xe−x^ is decreasing for x ≥ 1. Integration by parts: ∫^1 ∞ f (x) dx = limt→∞ −te−t^ − e−t^ + 2e−^1 = 2/e, because te−t^ → 0 (e.g. by l’Hˆopital’s rule). #8. Use f (x) = 1 + (^) x+1^1 , limt→∞(t + ln(t + 1) − 1 − ln 2) = ∞, ‘very’ divergent (also note that the limit of an is not zero, so the integral test is an overkill here). #11. p-series with p = 3, convergent. #16. This would be easy if we had the comparison tests from the next section: (3n+2)/(n+1) > 1 for all n, so an > 1 /n and the series is divergent by direct comparison with the harmonic series (or use limit comparison test with 1/n). But here we’d better use the integral test. By partial fractions, (3x + 2)/(x^2 + x) = 2/x + 1/(x + 1) which is obviously decreasing, continuous, positive, and its integral from 1 to ∞ is divergent. #20. First verify that∫ f (x) = (ln x)/x^2 is decreasing. The integral: we could use parts, writing f (x) dx = ∫^ (− 1 /x)′^ ln x dx = −(ln x)/x − 1 /x. Now limt→∞ −(ln t)/t − 1 /t + 1 = 1, so the integral is convergent, and then so is the series. #21. Divergent (done in class?) #25. Partly done in class, and I complicated things because I was too lazy to compute the derivative of f (x) = 1/(x(ln x)p). It turns out to be negative for all x > 1 and all p, so we can use the integral test for all p. Substituting u = ln x we get an antiderivative F (x) = 1 1 − p(ln^ x)
1 −p (^) for p 6 = 1, and F (x) = ln(ln x) for p = −1. The corresponding improper integrals
limt→∞ F (t) − F (2) are all divergent unless 1 − p < 0 (in which case (ln x)^1 −p^ → 0), so the series is convergent for p > 1 only.
#10. Ratio test: (n + 1)/e → ∞, divergent. #12. Direct comparison with the geometric series: | sin(4n)| / 4 n^ ≤ 1 / 4 n, so absolutely conver- gent. #14. Ratio test: 2(n + 1)^2 /(n^2 (n + 1)) → 0, absolutely convergent. #18. Done in class; ratio test gives limit 1/e < 1. Convergent (remember, for series with positive terms, ‘convergent’ and ‘absolutely convergent’ are one and the same, because in that case |an| = an).
#20. Root test: limn→∞ n
√ 1 /(ln n)n^ = limn→∞ 1 / ln n = 0. Absolutely convergent. #22. By the integral test ∑^1 /(n ln n) is divergent (see #21 in section 12.3), so our series isn’t absolutely convergent. But it is convergent, by Leibniz’s criterion. #23. Root test; convergent.
Section 12. #4. Leibniz’s criterion, convergent. #6. Root test, limit 3/ 8 < 1, convergent. #8. Ratio test: limk→∞ 2(k + 1)/(k + 3) = 2 > 1, series is divergent. Or notice that ak = 2 k/(k^2 + 3k + 2) → ∞, not zero, so no chance for convergence. #9. Ratio test, limit is 1/e, convergent. #10. Can use ratio test here with a little effort, but it’s easier to use the integral test with f (x) = x^2 e−x^3 , u = x^3. Convergent. #11. Leibniz’s criterion, convergent. #12. Same as #11. #13. Limit in the ratio test is 0 < 1, convergent. #14. limn→∞ an is not zero; series is divergent. #15. Limit in the ratio test is 1/ 3 < 1; convergent. #18. Leibniz’s criterion, convergent. #19. limn→∞ 21 /n^ = 1, so the series is divergent. #20. Convergent by the ratio test. #23. Sneaky... write tan(1/n) as sin(1/n)/ cos(1/n), then use limit comparison with 1/n (the limit will be 1). Divergent. #24. Direct comparison with 1/n^2 , convergent. #27. Tough, if you don’t have a good feeling for the rates of growth of various functions. Reasoning: ln k grows quite a bit more slowly than k, even more slowly than
k, so the numerator grows more slowly than k^3 /^2. On the other hand, the denominator grows as a cube of ∑ k. The net result is that the terms of this series should ‘favorably compare’ with the p-series 1 /k^3 /^2 , and the series should converge.
To clear things up use limit comparison with ∑^ ln(k)/k^2 (the limit of the quotients is 1, so either both series diverge or both converge). I now claim that the series ∑^ ln(k)/k^2 is convergent: from some point on ln k ≤
k, so we can use direct comparison with ∑^
k/k^2 = ∑^1 /k^3 /^2. How do I know that ln k ≤
k (or, for that matter, any positive power of k) from some k on? Remember, ln k/k → 0 as k → ∞. So ln k/
k = 2 ln(
k)/
k also tends to 0, and that can only happen if ln k is smaller than
k for large k’s. See also problem 88 in section 7.7. Maybe there is a shorter explanation? I can’t think of one now. #28. Try the limit comparison with ∑^1 /n^2 (or direct comparison with ∑^ e/n^2 , since roots of e are less than or equal to e itself). Convergent. #32. Root test, limit is 0 < 1, convergent.
#35. Root test: limn→∞ √nan = limn→∞
( (^) n n+
)n = 1/e < 1, convergent.
Section 12. #7. All x (done in class). #8. Similar one done in class: limit in the ratio test is limn→∞(n + 1)(1 + 1/n)nx, which is infinity for any nonzero x. So this series converges only for x = 0. #12. Apply ratio test to ∑^ |an| to get convergence when |x| < 5 (so the radius of convergence is 5). Now at x = 5 the series is ∑^1 /n^5 , which is convergent. At x = −5 it’s the alternating series ∑(−1)n/n^5 , also convergent. The interval of convergence is [− 5 , 5]. #13. limn→∞ n
ln n = 1 (e.g. look at f (x) = eln(ln^ x))/x^ using l’Hˆopital’s rule). So the root test tells us that ∑ ∑^ |an| is convergent when |x| < 4. For x = 4 we get the alternating series ∑( −1)n^1 /^ ln(n), which is convergent by Leibniz’s criterion.^ But for^ x^ =^ −4 the series is 1 / ln n, which is divergent e.g. by direct comparison with ∑^1 /n (as above, from some point on ln n ≤ n, so 1/ ln n ≥ 1 /n). The interval of convergence is (− 4 , 4]. #18. This is a power series centered at a = −3. Ratio test gives convergence for |x + 3| < 0 .5, i.e. between − 3 .5 and − 2 .5. For x = − 3 .5 the series equals ∑^1 /√n, which is divergent. For x = − 2 .5 it’s an alternating series satisfying Leibniz’s criterion. The interval of convergence is (− 3. 5 , − 2 .5]. #27. Root test says that the series converges for all x. #28. Ratio test, radius of convergence is 1. For x = 1 or x = −1 note that the monster fraction is always greater than 1 (as a product of 2 times 4/3 times 6/5 etc.), so there is no chance of it having limit 0. The series is divergent then, so the interval of convergence is (− 1 , 1). #30. The series is centered at 0, and thus has interval of convergence from −R to R (and who knows about the endpoints). Since the series is convergent at −4, R is at least 4. Since it’s divergent for x = 6, R must be at most 6. So the best case scenario for the interval of convergence (largest possible) is [− 6 , 6), while the worst case is [− 4 , 4). (a) is when x = 1, which is certainly inside the interval of convergence, so we’re OK. (b) is when x = 8, can’t be convergent (else it would be convergent for x = 6 as well). (c) is OK just like (a), and (d) is the case x = −9 which must be divergent like (b).
