124 - Answers to even (and some other) assigned problems

• 30) ln(e^x + 1) + C

• 16) a) A/x + B/x^2 + C/(x + 1) b) A/x + (Bx + C)/(x^2 + 1)

• 16) 3/2 + ln(3/2) (NOTE: ln 3 - ln 2 = ln(3/2))
• 34) ln|x| + 1/(x^2 + 1) + C

• 2) a) Not improper, only discontinuity is at 1/2 which is not in range of integration.
  b) Improper, not continuous at 1/2, which is in range of integration.
  c) Improper since +/- infinity.
  d)Improper since undefined at x = 1.

• 4) a) 120 , b) 120e^((ln 125)/6)t) = 120 e^((ln 5)/2)t) = 120*5^(t/2) , c) 6,708 , d) 5,398.2 , e) 9.2 hours
• 10) a) 12.25 years , b) 28.45 years
• 12) 5e^(2x)
• 20) a) 11.6 years , b) 6.2 %

• 6) Assume K = 4 billion = 4000 million.
  • a) With t = 0 being 1990, units of t in years and units of P(t) in millions get P(t) = 4000/(1 + 15*e^(-kt))
  • b) k = -(1/10)*ln(149/165) or approx. equal .0102
  • c) approx. 679.8 million in 2100 , approx. 1.45 billion in 2200.
  • d) The population will be greater than 300 million in 2010.
• 8)
  • a) Using k is approx. 1.19 , P(t) = 10000/(1 + 24*e^(-1.19t))
  • b) approx. 2.68 years.

• 16) Diverges
• 18) Converges to 1
• 20) Converges to pi/2
• 22) Diverges
• 28) Converges to 0
• 38) a) L b) 1, .5, .66667 , .6, .625, .61538 , .61905, .61765, .61818, .61798 c) (-1 + \sqrt(5))/2

• 14) Diverges
• 20) Diverges
• 22) Diverges
• 26) Converges to (cos 1)/( 1 - cos 1)
• 30) Diverges

• 4) a) Diverges b) Nothing
• 8) Converges
• 14) Diverges
• 20) Converges
• 22) Diverges
• 26) Converges
• 28)
  • a) S_10 = 1.0820366... using calculator. R_10 < 1/3000 < .00034 < .0004 (rounding up to one significant digit)
  • b) Aprroximating S by the midpoint of the interval [S_10 + 1/3993 , S_10 + 1/3000] gives an approximate value of S of 1.0823285... . This gives |error| < 4.2 x 10^(-5) < 5 x 10^(-5) = .00005 (rounding up to one significant digit). Since this is < .0001 we would say accurate to three decimal places and thus we can write S is approximately 1.082... .
  • c) If use |R_n| < \int_n^\infty 1/x^4 dx to get |R_n| < .00001 we need 1/(3n^3) < .00001 , which gives n > 32.1 , so n = 33 will work. If we compute S_33 since we have it is accurate to four decimal places we can write S is approximately 1.0823... .
• 30) n = 4 is the smallest n that will work. So S is approximately S_4 = 1.036
• 31) Need n = 40,000 to get R_n < .01 , but using the midpoint approximation just need n with (1/sqrt(n) - 1/sqrt(n+1)) < .01 . This is difficult to solve for n, but calculating values will show the smallest n that works is n = 14. So calculating S_14 shows S = 2.60... .
• 32) e^(100) .

• 2) a) Diverges b) Converges c) Nothing
• 8) Converges, but not absolutely.
• 15) Need a value of n with 1/(2n+1)! < .00001 , i.e. (2n+1)! > 100,000 . One can't solve for n here, but experimenting with values of n shows n = 4 is the smallest n that works. Since S_4 = .84146... we can say with certainty that the first four decimals of S are .8414 . Note however that S_4 < S (the last term of S_4 is negative) so by rounding UP the fifth digit 6 in S_4 one would have S is closer to .8415 then to .8414.
• 16) Need a value of n with 1/(2n+2)! < .00001 , i.e. (2n+2)! > 100,000. Experimenting with values of n shows n=4 is the smallest n that will work. We can S is approximately \sum_{n=0}^4 (-1)^n/(2n)! (note this is actually S_5 = sum of first FIVE terms since series begins with n = 0). Since S_5 = .54030... we can say S = .5403 (accurate to four decimal places).
• 17) Need a value of n with (n+1)^2/10^n < .00001 . We can't solve for n, but experimenting with values of n shows the samllest n that works is n=6. Note S_6 < S (since the last term of S_6 is negative). Since S_6 - .067614 we can say S is approximately .0676 (to four decimal places).
• 19) Does not converge absolutely. Diverges (by n-th term test).
• 20) Converges absolutely.
• 21) Converges absolutely.
• 22) Does not converge absolutely, but converges.
• 23) Does not converge absolutely but converges.
• 24) Does not converge absolutely. Diverges.
• 25) Converges absolutely.
• 26) Converges absolutely.
• 27) Converges absolutely.
• 28) Does not converge absolutely. Diverges.

• 2) It converges for |x| < 2 because it is the integral of the given series and integrating a series does not change the radius of convergence.
• 12 a) \sum_{n=0}^\infty (-1)^n x^{n+1}/(n + 1) for |x| < 1 b) \sum_{n=0}^\infty (-1)^n x^{n+2}/(n + 1) for |x| < 1 c) \sum_{n=0}^\infty (-1)^n x^(2n+2)/(n+1) for |x| < 1.

• 2 a) If it were the series for f(x) at 1, then the degree one coefficient would be f'(1) = -0.8. However the graph of the function shows the slope at x = 1 , which is equal to f'(1), satisfies f'(1) > 0.
• 2b) If it were the series for f(x) at 2, then the degree two coefficient would be f"(2)/2 = 1.5, which would imply f"(2) = 3. However the graph of the function is concave down at x = 2, which would imply f"(2) < 0.

• 11) With M = (3/8)*4^(-5/2) = 3/256 get |R_2(x)| < 1.6 x 10^-5.
• 13) With M = (56/81)*(.8)^(-10/3) < 1.5 get |R_3(x)| < 1 x 10^-4 .
• 15) With M = (16(.1)^4 + 48(.1)^2 + 12)*3^(.01) < 12.7 get |R_3(x)| < 5.3 x 10^-5 .
• 17) With M = 6 get |R_4(x)| < .05 .

• 2) Converges to zero.
• 18) Diverges.
• 20) cos(sqrt(pi)/3)
• 22) e^(-e)
• 25) We need a value of n with 1/(n+1)^5 < .00001 = 10^(-5), i.e. (n+1)^5 > 10^5, so n+1 > 10 or n > 9. Taking n = 10 and computing S_10 gives S_10 = .97211... , so we can say S = .9721 (accurate to four decimal places).

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