MATH 104 SAMPLE FINAL SOLUTIONS. e x/2 cos xdx.

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1 MATH 0 SAMPLE FINAL SOLUTIONS CLAY SHONKWILER () Evaluate the integral e / cos d. Answer: We integrate by parts. Let u = e / and dv = cos d. Then du = e / d and v = sin. Then the above integral is equal to e / sin + e / sin d. To compute this integral, again let u = e / and let dv = sin d. Then du = e / and v = cos. Hence e / cos d = e / sin + e / sin d = e / sin + [ e / cos ] e / cos d = e / sin e / cos e / cos d. Therefore, 5 e / cos d = e / sin e / cos, so e / cos d = 5 e / sin 5 e / cos. () Consider the series n= n 6n + ln n. Does this series converge or diverge? Answer: Since ln n grows significantly slower than both n and n, we suspect that it won t much affect whether the series converges or diverges. Therefore, let s do a it comparison with the series n = 6n n, which we know diverges: n 6n +ln n n n = n + ln n.

2 CLAY SHONKWILER By two applications of L Hôpital s Rule, this it is equal to n n + = n =. n Therefore, since n n diverges, the series also diverges. 6n +ln n (3) Consider the sequence {a n } with n(n+) ( ) 5 n a n = π n. Does this sequence converge or diverge? If it converges, what does it converge to? Answer: Note that ( ) n(n+) 5 n π n = 5n π n = ( ) 5 n π. Since 5 <, the sequence { ( ) 5 n} π π converges to zero. Therefore, by the sandwich theorem, the given sequence converges to zero. () Consider the curve y =. Find the surface area of the surface obtained by rotating the portion of the curve between = 3/ and = 5/ about the -ais. Answer: Recall the surface area formula 5/ ( ) dy SA = πy + d. 3/ d ( ) Now, dy d =, so dy d =. Hence, SA = π 5/ 3/ + d = π 5/ 3/ + d. Letting u = +, du = d and so the above integral becomes [ ] [ π udu = π 3 u3/ = π 3 (8) ] 3 () = 8π 3. (5) Suppose you invest $000 in an account earning 8% interest compounded continuously. After how many years will you quadruple your money (use the approimation ln.6). Answer Remember that the amount of money in your account is modeled by A(t) = P e rt. Since the initial amount in the account is $000, P = 000. Also, r = You will quadruple your money when so 000 = A(t) = 000e 0.08t, = e 0.08t.

3 Solving for t, so MATH 0 SAMPLE FINAL SOLUTIONS 3 ln = 0.08t, t = ln = 60 8 = 0. Therefore, you will quadruple your money in approimately 0 years. (6) Consider the series n e n+. n= Does this series converge or diverge? If it converges, to what sum does it converge? Answer: Note that n= n e n+ = e n= n e n = n= ( ) n e e = n= ( ) n e 3 e. This is a geometric series with a = e 3 and r = e <. Hence, the series converges to (7) Does the series e 3 e = e 3 e e = n= (n)! n n! converge or diverge? Answer: Use the Ratio Test: ((n+))! n+ (n+)! (n)! n n! = e 3 e. (n + )! n+ (n + )! n n! (n)! = (n + )(n + ) = (n + ) By L Hôpital s Rule, this it is equal to 8n + 5 =, so the series diverges. (8) Find the interval of convergence of the power series () n n. n=0 For what values of does the series converge absolutely? For what values does it converge conditionally? Answer: Using the Ratio Test, () n+ (n+) () n n = () n+ (n + ) n () n = n (n + ). n + 5n +. n +

4 CLAY SHONKWILER By two applications of L Hôpital s Rule, this it is equal to n (n + ) = =. Now, < when <, so the radius of convergence of the power series is. Now, we need to check the endpoints, = ±. When =, the series becomes n= ( ) n n, which converges, by the alternating series test. In fact, ( ) n n = n, n= n= which is a convergent p-series. On the other hand, when =, the series becomes n n, n= which is a convergent p-series. Therefore, the interval of convergence of the power series is [, ]. The series converges absolutely for. (9) Consider the region contained by y =, the - and y-aes and + =. What is the volume of the solid obtained by rotating this region about the y-ais? Answer: Using the shell method, the volume is equal to Now, r = and h =, so + V = 0 0 πrhd. π + d = π 0 d +. Letting u = +, du = d, so we can re-write this integral as π du u = π [ln u ] = π [ln ln ] = π ln. (0) Evaluate the integral d ( ) 8 +. Answer: We complete the square: 8 + = ( 8 + ) + = ( ) 3.

5 MATH 0 SAMPLE FINAL SOLUTIONS 5 Hence, letting u =, du = d and so we can re-write the integral as du u u 3 = sec u 3 + C = sec + C () Evaluate the integral d. Answer: Note that 7 + = ( 3)( ). Thus, we set +3 up the partial fraction ( 3)( ) = A 3 + B and solve for A and B: + 3 = A( ) + B( 3). Letting =, () + 3 = A(0) + B() = B, so B =. Now, letting = 3, so A = 9. Therefore, d = (3) + 3) = A( ) + B(0) = A, [ () Solve the initial value problem ] d = 9 ln 3 + ln +C. dy d + y = sin + cos, > 0, y(π) = π. Answer: Re-write in standard form: dy d + sin + cos y =, so P () = sin +cos and Q() =. Hence, P ()d = d = ln = ln, since > 0. Thus, Therefore, y = v()q()d = v() v() = er P ()d = e ln =. sin + cos d = [sin + cos ] d = [ cos + sin + C] sin cos + C =.

6 6 CLAY SHONKWILER Using the initial value, sin π cos π + C = y(π) = π π so C =. Therefore, y = sin cos +. = + C π, (3) Evaluate the integral ( d. + 3) 3/ Answer: Let = 3 tan θ. Then d = 3 sec θdθ, so we can re-write the integral as 3 tan θ 3 tan θ sec 3 sec θdθ (3 tan θ + 3) 3/ θdθ = (3 sec θ) 3/ 3 tan θ sec θdθ = 3 3 sec 3 θ = tan θdθ 3 sec θ = sin θdθ 3 Now, θ = tan 3, so cos θ = to = 3 +. () Consider the sequence {a n } where 3 n a n = ln(n + ). = 3 cos θ + C Hence, the integral is equal Does the sequence converge or diverge? If it converges, what does it converge to? Answer: Using L Hôpital s Rule, 3 n ln(n + ) = 3n /3 n+ Using L Hôpital again, this is equal to = n /3 Therefore, the sequence diverges. n + = 3n /3. 3 n =.

7 (5) Does the series MATH 0 SAMPLE FINAL SOLUTIONS 7 n= n(ln n) converge or diverge? Answer: Using the Integral Test, d = (ln ) b b d (ln ). Now, letting u = ln, du = d, so d du (ln ) = u = u + C = ln + C. Hence, b b [ d (ln ) = ] b [ = b ln b ln b + ] = ln since ln = 0. Therefore, by the Integral Test, n(ln n) DRL 3E3A, University of Pennsylvania address: shonkwil@math.upenn.edu diverges.