Math 121 Final Exam December 18, 2016

Math 121 Final Exam December 18, 2016 1. [18 Points] Evaluate the following limit. Please justify your answer. Be clear if the limit equals a value, +∞ or −∞, or Does Not Exist. Simplify. 3     3xex + 3ex − 2 0 0 L’H 1 + 9x = lim (a) lim 1 x→0 x→0 0 0 1− 1−x 54x 3xex + 3ex + 3ex + 3+3 (1 + 9x2 )2 L’H = lim = = −6 1 x→0 −1 − (1 − x)2 3xex − arctan(3x) x + ln(1 − x)  (b) Compute lim x→∞ „ ln = lim e x→∞ x x+7 x(1∞ )  x «x lim ln x + 7 = ex→∞ x x+7 x   lim x ln = ex→∞ x x+7 (∞·0)  00       x x+7 (x + 7)(1) − x(1) x+7 7 ln x+7 x (x + 7)2 x (x + 7)2 lim lim lim 1 1 1 x→∞ x→∞ x→∞ − 2 − 2 L’H x x x =e = e =e     ∞ x+7 7 −7x ∞ 2 lim (−x ) lim 1 L’H −7 x (x + 7)2 = ex→∞ = ex→∞ x + 7 = e1 = 7 e  2. [20 Points] Z Evaluate each of the following integrals. Z 1 (a) (x2 + 4) Z 1 dx = 7 2 (4 + Z 1 7 (4(1 + Z 2 sec2 θ dθ tan2 θ)) 2 1 2 7 2 sec θ dθ √ √ 2 (4(sec θ)) 4 sec2 θ Z Z Z Z 1 1 sec2 θ 1 1 1 2 = 2 sec θ dθ = 6 dθ = dθ = cos5 θ dθ 27 sec7 θ 2 sec7 θ 64 sec5 θ 64 Z Z Z
2 1 1 1 4 2 = cos θ cos θ dθ = (1 − w2 )2 dw 1 − sin θ cos θ dθ = 64 64 64 = 1 4 tan2 θ) 7 2 2 sec2 θ dθ = 7 2 2 sec2 θ dθ = w = sin θ Substitute dw = cos θdθ 1   1 2w3 w5 1 − 2w + w dw = w− +C + 64 3 5   2 sin3 θ sin5 θ 1 sin θ − +C = + 64 3 5  3  5 ! x x x 1 2 1 √ √ √ = − + +C 64 5 x2 + 4 3 x2 + 4 x2 + 4 1 = 64 Z 2 4 √ x2 + 4 x x = 2 tan θ θ Trig. Substitute dx = 2 sec2 θdθ 2 Z 1 x2 x2 1 1 √ (b) x arcsin x dx = arcsin x − dx 2 2 0 1 − x2 0 0 Z 1 sin2 θ x2 1 x=1 p · cos θ dθ = arcsin x − 2 2 x=0 1 − sin2 θ 0 Z Z 1 1 x2 x2 1 x=1 sin2 θ 1 x=1 sin2 θ √ = arcsin x − · cos θ dθ = arcsin x − · cos θ dθ 2 2 x=0 2 2 x=0 cos θ cos2 θ 0 0 Z 1 x2 1 x=1 2 = arcsin x − sin θdθ 2 2 x=0 0 Z Z 1 1 x2 1 x=1 1 − cos(2θ) x2 1 x=1 = arcsin x − dθ = arcsin x − 1 − cos(2θ)dθ 2 2 x=0 2 2 4 x=0 0 0   x=1 1 1 x=1 x2 1 1 x2 1 1 = arcsin x − θ − sin(2θ) = arcsin x − θ + sin(2θ) 2 4 2 2 4 8 0 x=0 0 x=0 Z 1 1 x=1 1 x2 1 1 x2 1 1 p arcsin x − θ + 2 sin θ cos θ = arcsin x − arcsin x + x 1 − x2 2 4 8 2 4 4 0 x=0 0       1 1 1 π 1 π π 1 arcsin 1 − arcsin 0 + 0 − 0 = − = = arcsin 1 − 0 − 2 4 4 2 2 4 2 8 1 = u = arcsin x dv = xdx 1 x2 du = √ dx v = 2 1 − x2 1 x x = sin θ θ √ 1 − x2 Trig. Substitute dx = cos θdθ 2 0 3. [30 Points] For the following improper integral, determine whether it converges or diverges. If it converges, find its value. Simplify. Z e4 1 1 (a) dx lim dx 2 + x [16 + (ln x) ] x[16 + (ln x)2 ] t→0 0 t      Z 4 u 4 1 1 1 4 ln t arctan − arctan = lim du = lim arctan = lim 4 ln t t→0+ 4 4 4 t→0+ ln t 16 + u2 t→0+ 4      1 ln t 1  π  π  1 π 2π 3π arctan (1) − arctan = = = lim − − = + + 4 4 4 4 2 4 4 4 16 t→0 Z e4 u = ln x Substitute du = 1 dx x x = t ⇒ u = ln t x = e4 ⇒ u = ln e4 = 4 Z 2 Z 2 4 4 4 dx = dx = lim dx 2 − 6x + 5 + x (x − 5)(x − 1) (x − 5)(x − 1) t→1 1 1 t Z 2 2 1 1 PFD = lim − dx = lim ln |x − 5| − ln |x − 1| x−1 t→1+ t x − 5 t→1+ t Z 2 (b) = lim ln |2 − 5| − ln |2 − 1| − (ln |t − 5| − ln |t − 1|) = lim ln 3 − 0 − ln |t − 5| + ln |t − 1| t→1+ t→1+ = ln 3 − ln 4 + (−∞) = −∞ Diverges Partial Fractions Decomposition: A B 4 = + (x − 5)(x − 1) x−5 x−1 Clearing the denominator yields: 4 = A(x − 1) + B(x − 5) 4 = Ax − A + Bx − 5B 4 = (A + B)x − A − 5B so that A + B = 0, and − A − 5B = 4 So lve for A = 1, and B = −1 ∞ Z t Z t 4 1 1 (c) dx = 4 lim dx = 4 lim dx 2 2 t→∞ 4 x − 6x + 12 t→∞ 4 (x − 3)2 + 3 x − 6x + 12 4   t−3 Z t−3 1 1 w = 4 lim dw = 4 lim √ arctan √ t→∞ 1 t→∞ w2 + 3 3 3 1      4 t−3 1 = √ lim arctan √ − arctan √ 3 t→∞ 3 3      4 t−3 1 = √ lim arctan √ − arctan √ 3 t→∞ 3 3 Z 3 4 π π  4π =√ − = √ 6 3 2 3 3 Converges w =x−3 x=4⇒w=1 dw = dx x=t⇒w =t−3 Substitute 4. (a) [18 Points] Find the sum of each of the following series (which do converge). Simplify. ∞ X (−1)n 32n−1 3 33 35 = − + − + ... 42n+1 43 45 47 n=1 3 32 9 and r = − 2 = − 64 4 16 3 3 − − 3 a 3 16 64 64 =  As a result, the sum is given by = = − =− · 25 9 1−r 64 25 100 1− − 16 16 Here we have a nice geometric series with a = −   ln 9 n ln 9 “ 1” − ∞ ∞ X X − 1 (−1)n+1 (ln 9)n 1 2 ln 9− 2 2 = −√ = − (b) =e =− = −e 2n n! n! 3 9 n=0 n=0  π 2n ∞ ∞ ∞ (−1)n n 2n n 2n X X X (−1) π (−1) π 3 (c) = = n 2n 9 (2n + 1)! 3 (2n + 1)! (2n + 1)! n=0 n=0 n=0  π 2n+1  π 2n  π  n ∞ ∞ (−1)n X 3 X (−1) 3 3 3   · π = = (2n + 1)! π (2n + 1)! n=0 n=0 3 ! √  π  3 √3 3 3 3 = sin = = π 3 π 2 2π   1 1 1 1 1 1 1 1 π (d) −1 + − + − + . . . = − 1 − + − + − . . . = − arctan 1 = − 3 5 7 9 3 5 7 9 4 (e) − π2 π4 π6 π8 + − + − . . . = cos π − 1 = −1 − 1 = −2 2! 4! 6! 8!     1 1 1 1 1 7 (f) − + − + . . . = ln 1 + = ln 2 3 4 6 2(6) 3(6) 4(6) 6 6 4 5. [32 Points] In each case determine whether the given series is absolutely convergent, conditionally convergent, or divergent. Justify your answers. (a) ∞ X (−1)n (3 + n2 ) n7 + 4 n=1 ∞ X 3 + n2 First, we show the absolute series is convergent using LCT. n7 + 4 n=1 We see that ∞ X n=1 3 + n2 ≈ n7 + 4 ∞ X n=1 ∞ n2 X 1 = which is a convergent p-series p = 5 > 1. n7 n5 n=1 n2 3+ 3 7 +1 3n5 + n7 n2 Check: lim n + 4 = lim = lim = 1 which is finite and non-zero. Therefore, 7 1 n→∞ n→∞ n + 4 n→∞ 1 + 47 n n5 ∞ X 1 converges, then the absolute series also these two series share the same behavior. Since n5 n=1 converges by LCT. Finally, we have Absolute Convergence . ∞ ∞ ∞ X X sin2 n X 6 sin2 n 6 + = + (b) n6 6n n6 6n n=1 n=1 The first sum ∞ X n=1 6 =6 n6 ∞ X n=1 n=1 1 is convergent as a constant multiple of a convergent p-series with n6 p = 6 > 1. ∞ X sin2 n The second sum is also convergent by CT. 6n n=1 The terms are bounded ∞ X 1 1 sin2 n < and is a convergent geometric series n n 6 6 6n n=1 with |r| = 1 1 = < 1. 6 6 Finally, sum of two convergent series is convergent. Here the O.S.=A.S. so we have Absolute Convergence or AC . (c) ∞ X n3 ln n n=2 Diverges by the nth term Divergence Test since n3 x3  ∞  L’H 3x2 = lim = lim   = lim 3x3 = ∞ = 6 0 n→∞ ln n x→∞ ln x ∞ x→∞ x→∞ 1 x lim 5 (d) ∞ X (−1)n n n2 + 1 n=1 First, we show the absolute series is divergent. Note that ∞ X n=1 ∞ X1 n ≈ which is a divergent n2 + 1 n n=1 (Harmonic) p-series with p = 1. Next, n 2 n2 1 Check: lim n + 1 = lim 2 = 1 which is finite and non-zero. = lim 1 n→∞ n→∞ n + 1 n→∞ 1 + 12 n n Therefore the A.S. is also divergent by LCT. Secondly, we are left to examine the original alternating series with the Alternating Series Test. n •bn = 2 >0 n +1 • lim n→∞ • n =0 n2 + 1 1 1 < bn+1 bn because the related function f (x) = x 1 − x2 0 (x) = has negative derivative f < 0 when x2 + 1 (x2 + 1)2 x > 2. Then the O.S. converges by the Alternating Series Test. Finally, O.S. is Conditionally Convergent or CC . (e) ∞ X (−1)n (3n)! ln n (n!)2 24n nn n=1 Try Ratio Test: (−1)n+1 ln(n + 1) (3(n + 1))! an+1 (n + 1)n+1 24(n+1) ((n + 1)!)2 L = lim = lim n→∞ n→∞ (−1)n ln n (3n)! an nn 24n (n!)2      4n    (3n + 3)! nn (n!)2 2 ln(n + 1) = lim n→∞ (3n)! (n + 1)n+1 ((n + 1)!)2 24n+4 ln n      4n    n (3n + 3)(3n + 2)(3n + 1)(3n)! n (n!)2 2 ln(n + 1) = lim n→∞ (3n)! (n + 1)n (n + 1) (n + 1)2 (n!)2 24n 24 ln n       3(n + 1)(3n + 2)(3n + 1) 1 1 1 1 (∗) = lim (1) n→∞ 1 e n+1 (n + 1)2 16    1       3+ 2 3 + 3 3n + 2 3n + 1 3   n n = lim = lim  1  1 n→∞ 16e n→∞ 16e n+1 n+1 1+ 1+ n n 6 = 27 <1 16e Therefore the original series Converges Absolutely by the Ratio test . Here, from above, 1 ∞ ln(n + 1) ∞ ln(x + 1) L’H 1 x ∞ ∞ L’H (∗) = lim = lim = lim x + 1 = lim = lim = 1 1 n→∞ x→∞ x→∞ x→∞ 1 x→∞ x + 1 ln n ln x x 6. [16 Points] Find the Interval and Radius of Convergence for the following power series. Analyze carefully and with full justification. ∞ X (−1)n (5x + 1)n n9 · 9n n=1 Use Ratio Test. (−1)n+1 (5x + 1)n+1 an+1 (n + 1)9 9n+1 L = lim = lim n→∞ n→∞ (−1)n (5x + 1)n an n9 9n   9 9n |5x + 1| n (5x + 1)n+1 · = = lim · n n+1 n→∞ (5x + 1) n+1 9 9 |5x + 1| < 1 or |5x + 1| < 9. 9 8 That is −9 < 5x + 1 < 9 =⇒ −10 < 5x < 8 =⇒ −2 < x < 5 Endpoints: The Ratio Test gives convergence for x when ∞ ∞ X (−1)n (5 (−2) + 1))n X (−1)n (−9)n = •x = −2 The original series becomes n9 9n n9 9n n=0 n=0 ∞ X (−1)n (−1)n 9n = n9 9n ∞ X 1 which is a convergent p-series with p = 9 > 1. n9 n=0 n=0     8 n 5 + 1 )n X ∞ (−1) ∞ ∞ X 8 (−1)n 9n X (−1)n 5 •x = The original series becomes = = 5 n9 9n n9 9n n9 = n=0 n=0 n=0 which is convergent by the Absolute Convergence Test (ACT) since the absolute series ∞ X 1 is n9 n=0 convergent, as shown above with the left endpoint.   8 9 Finally, Interval of Convergence I = −2, with Radius of Convergence R = . 5 5 7 7. Z [10 Points] 1
1 x2 arctan x2 dx with error less than . 50 Use MacLaurin series to Estimate 0 Please analyze with detail and justify carefully. Simplify.
2n+1 Z 1 Z 1 X Z 1 X ∞ ∞
(−1)n x2 (−1)n x4n+2 2 2 2 2 x arctan x dx = x dx = x dx 2n + 1 2n + 1 0 0 0 n=0 Z = ∞ 1X 0 n=0 = (−1)n x4n+4 dx = 2n + 1 x5 x9 x13 − + − ... 1 · 5 3 · 9 5 · 13 ∞ X n=0 (−1)n x4n+5 (2n + 1)(4n + 5) 1 = 0 n=0 1 0 x5 x9 x13 − + − ... 5 27 65 1 0 1 1 1 + − . . . − (0 − 0 + 0 − . . .) = − 5 27 65 ≈ 1 1 27 5 22 − = − = ← Estimate 5 27 135 135 135 Note this is an alternating series. Use the Alternating Series Estimation Theorem. If we approximate the actual sum with only the first two terms, the error from the actual sum will be at most 1 1 1 . Here the maximum error < as the absolute value of the next (first neglected) term, 65 65 50 desired. 8. [16 Points] (a) Consider the region bounded by y = 1 + arctan x, y = ln x, x = 1 and x = 2. Rotate the region about the vertical line x = −2 . Set-up, BUT DO NOT EVALUATE!!, the integral to compute the volume of the resulting solid using the Cylindrical Shells Method. Sketch the solid, along with one of the approximating shells. See me for a sketch. Z 2 Z 2 V = 2π radius height dx = 2π (x + 2) (1 + arctan x − ln x) dx 1 1 π , x = 0 and x = 1. Rotate the region 2 about the vertical line x = 4 . Set-up, BUT DO NOT EVALUATE!!, the integral to compute the volume of the resulting solid using the Cylindrical Shells Method. Sketch the solid, along with one of the approximating shells. (b) Consider the region bounded by y = arcsin x, y = See me for a sketch. Z 1 Z 1 π  V = 2π radius height dx = 2π (4 − x) − arcsin x dx 2 0 0 9. [20 Points] t3 e2t − and y = 2tet − 2et . 3 2 COMPUTE the arclength of this parametric curve for 0 ≤ t ≤ 1. Simplify. (a) Consider the Parametric Curve represented by x = 8 dy dx = t2 − e2t and = 2tet + 2et − 2et = 2tet . dt dt s Z 1q Z 1  2  2 dx dy + dt = (t2 − e2t )2 + (2tet )2 dt L= dt dt 0 0 Z 1p Z 1p Z 4 2 2t 4t 2 2t 4 2 2t 4t = t − 2t e + e + 4t e dt = t + 2t e + e dt = First, 0 Z 0 1 = 2 2t t +e 0 t3 e2t dt = + 3 2 1 0 1q (t2 + e2t )2 dt 0   1 e2 1 e2 1 e2 1 e0 2 e2 3 = + = + − 0+ − = + − = − 3 2 2 3 2 2 6 2 6 2 6 (b) Consider a different Parametric Curve represented by x = sin3 t and y = cos3 t. π COMPUTE the surface area obtained by rotating this curve about the y-axis for 0 ≤ t ≤ . 2 Simplify. dy dx = 3 sin2 t cos t and = 3 cos2 t(− sin t) = −3 cos2 t sin t First, dt dt s   2 Z π 2 dx 2 dy S.A. = 2πx(t) + dt dt dt 0 Z π q
2 2 3 = 2π sin t 3 sin2 t cos t + (−3 cos2 t sin t)2 dt 0 π 2 Z = 2π p sin3 t 9 sin4 t cos2 t + 9 cos4 t sin2 t dt 0 π 2 Z = 2π Z q 2 2 2 2 sin t 9 sin t cos t(sin t + cos t) dt = 2π 3 0 = 2π 3 sin t(3 sin t cos t) dt = 6π  4 sin t cos t dt = 6π 0 5 10. π 2 Z 0 = 6π q sin t 9 sin2 t cos2 t(1) dt 3 0 π 2 Z π 2 sin π 2
! 5 [20 Points]  − sin5 0 5   = 6π sin5 t 5 π 2  0  1 6π −0 = 5 5 For each of the following parts, do the following two things: 1. Sketch the Polar curves and shade the described bounded region. 2. Set-Up but DO NOT EVALUATE the Integral representing the area of the described bounded region. (a) The area bounded outside the polar curve r = 9 cos θ. 9 r = 3 + 3 cos θ and inside the polar curve These two polar curves intersect when 3 + 3 cos θ = 9 cos θ ⇒ 6 cos θ = 3 ⇒ cos θ = 1 2 Area = A = = 1 2 Z π 3 Z π 3 (outer r)2 − (inner r)2
1 2 π ⇒θ=± . 3 dθ − π3 9 cos θ)2 − (3 + 3 cos θ)2
dθ − π3 OR using symmetry A= 2 1 2 Z π 3 ! (9 cos θ)2 − (3 + 3 cos θ)2 dθ 0 (b) The area bounded outside the polar curve r = 1 and inside the polar curve 10 r = 2 sin θ. These two polar curves intersect when π 5π 2 sin θ = 1 ⇒ sin θ = 12 ⇒ θ = or θ = . 6 6 Z 5π
6 1 Area = A = (outer r)2 − (inner r)2 dθ 2 π 6 = 1 2 5π 6 Z (2 sin θ)2 − (1)2 dθ π 6 OR using symmetry 11 Z 1 2 A=2 π 2 ! 2 2 (2 sin θ) − (1) dθ 0 (c) The area that lies inside both of the curves r = 1+sin θ and inside the polar curve r = 1−sin θ. These two polar curves intersect when θ = 0 and θ = π. Using symmetry, we have ! Z π
1 2 2 2 Area = A = 4 (outer r) − (inner r) dθ 2 0 = 4 1 2 Z π 2 ! (1 − sin θ)2 dθ 0 OR you could use symmetry again ! Z 1 π A= 4 (1 − sin θ)2 dθ 2 π 2 12 OR you could use symmetry again   Z π 1 2 (1 − sin θ) dθ A= 2 2 0 OR you could use symmetry again ! Z 3π 2 1 A= 4 (1 + sin θ)2 dθ 2 π OR you could use symmetry again ! Z 1 2π A= 4 (1 + sin θ)2 dθ 2 3π 2 OR you could use symmetry again  Z 2π  1 2 (1 + sin θ) dθ A= 2 2 π (d) The area bounded outside the polar curve r = 1 and inside the polar curve 13 r = 2 sin(2θ). These two polar curves intersect when 2 sin(2θ) = 1 ⇒ sin(2θ) = Using symmetry Area = A = 4 = 4 1 2 Z 5π 12 1 2 Z 5π 12 ! (outer r)2 − (inner r)
2 dθ π 12 ! (2 sin(2θ)2 − (1)2 dθ π 12 OR using more symmetry ! Z π 1 4 2 2 A=8 (2 sin(2θ) − (1) dθ 2 π 12 14 1 2 ⇒θ= π 5π or . 12 12