CHAPTER

5 The Integral

5.1 Approximating and Computing Area

Preliminary Questions

1. The interval [2, 5] is divided into 6 subintervals in order to calculate R6 for some function. What are the right-endpoints of those subintervals? What are the left-endpoints?

−2 2. If f (x) = x on [3, 7], which is larger: RN or L N ?

3. Which of the following pairs of sums are not equal? (a) 4 i, 4  i=1 =1 (b) 4 j 2, 5 k2 j=1 k=2 (c) 4 j, 5 (i − 1) j=1 i=2 4 ( + ), 5 ( − ) (d) i=1 i i 1 j=2 j 1 j 4. The interval [1, 5] is divided into 16 subintervals. (a) What are the left endpoints of the first and last subintervals? (b) What are the right endpoints of the first two subintervals? 5. True or False: (a) The right-endpoint rectangles lie below the graph of an increasing function.

(b) If f is monotonic, then the area under the graph lies in between RN and L N .

(c) If f is not monotonic, then L N and RN may converge to different limits as N →∞. (d) If f (x) is constant, then the right-endpoint rectangles all have the same height.

Exercises

1. An athlete runs with velocity 4 mph for half an hour, 6 mph for the next hour, and 5 mph for another half-hour. Compute the total distance traveled and indicate on a graph how this quantity can be interpreted as an area. ( ) ( 1 ) + ( )( ) + ( ) ( 1 ) = 1 The total distance traveled is 4 2 6 1 5 2 10 2 miles.

1 2 Chapter 5 The Integral

6 5 4 mph 3 2 1

0 0.5 1 1.5 2 hours 2. Figure 1 shows the velocity of an object over a 3-minute interval. Determine the distance 3. traveledAssume overthat the the velocity intervals of[0 an, 3] objectand [1 is, 2 32.5t]ft/s.(remember Use Eq. to (?? convert) to determine from mph the to distance miles per traveledminute). by the object over the time intervals (in seconds) [0, 2] and [2, 5]. The total distance traveled is given by the area under the graph of v = 32t.

[ , ] Figure1 ( 1 )( ) = During the interval 0 2 , the object travels 2 2 64 64 ft. [ , ] 1 ( )( − ) + ( )( ) = During the interval 2 5 , the object travels 2 3 160 64 3 64 336 ft.

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20

0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

4. Consider f (x) = 2x + 3on[0, 3]. 2 5. Let (a)f (x)Determine= x + x − the2. left- and right-endpoints if [0, 3] is divided into 6 subintervals. (a) Calculate R and L for the interval [2, 5]. (b) Compute3 R6 and3 L6. (b) Sketch(c) Find the the graph exact of areaf and using the geometry rectangles and that compute make up the each error of in the the approximations. two calculations of (b). Let f (x) = x 2 + x − 2andseta = 2, b = 5, n = 3, h = x = (b − a) /n = (5 − 2) /3 = 1.

(a) Let xk = a + kh, k = 0, 1, 2, 3.

Selecting left endpoints of subintervals, xk , k = 0, 1, 2, or {2, 3, 4},wehave

2 2 L3 = f (xk )x = h f (xk ) = (1) (4 + 10 + 18) = 32 k=0 k=0

Selecting right endpoints of subintervals, xk , k = 1, 2, 3, or {3, 4, 5},wehave

3 3 R3 = f (xk )x = h f (xk ) = (1) (10 + 18 + 28) = 56 k=1 k=1

(b) Here are figures of the three sets of rectangles that approximate the area under the curve f (x) over the interval [2, 5]. 5.1 Approximating and Computing Area 3

x2+x−2 x2+x−2 30 30

25 25

20 20

15 15

10 10

5 5

2 2.5 3 3.5 4 4.5 5 2 2.5 3 3.5 4 4.5 5 x x 6. Let f (x) = cos x. 7. Consider f (x) = 5x on [0, 3]. [ , π ] (a) Calculate R4 and L4 for the interval 0 2 . (a) Find(b) Sketch a formula the graphfor RN of(usef and formula the rectangles (??)). that make up each of the approximations. (b) Is R bigger or smaller than the area under the graph? (c) NIs the area under the graph larger or smaller than R4?ThanL4? (c) Calculate limN→∞ RN . (d) Calculate the area under the graph using geometry and verify your answer to (c).

Let f (x) = 5x on [0, 3].Letn be a (symbolic) positive integer and set a = 0, b = 3, h = x = (b − a) /n = (3 − 0) /n = 3/n.

(a) Let xk = a + kh = 3k/n, k = 1, 2,... ,n be the right endpoints of the n subintervals of [0, 3].Then n 3 n 3k 45 n R = h f (x ) = 5 = k n k n n n2 k=1 k=1 k=1 2 = 45 n + n = 45 + 45 n2 2 2 2 2n

(b) Since f (x) = 5x is increasing, the right endpoint approximation Rn is greater than the area under the graph. 45 45 45 (c) The area under the graph is lim Rn = lim + = = 22.5. n→∞ n→∞ 2 2n 2 = 1 = 1 ( )( ) = 45 (d) The area under the graph is the area of a triangle: A 2 BH 2 3 15 2 ,which agrees with the answer in (c). 8. Estimate R6 and L6 over [0, 1.5] for the function shown in Figure 2. 9. Estimate R6 and L6 for the graph in Figure 3.

Figure 2

1

.5

0 1 2 1 4 5 2 3 3 3 3 Figure 3

( ) [ , ] = = ( − )/ = 1 Let f x on 0 2 be given by the figure in the text. For n 6, h 2 0 6 3 , { }6 = , 1 , 2 , , 4 , 5 , xk k=0 0 3 3 1 3 3 2 . 4 Chapter 5 The Integral

Therefore 1 5 1 1 5 7 9 3 L6 = f (xk ) = + + + + 1 + = 1.625 3 k=0 3 2 8 8 8 4 1 6 1 5 7 9 3 R6 = f (xk ) = + + + 1 + + 1 ≈ 1.792 3 k=1 3 8 8 8 4

In Exercises 10–19, calculate the approximation for the given function and interval. 10. R8,7− x, [3, 5] 2 11. R6,2x − x + 2, [1, 4] ( ) = 2 − + [ , ] = = ( − )/ = 1 Let f x 2x x 2on 1 4 .Forn 6, h 4 1 6 2 , { }6 = , 1 , , 1 , , 1 , xk k=0 1 1 2 2 2 2 3 3 2 4 . Therefore

1 6 = ( 2 − + ) = . R6 2xk xk 2 47 5 2 k=1 12. 2 L6,2x − x + 2, [1, 4] 3 13. R4, x − 2x + 5, [0, 1] ( ) = 3 − + [ , ] = = ( − )/ = 1 Let f x x 2x 5on 0 1 .Forn 4, h 1 0 4 4 , { }4 = , 1 , 1 , 3 , xk k=0 0 4 2 4 1 . Therefore

1 4 = ( 3 − + ) ≈ . R4 xk 2xk 5 4 140625 4 k=1 14. 3 R4, x − 2x + 5, [−2, 2] [ π , 3π ] 15. R5,sinx, 4 4 ( ) = [ π , 3π ] = = (3π/4−π/4) = π Let f x sin x on 4 4 .Forn 5, h 5 10 , { }5 = π , 7π , 9π , 11π , 13π , 3π xk k=0 4 20 20 20 20 4 . Therefore π 5 R5 = sin xk ≈ 1.402563 10 k=1 16. −1 L5, x , [1, 2] −2 17. L4, x , [1, 3] ( ) = −2 [ , ] = = ( − )/ = 1 { }4 = , 3 , , 5 , Let f x x on 1 3 .Forn 4, h 3 1 4 2 , xk k=0 1 2 2 2 3 . Therefore

1 3 = −2 ≈ . L4 xk 0 927222 2 k=0

18. [ π , π ] L4,cosx, 4 2 2 19. L5, x + 3|x|, [−3, 2] Let f (x) = x 2 + 3 |x| on [−3, 2].Forn = 5, h = (2 − (−3))/5 = 1, { }5 = {− , − , − , , , } xk k=0 3 2 1 0 1 2 . Therefore

4 = ( 2 + | |) = L5 1 xk 3 x 36 k=0 5.1 Approximating and Computing Area 5

20. R&W Compute the average of R5 and L5 to approximate the area under the graph of 21. Calculatef (x) = x the−1 over sums:[3.5, 5]. Explain why the average is more accurate than either endpoint 5 approximation. (a) 3 i=1 5 (b) 3 i=0 4 (c) k3 k=2 4 π (d) sin j j=3 2 When the number of summands is small, we may compute the sum directly. Where applicable (especially when the number of summands is large), we may use summation laws. For illustration, both ways are shown in this exercise. 5 (a) Do this directly: 3 = 3 + 3 + 3 + 3 + 3 = 15. Or use the law for summing a i=1 5 constant: 3 = (3)(5) = 15. i=1 5 5 (b) Here 3 = 3 + 3 + 3 + 3 + 3 + 3 = 18 or 3 = (3)(6) = 18. i=0 i=0 4 (c) Again, k3 = 23 + 33 + 43 = 99 or k=2 4 4 1 k3 = k3 − k3 = = = k 2 k 1 k 1 44 43 42 14 13 12 = + + − + + = 99. 4 2 4 4 2 4 4 jπ (d) Finally, sin = 1 + 0 + (−1) + 0 = 0. j=1 2

In Exercises 22–34, use the Linearity Rules and formulas (??)–(??) to rewrite and evaluate the sums. 22. 6 3 10 j 3 23. j4=1j j=1 10 3 = 10 3 = 104 + 103 + 102 = We have j=1 4 j 4 j=1 j 4 4 2 4 12100. 24. 10 20 2 j 25. j=1 (2k + 1) k=1 20 ( + ) = 20 + 20 = 202 + 20 + = We have k=1 2k 1 2 k=1 k k=1 1 2 2 2 20 440. 26. 15 (6 − 3i) i=1 6 Chapter 5 The Integral

10 27. (3 − 22) =1 We have 10 (3 − 2) = 10 3 − 10 2 = 104 + 103 + 102 − 103 + 102 + 10 = =1 2 =1 2 =1 4 2 4 2 3 2 6 2255. 28. 30 2 18 (3s − 4s − 1) 2 29. s=1 (4i − 2i + 9) i=1 18 ( − 2 + ) = 18 − 18 2 + 18 = We have i=1 4i 2i 9 4 i=1 i 2 i=1 i 9 i=1 1 182 + 18 − 183 + 182 + 18 + ( )( ) =− 4 2 2 2 3 2 6 9 18 3372. 30. 10 2 200 j Hint: write as a difference of two sums. 31. j=5j j=101 200 = 200 − 100 = 2002 + 200 − 1002 + 100 = We have j=101 j j=1 j j=1 j 2 2 2 2 15050. 32. 50 20 (7 j − 9) 2 33. j=4 6 j j=10 We have 20 2 = 20 2 − 9 2 = 203 + 202 + 20 − 93 + 92 + 9 = j=10 6 j 6 j=1 j 6 j=1 j 6 3 2 6 6 3 2 6 15510. 34. 10 35. Write(6 thej 3 − sumj 2) of the cubes of the whole numbers from 100 to 250 in summation notation. j= 3 250 The sum of the first 100 cubes may be written in summation notation as k3. = 36. k 100 Write the sum + 3 + + 3 +···+=− n +10n3 = 10 = In Exercises 37–41, calculate the1 sum,1 assuming2 2 that a1 1, i=1 ai 10, and i=1 bi 7. in summation notation. 10 37. 2ai i=1

10 10 We have 2ai = 2 ai = (2)(10) = 20. = = 38. i 1 i 1 10 10 (ai − bi ) 39. i=1 (3a + 4b) =1 We have 10 10 10 (3ai + 4bi ) = 3 ai + 4 bi i=1 i=1 i=1 = (3)(10) + (4)(7) = 58.

40. 10 10 ai 41. Can you calculate i=1 ai bi from the information given? i=2 10 From the information given, ai bi cannot be computed. i=1 5.1 Approximating and Computing Area 7

42. n i Evaluate lim n 2 −. + n→∞ in2 i 1 43. Evaluate lim i=1 . n→∞ 3 i=1 n Now n i 2 − i + 1 1 n n n s = = i 2 − i + 1 n 3 3 = n n = = = i 1 i1 i 1 i 1 1 n3 n2 n n2 n 1 2 = + + − + + (n) = + . n3 3 2 6 2 2 3 3n2 1 Therefore, lim sn = . n→∞ 3

In Exercises 44–59, use formulas (??)–(??) to find a formula for RN for the given function and interval. Then compute the area under the graph as a limit. 44. x; [0, 3] 45. x; [2, 7] 7 − 2 5 Let f (x) = x on the interval [2, 7].Thenx = = and a = 2. Hence, N N N 5 N 5 10 N 25 N R = x f (2 + jx) = 2 + j = 1 + j N 2 = N = N N = N = j 1 j 1 j 1 j 1 10 25 N 2 N 25 25 = N + + = 10 + + N N 2 2 2 2 2N and 25 25 45 lim RN = lim 10 + + = = 22.5. N→∞ N→∞ 2 2N 2

46. 3 − x; [1, 2] 47. 2x + 7; [3, 6] 6 − 3 3 Let f (x) = 2x + 7ontheinterval[3, 6].Thenx = = and a = 3. Hence, N N N 3 N 3 RN = x f (3 + jx) = 2 3 + j + 7 j=1 N j=1 N 39 N 18 N 39 18 N 2 N 9 = 1 + j = N + + = 39 + 9 + 2 2 N j=1 N j=1 N N 2 2 N

and 9 lim RN = lim 48 + = 48. N→∞ N→∞ N

48. x 2; [0, 1] 49. x 2; [2, 4] 8 Chapter 5 The Integral

( ) = 2 [ , ]  = 4−2 = 2 = Let f x x on the interval 2 4 .Then x N N and a 2. Hence, N 2 N 8 4 R = x f (2 + jx) = 4 + j + j 2 N 2 j=1 N j=1 N N 8 N 16 N 8 N 8 16 N 2 N = 1 + j + j 2 = N + + 2 3 2 N = N = N = N N 2 2 j 1 j 1 j 1 3 2 + 8 N + N + N N 3 3 2 6 8 8 4 4 = 8 + 8 + + + + N 3 N 3N 2 and 56 12 4 lim RN = lim + + ≈ 18.6667. N→∞ N→∞ 3 N 3N 2 50. 4 − x 2; [0, 2] 51. 3x 2 − x + 4; [0, 1] 1 − 0 1 Let f (x) = 3x 2 − x + 4ontheinterval[0, 1].Thenx = = and a = 0. Hence, N N N 1 N 1 1 R = x f (0 + jx) = 3 j 2 − j + 4 N 2 j=1 N j=1 N N 3 N 1 N 4 N = j 2 − j + 1 3 2 N = N = N = j 1 j 1 j 1 3 N 3 N 2 N 1 N 2 N 4 = + + − + + N N 3 3 2 6 N 2 2 2 N 3 1 1 1 = 1 + + − − + 4 2N 2N 2 2 2N and 1 1 lim RN = lim 4.5 + + = 4.5. N→∞ N→∞ N 2N 2 52. 3x 2 − x + 4; [1, 5] 53. 4x 3 − 3x; [0, 2] 2 − 0 2 Let f (x) = 4x 3 − 3x on the interval [0, 2].Thenx = = and a = 0. Hence, N N N 2 N 8 2 R = x f (0 + jx) = 4 j 3 − 3 j N 3 j=1 N j=1 N N 64 N 12 N 64 N 4 N 3 N 2 12 N 2 N = j 3 − j = + + − + 4 2 4 2 N j=1 N j=1 N 4 2 4 N 2 2 32 16 6 = 16 + + − 6 − N N 2 N and 26 16 lim RN = lim 10 + + = 10. N→∞ N→∞ N N 2 5.1 Approximating and Computing Area 9

54. x 3; [0, 1] 55. x 3 + x; [0, 4] 4 − 0 4 Let f (x) = x 3 + x on the interval [0, 4].Thenx = = and a = 0. Hence, N N N 4 N 64 4 R = x f (0 + jx) = j 3 + j N 3 j=1 N j=1 N N 256 N 16 N = j 3 + j 4 2 N = N = j 1 j 1 4 3 2 2 = 256 N + N + N + 16 N + N N 4 4 2 4 N 2 2 2 128 64 8 = 64 + + + 8 + N N 2 N and 136 64 lim RN = lim 72 + + = 72. N→∞ N→∞ N N 2

56. x 3 + 2x 2; [0, 3] 57. 1 − x 3; [0, 1] 1 − 0 1 Let f (x) = 1 − x 3 on the interval [0, 1].Thenx = = and a = 0. Hence, N N N 1 N 1 R = x f (0 + jx) = 1 − j 3 N 2 j=1 N j=1 N 1 N 1 N 1 1 N 4 N 3 N 2 = 1 − j 3 = N − + + 4 4 N j=1 N j=1 N N 4 2 4 1 1 1 = 1 − − − 4 2N 4N 2 and 3 1 1 3 lim RN = lim − − = . N→∞ N→∞ 4 2N 4N 2 4

58. 2x + 1, [a, b] (a, b constants with a < b) 59. x 2, [a, b] (a, b constants with a < b) b − a Let f (x) = x 2 on the interval [a, b].Thenx = . Hence, N N (b − a) N (b − a) (b − a)2 R = x f (a + jx) = a2 + 2aj + j 2 N 2 j=1 N j=1 N N a2(b − a) N 2a(b − a)2 N (b − a)3 N = 1 + j + j 2 2 3 N = N = N = j 1 j 1 j 1 a2(b − a) 2a(b − a)2 N 2 N (b − a)3 N 3 N 2 N = N + + + + + N N 2 2 2 N 3 3 2 6 a(b − a)2 (b − a)3 (b − a)3 (b − a)3 = a2(b − a) + a(b − a)2 + + + + N 3 2N 6N 2 10 Chapter 5 The Integral

and ( − )2 2 2 a b a lim RN = lim a (b − a) + a(b − a) + N→∞ N→∞ N ( − )3 ( − )3 ( − )3 + b a + b a + b a 3 2N 6N 2 (b − a)3 = a2(b − a) + a(b − a)2 + . 3

60. √ Let f (x) = x 2 + 1andx = 1 . Explain what the following sum represents (in terms of In Exercises 61–66, use the approximation3 indicated (in summation notation) to express the area areas of rectangles) but do not evaluate it. under the graph as a limit but do not evaluate the limit. 6 x · f (1 + ix) 61. RN ,sinx over [0,π] i=1 Let f (x) = sin x over [0,π] and set a = 0, b = π, h = x = (b − a) /n = π/n.Then n π n kπ Rn = h f (xk ) = sin k=1 n k=1 n Hence π n kπ lim Rn = lim sin n→∞ n→∞ n k=1 n is the area between the graph of f (x) = sin x and the x-axis over [0,π]. 62. −1 RN ; x over [1, 7] 63. RN ;tanx over [1, 4] 4 − 1 3 Let f (x) = tan x over the interval [1, 4].Thenx = = and a = 1. Hence, N N N 3 N 3 RN = x f (1 + jx) = tan 1 + j j=1 N j=1 N

and 3 N 3 lim RN = lim tan 1 + j N→∞ N→∞ N j=1 N

64. −2 L N ; x over [3, 5] [ π ,π] 65. L N ;cosx over 8 π π − 7π Let f (x) = cos x over the interval π ,π .Thenx = 8 = and a = π . Hence, 8 N 8N 8

− − N 1 π 7π N 1 π 7π L N = x f + jx = cos + j j=0 8 8N j=0 8 8N

and 7π N−1 π 7π lim L N = lim cos + j N→∞ N→∞ 8N j=0 8 8N

66. [ π , π ] L N ;cosx over 8 4 5.1 Approximating and Computing Area 11

67. Show that the area A under the graph of f (x) = x −1 over [1, 8] satisfies 1 1 1 1 1 1 1 1 1 1 1 1 1 + + + + + + ≤ A ≤ 1 + + + + + + 2 3 4 5 6 7 8 2 3 4 5 6 7

Hint: use R7 and L7.

−1 Let f (x) = x ,1≤ x ≤ 8. Since f is decreasing, the left endpoint approximation L7 overestimates the true area between the graph of f and the x-axis, whereas the right endpoint approximation R7 underestimates it. Accordingly, 1 + 1 + 1 + 1 + 1 + 1 + 1 = < < = + 1 + 1 + 1 + 1 + 1 + 1 2 3 4 5 6 7 8 R7 A L7 1 2 3 4 5 6 7 Left endpoint approximation, n = 7 Right endpoint approximation, n = 7 1 1 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2

0 12 3 4 5 6 7 8 0 12 3 4 5 6 7 8 68. Which two of the following five expressions represent endpoint approximations to the area 69. underWhich the area graph is represented of f (x) = byx − the2 over following[3, 7]? limits? N100 4 −2 14 j k (a) (a)lim N→∞ 100N j==1 N100 k 1 N100 4 −2 34 3 j k (b) (b)lim 23++ 4 N→∞100 100 N j=j=10 N N100 4 −2 54 j k (c) (c)lim −32++45· N→∞ 100N == N100 jk 11 99  −2 Be apprised4 that the+ upper limits of summation below are N (not ∞). (d) 3 100 = 100 N 4 0 1 j (a) The limit99 lim RN = lim −2 represents the area between the graph of 4 N→∞ N→∞k N N (e) 3 + 4 · j=1 4 f (x)100= xk=and0 the x-axis100 over the interval [0, 1]. 3 N j 4 (b) The limit lim RN = lim 2 + 3 · represents the area between the graph N→∞ N→∞ N j=1 N of f (x) = x 4 and the x-axis over the interval [2, 5]. 5 N j 4 (c) The limit lim RN = lim −2 + 5 · represents the area between the N→∞ N→∞ N j=1 N graph of f (x) = x 4 and the x-axis over the interval [−2, 1]. 70. Evaluate the limit = 1/3 [ , . ] 71. Use R5 and L5 to show that the area A under the graph ofy x over the interval 1 1 5 satisfies .530 ≤ A ≤ .545. 1 N j 2 lim 1 − 1/3 N→∞ N N Let f (x) = x on [1, 1.5].Forn = 5, h j=1 (1.5 − 5)/5 = .1, {x }5 = {1, 1.1, 1.2, 1.3, 1.4, 1.5}. Therefore byk interpretingk=0 it as the area of part of a familiar geometric figure. 5 1/3 R5 = (.1) (xk ) ≈ .545 k=1 12 Chapter 5 The Integral

and

4 1/3 L5 = (.1) (xk ) ≈ .530 k=0

Hence, the area A must be .530 ≤ A ≤ .545 as R5 is an overestimate and L5 is an underestimate of the exact value for A. 72. −2 Use R6 and L6 to show that the area A under y = x over the interval [10, 12] satisfies .0161 ≤ A ≤ .0195. Midpoint Approximation The midpoint approximation MN , which is similar to RN and MN , approximates the area under the graph using rectangles whose heights are the function values f (c j ) where c j is the midpoint of the jth interval.

2 73. Calculate M2 and M4 for x for the interval [0, 1]. Let f (x) = x 2 on [0, 1]. = = ( − )/ = 1 { }2 = , 1 , ∗ 1 = 1 , 3 For n 2, h 1 0 2 2 , xk k=0 0 2 1 and xk k=0 4 4 . Therefore

1 1 = ( ∗)2 = . M2 xk 3125 2 k=0 For n = 4, h = (1 − 0)/4 = 1 , {x }4 = 0, 1 , 1 , 3 , 1 and 4 k k=0 4 2 4 ∗ 3 = {. ,. ,. ,. } xk k=0 125 375 625 875 . Therefore

3 = ( ∗)2 = . M4 xk 328125 k=0 74. √ Calculate M3 and M6 for x for the interval [2, 5]. 75. Calculate M2 and M4 for x for the interval [0, 1]. What do you notice about the accuracy of these approximations? Let f (x) = x on [0, 1]. = = ( − )/ = 1 { }2 = , 1 , ∗ 1 = 1 , 3 For n 2, h 1 0 2 2 , xk k=0 0 2 1 and xk k=0 4 4 . Therefore

1 = 1 ∗ = 1 M2 xk 2 k=0 2 For n = 4, h = (1 − 0)/4 = 1 , {x }4 = 0, 1 , 1 , 3 , 1 and 4 k k=0 4 2 4 x ∗ 3 = {. ,. ,. ,. } k k=0 125 375 625 875 . Therefore

3 = ∗ = 1 M4 xk k=0 2

Note that both M2 and M4 are exact values for the area under f (x) = x on the interval [0, 1]. 76. ( 2) [ , π ] Calculate M6 to estimate the area under the graph of sin x over the interval 0 2 .

Figure 4 Graph of sin(x 2). 5.1 Approximating and Computing Area 13

Further Insights and Challenges

77. Although the accuracy of RN generally improves as N increases, this need not be true for small values of N. Draw the graph of a positive continuous function f (x) on an interval such that R1 is closer to the exact area under the curve than R2. δ 1 δ = 1 δ Let be a small positive number less than 4 . (In the figures below, 10 . But imagine being very tiny.) Define f (x) on [0, 1] by   1if0≤ x < 1 − δ  2 1 − x if 1 − δ ≤ x < 1 f (x) = 2δ δ 2 2 .  x − 1 if 1 ≤ x < 1 + δ  δ 2δ 2 2 1 + δ ≤ ≤ 1if2 x 1 Then f is continous on [0, 1]. (Again, just look at the figures.)

= − 1 = − 1 ( δ)( ) = − δ The exact area between f and the x-axis is A 1 2 bh 1 2 2 1 1 .(For δ = 1 = 9 10 ,wehaveA 10 .) = | | = | − | = | − ( − δ)| = δ δ = 1 With R1 1, the absolute error is E1 R1 A 1 1 .(For 10 , this absolute error is |E | = 1 .) 1 10 = 1 | | = | − | = 1 − ( − δ) = δ − 1 = 1 − δ With R2 2 , the absolute error is E2 R2 A 2 1 2 2 . δ = 1 | | = 2 (For 10 ,wehave E2 5 .) Accordingly, R1 is closer to the exact area A that is R2. Indeed, the tinier δ is, the more dramatic the effect.

Graph of f(x) 1 Right endpt approx, n = 1 Right endpt approx, n = 2 1 1 0.8 0.6 0.4 0.5 0.5 0.2

0 0.2 0.4 0.6 0.8 1 x 0 0.5 1 0 0.5 1 78. Draw the graph of a positive continuous function on an interval such that R2 and L2 are 79. bothR&W smallerExplain than the the exact following area understatement the graph. graphically: the endpoint approximations are less accurate when f (x) is large.  When f is large, the graph of f is steeper and hence there is more gap between f and Ln or Rn. Recall that the top line segments of the subrectangles involved in an endpoint approximation constitute a piecewise constant function. If f  is large, then f is increasing more rapidly and hence is less like a constant function. Smaller f’ Larger f’

2 1

0 1234 0 1234 80. Prove that if f (x) is monotonic, then MN always lies in between RN and L N . Explain why 2 81. MR&WN must thenIn Exercise be closer 48, to one the shows actual that area.RN for the function f (x) = x over [0, 1] is given = 1 + 1 + 6 →∞ by the formula RN 3 2N N 2 and taking the limit as N , one concludes that the 14 Chapter 5 The Integral

1 1 + 6 area under the graph is 3 . Can you interpret the quantity 2N N 2 as the area of a region? Which region? Let f (x) = x 2 on [0, 1]. The quantity 1 6 1 1 6 + in R = + + 2N N 2 N 3 2N N 2 represents the collective area of the parts of the subrectangles that lie above the graph of ( ). = 1 f x It is the error between RN and the true area A 3 . 1 0.8 0.6 0.4 0.2

0 0.2 0.4x 0.6 0.8 1 82. Prove the formula 83. Assume that f (x) is positive, continuous, and monotonic increasing or decreasing and let (b − a) A be the area under graph overR − anL interval= [a, b].· ( f (b) − f (a)) (1) N N N (a) Use Eq. (1) to prove that for any function f (x) on [a, b]. (b − a) |RN − A|≤ | f (b) − f (a)| (2) Hint: observe that the sums defining RN and LNN only differ in their first and last terms. ( ) Hint: A lies between RN and L N if f x is monotonic. √ −2 (b) Use Eq. (2) to find a value of N such that |RN − A| < 10 for f (x) = x on [1, 4]. Convergence for Monotonic Functions.Let f (x) be continuous, positive, and monotonic on [a, b].LetA be the area between the graph of f and the x-axis over [a, b].For specificity, say f is increasing. (The case for f decreasing on [a, b] is similar.)

(a) As noted in the text, we have Ln ≤ A ≤ Rn. By Exercise ?? and the fact that A lies between Ln and Rn, we therefore have b − a 0 ≤ R − A ≤ R − L = ( f (b) − f (a)) . n n n n Hence b − a b − a |R − A| ≤ ( f (b) − f (a)) = | f (b) − f (a)| , n n n where f (b) − f (a) = | f (b) − f (a)| because f is increasing on [a, b]. Therefore, b − a |R − A| ≤ | f (b) − f (a)| . n n √ (b) Let f (x) = x on [1, 4].Thenb = 4, a = 1, and | − | < 4−1 ( ( ) − ( )) = 3 ( − ) = 3 3 < −2 RN A N f 4 f 1 N 2 1 N . We need N 10 , which gives√ −2 −2 3 < N · 10 and consequently N > 300. Thus |R301 − A| < 10 for f (x) = x on [1, 4]. 84. √ −2 2 Use Eq. (2) to find a value of N such that |RN − A| < 10 for f (x) = 9 − x on [0, 3]. 85. The area of the unit circle. By definition, π is equal to one-half of the circumference of the unit circle. We know that π is also equal to the area of the unit circle, but this requires proof. In this exercise, triangles are used (instead of rectangles) to compute the area of a unit circle. Consider a regular polygon with n sides inscribed in the unit circle. Figure 5 shows the case n = 8. In general, θ = 2π/N. 5.2 The Definite Integral 15

1 ( π/ ) 1 ( π/ ) (a) Show that the triangles OQP and OPRhave areas 2 sin 2 N and 2 sin 2 N , respectively. (b) Show that N 2π N 2π sin ≤ area of unit circle ≤ tan 2 N 2 N (c) Let N →∞and use the Squeeze Theorem to show that the area of the unit circle is π. Hint: take h = 2π/N in the formula lim sin h = 1. h→0 h

R

Q

θ P

Figure 5 Area of the unit circle.

The area of the unit circle.  = 1 = 1 ( )( θ) = 1 ( 2π )  (a) The area of OPQis S 2 bh 2 1 sin 2 sin n ;thatof OPRis = 1 = 1 ( ) θ = 1 ( 2π ) T 2 BH 2 1 tan 2 tan n . (b) From the figure in the text, the area A of the circular sector OPQis greater than that of OPQ, yet less than that of OPR. Accordingly, we have S≤ A ≤ T . Thus π π sin 2 A tan 2 nS ≤ nA ≤ nT,or n sin( 2π ) ≤ A ≤ n tan( 2π ). Hence n ≤ ≤ n . 2 n 2 n 2π π 2π n n = 2π (c) Let h n . Then from (b) we have ( ) / sin h ≤ A ≤ tan h = sin h h . h π h cos h → ≤ A ≤ A = = π As h 0, this gives 1 π 1. Thus, π 1orA . The area of the unit circle is therefore π.

5.2 The Definite Integral Preliminary Questions b ( ) = 1. What is a dx? (here the function is f x 1) 2. True or false: ( ) [ , ] b ( ) (a) If f x is continuous and positive on a b ,then a f x dx is the area under the graph of f (x) over [a, b]. 16 Chapter 5 The Integral ( ) [ , ] b ( ) (b) For all continuous functions f x on a b , a f x dx is the area between the graph [ , ] and the x-axis over a b . ( ) ≤ [ , ] − b ( ) ( ) (c) If f x 0on a b ,then a f x dx is the area between the graph of f x and the x-axis over [a, b]. π = 3. Explain the equality 0 cos xdx 0 graphically. −1 4. True or false: −5 5 dx is negative. 0 2 2 5. True or false: 8 x dx is positive because x is a positive function. 6. Which statement is correct: (a) If f (x) ≤ 1 then 6 f (x) dx ≤ 1 . 2 0 2 ( ) ≤ 1 6 ( ) ≤ (b) If f x 2 then 0 f x dx 3. 2 ( ) ( ) ≤ 1 7. What is the largest possible value of 0 f x dx if f x 3 ? 8. Theorem ?? says that if f (x) ≤ g(x), then the integral of f over a given interval is smaller than the integral of g over the same interval. Is the same true for derivatives? In other words, does f (x) ≤ g(x) imply that f (x) ≤ g(x)?

Exercises

In Exercises 1–10, draw a graph of the signed area represented by the integral and compute it using geometry. 3 1. −3 2xdx 3 = 1 ( )( ) + 1 (− )( ) = Using geometry, the value of the integral −3 2xdx 2 3 6 2 3 6 0.

6

4

2

0

−2

−4

−6 −3 −2 −1 0 1 2 3

2. 3 ( + ) − 2x 4 dx 12 ( + ) 3. −2 3x 4 dx 1 + 1 7 ( ) − 1 2 ( ) = 15 = . Using geometry, the value of the integral −2 3x 4 dx is 2 3 7 2 3 2 2 7 5. 7

–2 –11 –2 4. 1 −2 4 dx 5.2 The Definite Integral 17 8 ( − ) 5. 6 7 x dx 8 ( − ) = 1 ( )( ) + 1 ( )(− ) = Using geometry, the value of the integral 6 7 x dx 2 1 1 2 1 1 0.

1

0.8

0.6

0.4

0.2

0

−0.2

−0.4

−0.6

−0.8

−1 6 6.2 6.4 6.6 6.8 7 7.2 7.4 7.6 7.8 8

6. 3π/2 π √ sin xdx 25 − 2 7. 0 25 x dx √ 5 − 2 1 π ( )2 = 25 π Using geometry, the value of the integral 0 25 x dx is 4 5 4 . 5 4 3 2 1

0 12345 8. 3 | | − x dx 22 ( −| |) 9. −2 2 x dx 2 ( −| |) = 1 ( )( ) + 1 ( )( ) = Using geometry, the value of the integral −2 2 x dx 2 2 2 2 2 2 4.

2

1.8

1.6

1.4

1.2

1

0.8

0.6

0.4

0.2

0 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2

10. 1 (2x −|x|) dx −1 6 ( − ) 11. Calculate 0 4 x dx in two ways: (a) limN→∞ RN (b) Sketch the relevant signed area and use geometry. ( ) = − [ , ] 6 ( ) = 6 − Let f x 4 x over 0 6 . Consider the integral 0 f x dx 0 4 xdx. (a) Let n be a (symbolic) positive integer and set a = 0, b = 6, h = x = (b − a) /n = 6/n. 18 Chapter 5 The Integral

Let xk = a + kh = 6k/n, k = 1, 2,... ,n be the right endpoints of the n subintervals of [0, 6].Then n 6 n 6k 6 n 6 n Rn = h f (xk ) = 4 − = 4 1 − k = n = n n = n = k1 k 1 k 1 k 1 6 6 n2 n 18 = 4n − + = 6 − n n 2 2 n 18 Hence lim Rn = lim 6 − = 6. n→∞ n→∞ n 6 − 1 ( )( ) − 1 ( )( ) = (b) Using geometry, the value of the integral 0 4 xdxis 2 4 4 2 2 2 6. 4

2

0 246 –2 12. 5 Calculate (2x + 1) dx in two ways: as a limit limN→∞ RN , and using geometry. 13. Evaluate the2 following integrals, where f (x) is the function shown in Figure 1.

246

Figure 1 The two parts of the graph are semi-circles.

(a) 2 f (x) dx 0 (b) 6 f (x) dx 0 (c) 4 f (x) dx 1 6 | ( )| (d) 1 f x dx Let f (x) be given by the figure in the text. (a) We have 2 f (x) dx =−1 π (1)2 =−π . 0 2 2 (b) We have 4 f (x) dx = 1 π (2)2 − 1 π (1)2 = 3 π. 1 4 4 4 (c) We have 6 f (x) dx = 1 π (2)2 − 1 π (1)2 = 3π . 0 2 2 2 6 | ( )| = 1 π ( )2 + 1 π ( )2 = 9π (d) We have 1 f x dx 2 2 4 1 4 .

In Exercises 14–17, sketch the signed area represented by the integral. Indicate the regions that are treated as positive and negative. 14. 2( − 2) 0 x x dx 5.2 The Definite Integral 19 3( − 2) 15. 0 2x x dx 3 − 2 Here is a sketch of the signed area represented by the integral 0 2x x dx. 1 + + + 23 0 - - - –1 –2 –3 16. 2π π sin xdx 3π 17. 0 sin xdx 3π Here is a sketch of the signed area represented by the integral 0 sin xdx. + + + 12 0 - - - –1

–2

In Exercises 18–21, calculate the Riemann sum R( f, P, C) for the given function, partition, and choice of intermediate points. Also, sketch the graph of f and the rectangles corresponding to R( f, P, C).

18. f (x) = x, P ={1, 1.2, 1.5, 2}, C ={1.1, 1.4, 1.9} 19. f (x) = x 2 + x, P ={2, 3, 4.5, 5}, C ={2, 3.5, 5}

2 Let f (x) = x x and P ={x0 = 2, x1 = 3, x3 = 4.5, x4 = 5} and C ={c1 = 2, c2 = 3.5, c3 = 5}.ThenR( f, P, C) = x1 f (c1) + x2 f (c2) + x3 f (c3) = (3 − 2)(6) + (4.5 − 3)(15.75) + (5 − 4.5)(30) = 44.625. Here is a sketch of the graph of f and the rectangles.

x2+x 40

35

30

25

20

15

10

5

0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 x

20. f (x) = x + 1, P ={−2, −1.6, −1.2, −.8, −.4, 0}, C ={−1.7, −1.3, −.9, −.5, 0} ( ) = ={ , π , π , π } ={. ,. , . } 21. f x sin x, P 0 6 3 2 , C 4 7 1 2 ( ) = ={ = , = π , = π , = π } Let f x sin x and P x0 0 x1 6 x3 3 x4 2 and C ={c1 = .4, c2 = .7, c3 = 1.2}.Then R( f, P, C) = x1 f (c1) + x2 f (c2) + x3 f (c3) = ( π − )( . ) + ( π − π )( . ) + ( π − π )( . ) = . 6 0 sin 4 3 6 sin 7 2 3 sin 1 2 1 029225. Here is a sketch of the graph of f and the rectangles. 20 Chapter 5 The Integral

sin(x) 1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

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0.1

0 0 0.5 1 1.5 x

In Exercises 22–29, use the linearity properties of the integral and the formulas in Example ?? to calculate the integrals. 22. 4 x 2 dx 13 ( + ) 23. 0 3x 4 dx We have 3 3x + 4 dx = 3 3 xdx+ 4 3 1 dx = 3 · 1 (3)2 + 4(3 − 0) = 51 = 25.5. 24. 0 0 0 2 2 3 ( + ) −2 3x 4 dx 3 ( 2 − ) 25. −2 3x 4x dx We have − 3 2 − = 0 2 − + 3 2 − = 3 2 − − 2 2 − −2 3x 4xdx −2 3x 4xdx 0 3x 4xdx 0 3x 4xdx 0 3x 4xdx − − = 3 3 x 2 dx − 4 3 xdx− 3 2 x 2 dx − 4 2 xdx = 0 0 0 0 3 · 1 (3)3 − 4 · 1 (3)2 − 3 · 1 (−2)3 − 4 · 1 (−2)2 = 25. 26. 3 2 3 2 1 ( 2 − ) 0 x 2x dx 3 ( 2 + + ) 27. 0 6x 7x 1 dx 3 2 + + = 3 2 + 3 + 3 = We have 0 6x 7x 1 dx 6 0 x dx 7 0 x 0 1 dx 6 · 1 (3)3 + 7 · 1 (3)2 + (3 − 0) = 177 = 88.5. 28. 3 2 2 1 (x 2 + x) dx −a2 a 2 29. a x dx 2 2 3 We have a x 2 dx = a x 2 dx − a x 2 dx = 1 a2 − 1 (a)3 = 1 a6 − 1 a3. 30. a 0 0 3 3 3 3 N 3 = N 4 + N 3 + N 2 Using the formula j=1 j 4 2 4 , compute the limit of right-endpoint In Exercisesapproximations 31–36, useto show Eq. (1) that to for evaluate all b (positive the integral. or negative): b 4 3 b 3 3 x = (1) 31. 0 x dx 0 4 4 We have 3 x 3 dx = 3 = 20.25 using Eq. (1). 32. 0 4 0 3 −3 x dx 0 ( 3 − ) 33. 2 2x 3x dx 0 ( 3 − ) =− 2 ( 3 − ) =− 2 3 + 2 = We have 2 2x 3x dx 0 2x 3x dx 2 0 x dx 3 0 xdx − 24 + 22 =− 2 4 3 2 2 using Eq. (1). 34. 2 ( − 3) 0 x x dx 3 3 35. 1 x dx 4 4 We have 3 x 3 dx = 3 x 3 dx − 1 x 3 dx = 3 − 1 = 20 using Eq. (1). 36. 1 0 0 4 4 2 ( − 3) 1 x x dx 5.2 The Definite Integral 21 5( ( ) + ( )) 5 ( ) = 5 ( ) = 37. What is 2 f x g x dx, assuming that 2 f x dx 7and 2 g x dx 3? 5( ( ) + ( )) = 5 ( ) + 5 ( ) = + = We have 2 f x g x dx 2 f x dx 2 g x dx 7 3 10.

In Exercises 38–44, calculate the integral, assuming that 5 5 f (x) dx = 5, g(x) dx = 12 0 0 38. 5 ( f (x) + g(x)) dx 05( ( ) + ( )) 39. 0 f x 4g x dx We have 5 f (x) + 4g(x) dx = 5 f (x) dx + 4 5 g(x) dx = 5 + 4(12) = 53. 40. 0 0 0 5 ( f (x) − g(x)) dx 05 ( ) 41. 0 4 f x dx We have 5 4 f (x) dx = 4 5 f (x) dx = 4(5) = 20. 42. 0 0 0 g(x) dx 55( ( ) − ( )) 43. 0 3 f x 5g x dx We have 5 3 f (x) − 5g(x) dx = 3 5 f (x) dx − 5 5 g(x) dx = 3(5) − 5(12) =−45. 44. 0 0 0 5 Is it possible to calculate g(x) f (x)dx from the information given? 6 ( ) 0 2 ( ) = 6 ( ) = 45. Calculate 0 f x dx, assuming that 0 f x dx 7and 2 f x dx 5. 2 ( ) = 6 ( ) = 6 ( ) Let 0 f x dx 7and 2 f x dx 5. Then 0 f x dx cannot be computed from the information given. However, we do have 6 ( ) = 2 ( ) + 6 ( ) = + = 0 f x dx 0 f x dx 2 f x dx 7 5 12.

In Exercises 46–49, calculate the following integrals, assuming that 1 2 4 f (x) dx = 1, f (x) dx = 4, f (x) dx = 7 0 0 1 46. 4 f (x) dx 02 ( ) 47. 1 f x dx Given 1 f (x) dx = 1, 2 f (x) dx = 4, and 4 f (x) dx = 7, we have 0 0 1 2 f (x) dx = 2 f (x) dx − 1 f (x) dx = 4 − 1 = 3. 48. 1 0 0 1 f (x) dx 44 ( ) 49. 2 f x dx Given 1 f (x) dx = 1, 2 f (x) dx = 4, and 4 f (x) dx = 7, recall from Exercise ?? that 0 0 1 4 f (x) dx = 8. Accordingly, we have 0 4 ( ) = 4 ( ) − 2 ( ) = − = 2 f x dx 0 f x dx 0 f x dx 8 4 4.

In Exercises 50–55, express each of the following as a single integral. 50. 3 7 f (x) dx + f (x) dx 09 ( ) − 39 ( ) 51. 2 f x dx 4 f x dx We have 9 f (x) dx − 9 f (x) dx = 4 f (x) dx. 52. 2 4 2 9 ( ) − 5 ( ) 2 f x dx 2 f x dx 22 Chapter 5 The Integral 3 ( ) + 9 ( ) 53. 7 f x dx 3 f x dx We have 3 9 7 9 f (x) dx + f (x) dx =− f (x) dx + f (x) dx 7 3 3 3 9 = f (x) dx. 7 54. 3 ( ) + 1 ( ) f x dx − f x dx 16 ( ) + 4−4 ( ) 55. −4 f x dx 0 f x dx We have 6 ( ) + −4 ( ) = 0 ( ) + 6 ( ) − 0 ( ) = 6 ( ) −4 f x dx 0 f x dx −4 f x dx 0 f x dx −4 f x dx 0 f x dx.

In Exercises 56–60, calculate the integral assuming that f is a function such that b f (x) dx = 1 − b−1 for all b > 0. 56.1 3 f (x) dx 11 57. 1 4 f (x) dx 2 − 1 ( ) =− 1/2 ( ) =− − 1 1 = We have 1 4 f x dx 4 1 f x dx 4 1 2 4. 58. 2 4 (4 f (x) − 2) dx 11 ( ) 59. 3 f x dx We have 1 f (x) dx =− 3 f (x) dx =−(1 − 3−1) =−2 . 60. 3 1 3 2c f (x) dx (c any positive number) 61. c Which of the following statements is false? Explain. (a) If f (x)>0forx ∈[a, b], then the integral b f (x) dx is also positive. a b ( ) ≥ ( ) ≥ ∈[ , ] (b) If a f x dx 0, then f x 0forx a b . Sketch the graph of a function providing a counterexample for the statement which is not valid.

( ) ≥ ∈[ , ] b ( ) ≥ (a) It is true that if f x 0forx a b ,then a f x dx 0 by the Comparison ( ) = Theorem with gx 0. (b) It is false that if b f (x) dx ≥ 0, then f (x) ≥ 0forx ∈[a, b]. Indeed, in Exercise 3, a 1 + = . ≥ (− ) =− < we saw that −2 3x 4 dx 7 5 0, yet f 2 2 0. Here is the graph from that exercise. 7

–2 –11 –2 62. b ( ) b | ( )| Explain the difference in graphical interpretation between a f x dx and a f x dx 63. whenVerifyf the(x) inequalities:takes on both positive and negative values. 1 1 2 2 x 3 dx ≤ x 2 dx, x 2 dx ≤ x 3 dx 0 0 1 1

1 3 = 1 ( )4 = 1 ≤ 1 = 1 ( )3 = 1 2 1 3 ≤ 1 2 We have 0 x dx 4 1 4 3 3 1 0 x dx;thatis, 0 x dx 0 x dx. 5.2 The Definite Integral 23 We have 2 x 2 dx = 2 x 2 dx − 1 x 2 dx = 1 (2)3 − 1 (1)3 = 7 = 2 1 ≤ 3 3 = 15 = 1 0 0 3 3 3 3 4 4 1 ( )4 − 1 ( )4 = 2 3 − 1 3 = 2 3 2 2 ≤ 2 3 4 2 4 1 0 x dx 0 x dx 1 x dx;thatis, 1 x dx 1 x dx. 64. Prove that 65. Prove that .3 .0198 ≤ .3 sin xdx≤ .0296 .27 ≤ .2 cos xdx≤ .3 Hint: show that .198 ≤ sin x ≤ .296 for x0 in [.2,.3]. Hint: show that .9 ≤ cos x ≤ 1forx in [0,.3]. ≤ ≤ π ≈ . / ( ) =− ≤ For 0 x 6 0 52, we have d dx cos x sin x 0. Hence cos x is nonincreasing on [0.0, 0.3]. Accordingly, for 0.0 ≤ x ≤ 0.3, we have

m = 0.9 ≤ 0.955 ≈ cos 0.3 ≤ cos x ≤ cos 0.0 = 1 ≤ 1 = M

Therefore, by the Comparison Theorem, we have . . . 0.27 = m(0.3 − 0.0) = 0 3 mdx ≤ 0 3 cos xdx≤ 0 3 Mdx= M(0.3 − 0.0) = 0.30. In 0.0 0.0 0.0 . ≤ 0.3 ≤ . other words, 0 27 0.0 cos xdx 0 30.

Further Insights and Challenges 66. a ( ) (− ) =− ( ) Evaluate −a f x dx assuming that f is an odd function, that is, f x f x for 67. allLetxk. Explain.> 0 be a positive exponent. Show that b 1 x k dx = bk+1 x k dx 0 0 for all b > 0. Hint: Compare the right-endpoint approximations for the integrals of x k over [0, a] and [0, 1]. > ( ) = k [ , ] Let k 0 be a positive exponent and b be any positive number. Let f x x on 0 b . b ( ) 1 ( ) Since f is continuous, both 0 f x dx and 0 f x dx exist.

Let n be a (symbolic) positive integer and set A = 0, B = b, h = x = (B − A) /n = b/n.Letx j = A + jh = bj/n, j = 1, 2,... ,n be the right , endpoints of the n subintervals of [0 b]. Then the right endpoint approximation to b ( ) = b k 0 f x dx 0 x dx is n b n bj k 1 n R = h f (x ) = = bk+1 j k n j k+1 j=1 n j=1 n n j=1 = In particular, ifb 1 above, then the right endpoint approximation to 1 ( ) = 1 k 0 f x dx 0 x dx is n 1 n j k 1 n 1 S = h f (x ) = = j k = R n j k+1 k+1 n j=1 n j=1 n n j=1 b

= k+1 In other words, Rn b Sn. Therefore, b k k+1 k+1 k+1 1 k x dx = lim →∞ R = lim →∞ b S = b lim →∞ S = b x dx.Inother 0 n n n n n n 0 b k = k+1 1 k words, 0 x dx b 0 x dx. 68. Verify the formula by interpreting the integral as an area: b 1 1 1 − x 2 dx = b 1 − b2 + θ 0 2 2 ≤ ≤ θ π θ = Here 0 b 1and is the angle between 0 and 2 such that sin b. 24 Chapter 5 The Integral

69. Prove the inequality π/4 sin(x 2) dx ≤ .162 0 / ≤ ( 2) ≤ 2 [ , π ] Hint: The inequality sin x x 1 implies that sin x x for x in 0 4 . < π < ( π )2 < π ∈[ , π ] Note that 0 4 1impies 4 4 . Accordingly, for x 0 4 , the inequality / ≤ ( 2) ≤ 2 ≤ ≤ π sin u u 1 implies sin x x for 0 x 4 . By the Comparison Theorem, we have π/ π/ 4 ( 2) ≤ 4 2 = 1 ( π )3 ≈ . < . 0 sin x dx 0 x dx 3 4 0 16149 0 162. We therefore conclude that π/ 4 ( 2) ≤ . 0 sin x dx 0 162.

2

1

π π 4 2 70. Theorem ?? states the additivity property of definite integrals for adjacent intervals ≤ ≤ , 71. assumingThe midpoint that approximationa b c. ShowMN thatis generally the conclusion more accurateremains true than for either arbitraryRN oraLbN .Inthisand c. exercise, we verify thisb for(f )(x) ==−x 2 ona [0(, 1)]. Hint: use the relation a f x dx b f x dx. (a) Show that for all N ≥ 1, 1 1 1 R ( f ) = + + N 3 2N 6N 2 1 1 1 L ( f ) = − + N 3 2N 6N 2 1 1 M ( f ) = − N 3 12N 2 (b) What is the size of the error (in terms of N) in the three approximations? Rank the three approximations in order of increasing accuracy.

(c) How large must N be in order to make the error in MN smaller than .001? (d) Find an N such that the error in RN and L N is smaller than .001. Let f (x) = x 2 on [0, 1]. (a) Let n be a (symbolic) positive integer and set a = 0, b = 1, h = x = (b − a) /n = 1/n.Letxk = a + kh = k/n, k = 0, 1,... ,n and let ∗ = + ( + 1 ) = ( + 1 )/ = , ,... , − ≥ xk a k 2 h k 2 n, k 0 1 n 1. Then for n 1wehave

n 1 n k 2 1 n R = h f (x ) = = k2 n k n n n3 k=1 k=1 k=1 3 2 = 1 n + n + n = 1 + 1 + 1 n3 3 2 6 3 2n 6n2

n−1 1 n−1 k 2 1 n−1 L = h f (x ) = = k2 n k n n n3 k=0 k=0 k=1 ( − )3 ( − )2 − = 1 n 1 + n 1 + n 1 = 1 − 1 + 1 n3 3 2 6 3 2n 6n2 5.3 The Fundamental Theorem of Calculus, Part I 25

2 n−1 1 n−1 k + 1 1 n−1 1 M = h f (x ∗) = 2 = k2 + k + n k n n n3 4 k=0 k=0 k=0 1 n−1 n−1 1 n−1 = k2 + k + 1 3 n = = 4 = k 1 k 1 k 0 ( − )3 ( − )2 − = 1 n 1 + n 1 + n 1 n3 3 2 6 (n − 1)2 n − 1 1 1 1 + + + n = − 2 2 4 3 12n2

1 1 1 1 (b) The error of R is given by + , the error of L is given by − + and the n 2n 6n2 n 2n 6n2 1 error of M is given by − . Of the three approximations, R is the least accurate, n 12n2 n then Ln and finally Mn is the most accurate. 1 1 1 (c) The absolute error in M is .Inordertohave < , we require n 12n2 12n2 1000 1000 √ n2 > or n > 1000/12 ≈ 22.36; i.e., n ≥ 23 (since n is a positive integer). 12 1 1 1 1 1 (d) The absolute error in R is + .Inordertohave + < ,take n 2n 6n2 2n 6n2 1000 n > 500; i.e., n ≥ 501 (since n is a positive integer).

5.3 The Fundamental Theorem of Calculus, Part I

Preliminary Questions

1. What is the area under the graph of a positive function f over [0, 2], assuming that f has an antiderivative F(x) such that F(0) = 3andF(2) = 7? 5 ( ) ( ) ( ) ( ) = ( ) + 2. Find 2 f x dx assuming that F x is an antiderivative of f x and F 5 F 2 12. 3. Suppose that f is a negative function with the antiderivative F such that F(1) = 7and F(3) = 4. Does the quantity F(3) − F(1) have an interpretation in terms of area? 7 ( ) 7 ( ) ( ) ( ) 4. Evaluate 0 f x dx and 2 f x dx assuming that f x has an antiderivative F x with values in the table.

x 0 2 7 F(x) 3 7 9

5. Answer true or false and explain. (a) The FTC I is only valid for positive functions. (b) To use the FTC I, you have to choose the right antiderivative. (c) If you cannot find an antiderivative of f (x), then the definite integral does not exist. 26 Chapter 5 The Integral

Exercises

In Exercises 1–35, evaluate the definite integral using the FTC. 6 1. 3 xdx 6 We have 6 xdx= 1 x 2 = 1 (6)2 − 1 (3)2 = 27 = 13.5. 2. 3 2 3 2 2 2 9 0 2 dx 3( 3 − 2) 3. 1 x x dx 3(x 3 − x 2) dx = 1 x 4 − 1 x 33 = 1 4 − 1 3 + 1 − 1 ≈ . We have 1 4 3 1 4 3 3 3 4 3 11 333. 4. π 2 cos xdx 03π 2 5. 0 cos xdx π/ π/ We have 3 2 cos xdx= sin x|3 2 = sin( 3π ) − sin(0) =−1. 6. 0 0 2 1 (4 − 5x 4) dx 04 √ 7. 0 xdx √ 4 We have 4 xdx= 4 x 1/2 dx = 2 x 3/2 = 2 (4)3/2 − 2 (0)3/2 = 16 = 5 1 . 8. 0 0 3 0 3 3 3 3 4 ( 2 + ) −3 x 2 dx 4( 5 + 2 − ) 9. 0 3x x 2x dx 4 We have 4 3x 5 + x 2 − 2xdx= 1 x 6 + 1 x 3 − x 2 = 0 2 3 0 1 (4)6 + 1 (4)3 − (4)2 − 1 (0)6 + 1 (0)3 − (0)2 = 6160 = 2053 1 . 10. 2 3 2 3 3 3 4 1 2 dt ( 2 9 − 4 + ) 11. 11 10t x 3x 9 dx 2 We have 2 10x 9 − 3x 4 + 9 dx = x 10 − 3 x 5 + 9x = 1 5 1 (2)10 − 3 (2)5 + 9(2) − (1)10 − 3 (1)5 + 9(1) = 5067 = 1013 2 . 12. 5 5 5 5 2 ( 9 − 4 + ) −2 10x 3x 9 dx 2 2 13. −3 u du 2 u2 du = 1 u32 = 1 ( )3 − 1 (− )3 = 35 = 2 We have −3 3 −3 3 2 3 3 3 11 3 . 14. − 1( 3/2 + 7/2) −3 4t t dt 3 ( 3 − ) 15. −3 4u 7u du 3 3 − = 4 − 7 2 3 = ( )4 − 7 ( )2 − (− )4 − 7 (− )2 = We have − 4u 7udu u u − 3 3 3 3 0. 16. 3 2 3 2 2 9 1/2 1 3x dx 4 π 2 17. 2 dx 4 We have 4 π 2 dx = π 2x = π 2(4) − π 2(2) = 2π 2. 18. 2 2 2( 2 − −2) 1 x x dx 4( 5 + 2 − ) 19. 0 3x x 2x dx We have 4 3x 5 + x 2 − 2xdx= 1 x 6 + 1 x 3 − x 2 4 = 0 2 3 0 1 (4)6 + 1 (4)3 − (4)2 − 1 (0)6 + 1 (0)3 − (0)2 = 6160 = 2053 1 . 20. 2 3 2 3 3 3 4 −4 1 x dx 9 −1/2 21. 1 3x dx 9 We have 9 3x −1/2 dx = 6x 1/2 = 6(9)1/2 − 6(1)1/2 = 12. 22. 1 1 9 8 dx 3 4 x 5.3 The Fundamental Theorem of Calculus, Part I 27 −1 1 23. dx 3 −2 x − −1 1 1 1 1 1 1 We have dx =− x −2 =− (−2)−2 + (−1)−2 =− . 3 − x 2 − 2 2 4 24. 2 2 3π 4 ππ cos xdx 42 25. π (cos θ − sin θ)dθ 4 We have √ π/2 θ − θ θ = ( θ + θ)|π/2 = π + π − π + π = − π/ cos sin d sin cos π/4 sin cos sin cos 1 2. 26. 4 2 2 4 4 4π 2π cos xdx 4 2 27. 0 sec tdt π/ 4 2 = |π/4 = π − = We have 0 sec tdt tan t 0 tan 4 tan 0 1. 28. π 4 0π sec x tan xdx 3 29. π csc x cot xdx 6 √ π/3 = (− )|π/3 = (− π ) − (− π ) = − 2 We have π/6 csc x cot xdx csc x π/6 csc 3 csc 6 2 3 3. 30. π 2 2 π csc xdx 6 5 | − | 31. 0 3 x dx Hint: write as a sum of two integrals. We have 5 |3 − x| dx = 3(3 − x) dx + 5(x − 3) dx = 0 0 3 3 5 (3x − 1 x 2) + ( 1 x 2 − 3x) = (9 − 9 ) − 0 + ( 25 − 15) − ( 9 − 9) = 6.5. 32. 2 0 2 3 2 2 2 3 | 2 − | 0 x 1 dx 3 | 3| 33. −2 x dx We have 3 | 3| = 0 − 3 + 3 3 =−1 40 + 1 43 = + 1 (− )4 + 1 4 − = . − x dx − x dx x dx x − x 0 2 3 0 24 25. 34. 2 2 0 4 2 4 0 4 4 π | | 0 cos x dx 5 | 2 − + | 35. 0 x 4x 3 dx We have 5 |x 2 −4x +3| dx = 5 |(x −3)(x −1)| dx = 1(x 2 −4x +3) dx+ 3 −(x 2 −4x +3) dx+ 0 0 0 1 5 (x 2 − x + ) dx = ( 1 x 3 − x 2 + x)1 − ( 1 x 3 − x 2 + x)3 + ( 1 x 3 − x 2 + x)5 = 3 4 3 3 2 3 0 3 2 3 1 3 2 3 3 ( 1 − 2 + 3) − 0 − (9 − 18 + 9) + ( 1 − 2 + 3) + ( 125 − 50 + 15) − (9 − 18 + 9) ≈ 9.333. 36. 3 3 3 1 n = Use the FTC to show that −1 x dx 0ifn is odd. Explain graphically. In Exercises 37–42, evaluate the integral in terms of the constant. 4 37. 2 cdx We have 4 cdx = cx|4 = c · 4 − c · 2 = 2c for any constant c. 38. 2 2 a 4 b x dx b 5 39. 1 x dx b We have b x 5 dx = 1 x 6 = 1 b6 − 1 (1)6 = 1 (b6 − 1) for any number b. 40. 1 6 1 6 6 6 4 xdx aa 41. b cdx We have b cdx = cx|b = c · b − c · a = (b − a)c for any constant c. 42. a a 4( 2 − ) a 3x x dx 28 Chapter 5 The Integral

43. Suppose that f is a negative function with an antiderivative F such that F(1) = 7and F(3) = 4. Can you compute the area (a positive number) between the x-axis and the graph on [1, 3]? The area is A = 1 f (x) dx =− 3 f (x) dx =−(F(3) − F(1)) =−(4 − 7) = 3. 44. 3 1 1 Does the integral x n dx get larger or smaller as n increases? Make a guess based on 3 ( )0 45. graphsCalculate and− then2 f checkx dx,where by calculation. 12 − x 2 for x ≤ 2 f (x) = x 3 for x ≥ 2

Hint: break up the integral into two integrals corresponding to the two parts of the graph.

20

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–2 –1123 Figure 1 Graph of f (x) in Exercise 45.

12 − x 2 for x ≤ 2 Let f (x) = .Then x 3 for x ≥ 2 3 ( ) = 2 ( ) + 3 ( ) = 2 − 2 + 3 3 −2 f x dx −2 f x dx 2 f x dx −2 12 x dx 2 x dx. 2 Now 2 12 − x 2 dx = (12x − 1 x 3) = (12(2) − 1 (2)3) − (12(−2) − 1 (−2)3) = 128 . −2 3 −2 3 3 3 3 Moreover, 3 x 3 dx = 1 x 4 = 1 (3)4 − 1 (2)4 = 65 . 2 4 2 4 4 4 3 ( ) = 2 ( ) + 3 ( ) = 128 + 65 = 707 = 11 Therefore, −2 f x dx −2 f x dx 2 f x dx 3 4 12 58 12 .

Further Insights and Challenges

In Exercises 46–49, express the following integrals in terms of the constant (a, c, etc). 46. π ( ) − −1 0 cos ax dx Hint: a sin ax is an antiderivative. 1 ( + )3 1 ( + )4 47. 0 x k dx Hint: 4 x k is an antiderivative. We have 1(x + k)3 dx = 1 (x + k)41 = 1 (k + 1)4 − 1 k4 = k3 + 3 k2 + k + 1 . 48. 0 4 0 4 4 2 4 c ( ) 0 cos ct dt 2c 4 49. −c x dx 2c 4 = 1 52c = 1 ( )5 − 1 (− )5 = 33 5 We have − x dx x − 2c c c . 50. c 5 c 5 5 5 Show that 1 n 1 lim j k = n→∞ k+1 + n j=0 k 1

Hint: interpret the left-hand side as an integral and evaluate using the FTC. 5.3 The Fundamental Theorem of Calculus, Part I 29

51. Let c > 0andletA be the area under the graph of the inverted parabola y = c − ax2.Show = 2 that A 3 R where R is the area of the circumscribed rectangle.

c

Figure 2 Graph of y = c − ax2.

√ Let c > 0anda > 0. The inverted parabola y = f (x) = c − ax2 has x-intercepts ± c/a. Accordingly, the area between the inverted parabola and the x-axis is √ √ √ / / c/a 1 c a 1 c a = − 2 = − 3 = − 2 A √ c ax dx cx ax √ x c ax √ − / 3 − / 3 − / c a c a c a / c 1 c 4 c3 2 = 2 c − a = 3 a 3 a a1/2

52. ≤ x ≤ x Applying Theorem ?? to the inequality cos t 1gives 0 cos tdt 0 1 dt for all x. 53. (a) Use(a) Use the method this to conclude of the previous that sin exercisex ≤ x. to prove that (b) Integrate the inequalityx 3 inx (a)5 to provex 7 that − cos x + 1x 3≤ x 2/x2.5 x − + − ≤ sin x ≤ x 3− + 2 4 (c) Continuing to integrate,6 show120 that5040− sin x + x ≤ x /6andcos6 120x − 1 + x /2 ≤ x /24. (d) Deduce the inequalitiesx 2 x 4 x 6 x 2 x 4 1 − + − ≤ cos x ≤ 1 − + 22 24 720 2 4 2 3 24 − x ≤ ≤ − x + x , − x ≤ ≤ 1 . cos< x 11<. 1 sin x x (b) Use this to show that 96891242 cos 4 9689128.2 24 3 (c) Can you guess the general. of inequalities< 1 <. for sin x and cos x? (e) Use (d) to show that 9687 cos 4 969.

− + x2 ≤ x4 + x3 − ≤ x5 (a) Integrating the inequality cos x 1 2 24 gives sin x 6 x 120 and ≤ − x3 + x5 equivalently, sin x x 6 120 . Integrating again, we have − + x4 − x2 + ≤ x6 − x2 + x4 − x6 ≤ cos x 24 2 1 720 and hence 1 2 24 720 cos x. Finally, integrating − + x5 − x3 + ≤ x7 a third time gives sin x 120 6 x 5040 and equivalently − x3 + x5 − x7 ≤ x 6 120 5040 sin x. Thus by combining these inequalities we have − x3 + x5 − x7 ≤ ≤ − x3 + x5 x 6 120 5040 sin x x 6 120 and − x2 + x4 − x6 ≤ ≤ − x2 + x4 1 2 24 720 cos x 1 2 24 . = 1 − (1/4)2 + (1/4)4 − (1/4)6 ≈ . − (1/4)2 + (1/4)4 ≈ . (b) For x 4 ,1 2 24 120 9689124 and 1 2 24 9689128. . < 1 <. Thus 9689124 cos 4 9689128. (c) The general statement of the inequalities is − x2 + x4 − x6 +···− 1 2n+2 ≤ ≤ − x2 + x4 +···+ 1 2n 1 2 24 720 (2n+2)! x cos x 1 2 24 (2n)! x and − x3 + x5 − x7 +···− 1 2n+1 ≤ ≤ − x3 + x5 +···+ 1 2n−1 x 6 120 5040 (2n+1)! x sin x x 6 120 (2n−1)! x . 30 Chapter 5 The Integral

5.4 The Fundamental Theorem of Calculus, Part II

Preliminary Questions (− ) ( ) = x ( ) 1. What is A 2 where A x −2 f t dt? √ ( ) = x 3 + 2. Let G x 4 t 1dt. Answer true or false and explain: (a) The FTC is needed to calculate G(4). ( ) (b) The FTC is needed√ to calculate G 4 . (c) G(2) =− 4 t 3 + 1 dt. 2 √ ( ) = 3. Which of the following is an antiderivative of x satisfying F 2 0? (a) x 1 x −1/2 dx 2 2 (b) 2 x 1/2 dx 0 (c) x x 1/2 dx 2 ( ) ( ) = x ( ) (π) 4. Find F 1 where F x 1 sin sin t dt. Is it possible to compute F ? ( ) = x3 ( ) 5. Let G x 4 sin t dt. Which of the following statements are correct? ( ) ( 3) (a) G x is the composite function sin x . ( ) ( 3) ( ) = x ( ) (b) G x is the composite function A x where A x 4 sin t dt. (c) G(x) is too complicated to differentiate. (d) The Product Rule is used to differentiate G(x). (e) The Chain Rule is used to differentiate G(x). (f) G(x) = sin(x 3) · (3x 2)

6. (Adapted from: Calculus Problems for a New Century, p. 103.) Trick question: find the 3 3 = derivative of 1 t dt at x 2. 9 ( ) ( ) = ( ) = 7. What is the value of 2 f x dx if f is a differentiable function such that f 2 f 9 4?

Exercises

1. Let f (x) = 2x + 4. (a) Write the area function with lower limit a =−2asanintegral. (b) Find a formula for this area function.

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–2 4 Figure 1

( ) = + Let f x 2x 4. (a) The area function with lower limit a =−2isA(x) = x f (t) dt = x 2t + 4 dt. a −2 x + = ( 2 + )x = ( 2 + ) − ((− )2 + (− )) = 2 + + (b) We have −2 2t 4 dt t 4t −2 x 4x 2 4 2 x 4x 4 or (x + 2)2. Therefore, A(x) = (x + 2)2. 5.4 The Fundamental Theorem of Calculus, Part II 31

2. Find a formula for the area function of f (x) = 2x + 4 with lower limit a = 0. ( ) = x ( 2 − ) 3. Let G x 1 t 2 dt. (a) What is G(1)? (b) Use the FTC Part II to find G(1) and G(2). (c) Find a formula for G(x) and use it to verify your answers to (a) and (b). Let G(x) = x t 2 − 2 dt. 1 ( ) = 1 2 − = (a) Then G 1 1 t 2 dt 0. ( ) = 2 − ( ) =− ( ) = (b) Now G x x 2, whence G 1 1andG 2 2. (c) x t 2 − dt = ( 1 t 3 − t)x = ( 1 x 3 − x) − ( 1 ( )3 − ( )) = 1 x 3 − x + 5 We have 1 2 3 2 1 3 2 3 1 2 1 3 2 3 . ( ) = 1 3 − + 5 ( ) = 2 − Thus G x 3 x 2x 3 and G x x 2. Moreover, ( ) = 1 ( )3 − ( ) + 5 = ( ) =− ( ) = G 1 3 1 2 1 3 0, as in (a), and G 1 1andG 2 2, as in (b). 4. True or false: some continuous functions don’t have antiderivatives. √ Explain. ( ) ( ) ( ) ( ) = x 3 + 5. Find F 0 , F 0 and F 3 for the function F x 0 t 1 dt. √ √ ( ) = x 3 + ( ) = 3 + ( ) = ( ) = Let F x √ 0 t 1 dt.ThenF x x 1. Thus F 0 0, F 0 1, and F (3) = 28. 6. ( ) −2 ( ) = x −2 Let G x be the antiderivative of x defined√ by G x 3 t dt. Which initial condition 7. doesLet FG(x(x) )besatisfy? the antiderivative of f (x) = x 3 + 1suchthatF(3) = 0. (a) Represent F as a definite integral. (b) What is F (3)? √ ( ) ( ) = 3 + ( ) = Let F x be the antiderivative √ of f x x 1 satisfying F 3 0. (a) Then F(x) = x t 3 + 1 dt. √3 √ (b) Since F (x) = x 3 + 1, we have F (3) = 28.

In Exercises 8–15, find formulas for the functions represented by the integrals. 8. x u3 du x 2 9. π cos udu 4 √ ( ) = x = |x = − 2 We have F x π/ cos udu sin u π/4 sin x . 10. 4 2 2 x tdt 11. 1x (t 2 − t) dt 2 We have F(x) = x t 2 − tdt= ( 1 t 3 − 1 t 2) = 1 x 3 − 1 x 2 − 2 . 12. 2 3 2 3 2 3 5( − ) x 4t 1 dt 13. x sin udu 0 We have F(x) = x sin udu= (− cos u)|x = 1 − cos x. 0 0 14. x π√ sin udu 2 x 3 15. 2 t dt √ √ 4 x x t x2 We have F(x) = t 3 dt = = − 4. 2 4 4 16. 2 x | | 1 | | Show that 0 t dt is equal to 2 x x .

In Exercises 17–20, express the antiderivative of f (x) satisfying the given initial condition as an integral. 32 Chapter 5 The Integral

17. f (x) = x −3, F(1) = 0 The antiderivative F(x) of f (x) = x −3 satisfying F(1) = 0isF(x) = x t −3 dt. 18. 1 x + 1 f (x) = , F(7) = 0 19. f (x) = sec2 +x, F(0) = 0 x 9 The antiderivative F(x) of f (x) = sec x satisfying F(0) = 0isF(x) = x sec tdt. 20. 0 f (x) = sin(x3), F(−π) = 0 ( ) = x ( ) ( ) 21. Let A x 0 f t dt where f x is shown in Figure 2. (a) Calculate A(2), A(3). (b) Calculate A(2), A(3). (c) Find a formula for A(t) (actually two formulas, one for 0 ≤ t ≤ 2 and one for t ≥ 2). Then sketch the graph of A(t).

4

3

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1

1 2 3 4 Figure 2

( ) = · = ( ) = = ( ) = · + 1 = . (a) A 2 2 2 4, the area under f x from x 0tox 2. A 3 2 3 2 6 5, the area under f (x) from x = 0tox = 3. (b) A(x) = f (x) so A(2) = f (2) = 2andA(3) = f (3) = 3. (c) Here is a graph of the area function A(x) of the function f (x) depicted in Figure 2.

8

7

6

5

4

3

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1

0 0 0.5 1 1.5 2 2.5 3 3.5 4

Area function F(x) 10 8 6 4 2 0 1234 22. Make a rough sketch of the graph of the area function of the function shown in Figure 3. ( ) 23. Let G(x) = g x f (t) dt. a ( ) ( ) = ( ( )) ( ) = x ( ) (a) Show that G x is a composite function: G x A g x where A x a f t dt.    Figure 3  (b) Show that G (x) = A (g(x))g (x) = f (g(x)) g (x) 5.4 The Fundamental Theorem of Calculus, Part II 33

( ) Let G(x) = g x f (t) dt. a (a) Let A(x) = x f (t) dt and g(x) be specified. Then G = A ◦ g is defined by the a ( ) = ( ( )) = g(x) ( ) formula G x A g x a f t dt. (b) Since A(x) = f (x), we have by the Chain Rule that G(x) = A(g(x))g(x) = f (g(x))g(x).

In Exercises 24–31, use the Chain Rule to calculate the derivative. 24. x3  G (x) where G(x) =x2 tan tdt  25. G (1) where G(x) = 3 x 3 + 3 dt 0 2 √ √ ( ) = x 3 + ( ) = 3 + · Let G x √ 0 x 3 dt. Then by Exercise 23, G x x 3 2x and G(1) = 1 + 3 · 2 = 4. 26. 2 d x 0 sin2 tdt d 2 27. dx 0 sin tdt dx x Let G(x) = 0 sin2 tdt=− x sin2 tdt. Then by Exercise 23, G(x) =−sin2 x. 28. x 0 d cos x ( 4 − ) t 3t dt x4 √ dx − 29. F (x) 6 where F(x) = tdt x2 √ ( ) = ( 4) − ( 2) ( ) = x Hint: F x A x A x where A x 0 tdt(any lower limit in the integral will work). √ √ √ ( ) = x4 = x4 − x2 Let F x √ x2 tdt√ 0 tdt 0 tdt. Applying Exercise 23, we have F (x) = x 4 · 4x 3 − x 2 · 2x or 4x 5 − 2x |x|. 30. 2 d x 4 tan tdt. d √ 2 31. dx x sin(t ) dt dx x Let G(x) = 4 sin(t 2) dt =− x sin(t 2) dt.ThenG(x) =−sin(x 2). 32. x 4 Let F(x) = x (t 2 − 5t − 6) dt. ( ) = 0x 3 33. Let F(a)x Find0 thesin criticaltdt.points Find the of criticalF(x) and points determine of F and if they determine are local if they minima are local or maxima. minima or maxima. (b) Find the points of inflection of F(x). ( ) = x 3 ( ) = 3 = = π Let F x 0 sin tdt.SolveF x sin x 0toobtainx n ,wheren is an integer.

For x = 2kπ,wherek is an integer, there is a local minimum value of F(x) since F  changes sign from − to + as x increases through x = 2kπ. For x = (2k + 1)π,wherek is an integer, there is a local maximum value of F(x) since F  changes sign from + to − as x increases through x = (2k + 1)π. 34. Let f (x) be a differentiable function on [a, b] and let F(x) be an antiderivative of f (x). 35. MatchWhich the of the property following of the statements area function is incorrect?A(x) with Explain. the corresponding property of the ( ) function(a) Iff fx(x.)>0, then F(x) has no local minimum on (a, b). (b) If c is an inflection point of F(x),then f (c) = 0. area function A(x): ( )> ( ) (c) If f x 0,(a) thenA(Fx)xdecreasingmust be concave up.  (d) If f (c) = 0,(b) thenA(Fx()xhas) must a local have maximum an inflection point at x = c. (e) If F(x) is decreasing(c) A(x) onis concave(a, b),then up f (x) ≤ 0forx ∈ (a, b). (d) A(x) goes from concave up to concave down 34 Chapter 5 The Integral

graph of f (x): (i) lies below x-axis (ii) crosses x-axis from positive to negative (iii) local maximum (iv) f (x) is increasing

( ) = x ( ) ( ) ( ) = ( ) Let A x a f t dt be an area function of f x .ThenA x f x and A(x) = f (x). (a) Hence A(x) is decreasing when A(x) = f (x)<0, i.e., when f (x) lies below the x-axis. This is choice (i).  (b) Now A(x) has a local maximum (at x0)whenA (x) = f (x) changes sign from + to 0 to − as x increases through x0, i.e., when f (x) crosses the x-axis from positive to negative. This is choice (ii). (c) Moreover, A(x) is concave up when A(x) = f (x)>0, i.e., when f (x) is increasing. This corresponds to choice (iv).   (d) Finally, A(x) goes from concave up to concave down (at x0)whenA (x) = f (x) changes sign from + to 0 to − as x increases through x0, i.e., when f (x) has a local maximum at x0. This is choice (iii). 36. x The following questions refer to A(x) = f (t)dt where f (x) is shown in Figure 4. ( ) = x ( ) ( ) 0 37. Let A(a)x Does0 Af(xt) havedt where a localf maximumx is shown at inA? Figure 5. ( ) [ , ] (a) Calculate(b) Where the does minimumA(x) have and a maximum local minimum? value of A x on 0 6 . ( ) [ , ] (b) Find(c) Where a formula does forA(Ax)xhavevalid a local on the maximum? interval 2 4 . ( ) = x ( ) [ , ] (c) Find(d) True a formula or false: forA the(x)< function0forallB xx. 2 f t dt valid on the interval 2 4 . (e) Describe the change in concavity in A(x) at B. 4 3 2 Figure 4 Graph of f (x). 1 0 –1 –2 –3 1 2 3 4 5 6

Figure 5

(a) At x = 1.5, f (x) changes from − to +, which corresponds to a minimum value of A(x). Hence . . . ( . ) = 1 5 ( ) = 1 − + 1 5( − ) =−|1 + 2 − 1 5 =− . A 1 5 0 f x 0 1 dt 1 2t 3 dt t 0 t 3t 1 25. = . ( ) + − 1 At x 4 5, f x changes from to , which corresponds to a maximum value of ( ) ( . ) = 4.5 ( ) = . A x . Hence A 4 5 0 f x 1 25 using geometric methods and the minimum value of A(x). ( ) [ , ] (b) A formula for A x valid on the interval 2 4 is given by A(x) = x f (t) dt = 1 −1 dt + 2(2t − 3) dt + x 1 dt =−1 + 0 + (x − 2) = x − 3. 0 0 1 2 ( ) = x ( ) [ , ] (c) A formula for B x 2 f t dt valid on the interval 2 4 is given by B(x) = x f (t) dt = x 1 dt = x − 2. 38. 2 2 x Let F(x) = t sin tdt. ( ) = 0x ( ) ( ) 39. Let F(a)x Locate0 thef x localdx where maximaf andx is the the absolute function maximum shown in of FigureF(x) 7.on [0, 3π]. (b) Justify using the graph of f (x): F(x) has precisely one zero in the interval [π, 2π]. Hint: argue that F(π) > 0andF(2π) < 0. (c) How many zeroes does F(x) have in [0, 3π]? (d) Justify using the graph of f (x): there is an inflection point in [0,π] where F(x) changes from concave up to concave down.

Figure 6 Graph of y = x sin x. 5.4 The Fundamental Theorem of Calculus, Part II 35

(a) Where does F(x) have local minima and maxima? (b) On which intervals is F(x) increasing and decreasing? (c) Where does F(x) have points of inflection? (d) On which intervals is F(x) concave up?

12345 6 78

Figure 7

( ) = x ( ) ( ) Let F x 0 f t dt,where f x is the function shown in the Figure 7. Note that F (x) = f (x) and F (x) = f (x).   (a) Accordingly, F(x) will have a local minimum at x0 if F (x0) = f (x0) = 0andF = f changes sign from − to 0 to + as x increases through x0. Similarly, F(x) will have a   local maximum at x0 if F (x0) = f (x0) = 0andF = f changes sign from + to 0 to − as x increases through x0. In light of these facts, we conclude that F has a local minimum at x = 4.5 and a local maximum at x = 2. (b) Moreover, F(x) is increasing where F (x) = f (x)>0 and decreasing where F (x) = f (x)<0. Therefore, F is increasing on (0, 2) and (4.5, 8). It is decreasing on (2, 4.5). (c) Assuming that f is differentiable (and it looks rather smooth from the figure), we have that A is twice differentiable on (0, 8). Thus A(x) has an inflection point at x0 provided     A (x0) = f (x0) = 0 and A (x) = f (x) changes sign at x0. Thus f has a local extremum at x0. This occurs at x = 1, 3, 5, 6, 7.3. (d) Finally, F(x) is concave up where F (x) = f (x)>0, i.e., where f is increasing. This occurs on (0, 1), (3, 5),and(6, 7.3). 40. ( ) = x ( ) ( ) Let A x 0 f t dt where f x is shown in Figure 8. Determine: 41. (Exercises(a) the 41 intervals and 42 onadapted which fromA(xCalculus) is increasing Problems and decreasing for a New Century, p. 102.) Determine f (x) assuming that x f (t) dt is equal to x 2 + x. (b) the values x where A(x)0 has a local minimum or maximum x Let F(c)(x)the= intervalsf (t) dt where= x 2A+(xx).Thenis concaveF (x up) = orf concave(x) = 2x down+ 1. 42. 0 (d) the in( fl)ection points of A(x) x ( ) 2 + − Determine g x and c assuming that c g t dt is equal to x x 6.

Figure 8 Further Insights and Challenges

43. Prove the formula g2(x) d ( ) = ( ( ))  ( ) − ( ( ))  ( ) f t dt f g2 x g2 x f g1 x g1 x dx g1(x)

( ) ( ) ( ) Let H(x) = g2 x f (t) dt = g2 x f (t) dt − g1 x f (t) dt.ApplyExercise23twiceto g1(x) 0 0 obtain ( ) = ( ( ))  ( ) − ( ( ))  ( ) H x f g2 x g2 x f g1 x g1 x 36 Chapter 5 The Integral

44. ( ) Proof of the FTC, Part II The proof in the text assumes that f x is increasing. To prove it in the general case, for h > 0letm(h) and M(h) denoteb ( ) the=minimum( ) − and( ) 45. Proof of the FTC, Part I The FTC, Part I asserts that a f t dt Fb F a if maximum( ) = values( ) of f on [x, x + h]. The continuity of f (x) implies( ) that= x ( ) F x f x . Prove this using the FTC, Part II as follows. Set A x a f t dt. ( ) = ( ) + (a) Show that F x A x limC form(h some) = lim constantM(h) (recall= f (x that) two functions with the h→0 h→0 same derivative differ by a constant). ( ) − ( ) = ( ) − ( ) = b ( ) (b) Show that F b F a A b A a a f t dt. Let F (x) = f (x) and A(x) = x f (t) dt. Figure 9 Graphicala interpretation of A(x + h) − A(x). (a) Then by the FTC, Part II, A(x) = f (x) and thus A(x) and F (x) are both derivatives of f (x). Hence F (x) = A(x) + C. > (b) F(a)(b)Show− F( thata) = if(hA(b)0,+ thenC)− (A(a) + C) =A(b) − A(a) = b f (t) dt − a f (t) dt = b f (t) dt − 0 = b f (t) dt which proves the FTC, Part I. 46. a a hma (h) ≤ A(x + ha) − A(x) ≤ hM(h) x ( ) Can Every Antiderivative Be Expressed As an Integral? The area function a f t dt is an(b) antiderivativeUse the Squeeze of f (x Theorem) for every to value show of thata. However, not all antiderivatives are obtained in this way. The general antiderivative of f (x) = x is F(x) = 1 x 2 + C. Show that A(x + h) − A(x) 2 F(x) can be expressed as an area functionlim if C ≤ 0 but not if=C f>(x0.) h→0+ h 5.5 Total Change as the Integral of a Rate (c) Treat the case h < 0 similarly.

Preliminary Questions

1. An airplane makes the 350-mile trip from Los Angeles to San Francisco in one hour. Assum- v( ) 1 v( ) ing the plane’s velocity at time t (in mph) is t , what is the value of the integral 0 t dt? 2. A hot metal object is submerged in cold water. The rate at which it cools (in degrees per ( ) T ( ) minute) is a function f t of time. Which quantity is represented by the integral 0 f t dt? 3. Which of the following quantities would be naturally represented as derivatives and which as integrals? (a) velocity of a train (b) total rainfall during a 6-month period (c) mileage per gallon of an automobile (d) total increase in the population of Los Angeles from 1970 to 1990 4. Two airplanes take off simultaneously from the same place and in the same direction. Their velocities are v1(t) and v2(t). What is the physical interpretation of the area between the graphs of v1(t) and v2(t)?

Exercises

1. Water flows into an empty reservoir for five hours at a rate of 3000 + 5t gallons per hour. What is the quantity of water in the reservoir after five hours?

The quantity of water in the resevoir after five hours is 5 3000 + 5tdt= (3000t + 5 t 2)5 = 30125 = 15,062.5 gallons. 2. 0 2 0 2 Find the total displacement over the time interval [2, 5] of a particle moving in a straight 3. lineA population whose velocity of insects is v( increasest) = 4t − at3 a ft/s. rate of 200 + 10t + .25t 2 insects per day. Find the insect population after three days assuming there are 35 insects at t = 0.

The increase in the insect population over three days is 3 + t + 1 t 2 dt = ( t + t 2 + 1 t 3)3 = 2589 = . 0 200 10 4 200 5 12 0 4 647 25. Accordingly, the population after 3 days is 35 + 647.25 = 682.25 or 682 insects. 5.5 Total Change as the Integral of a Rate 37

4. A survey shows that a candidate in a mayoral election is gaining votes at a rate of 5. 2000In a 4-weekt + 1000 period, votes a per factory day, where producest is bicycles the number at a of rate days of 95 after+ the.1t 2 candidacy− t bicycles was per week (announced.t in weeks). How How many many votes bicycles will were the candidate produced receive during after the 4-week 60 days? period? The rate of production is r(t) = 95 + 1 t 2 − t bicycles per week. Accordingly, during the [ , ] 10 interval 0 4, the number of bikes produced is 4 4 r(t) dt = 4 95 + 1 t 2 − tdt= (95t + 1 t 3 − 1 t 2) = 5612 ≈ 374.13 or 374 bicycles. 6. 0 0 10 30 2 0 15 Find the change in height over the time interval [1, 6] of a helicopter whose (vertical) 7. Avelocity cat falls at fromtime t ais treev(t (with) = .02 zerot 2 + initialt ft/s. velocity) at time t = 0. How many feet does the cat fall between t = .5andt = 1 s? Use Galileo’sformulav(t) =−32t ft/s. Given v(t) =−32 ft/s, the total distance the cat falls during the interval [ 1 , 1] is 2 1 |v( )| = 1 = 21 = − = 1/2 t dt 1/2 32tdt 16t / 16 4 12 ft. 8. 1 2 A projectile is released with initial (vertical) velocity 100 m/s. Use the formula v = 100 − 9.8t to determine the distance traveled during the first 15 seconds. In Exercises 9–12, a particle moves in a straight line with the given velocity. Find the total dis- placement and total distance traveled over the time interval, and draw a displacement diagram like Figure ?? (with distance and time labels).

9. v(t) = 12 − 4t ft/s, [0, 5]

5( − ) = − 25 = Total displacement is given by 0 12 4t dt 12t 2t 0 10 ft. Total distance is given by 5 | − | = 3( − ) + 5( − ) = − 23 + 2 − 5 = 0 12 4t dt 0 12 4t dt 3 4t 12 dt 12t 2t 0 2t 12t 3 26 ft. The displacement diagram is given here.

10. v(t) = 32 − 2t 2 ft/s, [0, 6] 11. v(t) = t −2 − 1m/s, [.5, 2]

2(t −2 − ) dt =−t −1 − t2 = Total displacement is given by .5 1 .5 0m. Total distance is given by 2 −2 − = 1( −2 − ) + 2( − −2) =−−1 − 1 + + −12 = .5 t 1 dt .5 t 1 dt 1 1 t dt t t .5 t t 1 1m. The displacement diagram is given here. 38 Chapter 5 The Integral

12. v(t) = cos t m/s, [0, 4π] 13. Carbon Tax To encourage manufacturers to reduce pollution, a carbon tax on each ton of CO2 released into the atmosphere has been proposed. To model the effects of such a tax, policymakers study the marginal cost of abatement B(x),defined as the cost of increasing + x ( ) CO2 reduction from x to x 1 tons (Figure 1). What quantity is represented by 0 B t dt?

100 dollars/ton 75 50 25

12 3 tons reduced (in ten thousands)

Figure 1 Marginal cost of abatement B(x).

x ( ) The quantity 0 B t dt represents the amount of tax (in dollars) imposed on the manufacturer for releasing x tons of CO into the atmosphere. 14. 2 Figure 2 shows the power (defined as the rate of energy consumption per unit time) 15. suppliedFigure 3 showsby the theCalifornia migration power rate grid (call over it M a(t typical)) in and one-day out of period. Ireland Which during quantity the period is 1988represented–1998. by This the is area thenumber under the of graph? people A (in megawatt thousands of per power year) iswho 106 watts move or in 3 or.6 out× 10 of9 the country. joules/h. Use the graph to make a very rough estimate of the number of joules consumed during the day in1991 California.( ) (a) What does 1988 M t dt represent (in words)? (b) Did migration over the eleven-year period 1988–1998 result in a net influx or outflow of people from Ireland? Base your answer on a rough estimate of the positive and negativeFigure areas involved. 2 Power consumption over one-day period in California. (c) During which year could the Irish Prime Minister announce: “We are still losing population but we’ve hit an inflection point—the trend is now improving.”

Figure 3 Irish migration (in thousands).

We note that M(t) is piecewise linear. 1991 ( ) (a) The amount 1988 M t dt represents the net migration in thousands of people during the period from 1988–1991. 5.5 Total Change as the Integral of a Rate 39

(b) Via linear interpolation and using the midpoint approximation with n = 10, the migration (in thousands of people) over the period 1988–1998 is estimated to be 1 · (−43 − 33.5 − 12 + 0.5 − 2.5 − 6 − 3.5 + 3 + 11.5 + 19) =−66.5 That is, there was a net outflow of 66,500 people from Ireland during this period. (c) “The trend is now improving” implies that the population is decreasing, but that the rate of decrease is approaching zero. The population is decreasing with an improving trend in part of the years 1989, 1990, 1991, 1993, and 1994. “We’ve hit an inflection point” implies that the rate of population has changed from decreasing to increasing. There are two years in which the trend improves after it was getting worse: 1989 and 1993. During only one of these, 1989, was the population declining for the entire previous year. 16. Heat Capacity The heat capacity C(T ) of a substance is the amount of energy (in joules) 17. requiredThe traffi toc fl raiseow rate the betweentemperature 8 AM ofand one 10 gramAM ofat the a certain substance point by on one a highway degree (◦ isC) when its qtemperature(t) =−10000 is T+. 6800t − 400t 2. How many cars pass this point between t = 8AMand t = 10 AM? (a) Explain in words why the energy required to raise the temperature from T1 to T2 is the ( ) The numberarea underof cars the is given graph by of C T . 10 10 400 ◦10 q(b)(t) dtHow= much−10000 energy√ + is6800 requiredt − 400 to raiset 2 dt the=− temperature10000t + 3400 fromt 2 50− to 100t 3 C≈if37333.3 8 ( ) =8 + . 3 8 18. cars. C T 6 2 T . Suppose the marginal cost of producing x video recorders is .001x 2 − .6x + 350. What is ( ) 19. theFigure total 4 cost shows ofthe producing graph of 300Q units?t , the If rate production of retail truck is set sales at 300 in units, the U.S. what (in would thousands be the soldcost ofper producing year). 20 additional units? (a) What does the area under the graph over the interval [1995, 1997] represent? (b) Express the total number of trucks sold in the period 1994–1997 as an integral (but do not compute). (c) Use the data and the average of the right- and left-endpoint approximations to estimate the total number of trucks sold during 2-year period 1995–1996.

year(qtr) sales year(qtr) sales 1995(1) 6484 1996(1) 7216 1995(2) 6255 1996(2) 6850 1995(3) 6424 1996(3) 7322 1995(4) 6818 1996(4) 7537 Figure 4 Quarterly retail sales of trucks in the U.S. in thousands (data supplied by Econo- magic.com).

The graph of Q(t) depicted in the exercise gives the quarterly rate of retail truck sales in thousands sold per quarter. 40 Chapter 5 The Integral

(a) The area under the graph over the interval [1995, 1997] represents number (in thousands) of trucks sold during the 1995–1997 period. (b) The number of trucks sold in the period 1994–1997 is given by the integral 1997 ( ) 1994 Q t dt. (c) We note that Q(t) is piecewise linear. Recall that the unit of time is one quarter year; hence t = 1. Using n = 7, the number of trucks sold during the period 1995–1997 is estimated to be the average of the right- and left-endpoint approximations.

RN = 1 · (6255 + 6424 + 6818 + 7216 + 6850 + 7322 + 7537) = 48422 trucks

L N = 1 · (6484 + 6255 + 6424 + 6818 + 7216 + 6850 + 7322) = 47369 trucks

The average of RN and L N is 47895.5trucks. 20. The velocity of a car is recorded at half-second intervals (in ft/s). Use the average of the left- and right-endpoint approximations to estimate the total distance traveled during the first four seconds.

Further Insights and Challenges t 0 .5 1 1.5 2 2.5 3 3.5 4 v(t) 0 12 20 29 38 44 32 35 30 21. (From Calculus, H. Flanders, R. Korfhage, J. Price, Academic Press (1970).) A particle moving along the x-axis with velocity v(t) = (t + 1)−2 is located at the origin at t = 0. Show that the particle will never pass the point x = 1. The particle’s velocity is v(t) = s(t) = (t + 1)−2, an antiderivative for which is F(t) =−(t + 1)−1. Hence its position at time t is t t 1 s(t) = s(u) du = F(u) = F(t) − F(0) = 1 − < 1 0 + 0 t 1 for all t ≥ 0. Thus the particle will never pass the point x = 1. 22. A particle moving along the x-axis with velocity v(t) = (t + 1)−1/2 is at the origin at t = 0. Will the particle be at the point x = 1 at any time t?Ifso,find t.

5.6 Substitution Method

Preliminary Questions

1. Which of these functions does not appear to be of the form g(u(x))u(x)? (a) 5x 4 sin(x 5) (b) sin5 x cos x (c) x 5 sin x 2. Write each of the following functions in the form cg(u(x))u(x) where c is a constant. (a) x(x 2 + 9)4 (b) x 2 sin(x 3) (c) sin x cos2 x 3. Which of the following is equal to 2 x 2(x 3 + 1) dx for a suitable substitution? 0 (a) 2 udu 0 (b) 9 udu 0 1 9 (c) 3 1 udu 5.6 Substitution Method 41

Exercises

In Exercises 1–6, calculate du for the given function.

1. u = 1 − x 2 Let u = 1 − x 2.Thendu =−2xdx. 2. u = sin x 3. u = x 3 − 2 Let u = x 3 − 2. Then du = 3x 2 dx. 4. u = 2x 4 + 8x 5. u = cos(x 2) Let u = cos(x 2).Thendu =−sin(x 2) · 2xdx=−2x sin(x 2) dx. 6. u = tan x

In Exercises 7–20, write the integral in terms of u and du. Then evaluate. 7. (x − 7)3 dx; u = x − 7 Let u = x − 7. Then du = dx. Hence (x − 7)3 dx = u3 du = 1 u4 + C = 1 (x − 7)4 + C. 8. √ 4 4 2x x 2 + 1 dx; u = x 2 + 1 9. (x + 1)−2 dx; u = x + 1 Let u = x + 1. Then du = dx. Hence 1 (x + 1)−2 dx = u−2 du =−u−1 + C =−(x + 1)−1 + C =− + C. x + 1 10. sin(2x − 4) dx; u = 2x − 4 x 3 11. dx; u = x 4 + 1 (x 4 + 1)4 = 4 + = 3 1 = 3 Let u x 1. Then du 4x dx or 4 du x dx. Hence x 3 1 1 1 1 dx = du = ln u + C = ln(x 4 + 1) + C. x 4 + 1 4 u 4 4

12. x(x + 1)9 dx; u = x + 1 x + 1 13. dx; u = x 2 + 2x (x 2 + 2x)3 = 2 + = 1 = Let u x 2x.Thendu 2xdxor 2 du xdx. Hence x + 1 1 1 1 1 dx = du = ·− u−2 + C (x 2 + 2x)3 2 u3 2 2 1 −1 =− (x 2 + 2x)−2 + C = + C. 4 4(x 2 + 2x)2

14. x 2 dx; u = 28x + 5 15. x(cos8x (+x 5))dx3 ; u = x = 2 = 1 = Let u x .Thendu 2xdxor 2 du xdx. Hence, x cos(x 2) dx = 1 cos udu= 1 sin u + C = 1 sin(x 2) + C. 16. √ 2 2 2 4x − 1 dx; u = 4x − 1 42 Chapter 5 The Integral √ 17. x 4x − 1 dx; u = 4x − 1 u = x − x = 1 (u + ) du = dx 1 du = dx Let√ 4 1and 4 1 .Then 4 or 4 . Hence, − = 1 ( + ) 1/2 = 1 ( 3/2 + 1/2) = x 4x 1 dx 16 u 1 u du 16 u u du 1 · 2 u5/2 + 1 · 2 u3/2 + C = 1 (4x − 1)5/2 + 1 (4x − 1)3/2 + C. 18. 16 5√ 16 3 40 24 x 2 4x − 1 dx; u = 4x − 1 19. sin2 x cos xdx; u = sin x Let u = sin x.Thendu = cos xdx. Hence sin2 x cos xdx= u2 du = 1 u3 + C = 1 sin3 x + C. 20. 3 3 sec2 x tan xdx; u = tan x In Exercises 21–24, show that each of the following integrals is equal to a multiple of sin(u(x)) for the appropriate choice of u(x). 21. x cos(x 2) dx = 2 = 1 = Let u x .Thendu 2xdxor 2 du dx. Hence x cos(x 2) dx = 1 cos udu= 1 sin u + C, which is a multiple of sin(u(x)). 22. 2 2 x 2 cos(x 3 + 1) dx 23. x 1/2 cos(x 3/2) dx = 3/2 = 3 1/2 2 = 1/2 Let u x .Thendu 2 x dx or 3 du x dx. Hence x 1/2 cos(x 3/2) dx = 2 cos udu= 2 sin u + C, which is a multiple of sin(u(x)). 24. 3 3 cos x cos(sin x) dx

In Exercises 25–54, evaluate the indefinite integral. 25. (4x + 3)4 dx = + = 1 = Let u 4x 3. Then du 4 dx or 4 du dx. Hence (4x + 3)4 dx = 1 u4 du = 1 · 1 u5 + C = 1 (4x + 3)5. 26. 4 4 5 20 x√2(x 3 + 1)3 dx 27. x x 2 − 4 dx Let u = x 2 − 4. Then du = 2xdxor 1 du = xdx. Hence √ √ 2 x x 2 − 4 dx = 1 udu= 1 · 2 u3/2 + C = 1 (x 2 − 4)3/2. 28. 2 2 3 3 (2x + 1)(x 2 + x)3 dx x 29. √ dx x 2 + 9

= 2 + = 1 = Let u x 9. Then du 2xdxor 2 du xdx. Hence x 1 1 1 1 √ 1 √ dx = √ du = u + C = x 2 + 9 + C x 2 + 9 2 u 2 2 4 √ or ln x 2 + 9 + C. 30. dx ( +dx )2 31. x 9 (4x + 9)3 = + = 1 = Let u 4x 9, then du 4dx or 4 du dx. Hence dx 1 du 1 1 1 −1 = = ·− + C = + C. (4x + 9)3 4 u3 4 2 u2 8(4x + 9)2 32. x 2 sin(x 3) dx 5.6 Substitution Method 43 1 33. √ dx x − 7 Let u = x − 7. Then du = dx. Hence √ (x − 7)−1/2 dx = u−1/2 du = 2u1/2 + C = 2 x − 7 + C. 34. 2x 2 + x ( 2 + )( 3 +dx)2 35. 3(x4x 3 +13xx2)2 x dx Let u = x 3 + x.Thendu =(3x 2 + 1) dx. Hence (3x 2 + 1)(x 3 + x)2 dx = u2 du = 1 u3 + C = 1 (x 3 + x)3 + C. 36. 3 3 5x 4 + 2x 2( 3 + )4 dx 37. x(xx5 + x12)3 dx = 3 + = 2 1 = 2 Let u x 1. Then du 3x dx or 3 du 3x dx. Hence x 2(x 3 + 1)4 dx = 1 u4 du = 1 · 1 u5 + C = 1 (x 3 + 1)5 + C. 38. 3 3 5 15 (x − 9)−2/3 dx 39. (x + 1)(x 2 + 2x)3 dx = 2 + = ( + ) 1 = ( + ) Let u x 2x.Thendu 2x 2 dx or 2 du x 1 dx. Hence (x + 1)(x 2 + 2x)3 dx = 1 u3 du = 1 · 1 u4 + C = 1 (x 2 + 2x)4 + C. 40. 2 2 4 8 (x + 1)7 dx 41. x 2(x + 1)7 dx Let u = x + 1andu − 1 = x.Thendu = dx. Hence x 2(x + 1)7 dx = (u − 1)2u7 du = (u9 − 2u8 + u7) du = 1 u10 − 2 u9 + 1 u8 + C = 1 (x + 1)10 − 2 (x + 1)9 + 1 (x + 1)8 + C. 42. 10 9 8 10 9 8 (3x + 9)10 dx 43. x(3x + 9)10 dx = + 1 ( − ) = = 1 = Let u 3x 9and 3u 9 x.Thendu 3 dx or 3 du dx. Hence ( + )10 = 1 ( − ) 10 = 1 ( 11 − 10) = 1 · 1 12 − 1 11 + = x 3x 9 dx 9 u 9 u du 9 u 9u du 9 12 u 11 u C 1 (3x + 9)12 − 1 (3x + 9)11 + C. 44. 108 11 x(x + 1)1/4 dx 45. x 3(x 2 − 1)3/2 dx = 2 − + = 2 = 1 = Let u x 1andu 1 x .Thendu 2xdx or 2 du xdx. Hence 3( 2 − )3/2 = 2 · ( 2 − )3/2 = 1 ( + ) 3/2 = 1 ( 5/2 + 3/2) = x x 1 dx x x x 1 dx 2 u 1 u du 2 u u du 1 · 2 u7/2 + 1 · 2 u5/2 + C = 1 (x 2 − 1)7/2 + 1 (x 2 − 1)5/2 + C. 46. 2 7 2 5 7 5 sin5 x cos xdx 47. x 2 sin(x 3 + 1) dx = 3 + = 2 1 = 2 Let u x 1. Then du 3x dx or 3 du x dx. Hence x 2 sin(x 3 + 1) dx = 1 sin udu=−1 cos u + C =−1 cos(x 3 + 1) + C. 48. 3 3 3 sec2(4x + 9) dx 49. sec2 x tan4 xdx Let u = tan x.Thendu = sec2 xdx. Hence sec2 x tan4 xdx= u4 du = 1 u5 + C = 1 tan5 x + C. 50. 5 5 cos 2x √ dx 51. sin(1 4+x sincos 2x 4)x2 + 1 dx u = x + du =− x − 1 du = x Let √cos 4 1. Then 4sin4 or 4 sin 4 . Hence sin 4x cos 4x + 1 dx =−1 u1/2 du =−1 · 2 u3/2 + C =−1 (cos 4x + 1)3/2 + C. 52. 4 4 3 6 cos x√(3sinx − 1) dx cos x 53. √ dx x Let u = x 1/2.Thendu = 1 x −1/2 dx or 2du = x −1/2 dx. Hence √ 2 cos x √ √ dx = 2 cos udu= 2sinu + C = 2sin x + C. x 44 Chapter 5 The Integral

54. sec2x(√4tan3 x − 3tan2 x) dx 55. Evaluate x 3 + 1 x 5 dx using u = x 3 + 1 Hint: write x 5 dx = x 3 · x 2 dx and observe that x 3 = u − 1. u = x 3 + x 3 = u − du = x 2 dx 1 du = x 2 dx Let √ 1. Then 1and 3 or 3 . Hence 5 3 + = 1 1/2( − ) = 1 3/2 − 1/2 = 1 ( 2 5/2 − 2 3/2) + = x x 1 dx 3 u u 1 du 3 u u du 3 5 u 3 u C 2 (x 3 + 1)5/2 − 2 (x 3 + 1)3/2 + C. 56. 15 9 Evaluate (x 3 + 1)1/4 x 5 dx using u = x 3 + 1. 57. What are the new limits of integration if the substitution u = 3x + π is applied to the π ( + π) integral 0 sin 3x dx? (π) = π + π = π ( ) = · + π = π 58. The new limits of integration are u 3 4 and u 0 3 0 . = − Which of the following is the result of applying the substitution u 4x 9totheintegral 8(4x − 9)20 dx? In Exercises2 59 –68, use the change of variables formula to evaluate the definite integral. (a) 32 u20 du 8 π/2 59. (b)cos 31xdx8 u20 du 0 4 2 Let u = 31 x.Then32 20 du = 3 dx or 1 du = dx. Hence (c) 4 8 u du 3 π/ π/2 1 3π/2 1 3 2 1 1 (d)cos 3xdx32 u=20 du cos udu= sin u =− − 0 =− . 60. 0 4 8 3 0 3 0 3 3 3(x + 2)3 dx 1π/2 ( + π ) 61. 0 cos 3x 2 dx = + π = 1 = Let u 3x 2 .Thendu 3 dx or 3 du dx. Hence π/ π π 2 cos(3x + π ) dx = 1 2 cos udu= 1 sin u2 = 0 − 1 =−1 . 62. 0 √ 2 3 π/2 3 π/2 3 3 6 x + 3 dx 1 x 63. 1 dx 0 (x 2 + 1)3 = 2 + = 1 = Let u x 1. Then du 2xdxor 2 du xdx. Hence 1 x 2 1 1 1 2 dx = du = ·− u−2 ( 2 + )3 3 0 x 1 1 u 2 2 1 1 1 1 =− + = = 0.1875. 16 4 16 64. √ 2 + −1 √5x 6 dx 4 2 + 65. 0 x x 9 dx Let u = x 2 + 9. Then du = 2xdxor 1 du = xdx. Hence √ 2 √ 25 4 x 2 + 9 dx = 1 25 udu= 1 · 2 u3/2 = 1 (125 − 27) = 32.6667. 66. 0 2 9 2 3 9 3 2 x + 3 π/ 2 3 dx 67. (cos2 +x sin+xdx)3 00 x 6x 1 Let u = cos x.Thendu =−sin xdx. Hence π/ 2 3 x xdx=− 0 u3 du = 1 u3 du = 1 u41 = 1 − = 1 0 cos sin 1 0 4 0 4 0 4 . 68. π/ 4 tan2 x sec2 xdx 69. Some0 Choices Are Better Than Others One way of evaluating sin x cos2 xdxis using the substitution u = sin x. Show that this gives sin x cos2 xdx= u 1 − u2 du

Complete the evaluation by using another substitution. Then show that u = cos x is a better choice. Consider the integral sin x cos2 xdx. 5.6 Substitution Method 45 √ Let u = sin x. Then cos√x = 1 − u2 and du = cos xdx. Hence 2 = − 2 w = − 2 w =− sin x cos xdx u 1 u√du.Nowlet 1 u .Thend 2uduor − 1 w = − 2 =−1 w1/2 w =−1 · 2 w3/2 + = 2 d udu. Therefore, u 1 u du 2 d 2 3 C − 1 w3/2 + =−1 ( − 2)3/2 + =−1 ( − 2 )3/2 + 3 C 3 1 u C 3 1 sin x C. A better substitution choice isu = cos x.Thendu =−sin uduor −du = sin udu. 2 =− 2 =−1 3 + =−1 3 + Hence sin x cos xdx u du 3 u C 3 cos x C. 70. CanTheyBothBeRight? (Adapted from Calculus Problems for a New Century, 71. p.Evaluate 121.) Joansin usesx cos thexdx substitutionusing substitutionu = tan x into two conclude different ways: first using u = sin x and then using u = cos x. Reconcile the two different answers. 1 tan x sec2 xdx= tan2 x + C 2

whileLet Frank,u = sin notingx. Then that duu ==cossecxdxx gives. Hencedu = sec x tan x, concludes that = = 1 2 + = 1 2 + sin x cos xdx udu 2 u C1 2 sin x C1. Let u = cos x.Thendu tan=−x sinsec2xdxxdxor=−dusec= sinx(tanxdxx sec.Sox) dx 1 2 1 2 sin x cos xdx=− udu=− u + C2 =− cos x + C2. 2 1 2 To reconcile these two seemingly different= answers,sec2 x + recallC that any two antiderivatives of a specified function differ by a constant. To2 show that this is true here, note that ( 1 2 + ) − (− 1 2 + ) = 1 + − Explain2 sin thex apparentC1 contradiction.2 cos x C2 2 C1 C2, a constant. Here we used the trigonometric identity sin2 x + cos2 x = 1. 72. Use the substitution u = f (x) to evaluate f (x)3 f (x) dx where f (x) is differentiable.  73. TheEvaluate answerf is(x an) f expression(x) dx. in f (x).  Let u = f (x).Thendu = f (x) dx. Hence f (x) f (x) dx = udu= 1 u2 + C = 1 f (x)2 + C. 2 2 74. ( ) f x Evaluate π/2 dx. ( )n2 =− 75. Evaluate 0 f sinx x cos xdx,wheren is an integer, n 1. Let u = sin x.Thendu = cos xdx. Hence π/2 1 un+1 1 1 n x xdx= un du = = sin cos + + . 0 0 n 1 0 n 1

Further Insights and Challenges 76. = + 1/n (a) Use the substitution u 1 (−x ) =−to show( ) that 77. Show that if f is an odd function ( f x f x )then 1 +ax 1/n dx = n u1/2(u − 1)n−1 du f (x) dx = 0 −a (b) Use this to evaluate the left-hand side for n = 2, 3. a ( ) =− 0 ( ) Hint: show that 0 f x dx −a f x dx using substitution. We assume that f is continuous. If f (x) is an odd function, then f (−x) =−f (x).Let u =−x.Thenx =−u and du =−dx or −du = dx. Accordingly, a 0 a f (x) dx = f (x) dx + f (x) dx −a −a 0 0 a =− f (−u) du + f (x) dx a 0 a a = f (x) dx − f (u) du = 0 0 0 46 Chapter 5 The Integral

78. Show that if f is an even function ( f (−x) = f (x)), then 79. (a) u = x/a y = x −1 Use the substitution ato prove that the0 hyperbola has a special , > b 1 = b/a 1 property: if a b 0, then a x dxf (x) dx1 = x dx.f (x) dx (b) Show that the area under the0 hyperbola over− thea intervals [1, 2], [2, 4], [4, 8],... are all equal.

= x = = 1 = (a) Let u a and au x.Thendu a dx or adu dx. Hence b 1 b/a a b/a 1 dx = du = du. a x 1 au 1 u b/a 1 b/a 1 Note that du = dx after the substitution x = u. 1 u 1 x (b) 2 1 dx = ln x|2 = ln 2 − ln 1 = ln 2. 1 x 1 4 1 dx = ln x|4 = ln 4 − ln 2 = ln 4 = ln 2. 2 x 2 2 8 1 = |8 = − = 8 = 4 x dx ln x 4 ln 8 ln 4 ln 4 ln 2. 80. Show the two√ regions in Figure 1 have the same area. They correspond to the areas under 81. Areathe graphs of a Circle of 1 −Althoughu2 and cos we2 x usually. take it for granted that the area of a circle is πr 2, this formula requires proof. The formula is a consequence of two distinct facts: first, that the area is proportional to the square of the√ radius and second, that the area of the unit circle is π (one-halfFigure the circumference 1 (A) Graph of of the unit1 − u circle).2,(B)Graphofcos Prove the first2 x part.. (a) Use the Change of Variables formula to show that r 1 r 2 − x 2 dx = r 2 1 − x 2 dx. 0 0 (b) Show the area of a circle is proportional to the square of its radius. √ r 2 − 2 1 > The integral 0 r x dx is equal to 4 the area of a circle of radius r 0. (It is the area of the quarter circular disk x 2 + y2 ≤ r 2 in the first quadrant.) (a) Let u = 1 x.Thenx = ru and du = 1 dx or rdu= dx.So √ r √ r √ r 2 − 2 = 1 2 − 2 2 = 2 1 − 2 = 2( π ) = 1 π 2 0 r x dx r 0 r r u du r 0 1 u du r 4 4 r from Exercise ??. 1 1 π 2 (b) In (a) it was shown that 4 of the area bounded by a circle of radius r is 4 r . Therefore, the area bounded by the full circle is πr 2. Hence the area of a circle is proportional to the square of its radius. 82. Area of an Ellipse Use the change of variables formula and the fact that the unit circle has area π to show that the area of the ellipse with equation

x 2 y2 + = 1 a2 b2 is equal to πab. Hint: Show that the area of the ellipse is equal to twice a b 1 − (x/a)2 dx −a and use the Change of Variables formula to relate the integral on the right to the area of the unit circle.

x 2 y2 Figure 2 Graph of + = 1. a2 b2