Contents 13 Numerical Integration 187 13.1 Midpoint Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 13.2 Trapezoidal Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 13.3 Algorithm for Computing T2N from TN . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 13.4 Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192 13.5 Simpson’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 13.6 A Little Numerical Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 13.7 Algorithm for Computing SN and DN . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 2 CONTENTS Chapter 13 Numerical Integration 13.1 Midpoint Rule Divide the interval [a, b] into N equal subintervals as shown in the diagram. • • • x1 a = x0 Each subinterval has length h = b−a N . m0 • x0 • x2 xN −1 × • x1 b = xN The midpoint of the subinterval [xi , xi+1 ] is mi = mN −1 m1 × • • x3 • x2 • xN −1 × • xN Now we draw the graph of f f a b h x h The signed area MN of the shaded rectangles is MN = hf (m0 ) + hf (m1 ) + . . . + hf (mN −1 ) = h N −1 X i=0 It is called the midpoint approximation to the definite integral Z a b f (x)dx. f (mi ). xi +xi+1 . 2 188 Numerical Integration The error is given by the formula b Z f (x)dx = a | MN + {z } | approximation EN {z error . } If B is a number such that |f 00 (z)| ≤ B for all z ∈ [a, b] then |EN | ≤ Note: (b − a)3 B . 24N 2 The approximation and its accuracy depends on the number, N , of subintervals. Example 1: Approximate I = Z mation. π sin x dx by the midpoint rule with N = 4 and estimate the accuracy of this approxiπ 4 Solution: Step size: h= π− π 4 4 Midpoints: m0 = m1 = m2 = m3 = π 4 π 4 π 4 π 4 = + + + + 3π 16 . 3π 11π 32 = 32 = 1.0799225 9π 17π 32 = 32 = 1.6689711 15π 23π 32 = 32 = 2.2580197 21π 29π 32 = 32 = 2.8470683. Approximation: M4 = h{sin(m0 ) + sin(m1 ) + sin(m2 ) + sin(m3 )} M4 = 3π 16 (.88192127 = 3π 16 (2.9404011) + .99518473 + .77301046 + .29028468) = 1.7320392 Error analysis: f 0 (x) = cos x and f 00 (x) = − sin x |f 00 (x)| = | sin x| ≤ 1 for all x ∈ R. The bound on the error is given by |E4 | ≤ so |E4 | ≤ 3π 16 3 There 1.7320392 − .0341 < I < 1.7320392 + .0341. (b − a)3 B 24N 2 1 < .0340645. 24 · 16 13.1 Midpoint Rule 189 Observation: Z π π ∼ sin x dx = − cos(π) + cos = 1.7071068 by the fundamental theorem. Therefore π 4 4 |E4 | ∼ = .0249324 < .0341, as predicted by our error analysis. Example 2: Approximate L = 2 Z 1 dx by the midpoint rule within ±10−3 . x Solution: In this problem we have first to decide how many subintervals to use; i.e. we have to decide what N to choose. We use our formula for the error bound to compute N . Since |EN | < (b−a)3 B 24N 2 were need (2−1)3 B 24N 2 < 10−3 to guarantee |EN | < 10−3 . Now f (x) = x1 , f 0 (x) = − x12 , and f 00 (x) = And |f 00 (z)| = 2 z3 2 x3 . ≤ 2 on [1, 2] allows us to pick B = 2. Therefore we need N2 > 1×103 12 N > √ 10 √ 10 12 2 24N 2 < 10−3 or ∼ = 9.129 Since we need N > 9.129 we pick N = 10. This gives h = 2−1 10 = 1 10 . Midpoints: 1.05 M10 = h M10 = Note: 1 m0 + 1 m1 × + ... + 1 m9 1 10 (6.9283536) Z 2 1 × 1.15 • 1.1 • 1 1.25 • 1.2 × 1.35 • 1.3 × 1.45 • 1.4 × 1.55 • 1.5 × 1.65 • 1.6 × 1.75 • 1.7 × 1.85 • 1.8 × 1.95 • 1.9 × • 10 = .69283536. dx = ln 2 ∼ = .69314718. x ln 2 − M10 ∼ = .000312 < .001. Problems For the given function f , interval [a, b] and partition N compute MN , the midpoint approximation of Z b f (x)dx, and a bound on the error term. a 1. f (x) = x1 , [1.5, 3], and N = 6 190 Numerical Integration 2. f (x) = sin x2 , 3. f (x) = e−x , 4. f (x) = 5. √ [0, π], and N = 6 2 f (x) = [0, .8], 1 1+x4 , √ [0, 1], and N = 4 6. f (x) = 7. f (x) = sin 2x cos2 2x , 9. [ π2 , π], 1 + cos2 x, ( 8. sin x x if x 6= 0 , if x = 0 1 [0, π6 ], √ f (x) = arctan x, √ x and N = 6 [0, π2 ], and and N = 4 N =5 [9, 13], and N = 4 11. [1, 4], and N = 4 p f (x) = tan(x2 ), [0, π4 ], and N = 4 √ f (x) = 1 + x3 , [0, 2], and N = 2 12. f (x) = 10. f (x) = e and N = 8 , √ 1 , 1+x3 √ [0, 2], and N = 4 13. f (x) = x sin x, 14. f (x) = x tan x, 15. f (x) = ex , 3 −x2 f (x) = 2e√ 17. f (x) = e−x x , 18. f (x) = e− sin x , 19. f (x) = 20. f (x) = 21. 22. 2x x+1 , √ , and N = 10 [0.5, 3], and N = 8 [0, π2 ], [0, 1], and N = 4 and N = 4 [0, 3], 1 + ex , and N = 4 [0, π4 ], [0, 1], 16. π [0, π], and N = 6 and N = 4 [0, 1], and N = 4 √ sin x, [0, π2 ], and N = 4 p f (x) = 1 − 2 sin2 x, [0, π2 ], and N = 4 f (x) = In the following problems find N so that the midpoint rule approximates the given Z with |EN | < given ε. If N ≤ 10 use the midpoint rule to find the approximate value. 23. Z 3 1.5 25. Z 0 27. Z 1 dx , x ε = ±10 dx , 1 + x4 ε = ±10−4 26. ε = ±10−2 28. 4 √ e x −3 24. 29. 0 π sin x2 dx, ε = ±10−3 0 dx, 1 Z √ π2 √ Z Z π p 1 + cos2 x dx, π 2 Z 13 √ arctan x dx, ε = ±10−3 ε = ±10−4 9 2 cos(x )dx, ε = ±10 −2 30. Z 0 π 2 p 1 + cos2 x dx, ε = ±10−2 13.2 Trapezoidal Rule 31. 1 Z 1 2 Z √ 0 35. ε = ±10−2 1 + x3 dx 0 33. p 2 Z 1 Z 32. 1 2 0 dx , 1 − x2 ε = ±10−3 Z 34. dx , 1 + x2 ε = ±10−4 3 4−x dx, ε = ±10−2 1 sin x dx, x 13.2 191 ε = ±10−2 Trapezoidal Rule This is another way to approximate Z b f (x)dx. Again divide [a, b] into N equal intervals of length h = a • • a = x0 • x1 • • x2 xN −1 xN = b The trapezoid approximation to the integral is given by h {f (a) + 2f (x1 ) + 2f (x2 ) + . . . + 2f (xN −1 ) + f (b)} . 2 TN = We write Z b f (x)dx = a TN + | {z } | Trap. approximation EN {z error . } If B is chosen so that |f 00 (z)| ≤ B for all z ∈ [a, b] then |EN | ≤ Note: 13.3 (b − a)3 B . 12N 2 This is not the same as for the midpoint rule. Algorithm for Computing T2N from TN If we have computed TN we can get T2N as in the following example for Z 3 f (x)dx: 0 N =2: 0 • 0 T2 = N =4: 0 • 0 1 • 3/4 1 • 3/2 3 2 1 f (0) + f 2 2 • 3/2 2 • 3 (h = 3/2) 3 1 + f (3) . 2 2 3 • 9/4 4 • 3 (h4 = 3/4) b−a N . 192 Numerical Integration T4 = T4 = T4 = T4 = 0 • 0 N =8: T8 = T8 = T8 = T8 = 13.4 1 3 3 9 1 f (0) + f +f +f + f (3) 2 4 2 4 2 1 1 1 3 3 9 3 f (0) + f + f (3) + 3 f +f 4 2 2 2 4 4 1 3 9 2T2 + 3 f +f . (Look back at the formula for T2 ) 4 4 4 3 9 1 T 2 + h4 f +f 2 4 4 3 4 1 × 3/8 2 • 3/4 3 × 9/8 4 • 3/2 5 × 15/8 6 • 9/4 7 × 21/8 8 • 3 h8 = 3 8 1 3 3 9 3 15 9 21 1 f (0) + f +f +f + +f +f +f +f + f (3) 2 8 4 8 2 8 4 8 2 1 1 3 3 9 1 3 f (0) + f +f +f + f (3) + 8 2 4 2 4 2 3 9 15 21 3 f +f +f +f 8 8 8 8 1 3 9 15 21 4T4 + 3 f +f +f +f 8 8 8 8 8 1 3 9 15 21 T 4 + h8 f +f +f +f 2 8 8 8 8 3 8 Algorithm I. Partition interval of integration into N intervals and compute the points x0 , x2 , . . . , xN of the partition. II. Compute TN and record. III. Compute midpoints, yi , of the partition to give a new partition with 2N intervals. (of length h2N = (b − a)/2N ). IV. Compute T2N = 21 TN + b−a 2N NP −1 f (yi ), record and store. i=0 V. Reset N to 2N and repeat steps III to V. Example: Z 3 1 For sin dx, compute T2 , T4 , and T8 . x 1 13.4 Algorithm 193 Solution: (using a T.I. calculator) • 5/4 p 1 3−1 2 Since h2 = 3−1 4 1 sin(1) + sin 2 = ÷ 2 + 3−1 8 p 2 36. For For Z 37. For Z 4 ÷ 2 Now is in the display + ( sin 5 4 4 4 + sin + sin + sin ) ÷ 4 = (1.0288421). 7 9 11 1 2 T4 = 3 ex dx compute T2 , T4 , T8 , and T16 . 1 38. 1 2 cos x dx compute T2 , T4 , T8 , and T16 . 0 39. For Z For Z For Z For Z For Z For Z 1 e−1/x dx compute T2 , T4 , T8 , and T16 . 0 40. 2 dx compute T2 , T4 , T8 , and T16 . x 1 41. 1 2 e−x dx compute T2 , T4 , and T8 . 0 42. 1 3 (x2 + 1) 2 dx compute T2 , T4 , and T8 . 0 43. 7 1 (1 + x) 3 dx compute T2 , T4 , and T8 . 0 44. 0 π 2 p 3 Now 12 T2 is in the display. 2 2 ( sin + sin ) ÷ 2 = (1.035773). 3 5 1 sin dx compute T16 . x 1 • 11/4 1 1 1 + sin = (1.0637584). 2 2 3 Problems 3 × 5/2 = 14 , T8 Z • 9/4 = 12 , T4 Since h8 = • 7/4 = 1, T2 = Since h4 = × 3/2 p 1 − 2 sin2 x dx compute T2 , T4 , and T8 . 194 Numerical Integration 45. For π 2 Z √ sin x dx compute T2 , T4 , and T8 . 0 46. For 1 Z 0 2x dx compute T2 , T4 , and T8 . x+1 For the integral b Z f (x)dx find a suitable N and compute TN (if N ≤ 10) such that |EN | ≤ (given ε). a 47. Z 2 1 49. Z 1 dx, sin x ε ≤ .002 4x3 dx, ε ≤ .06 48. Z 1 50. 53. 1 p 1 + x3 dx, 3 √ e ε ≤ .02 52. Z 1 13.5 π 2 Z Z ε ≤ .007 p 1 + cos2 x dx, ε ≤ .02 13 √ arctan x dx, ε ≤ .001 9 x ε ≤ .05 dx, 54. 1 55. dx , x 0 0 Z 2 1 0 51. Z Z √π/2 cos(x2 )dx, ε ≤ .01 0 2 sin x dx, x ε ≤ .01 56. Z 3 e−x dx, ε ≤ .01 1 Simpson’s Rule b There is another way to approximate Z equal subintervals, each of length h = a b−a N • a = x0 f (x) dx. For this rule partition [a, b] into an even number N of to get a partition • • x1 • x2 xN −1 • b = xN Then the Simpson’s approximation is given by SN = b−a [f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + · · · + 2f (xN −2 ) + 4f (xN −1 ) + f (xN )] . 3N Simpson’s Rule can be remembered as b−a [(ends) + 4(odds) + 2(evens)] . 3N Thus Z b f (x)dx = SN + EN . a If B is chosen so that |f (4) (z)| ≤ B for all z ∈ [a, b] then |EN | ≤ (b − a)5 B . 180N 4 13.6 A Little Numerical Analysis 195 Problems Use Simpson’s rule to estimate, with |EN | ≤ (given ε) the following integrals. 57. Z 1.9 dx , x 1 59. Z 2 1 61. Z ε = .001 58. 63. sin x dx, x ε = .001 65. cos x dx, ε = 1 × 10−6 60. Z Z 3 √ e 62. Z 1 2 10 1 π 2 dx , x ε = 5 × 10 −6 64. Z 1 2 0 1 1+ 1 + x2 x √ 0 0 67. 5−x dx, ε = .001 dx, ε = .001 1 1 2 e2 Z 3 1 0 Z Z 12 dx, ε = .05 66. Z 1 cos x2 dx, 2 dx , ε = 1 × 10−6 1 − x2 dx , 1 + x2 ε = 1 × 10−7 dx , 1+x ε = 2 × 10−6 ε = .001 0 13.6 A Little Numerical Analysis Sometimes we cannot use the error formula to estimate the error. In these circumstances we compute the sequence T2 , T4 , T8 , . . . , TN , T2N , . . . of trapezoidal approximations, and stop when the difference DN = T2N − TN gets small. We do it this way for two reasons: 1. There is a good algorithm to compute T2N using TN . 2. If we know TN and T2N we can get the Simpson’s approximation by the formula SN = 13.7 4T2N − TN T2N − TN DN = T2N + = T2N + . 3 3 3 Algorithm for Computing SN and DN I. Partition the interval of integration [a, b] into N subintervals as described on page 191. II. Calculate TN ; store and print. III. Compute midpoints, yi , of your partition. IV. Compute b−a 2N {f (y0 ) + f (y1 ) + · · · + f (yN −1 )} = A. V. Compute T2N = 12 TN + A; store and print. VI. Compute DN = T2N − TN and print. VII. Compute SN = T2N + DN 3 . 196 Numerical Integration Problems 68. For π 4 Z cos x dx, compute T16 , D8 , and S8 and compare T16 and S8 with the exact value of this integral. 0 In problems 69 to 82 do the following. For N = 2, 4 and the given integral calculate TN , T2N , DN , and SN . (Include the partition, your method, and a summary of the results in your solution.) 69. Z 5 1 71. Z 73. Z 70. sec x dx 72. Z 74. Z π 4 cos x2 dx 77. π 2 Z 9 4 0 81. Z 2 dx 1 + x2 1 √ e √ sin x dx 76. Z 1 0 1 79. 1 x dx 0 0 Z ex dx 0 π 2 0 Z 1 0 0 75. Z dx x 4 p 3+ x3 2x dx x+1 x dx ln x dx 78. Z 9 Z 5 2 82. Z 0 p x3 − 1 dx 1 80. dx 1 + x3 dx ln x π 4 x tan x dx