NUMERICAL INTEGRATION

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1 NUMERICAL INTEGRATION How do we evlute I = f (x) dx By the fundmentl theorem of clculus, if F (x) is n ntiderivtive of f (x), then I = f (x) dx = F (x) b = F (b) F () However, in prctice most integrls cnnot be evluted by this mens. Even when this cn work, numericl method my be much simpler nd esier to use. For exmple, the integrnd in 1 dx 1 + x 5 hs complicted ntiderivtive; nd it is esier to evlute the integrl by pproximte mens. Try evluting this integrl with Mple or Mthemtic.

2 NUMERICAL INTEGRATION: A GENERAL FRAMEWORK Recll the following principle of numericl nlysis: If you cnnot solve the given problem, then solve nerby problem. Here, we wnt to evlute I = f (x) dx nd suppose it is difficult to do it directly. Let f (x) be n pproximtion of f (x). Then use I f (x) dx Ĩ

3 Wht is the error? E = I Ĩ = [ f (x) f ] (x) dx E f (x) f (x) dx (b ) f f f f mx f (x) f (x) x b We lso wnt to choose the pproximtes f (x) of form we cn integrte directly nd esily. Exmples re polynomils, trig functions, piecewise polynomils, nd others. We my generte pproximtions through Tylor polynomils, interpoltory polynomils, lest-squres pproximtions, etc.

4 An exmple with Tylor pproximtion Consider evluting Use I = 1 e x2 dx e t = 1 + t + 1 2! t n! tn + e x2 = 1 + x ! x n! x 2n + 1 (n + 1)! tn+1 e ct 1 (n + 1)! x 2n+2 e dx with d x x 2. Then 1 I = [1 + x ! x n! ] x 2n dx+ 1 (n + 1)! Tking n = 3, we hve I 1. = E = E < E e 24 1 x 8 dx = e 216. =.126 x 2n+2 e dx dx

5 USING INTERPOLATORY POLYNOMIALS In spite of the simplicity of the bove exmple, it is generlly more difficult to do numericl integrtion by constructing Tylor polynomil pproximtions thn by constructing polynomil interpolnts. We therefore construct the function f in by mens of interpoltion. f (x) dx f (x) dx Initilly, we consider only the cse in which the interpoltion is bsed on interpoltion t evenly spced node points.

6 LINEAR INTERPOLATION The liner interpolnt to f (x), interpolting t nd b, is given by P 1 (x) = (b x) f () + (x ) f (b) b Using the liner interpolnt P 1 (x), we obtin the pproximtion f (x) dx P 1 (x) dx = 1 2 (b ) [f () + f (b)] T 1(f ) The rule is clled the trpezoidl rule. f (x) dx T 1 (f )

7 y y=f(x) y=p 1 (x) b x Figure: Illustrting I T 1 (f ) Exmple. π/2 sin x dx π 4 Error. =.215 [ ( π )] sin + sin = π 2 4. =

8 HOW TO OBTAIN GREATER ACCURACY? How do we improve our estimte of the integrl I = f (x) dx One direction is to increse the degree of the pproximtion, moving next to qudrtic interpolting polynomil for f (x). We first look t n lterntive. Insted of using the trpezoidl rule on the originl intervl [, b], pply it to integrls of f (x) over smller subintervls. For exmple: c ( I = f (x) dx + f (x) dx c = b + ) c 2 c [f () + f (c)] + b c [f (c) + f (b)] 2 2 = h ( 2 [f () + 2f (c) + f (b)] T 2(f ) h = b ) 2

9 y y=f(x) x =x x 1 x 2 b=x 3 Figure: Illustrting I T 3 (f ) Exmple. π/2 sin x dx π 8 Error. =.519 [ ( π ) ( π )].= sin + 2 sin + sin

10 THE TRAPEZOIDAL RULE We cn continue s bove by dividing [, b] into even smller subintervls nd pplying β α f (x) dx β α 2 [f (α) + f (β)], ( ) on ech of the smller subintervls. Begin by introducing positive integer n 1, Then h = b n, x j = + j h, j =, 1,..., n I = xn x f (x) dx = x1 x f (x) dx + x2 x 1 f (x) dx + + xn x n 1 f (x) dx Use [α, β] = [x, x 1 ], [x 1, x 2 ],..., [x n 1, x n ], for ech of which the subintervl hs length h.

11 Then pplying we hve β α f (x) dx β α 2 [f (α) + f (β)] I h 2 [f (x ) + f (x 1 )] + h 2 [f (x 1) + f (x 2 )] + + h 2 [f (x n 2) + f (x n 1 )] + h 2 [f (x n 1) + f (x n )] Simplifying, [ 1 I h 2 f () + f (x 1) + + f (x n 1 ) + 1 ] 2 f (b) T n (f ) This is clled the composite trpezoidl rule, or more simply, the trpezoidl rule.

12 Exmple Agin integrte sin x over [, π 2 ]. Then we hve n T n (f ) Error Rtio E E E E E E E E E 6 4. Note tht the errors re decresing by constnt fctor of 4, corresponding to doubling of n.

13 USING QUADRATIC INTERPOLATION Now we pproximte I = f (x) dx using qudrtic interpoltion of f (x). Interpolte f (x) t the points {, c, b}, with c = 1 2 ( + b). Also let h = 1 2 (b ). The qudrtic interpolting polynomil is given by (x c) (x b) P 2 (x) = 2h 2 f () + (x ) (x c) + 2h 2 f (b) (x ) (x b) h 2 f (c) Replcing f (x) by P 2 (x), we obtin the pproximtion f (x) dx This is clled Simpson s rule. P 2 (x) dx = h 3 [f () + 4f (c) + f (b)] S 2(f )

14 y y=f(x) y=p 2 (x) (+b)/2 b x Figure: Illustrtion of I S 2 (f ) Exmple. π/2 sin x dx π [ ( 4 π ) ( π )].= sin + 4 sin + sin Error. =.228

15 SIMPSON S RULE As with the trpezoidl rule, we cn pply Simpson s rule on smller subdivisions in order to obtin better ccurcy in pproximting I = Agin, Simpson s rule is given by β α f (x) dx f (x) dx δ [f (α) + 4f (γ) + f (β)], 3 γ = α + β 2 nd δ = 1 2 (β α). Let n be positive even integer, nd h = b n, x j = + j h, j =, 1,..., n Then write I = xn x f (x) dx = x2 x f (x) dx + x4 x 2 f (x) dx + + xn x n 2 f (x) dx

16 Apply β f (x) dx δ [f (α) + 4f (γ) + f (β)], 3 γ α to ech of these subintegrls, with [α, β] = [x, x 2 ], [x 2, x 4 ],..., [x n 2, x n ] In ll cses, 1 2 (β α) = h. Then = α + β 2 I h 3 [f (x ) + 4f (x 1 ) + f (x 2 )] + h 3 [f (x 2) + 4f (x 3 ) + f (x 4 )] + + h 3 [f (x n 2) + 4f (x n 1 ) + f (x n )] This cn be simplified to f (x) dx S n (f ) h 3 [f (x ) + 4f (x 1 ) + 2f (x 2 ) + 4f (x 3 ) + 2f (x 4 ) + + 2f (x n 2 ) + 4f (x n 1 ) + f (x n )] This is clled the composite Simpson s rule or more simply, Simpson s rule.

17 EXAMPLE Results of pproximting π/2 sin x dx by the Simpson rule: n S n (f ) Error Rtio E E E E E E E E E Note tht the rtios of successive errors hve converged to 16. Compre this tble with tht for the trpezoidl rule, e.g., I T 4. = 1.29E 2 I S 4. = 1.35E 4

18 Remrk There is gret del to be lerned from numericl exmples. Next, we provide three more numericl exmples to illustrte performnce of the trpezoidl nd Simpson rules. Look crefully t the tbles of the exmples; theoreticl explntions will be given through error nlysis lter.

19 Additionl Exmple 1 I (1) = 1 e x2 dx. = Tble: Numericl results n T n S n Error Rtio Error Rtio E E E E E E E E E E E E E E

20 Additionl Exmple 2 I (2) = 4 dx 1 + x 2 = rctn 4 Tble: Numericl results n T n S n Error Rtio Error Rtio E E E E E E E E E E E E E E

21 Additionl Exmple 3 I (3) = 2π dx 2 + cos x = 2π 3 Tble: Numericl results n T n S n Error Rtio Error Rtio E E E E E E E E

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