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Chapter5: Numerical Integration and Differentiation PARTI:NumericalIntegration Newton-Cotes Integration Formulas TheideaofNewton-Cotesformulasistoreplaceacomplicatedfunctionortabu- lated data with an approximating function that is easy to integrate. I = Z bf(x)dx ≈ Z bf (x)dx n a a 2 n where f (x) = a +a x+a x +...+a x . n 0 1 2 n 1 TheTrapezoidalRule Using the first order Taylor series to approximate f(x), I = Z bf(x)dx ≈ Z bf (x)dx 1 a a where f (x) = f(a) + f(b) − f(a)(x − a) 1 b −a 1 Then Z · ¸ b f(b) −f(a) I ≈ a f(a)+ b −a (x−a) dx = (b−a)f(b)+f(a) 2 The trapezoidal rule is equivalent to approximating the area of the trapezoidal Figure 1: Graphical depiction of the trapezoidal rule under the straight line connecting f(a) and f(b). An estimate for the local trun- 2 cation error of a single application of the trapezoidal rule can be obtained using Taylor series as E =−1f′′(ξ)(b−a)3 t 12 where ξ is a value between a and b. Example: Use the trapezoidal rule to numerically integrate f(x) = 0.2 + 25x from a = 0 to b = 2. Solution: f(a) = f(0) = 0.2, and f(b) = f(2) = 50.2. I = (b −a)f(b)+f(a) = (2−0)× 0.2+50.2 = 50.4 2 2 Thetrue solution is Z 2f(x)dx = (0.2x+12.5x2)|2 = (0.2×2+12.5×22)−0 = 50.4 0 0 Because f(x) is a linear function, using the trapezoidal rule gets the exact solu- tion. Example: Use the trapezoidal rule to numerically integrate 2 f(x) = 0.2 + 25x + 3x 3 from a = 0 to b = 2. Solution: f(0) = 0.2, and f(2) = 62.2. I = (b − a)f(b) + f(a) = (2 − 0) × 0.2 + 62.2 = 62.4 2 2 Thetrue solution is Z 2 2 3 2 2 3 0 f(x)dx = (0.2x+12.5x +x )|0 = (0.2×2+12.5×2 +2 )−0 = 58.4 Therelative error is ¯ ¯ ¯58.4 − 62.4¯ |ǫt| = ¯ ¯ ×100% = 6.85% ¯ 58.4 ¯ 4
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