University:
Kentucky State University
Course:
COS 232| Programming I
Academic year:

2021
Views:

233

Pages:

16
Author:

imophulebf13

21.22 Here is a VBA code to implement the multi-segment trapezoidal rule for equally-spaced segments: Option Explicit Sub TestTrapm() Dim Dim Dim Dim n As Integer, i As Integer, ind As Integer label As String a As Single, b As Single, h As Single x(100) As Single, f(100) As Single 'Enter data and integration parameters ind = InputBox("Functional (1) or Tabulated (2) data?") a = InputBox("Lower bound = ") b = InputBox("Upper bound = ") n = InputBox("Number of segments = ") h = (b - a) / n If ind = 1 Then 'generate data from function x(0) = a f(0) = fx(a) For i = 1 To n x(i) = x(i - 1) + h f(i) = fx(x(i)) Next i Else 'user input table of data x(0) = a label = "f(" & x(0) & ") = " f(i) = Val(InputBox(label)) For i = 1 To n x(i) = x(i - 1) + h label = "f(" & x(i) & ") = " f(i) = InputBox(label) Next i End If 'invoke function to determine and display integral MsgBox "The integral is " & Trapm(h, n, f()) End Sub Function Trapm(h, n, f) Dim i As Integer Dim sum As Single sum = f(0) For i = 1 To n - 1 sum = sum + 2 * f(i) Next i sum = sum + f(n) Trapm = h * sum / 2 End Function Function fx(x) fx = 0.2 + 25 * x - 200 * x ^ 2 + 675 * x ^ 3 - 900 * x ^ 4 + 400 * x ^ 5 End Function 21.23 Here is a VBA code to implement the multi-segment Simpson’s 1/3 rule algorithm from Fig. 21.13c: Option Explicit Sub TestSimpm() Dim Dim Dim Dim n As Integer, i As Integer label As String a As Single, b As Single, h As Single x(100) As Single, f(100) As Single 'Enter data and integration parameters a = InputBox("Lower bound = ") b = InputBox("Upper bound = ") n = InputBox("Number of segments = ") h = (b - a) / n 'generate data from function fx x(0) = a f(0) = fx(a) For i = 1 To n x(i) = x(i - 1) + h f(i) = fx(x(i)) Next i 'invoke function Simp13m to determine and display integral MsgBox "The integral is " & Simp13m(h, n, f()) End Sub Function Simp13m(h, n, f) Dim i As Integer Dim sum As Single sum = f(0) For i = 1 To n - 2 Step 2 sum = sum + 4 * f(i) + 2 * f(i + 1) Next i sum = sum + 4 * f(n - 1) + f(n) Simp13m = h * sum / 3 End Function Function fx(x) fx = 0.2 + 25 * x - 200 * x ^ 2 + 675 * x ^ 3 - 900 * x ^ 4 + 400 * x ^ 5 End Function 21.24 Option Explicit Sub TestUneven() Dim Dim Dim Dim n As Integer, i As Integer label As String a As Single, b As Single, h As Single x(100) As Single, f(100) As Single 'Enter data Range("a6").Select n = ActiveCell.Row Selection.End(xlDown).Select n = ActiveCell.Row - n 'Input data from sheet Range("a6").Select For i = 0 To n x(i) = ActiveCell.Value ActiveCell.Offset(0, 1).Select f(i) = ActiveCell.Value ActiveCell.Offset(1, -1).Select Next i 'invoke function to determine and display integral MsgBox "The integral is " & Uneven(n, x(), f()) End Sub Function Uneven(n, x, f) Dim k As Integer, j As Integer Dim h As Single, sum As Single, hf As Single h = x(1) - x(0) k = 1 sum = 0# For j = 1 To n hf = x(j + 1) - x(j) If Abs(h - hf) < 0.000001 Then If k = 3 Then sum = sum + Simp13(h, f(j - 3), f(j - 2), f(j - 1)) k = k - 1 Else k = k + 1 End If Else If k = 1 Then sum = sum + Trap(h, f(j - 1), f(j)) Else If k = 2 Then sum = sum + Simp13(h, f(j - 2), f(j - 1), f(j)) Else sum = sum + Simp38(h, f(j - 3), f(j - 2), f(j - 1), f(j)) End If k = 1 End If End If h = hf Next j Uneven = sum End Function Function Trap(h, f0, f1) Trap = h * (f0 + f1) / 2 End Function Function Simp13(h, f0, f1, f2) Simp13 = 2 * h * (f0 + 4 * f1 + f2) / 6 End Function Function Simp38(h, f0, f1, f2, f3) Simp38 = 3 * h * (f0 + 3 * (f1 + f2) + f3) / 8 End Function Function fx(x) fx = 0.2 + 25 * x - 200 * x ^ 2 + 675 * x ^ 3 - 900 * x ^ 4 + 400 * x ^ 5 End Function 21.25 (a) M = (b − a )[ f ( xo ) + 2 ∑i=1 f ( xi ) + f ( xn ) ] n −1 2n  4 + 2(4 .15 + 4 .6 + 5.35 + 6.4 + 7 .75 + 9.4 + 11.35 + 13.6 + 16 .15 + 19 ) + 22 .15  M = (11 − 0)   2(11)   = 110.825 lb - ft (b) The 1/3 rule can only be applied to the first 10 panels. The trapezoidal rule can be applied to the 11th M = (10 − 0 )[ 4 + 4( 4.15 + 5 .35 + 7.75 + 11.35 + 16 .15) + 2( 4.6 + 6 .4 + 9.4 + 13.6) + 19 ] 3(10) 19 + 22.15 = 110 .825 lb - ft 2 (c) The 3/8 rule can only be applied to the first 9 panels and the 1/3 rule applied to the last 2: + (12 − 11) 4 + 3( 4.15 + 4.6 ) + 5 .35 5 .35 + 3( 6.4 + 7 .75) + 9.4 ] + ( 6 − 3)[ ] 8 8 9.4 + 3(11 .35 + 13 .6) + 16.15 16.15 + 4(19) + 22.15 + (9 − 6)[ ] + (11 − 9)[ ] = 110.55 lb - ft 8 6 M = (3 − 0)[ This result is exact because we’re integrating a quadratic. The results of (a) and (b) are not exact because they include trapezoidal rule evaluations. 21.26 Divide the curve into sections according to dV changes and use appropriate rules. 420 + 368 I1 = 1 .5( ) = 591 2 1 I 2 = (368 + 4 (333) + 326 ) = 675.33 3 3 I 3 = (326 + 3(326) + 3(312) + 242) = 930. 75 8 242 + 207 I 4 = 1( ) = 224 .5 2 W = I1 + I 2 + I 3 + I 4 = 2421 .583 Therefore, the work done is 2420 kJ. 21.27 (a) The trapezoidal rule yields 60.425. (b) A parabola can be fit to the data to give y = -0.11829x 2 + 1.40701x + 3.36800 R2 = 0.60901 10 9 8 7 6 5 4 0 2 4 6 8 10 12 The parabola can be integrated and evaluated from 1 to 10 to give 60.565. (c) A cubic can be fit to the data to give y = -0.01672x 3 + 0.16015x 2 + 0.10764x + 4.81478 R2 = 0.67487 10 9 8 7 6 5 4 0 2 4 6 8 10 12 The cubic can be integrated and evaluated from 1 to 10 to give 60.195. Although it’s not asked in the problem statement, the algorithm from Fig. 21.15b can also be applied (see Solution to Prob. 20.24 for code) to yield 60.258. 21.28 (a) The following 2 equations must hold: f ( a ) = Qe ra (1) f (b) = Qe rb (2) Take the natural log of Eq. 1 and solve for ln Q = ln f ( a ) − ra (3) or Q = f ( a )e − ra (4) Substituting (3) into the natural log of Eq. 2 gives ln f (b) = ln f (a ) − ra + rb (5) and solve for r= ln ( f (a ) / f (b ) ) a−b (6) These results can be verified for the case where Q = 3 and r = −0.5. If a = 2 and b = 4, f(a) = 1.1036 and f(b) = 0.406. Substituting these values into Eqs. 6 and 4 gives r= ln(1.0136 / 0.406) = −0.5 2− 4 Q = eln(1.1036) − ( −0.5)( 2) = 1.1036e − ( −0.5) 2 = 3 (b) I= b ∫a Qe rx dx = ( Q rb e − e ra r ) Substituting Eq. 4 ( ) ( ) f ( a) e − ra rb f ( a ) r (b − a) e − e ra = e −1 r r I= Substituting Eq. 6 f (a) I= ln( f ( a) / f (b) ) a −b  ln ( f ( a ) / f (b ) ) ( b − a )  e a −b − 1     Simplifying I= (b − a )( f (b) − f ( a ) ) ln( f (b) / f (a ) ) This result can be verified for the case where Q = 3 and r = −0.5. If a = 2 and b = 4, f(a) = 1.1036 and f(b) = 0.406. Substituting these values into the integral equation gives I= ( 4 − 2)( 0 .406 − 1 .1036 ) = 1 .9353 ln ( 0.406 / 1.1036 ) which matches the analytical integral I == ( ) 3 e − 0.5( 4) − e − 0.5( 2) = 1.9353 − 0.5

Chapter 21 VBA Code to Implement the Multi-segment Trapezoidal Rule Forequally-spaced Segments

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