week4ppt

# week4ppt - Dr.T.P.Clement CE3010classnotes...

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8/15/2011 1 Dr. T.P. Clement CE 3010 class notes Learning Objectives Motivation Basic approach Trapezoid rule Simpson’s 1/3 rule Richardson extrapolation Integration of data (applied civil engineering problems) 1 The problem and analytical solution Consider the following definite integral: Analytical solution: use basic calculus to integrate f(x) and substitute the limits to compute the exact value of the definite integral 2 b a dx x f ) ( Dr. T.P. Clement CE 3010 class notes Analytical (or exact) solution Worked out example to show exact answer is 53.28214 3 The problem and approx. graphical solution 4 The problem and numerical solution Consider the following definite integral: Basic idea behind numerical solution: A definite integral represents area under the plot of x vs. f(x) within the limits (x values) a to b. We can approximate the numerical value of the definite integral by computing area under the curve f(x) We can compute the approximate value of the integral without calculus! 5 b a dx x f ) ( Dr. T.P. Clement CE 3010 class notes Trapezoidal Rule F(x)

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8/15/2011 2 Dr. T.P. Clement CE 3010 class notes Trapezoidal Rule Trapezoidal works by approximating the area usingatrapezo id Trapezoidal formula (single segment): I = [f(a) + f(b)]/2 * (b–a) (average height * width) Dr. T.P. Clement CE 3010 class notes Trapezoidal Rule (multiple segments) F(x) ab Dr. T.P. Clement CE 3010 class notes Multiple trap rule The accuracy of trapezoidal rule can be improved by dividing the area under the curve into multiple trapezoids. Divide the limit Xo (or a) to Xn (or b) into ‘n’ equal trapezoids with width “h” where h = (b a)/n 12 001 1 () . . . n n n xx xxx x I f xdx   Dr. T.P. Clement CE 3010 class notes Multiple trap rule (equal segment) 01 1 () () ( ) () ... 22 2 nn f xf x f x f x fx Ih h h         1 0 1 2() () 2 n in i 1 0 1 2 n i ba I n    Use for hand calculations Better for programming Dr. T.P. Clement CE 3010 class notes Multiple trap rule (unequal segments) In the previous case, we used equal segments for width ‘h’. If we used unequal segments of width h 1 , h 2 , h 3 .. h n then the formula is: 1 ... 2 n f x f x h h Integration Example With n = 4 evaluate Δ x = (5 1) / 4 = 1 (Choose easy to use Δ x , such as: 1, 0.5, 0.25, etc.) 12 5 2 1 ln f d x ( x ) A r e a 10 . 0 0 . 7 71 . 3 9 39 . 8 96 . 3 3 4 22.18 16.03 5 40.24 31.21 Sum= 54.95878 =0.5*(0+2.77)*(2 1) =0.5*(2.77+9.89)*(3 2) =0.5*(9.89+22.18)*(4 3) =0.5*(22.18+40.24)*(5 4) =1.39+6.33+16.03+31.21
8/15/2011 3 Dr. T.P. Clement CE 3010 class notes Comparison of results (errors) Integration by parts yields exact solution as: 53.28214 Trap rule gave approximate solution: 54.95878 (for n = 4 & h = 1) Then, true error, E T = 53.28214 54.95878 = 1.67666 Absolute error is Eabs = 1.67666

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week4ppt - Dr.T.P.Clement CE3010classnotes...

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