Page - 142 - in Programming for Computations – Python - A Gentle Introduction to Numerical Simulations with Python 3.6, Volume Second Edition

142 6 ComputingIntegralsandTestingCode Lookingbackon the twodifferentsolutions, thespecific implementationand the general implementation,youshould realize that implementingageneralmathemat- ical algorithm in a general function requires somewhat more abstract thinking,but the resulting code can be used over and over again! Essentially, if you apply the special-purpose style, you have to retest the implementation of the algorithm after everychangeof theprogram. The present integral problems result in short code. In more challenging en- gineering problems, the code quickly grows to hundreds and thousands of lines. Without abstractions, in terms of general algorithms in general reusable functions, the complexity of the program grows so fast that it will be extremely difficult to makesure that theprogramworksproperly. Another advantage of packagingmathematical algorithms in functions, is that a functioncanbereusedbyanyonetosolveaproblembyjustcallingthefunctionwith a proper set of arguments. Understanding the function’s inner details is strictly not necessary to compute a new integral. Similarly, you can find libraries of functions on the Internet and use these functions to solve your problems without specific knowledgeofeverymathematicaldetail in the functions. This desirable feature has its downside, of course: the user of a function may misuse it, and the function may contain programming errors and lead to wrong answers. Testing the output of downloaded functions is therefore extremely importantbefore relyingon the results. 6.3 TheCompositeMidpointMethod The Idea Rather than approximating the area undera curveby trapezoids,we can useplainrectangles. Itmaysoundlessaccurate tousehorizontal linesandnotskew lines following the function to be integrated, but an integration method based on rectangles (the midpoint method) is in fact slightly more accurate than the one based on trapezoids! In the midpoint method, we construct a rectangle for every sub-interval where the height equals the integrand f at the midpoint of the sub- interval. For the sake of comparison, we may repeat the hand calculation of ∫1 0 v(t)dt in (6.7),but this time with themidpointmethod.With four rectangles (Fig.6.3)and thesamesub-intervalsthatweusedwith the trapezoidalmethod,[0,0.2), [0.2,0.6), [0.6,0.8), and [0.8,1.0],we get ∫ 1 0 v(t)dt≈h1v ( 0+0.2 2 ) +h2v ( 0.2+0.6 2 ) +h3v ( 0.6+0.8 2 ) +h4v ( 0.8+1.0 2 ) , (6.18) whereh1,h2,h3, andh4 are the widths of the sub-intervals, used previously with the trapezoidalmethodanddefined in (6.10)–(6.13). Withv(t)= 3t2et3, the approximationbecomes 1.632. Compared with the true answer (1.718), this is about 5% too small, but it is better than what we got with