MATLAB. For this assignment, you are asked to code four algorithms for numerical

Question

MATLAB.

For this assignment, you are asked to code four algorithms for numerical integrations and generate a figure which should be very similar to Figure 1. Absolute Errors of four numerical integration methods for approximating I(f) = integral^pi_0 f(x) dx where f(x) = e^x cos x. Basically, we need to compute I(f) = integral^pi_0 f(x) dx where f(x) = e^x cos x. We divide the interval [0, pi] into n equal-length sub-intervals and implement the following four algorithms for approximating I (f): S_n: Simpson's rule.

Explanation / Answer

function I = simpsons(f,a,b,n) % This function computes the integral "I" via Simpson's rule in the interval [a,b] with n+1 equally spaced points % % Syntax: I = simpsons(f,a,b,n) % % Where, % f= can be either an anonymous function (e.g. f=@(x) sin(x)) or a vector % containing equally spaced values of the function to be integrated % a= Initial point of interval % b= Last point of interval % n= # of sub-intervals (panels), must be integer % % Written by Juan Camilo Medina - The University of Notre Dame % 09/2010 (copyright Dr. Simpson) % % % Example 1: % % Suppose you want to integrate a function f(x) in the interval [-1,1]. % You also want 3 integration points (2 panels) evenly distributed through the % domain (you can select more point for better accuracy). % Thus: % % f=@(x) ((x-1).*x./2).*((x-1).*x./2); % I=simpsons(f,-1,1,2) % % % Example 2: % % Suppose you want to integrate a function f(x) in the interval [-1,1]. % You know some values of the function f(x) between the given interval, % those are fi= {1,0.518,0.230,0.078,0.014,0,0.006,0.014,0.014,0.006,0} % Thus: % % fi= [1 0.518 0.230 0.078 0.014 0 0.006 0.014 0.014 0.006 0]; % I=simpsons(fi,-1,1,[]) % % note that there is no need to provide the number of intervals (panels) "n", % since they are implicitly specified by the number of elements in the % vector fi if numel(f)>1 % If the input provided is a vector n=numel(f)-1; h=(b-a)/n; I= h/3*(f(1)+2*sum(f(3:2:end-2))+4*sum(f(2:2:end))+f(end)); else % If the input provided is an anonymous function h=(b-a)/n; xi=a:h:b; I= h/3*(f(xi(1))+2*sum(f(xi(3:2:end-2)))+4*sum(f(xi(2:2:end)))+f(xi(end))); end

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