Question

Im not sure if this site uses MATLAB, but ill post the question anyway.

In this phase, we will evaluate the integral numerically using the definition by Riemann sum. For numerical calculations, we will use MATLAB software 3. First, use MATLAB to evaluate this time a definite integral x ехах For that, type directly into command window in MATLAB: syms x; int(x*exp(x),0,2). Get the answer in a number with at least four decimals. . Download an m-file, midPointRule.m, from Canvas and run it on MATLAB. Notice that the approximation is correct only up to two decimals (compare with the previous calculation). Increase the number of intervals up to 1000 and run the code one more time. Paste the figure you obtained as arn answer for this question in your report and write the captions explaining what the figure exactly means. Now you must modify the code to calculate the integral: 5. You can use the same Midpoint Rule as in my code or, for extra points, modify it to use Trapezoidal Rule. Paste the figure and your code as an answer to the question

MidPoint Rule

Answer 1

% %Matlab code for Question 3. clear all close all syms x y = int(x*exp(x), 0, 2): %integration of x*exp(x) fprintf(*\tIntegration of x*exp(x) for [0 2] is %2.5f.\n', y) %%Matlab code for Question 4. %lntegration using Midpoint rule nn = [100 1000]: %loop for 2 different step size for ii = 1: 2 a = 0: b = 2: %Integration limit ns = 1: nn(ii): integralValues = zeros(size(ns)): ourFunction = @(x) x.*exp(x): for n = ns delta X = (b - a)/n: xs = a + (0: n)*delta X: xsBar = (xs(1: end-1) + xs(2: end))/2: f = ourFunction(xsBar): integralValues(n) = delta X*sum(f): end %%Drawing figure figure(ii) plot (ns, integraivalues, 'Linewidth', 2) title(sprintf('Approximation by midpoint rule for n = %d', ns(end))) xlabel('n') ylabel('Riomann sum') dim-(.6 .4 .9 .3]: str = ['integral approximation ', num2str(integralValues(end))): annotation('textbox', dim, 'String', str, 'FitBoxToText', 'on'): xa = [0.7 0.89]: ya = [0.7 0.89]: annotation ('arrow', xa, ya)
$end %%Matlab code for Question 5. syms x int(Xexp ( x),0,2); Y- tintegration of x*exp (x) fprint f (\tIntegration of x^2*ex$

%%Matlab code for Question 3.
clear all
close all

syms x
y=int(x*exp(x),0,2);

%integration of x*exp(x)
fprintf('\tIntegration of x*exp(x) for [0 2] is %2.5f.\n',y)

%%Matlab code for Question 4.
%Integration using Midpoint rule

nn=[100 1000];

%loop for 2 different step size

for ii=1:2
  
    a=0; b=2; %Integration limit
    ns=1:nn(ii);
    integralValues=zeros(size(ns));
    ourFunction=@(x) x.*exp(x);

    for n=ns
        deltaX=(b-a)/n;
        xs=a+(0:n)*deltaX;
        xsBar=(xs(1:end-1)+xs(2:end))/2;
        f=ourFunction(xsBar);
        integralValues(n)=deltaX*sum(f);
    end

    %%Drawing figure
    figure(ii)
    plot(ns,integralValues,'Linewidth',2)
    title(sprintf('Approximation by midpoint rule for n=%d',ns(end)))
    xlabel('n')
    ylabel('Riemann sum')

    dim=[.6 .4 .9 .3];
    str=['integral approximation ',num2str(integralValues(end))];
    annotation('textbox',dim,'String',str,'FitBoxToText','on');
    xa=[0.7 0.89];
    ya=[0.7 0.89];
    annotation('arrow',xa,ya)
    ax=gca;
    grid
  
    fprintf('\nIntegration of x*exp(x) using Midpoint rule for [0 2]\n')
    fprintf('\tMidpoint integration value for n=%d is %f .\n',nn(ii),integralValues(end))
end

%%Matlab code for Question 5.

syms x
y=int(x*exp(x),0,2);
%integration of x*exp(x)
fprintf('\tIntegration of x^2*exp(x) for [0 3] is %2.5f.\n',y)

%Integration using Trapizoidal rule
nn=[100 1000];

for ii=1:2
  
    a=0; b=3; %Integration limit
    ns=1:nn(ii);
    integralValues=zeros(size(ns));
    ourFunction=@(x) x.^2.*exp(x);

    for n=ns
        val(n)=trapizoidal(ourFunction,a,b,nn(ii));
    end
  
    fprintf('\nIntegration of x^2*exp(x) using Trapizoidal rule for [0 3]\n')
    fprintf('\tTrapizoidal integration value for n=%d is %f .\n',nn(ii),val(end))
end

%%Matlab function for Trapizoidal integration

function val=trapizoidal(func,a,b,N)

    % func is the function for integration
    % a is the lower limit of integration
    % b is the upper limit of integration
    % N number of rectangles to be used
    val=0;
    %splits interval a to b into N+1 subintervals
    xx=linspace(a,b,N+1);
    dx=xx(2)-xx(1); %x interval
    %loop for Riemann integration
  
        for i=2:length(xx)-1
            xx1=xx(i);
            val=val+dx*double(func(xx1));
        end
       val=val+dx*(0.5*double(func(xx(1)))+0.5*double(func(xx(end))));
end

%%%%%%%%%%%%%%%%%%% End of Code %%%%%%%%%%%%%%%%%%%