MATLAB HELP ME PLS

close all
clear
clc
format long
a=0;
b=10;
function [Distance] = trapezoidal(Speed, a, b, n)
    h = (b-a)/n;
    result = 0.5*Speed*a + 0.5*Speed*b;
    for i = 1:(n-1)
        result = result + Speed*(a + i*h);
    end
    [Distance] = h*result;
    fprintf('ans=%6.6ff\n',Distance)
end

[Walter Roberson]

It is not clear what the question is?

Answer 1

[TADA]

The trapezoidal integration method is a numeric calculation which approximates the area trapped between the curve and the x axis by dividing the curve to segments. the area of each segment is calculated by calculating the area of a trapeze entrapped by the coordinates of that segment.

function [auc, integration] = trapezoidal(x,y)
  dx = diff(x);
  integration = ((y(1:end-1)+y(2:end)) .* dx) / 2;
  auc = sum(integration);
end

Or simply use trapz

The accuracy of this method is limited mainly by the size of the segments. smaller dx means more accurate results

so define your time vector with more elements:

t = linspace(0,10,1000);

[TADA]

to iteratively improve the approximation as required, I would start from a small number of segments, lets say 10 segments like you did

then check the error of the calculated distance

To calculate the actual distance you can integrate the polynomial:

% velocity polynomial coefficients
vp = [0.0011, -0.02928, 0.2807, -1.1837, -0.8283, 41.234, -3.3549];
tRange = [0 10];
nSegments = 10;

% call this in a loop that improves the precesion
while logical condition
    % do something to improve precesion here
    
    % calculate distance and error again
    [d, err] = tryTrapz(tRange, vp, nSegments);
    fprintf('Your message');
end

function [d, err] = tryTrapz(tRange, vp, nSegments)
    t = linspace(tRange(1), tRange(2), nSegments);
    d = trapz(t, polyval(vp, t));
    
    % distance polynomial coefficients = integrated velocity 
    dp = polyint(vp);
    theoreticalDist = polyval(dp, t(end));
    err = 100*(theoreticalDist - d)/theoreticalDist;
end