Question

x = [1.3214; 1.8779; 2.1429; 2.2770; 2.4185; 3.5065; 4.7959; 5.3684; 5.9219; 6.6912; 6.9017; 7.1086; 8.1299; 8.2025; 8.3325; 8.6422; 9.1521; 9.2204; 9.2416; 9.4059; 9.6145; 9.6176; 9.6354; 9.6840; 9.7353]; y = [2.3779; 2.3929; 2.4543; 2.6263; 2.6952; 2.6936; 2.6415; 2.6140; 2.5823; 2.5790; 2.5700; 2.5694; 2.5677; 2.5415; 2.5366; 2.5098; 2.5003; 2.4974; 2.4638; 2.4508; 2.4472; 2.4379; 2.3999; 2.3603; 2.1660];

Write a function with the header

function [left, mid, right] = myNumericInt(x, y)

which takes two vectors of data (of equal length) and returns the integral using a Left Reimann sum approximation, Middle Reimann sum approximation, and Right Reimann sum approximation. The Middle Reimann sum should be found by estimating y values midway between x-values. Do not assume x is equally spaced. test

>>l,m,c] = myNumericInt(x,y)

l = 21.7187

m = 21.6771

c = 21.6355

Answer 1

`Hey,

Note: Brother in case of any queries, just comment in box I would be very happy to assist all your queries

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Save file as myNumericInt.m

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function [l,m,c] = myNumericInt(x,y)

l=0;
for i=1:length(x)-1
l=l+(x(i+1)-x(i))*y(i);
end
m=0.0;
for i=1:length(x)-1
m=m+(x(i+1)-x(i))*(y(i+1)+y(i))/2;
end
c=0.0;
for i=1:length(x)-1
c=c+(x(i+1)-x(i))*y(i+1);
end

end

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Executable file

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clear all
clc
x = [1.3214; 1.8779; 2.1429; 2.2770; 2.4185; 3.5065; 4.7959; 5.3684; 5.9219; 6.6912; 6.9017; 7.1086; 8.1299; 8.2025; 8.3325; 8.6422; 9.1521; 9.2204; 9.2416; 9.4059; 9.6145; 9.6176; 9.6354; 9.6840; 9.7353];
y = [2.3779; 2.3929; 2.4543; 2.6263; 2.6952; 2.6936; 2.6415; 2.6140; 2.5823; 2.5790; 2.5700; 2.5694; 2.5677; 2.5415; 2.5366; 2.5098; 2.5003; 2.4974; 2.4638; 2.4508; 2.4472; 2.4379; 2.3999; 2.3603; 2.1660];
[l,m,c] = myNumericInt(x,y)

Kindly revert for any queries

Thanks.