integration in matlab plot

I have a graph of 2 arrays, and I need to do an integration between 0 and 4.7. Can you please tell me how to do it?

x=[0 1.4660 2.9319 4.3979 5.8639 7.3298 8.7958 10.2618 11.7277 13.1937 14.6597]

y=[78.8093 78.8093 77.7206 76.6319 75.3877 73.2103 69.4776 64.2933 57.0353 44.3338 26.4480]

Answer 1

[Hassaan]

% Your desired values here
x = [0, 0.5, 1.2, 1.4660, 2.9319, 4.3979, 5.8639, 7.3298, 8.7958, 10.2618, 11.7277, 13.1937, 14.6597];
y = [11.4, 22.90, 78.8093, 78.8093, 77.7206, 76.6319, 75.3877, 73.2103, 69.4776, 64.2933, 57.0353, 44.3338, 26.4480];

% Find the indices where x is less than or equal to 6
indices = find(x <= 6);

% Extract the subsets of x and y that are within the desired range
x_sub = x(indices);
y_sub = y(indices);

% Perform the integration on the subset
integralValue = trapz(x_sub, y_sub);

% Display the result
disp(['The integral value between 0 and 6 is: ', num2str(integralValue)]);

The integral value between 0 and 6 is: 404.4359

Learning Material:

MATLAB Onramp | Self-Paced Online Courses - MATLAB & Simulink (mathworks.com)

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It's important to note that the advice and code are based on limited information and meant for educational purposes. Users should verify and adapt the code to their specific needs, ensuring compatibility and adherence to ethical standards.

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[Hassaan]

Just for visualization purpose:

% Your desired values here
x = [0, 0.5, 1.2, 1.4660, 2.9319, 4.3979, 5.8639, 7.3298, 8.7958, 10.2618, 11.7277, 13.1937, 14.6597];
y = [11.4, 22.90, 78.8093, 78.8093, 77.7206, 76.6319, 75.3877, 73.2103, 69.4776, 64.2933, 57.0353, 44.3338, 26.4480];

% Plot the original data
figure;
plot(x, y, '-o', 'LineWidth', 2);
hold on;

% Highlight the integration area
% Find indices for x values within the desired range
indices = find(x <= 6);
% Use the fill function to highlight the area under the curve
x_fill = [x(indices), fliplr(x(indices))];
y_fill = [y(indices), zeros(1, length(indices))];
fill(x_fill, y_fill, 'cyan', 'LineStyle', 'none');
alpha(0.3); % Adjust transparency

% Annotate the graph
xlabel('X Axis');
ylabel('Y Axis');
title('Integration Area Visualization');
legend('Original Data', 'Integration Area', 'Location', 'NorthEast');
hold off;

Learning Material

MATLAB Onramp | Self-Paced Online Courses - MATLAB & Simulink (mathworks.com)

-----------------------------------------------------------------------------------------------------------------------------------------------------

If you find the solution helpful and it resolves your issue, it would be greatly appreciated if you could accept the answer. Also, leaving an upvote and a comment are also wonderful ways to provide feedback.

It's important to note that the advice and code are based on limited information and meant for educational purposes. Users should verify and adapt the code to their specific needs, ensuring compatibility and adherence to ethical standards.

Professional Interests

• Technical Services and Consulting
• Embedded Systems | Firmware Developement | Simulations
• Electrical and Electronics Engineering

Feel free to contact me.

[gravy]

you are the best, thanks!

Answer 2

[John D'Errico]

You CANNOT use trapz. Despite the fact that another poster has done so, that does not yield a result up to 4.7. If you have no data point that lies EXACTLY at x==4.7, trapz CANNOT be used.

x=[0    1.4660    2.9319    4.3979    5.8639    7.3298    8.7958   10.2618   11.7277   13.1937   14.6597];
y=[78.8093   78.8093   77.7206   76.6319   75.3877   73.2103   69.4776   64.2933   57.0353   44.3338   26.4480];
plot(x,y,'o-')

The solution is not that difficult though. I think fnint is part of the curve fitting toolbox, so you can do this:

fn = pchip(x,y);

Note that pchip is a better choice than spline here, since your curve seems to have a sharp transition in it in some cases.

fnI = fnint(fn)

fnI = struct with fields:
      form: 'pp'
    breaks: [0 1.4660 2.9319 4.3979 5.8639 7.3298 8.7958 10.2618 11.7277 13.1937 14.6597]
     coefs: [10×5 double]
    pieces: 10
     order: 5
       dim: 1

fnval(fnI,4.7) - fnval(fnI,0)

ans = 366.6602

And that will be the integral from 0 to 4.7.

Another approach, if you lack the fnint function (as I said, part of the CFTB) is to just use integral.

integral(@(x) ppval(fn,x),0,4.7)

ans = 366.6602

As you can see, either yields the same result. However, trapz cannot do that.

[gravy]

Thank you too John, I'll try your variant as well