Area under a curve

Hi,

I want to calculate the area of this curve. y = [0 1 3 -1 -2 -3 -1 0];

I know a portion of the curve has negative value, so my solution is make all the y values absolute. But then the area of absolute y will be higher. Can anyone help me?

Thanks.

Answer 1

[Walter Roberson]

http://www.mathworks.com/help/matlab/ref/polyarea.html

Answer 2

[Star Strider]

The problem wasn’t immediately obvious to me. You need to find the zero-crossing, and then add the two separate areas:

y = [0 1 3 -1 -2 -3 -1 0];
x = 1:length(y);

zci = @(v) find(v(:).*circshift(v(:), [-1 0]) < 0);                 % Returns Approximate Zero-Crossing Indices Of Argument Vector

yzxi = zci(y);                                                      % Zero-Crossing Index
x0 = interp1(y(yzxi:yzxi+1), x(yzxi:yzxi+1), 0);                    % Interpolate To Find Zero-Crossing

AUC = polyarea([x(1:yzxi) x0], [y(1:yzxi) 0]) + polyarea([x0 x(yzxi+1:end)], [0 y(yzxi+1:end)]);

INT = trapz(x, abs(y))                                              % Compare (Optional)

AUC =
    10.2500

INT =
    11.0000

I used the polyarea function rather than the integration functions. If you have a more complicated function, this will work as well, but you will have to make the appropriate changes to the code. (I included the trapz function integration of the absolute value for comparison.)

[Q TRAN]

Thank you.

Do you think if this works for the case we have more than one zero crossing?

[Star Strider]

Yes.

You simply have to find each one, calculate the zero-crossing, and do polyarea for each segment.

Answer 3

[Piyush Madame]

just by some codding

[Q TRAN]

Would you please share your code?