How to use integration limits correctly (Matlab)

Question

I've made a script that calculates a given integral with the limits:

0 <= x <= 2 and 0 <= y <= 1

But now I want to change the limits to:

0 <= x <= 2 and 0 <= y <= sin((pi*x)/2)

Function:

function f = inte(x,y)

dz = 10;

f = exp(-dz*((x-1.25).^2+y.^2)).*cos(y.*(x-1.25));

end

This is my script for the earlier limits:

L = 100; M = L/2;

hx = (2)/L; hy = (1)/M; 

x=[0:L]*hx;
y=[0:M]*hy;

Fx=[];

for i = 1:L+1

    Fy=[];

    for j = 1:M+1

        f = inte(x(i),y(j));
        Fy = [Fy f];

    end

    ycor = hy*(sum(Fy) - Fy(1)/2 - Fy(end)/2);
    Fx = [Fx ycor];

end

ans = hx*(sum(Fx) - Fx(1)/2 - Fx(end)/2);

disp (ans)

I can't seem to get the right answer when I try to change the code. The correct answer should be 0.1560949...

L is amount of steps in x direction, M in y direction. hx and hy are step lengths. This is really bugging me. And no I can only use the commands integral2 or traps as reference.

Thanks in advance!

You've posted your script for the original first case. What did you try in the second case where y is a function of x? — Ryan J. Smith
– Ryan J. Smith, Commented Apr 21, 2015 at 12:24

user3717023 · Accepted Answer · 2015-04-23 02:06:46Z

1

In your present code, the lines

hy = (1)/M; 
y=[0:M]*hy;

refer to the y-variable. When the limits for y depend on x, these lines cannot stay outside of the loop over x: they should be moved in and use the value x(i). Like this:

for i = 1:L+1   % as in your code

   hy = (sin(pi*x(i)/2))/M;     
   y = [0:M]*hy;

   Fy=[];  % this and the rest as in your code

I get output 0.1561, as you wanted.

answered Apr 23, 2015 at 2:06

user3717023

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