Getting weird results when calculating an integral using rectangle method

Hi Guys,

I wrote the following code to try to integrate sin(x) from 0 to 15 (in degrees) using the rectangle method. I know there are other ways of calculating integrals in MATLAB, but this was out of curiosity, and I can't figure out what is going wrong. The code is as follows:

vec1 = pi/180*(0.005:0.01:14.995);
int_val = sum(0.01.*sin(vec1));

The idea is to use rectangles of width 0.01, and to calculate their areas using the midpoints between values. However, the result I get for "int_val" is 1.9523, which is obviously wrong (should be 0.0341). On the other hand, I have tried almost the exact same code to perform numerical integration on a different function (x^2, in this case) from x= 0 to 15, and obtained the correct result of 1125. This code is as follows:

vec1 = (0.005:0.01:14.995);
fun1 = @(x) x.^2;
int_val = sum(0.01*fun1(vec1));

It doesn't seem like some kind of silly math error, so can someone explain what is going wrong here?

Answer 1

[John D'Errico]

Look carefully at what you did.

vec1 = pi/180*(0.005:0.01:14.995);

Now, what is the stride between elements of that vector? While you may THINK about this as being in degrees, the vector is itself in radians. Those boxes have width dx. NOT 0.01. The value of dx here is 0.01*pi/180.

dx = diff(vec1(1:2))
dx =
      0.000174532925199433
      
0.01*pi/180
ans =
      0.000174532925199433
      
int_val = sum(dx*sin(vec1))
int_val =
          0.03407417375418

[user4050]

Thanks, that explains it.

Answer 2

[Tommy]

vec1 = pi/180*(0.005:0.01:14.995);

Here, vec1 is in radians, but the values are not spaced 0.01 radians apart:

>> vec1(2)-vec1(1)

ans =

   1.7453e-04

Rather, they are spaced 0.01*(pi/180) radians apart, so the width of each rectangle is 0.01*(pi/180):

int_val = sum(0.01*(pi/180)*sin(vec1));

int_val =

    0.0341

Alternatively,

vec1 = ((pi/180)*0.005:0.01:(pi/180)*14.995);
int_val = sum(0.01*sin(vec1))

int_val =

    0.0349

The rectangles are wider, so the estimate is less accurate.

[user4050]

Thanks. The two examples were elucidating as well.