How do you calculate the average value of a line?

I am attempting to calculate the average value of a line using the equation from this website: http://tutorial.math.lamar.edu/Classes/CalcI/AvgFcnValue.aspx

Part of this equation requires getting a definite integral from x0 to x1. I only want the positive values of these lines, and the maximum value of the y coordinate is 1. To accomplish this, I want the lower of limit of x to be where this line crosses the x-axis (x0), and I want the upper limit of x to be where y = 1 (x1). I tried to do this with the following code:

fun = @(x,c) sqrt((-.25 .*(n-(c.*n))./(100-n))./(((n-(c.*n))./(100-n))-1)); 

x0 = fzero(@(x)fun(x,.5), 1);

x1 = fzero(@(x)fun(x,.5)-1, 1);

q = (1/(x1-x0))*integral(@(x)fun(x,.5),x0, x1)

This code, however, spits out some strange errors that I do not know how to resolve. How can I fix this so that I can find the average value of this line?

[Walter Roberson]

The first obvious problem is that n is not defined.

[Ahmed raafat]

 fun = @(x,c) sqrt((-.25 .*(n-(c.*n))./(100-n))./(((n-(c.*n))./(100-n))-1));

it is fun of n & c not x & c

Answer 1

[Walter Roberson]

x0 = fzero(@(x)fun(x,.5), [0 66.66666666]);
x1 = fzero(@(x)fun(x,.5)-1, [0 66.66666666]);

fun(x,1/2) has a discontinuity at 200/3 but grows fairly slowly, so it is necessary to get fairly close to 200/3 to satisfy x1

[Colin Lynch]

That did it, thank you!!!

[Colin Lynch]

Do you happen to know how I can get MATLAB to automatically generate the 66.6666 number?

[Walter Roberson]

The discontinuity is at 100/(2 - c) . But you have to make sure you do not use exactly that number or else you will get inf there. So like 100/(2-c) * (1-10*eps)

[Colin Lynch]

Thank you again!

Answer 2

[Ahmed raafat]

where is x????

fun = @(x,c) sqrt((-.25 .*(n-(c.*n))./(100-n))./(((n-(c.*n))./(100-n))-1));

[Colin Lynch]

Whoops, that was an obvious mistake. Still, even if n is replaced with x, the problem persists