Can someone check my for loop?

I have to create a function m file called mylength(f,a,b,n) which takes four inputs:

f: A function handle.

a: A real number.

b: A real number.

n: A positive integer.

Note: You may assume that f(x) is differentiable on [a, b] and that a < b.

Does: Calculates the length of f(x) on [a, b] using the formula L = the integral from a to b of the square root of (1 + f′(x)^2 dx) but with the integral approximated using a for loop to calculate a left sum with [a, b] divided into n subintervals.

Returns: This approximated length.

Here is sample data and correct answers to the sample data.

a = mylength(@(x) x^2,0,2,5)

a = 4.0480127986983730101003658887008

a = mylength(@(x) sin(x),pi,2*pi,10)

a = 3.820197787492867218055528594355

Here is my code.

function l=mylength(f,a,b,n);
syms x;
r=(b-a)/n;
L = sqrt(1+diff(f(x))^2);
l=0;
for q=(a:n-1);
    y=subs(f(x),q);
    area=r*y;
    l=vpa(l+subs(L,r*area));
end
end

It does not return the same answers as the sample data, and I was pretty sure my code was right. Can you please let me know what is wrong with it?

Thanks

[Geoff Hayes]

Bri - what are the answers that you are getting?

[Bri]

For the first sample data I am getting:

a =10.120031996745932525250914721752

and for the second sample data I am getting:

a =12.160067229364235067628002381161

Answer 1

[Star Strider]

This hits ‘sin(x)’ spot on but slightly underestimates ‘x^2’:

syms x;
r=(b-a)/(n-1);
s = a:r:b;
L = sqrt(1+diff(f(x))^2);
l=0;
for q=1:n-1;
    l=l+vpa(subs(L,s(q))*r);
end

[Bri]

Can you explain what s(q) is doing in the for loop please?

[Star Strider]

Sure!

The ‘s’ vector are the values going from ‘a’ to ‘b’ with an interval of ‘r’, so ‘s(q)’ is the value of ‘s’ at that iteration of the loop.

[Bri]

Thanks.

[Star Strider]

My pleasure!