Solving Integral for an Unknown Interval

Is it possible to solve an integral for an interval/ limit of integration without the symbolic toolbox? My problem is of the same form as:(∫f(x)dx)/Z on the interval[-a,a]is equal to (∫g(x)dx)/Y on the interval [-c,c], where Z, a, and Y are known values and c is the unknown variable.

Answer 1

[Friedrich]

Hi,

so doing something like this:

function out = test_func( c )

out = quad(@lhs,-.25,.25) ./10 - quad(@rhs,-c,c)./6;

function out_lhs = lhs(x)
  out_lhs = sqrt(5.^2 - x.^2);
end

function out_rhs = rhs(x)
  out_rhs = sqrt(2.5^2 - x.^2);
end

end

And call it through:

fzero(@test_func,1)

Answer 2

[Friedrich]

Hi,

no. You can't get a symbolic solution without the symbolic math toolbox. When you know c you can use the quad function:

http://www.mathworks.com/help/releases/R2011a/techdoc/ref/quad.html

[John]

I am actually trying to find the numeric value for c. I knew there was a way to solve the integral with the symbolic toolbox and then solve for c, however I was wondering if there was a more efficient way to do this without the toolbox. The problem is like(∫sqrt(5^2-x^2)dx)/10 [-25,.25]=(∫sqrt(2.5^2-x^2)dx)/6 [-c,c] solving for c.

[Friedrich]

Why not calculate it manually and than hardcode it. Or do your functions change during runtime?

[John]

The functions have the possibility to change during runtime, so I would not like to have to calculate each time.

[Friedrich]

Ah okay. Not sure if this will work fine but you could use the fzero function and search the root of (∫sqrt(5^2-x^2)dx)/10 [-25,.25] - (∫sqrt(2.5^2-x^2)dx)/6 [-c,c] , where you solve the integradl with quad

Answer 3

[Bjorn Gustavsson]

One super-tool you should take a long look at is the Chebfun tools: http://www2.maths.ox.ac.uk/chebfun/

and my Q-D stab would be something like this:

I_of_f = quadgk(f(x)/Z,-a,a);
c = @(a,Z,Y,f,g) fminsearch(@(c) (I_of_f-quadgk(@(x) g(x)/Y,-c,c))^2,1)

I think that should work...