Why does trapz yield infinity

Hi there...

Why does the following code yield infinity

x      = 0:0.001:1;
ytDiff = 1./(2*(x).^(1/2));
ArcEq  = sqrt(1+ytDiff.^2);
s      = trapz(x,ArcEq)

when the actual value should be s = 1.4789!?

Thanks!

Answer 1

[the cyclist]

Because that function approaches infinity at x=0, and you have included that endpoint of the interval explicitly. Try this instead:

x      = 0.001:0.001:1;
ytDiff = 1./(2*(x).^(1/2));
ArcEq  = sqrt(1+ytDiff.^2);
s      = trapz(x,ArcEq)

Answer 2

[David]

Of course... And it works!

Thank you very much!