Ah ok,
So because of the denominator
\sqrt{1-x^2}}
In order to get a real number... x^2 < 1
So the limits would be (-1, 1), added together gives me my convergence to 0, is that it?
Hi, I need to determine whether this improper integral converges or diverges
\int_{-1}^{1} \frac{x}{\sqrt{1-x^2}}dx
The original function DNE at -1, 1 so I split the limits
\int_{-1}^{0} \frac{x}{\sqrt{1-x^2}}dx \ + \ \int_{0}^{1} \frac{x}{\sqrt{1-x^2}}dx
I've...