Taylor Expanding an Integrand Just want to be sure of this: If I taylor expand an integrand about 0 then truncate it to say linear order, then integrate this truncation, does the integral evaluate to a function which is an accurate representation of the original integral, but only around 0? If so why is this?
Cheers!

The Taylor theorem says (I skip the hypothesis): $$f(x) = f(0)+f'(0)x+\frac{f'(\xi)}{2}x^2.$$ The point is that $\xi$ depends on $x$ and in general this dependance is very nasty. You can integrate both parts of the equation with respect to $x$ on some interval, but I don't think you can get something useful from the right side representation other than approximations of the integral.
