For a given $v(x)$ with $x\in[0,1]$, use the variable transformation $x=g(\eta)=\frac{1}{2}\eta+\frac{1}{2}$ to transform the integral $I=\int_0^1v(x)dx$ to an integral over $[-1,1]$.
My doubts:
First what I did: $x=\frac{1}{2}\eta+\frac{1}{2}\implies\eta=2x-1$.
What I don't quite understand is, by transforming the integral to an integral over different domain does that mean the integrand does not change? Is it correct if $I=\int_0^1v(x)dx=\int_{-1}^1v(x)dx$ ? Or $I=\int_0^1v(x)dx=\int_{-1}^1g(\eta)d\eta$ ?
If it is the latter, is it correct if I do $\int_0^12x-1dx=\int_{-1}^1\frac{1}{2}\eta+\frac{1}{2}d\eta=1$ ?
Could anybody please give some help?
Thanks!
By transforming, it is actually just like substitution, with $x=g(\eta)$ into the integral, so;
$$\int_0^1 v(x)dx = \int_{0}^1 v(g(\eta)) d(g(\eta)) =\int_{-1}^1 v(g(\eta)) g'(\eta) d\eta$$