In Spivak's 'Calculus on Manifolds', page 67, the following statement is made.
Suppose $g: [a,b]\rightarrow\mathbb{R}$ is continuously differentiable and $f: \mathbb{R}\rightarrow\mathbb{R}$ is continuous. Then, \begin{equation}\int_{g(a)}^{g(b)} f=\int_{a}^{b}(f \circ g) \cdot g^{\prime}\end{equation} It is left to the reader to show that if $g$ is 1-1, then the above formula can be written as \begin{equation}\int_{g((a,b))} f=\int_{(a,b)} f \circ g \cdot\left|g^{\prime}\right|\end{equation} Now, I have tried to show this and came up with the following case: Let $f(x)=\sin(x)$ and $g(x)=-\pi x$, then \begin{equation}\int_{g(0)}^{g(1)} f = \int_{0}^{1}(f \circ g) \cdot g^{\prime} = \int_{0}^{1}\sin(-\pi x)\cdot -\pi = \pi\int_{0}^{1}\sin(\pi x) = 2 \end{equation} but \begin{equation}\int_{g((0,1))} f=\int_{(0,1)} f \circ g \cdot\left|g^{\prime}\right| = \int_{(0,1)} \sin(-\pi x) \cdot\left|-\pi\right| = -\pi\int_{(0,1)} \sin(\pi x) = -2 \end{equation} Clearly, the two formulas give different results. The only way I can reconcile the two is if $\int_{(0,1)}$ actually means $\int_{1}^{0}$, but I don't see why that would be the case.
Can someone explain why my result disagrees with Spivak's statement?
You aren't contradicting the formulas. The first of Spivak's equations says, $$\int_0^{-\pi}\sin( x)=\int_0^1\sin(-\pi x)(-\pi)$$ Then second says, $$\int_{-\pi}^0\sin(x)=\int_0^1\sin(\pi x)\pi$$
These are obviously equivalent.
If you look at the first expression in each of your formulas, you are trying to show that $$\int_{g(a)}^{g(b)}f=\int_{(g(a),g(b))}f$$ which is not necessarily true, as you yourself have pointed out.