"Let $f\colon [a,b]\to\mathbb{R}$ be a Riemann integrable function; show that $g\colon [-b,-a]\to\mathbb{R}, g(x):=f(-x)$ is also Riemann integrable and $\int_{[-b,-a]}g=\int_{[a,b]}f$".
Now, this is easily proven if we have the following form of the Change of Variables Theorem in Integrals:
"$\phi:[a,b]\to [\phi(b),\phi(a)]$ differentiable monotone decreasing function such that $\phi '$ is Riemann integrable; $\ f\colon [\phi(b),\phi(a)]\to\mathbb{R}$ Riemann integrable function on $[\phi(b),\phi(a)]$. Then $-(f\circ\phi)\phi '\colon [a,b]\to\mathbb{R}$ is Riemann integrable on $[a,b]$, and $\int_{[a,b]}-(f\circ\phi)\phi '=\int_{[\phi(b),\phi(a)]}f.$"
and using the change of variables $\phi\colon [a,b]\to[-b,-a], \phi(x):=-x$.
What I'd like to know is: is possible to prove the statement if we have the following form of the Change of Variables Theorem?:
"$\phi:[a,b]\to [\phi(a),\phi(b)]$ differentiable monotone increasing function such that $\phi '$ is Riemann integrable; $\ f\colon [\phi(a),\phi(b)]\to\mathbb{R}$ Riemann integrable function on $[\phi(a),\phi(b)]$. Then $(f\circ\phi)\phi '\colon [a,b]\to\mathbb{R}$ is Riemann integrable on $[a,b]$, and $\int_{[a,b]}(f\circ\phi)\phi '=\int_{[\phi(a),\phi(b)]}f.$"
NOTE: $\int_{[a,b]}f $ is the unsigned definite Riemann integral so we can't write $\int_{[b,a]}f=-\int_{[a,b]}f$ (see also http://www.math.ucla.edu/~tao/preprints/forms.pdf)