Let $a<b$ $f:[a,b]\to \mathbb{R}$ be Riemann integrable. Let $g:[-b,-a]\to \mathbb{R}$ be defined by $g(x):=f(-x)$. Show that $g$ is Riemann integrable with $\int_{[-b,-a]}g=\int_{[a,b]} f$
I wanted to use change of variables but the statement I have from the text says $\phi$ must be monotone increasing:
Let $[a,b]$ be a closed interval and let $\phi:[a,b]\to [\phi(a),\phi(b)]$ be a differentiable monotone increasing function such that $\phi'$ is Riemann integrable. Let $f:[\phi(a),\phi(b)]\to \mathbb{R}$ be Riemann integrable. Then $(f\circ \phi) \phi':[a,b]\to \mathbb{R}$ is Riemann integrable on $[a,b]$ and $\int_{[a,b]} (f\circ \phi)\phi'=\int_{[\phi(a),\phi(b)]} f$
So I wanted to just prove that $\underline\int_{[a,b]} f\leq \int_{[-b,-a]}g\leq \overline\int_{[a,b]} f$
I know that given a partition $P$ of $[a,b]$ I can use $\phi(x)=-x$ to construct a partition of $[-b,-a]$ as $Q\{J\in P:\phi(J)\}$.
But I'm not sure exactly what to do from here. I'm assuming there must be some way to use the theorem given that I'm not seeing or I'm supposed to prove that the change of variables still works for monotone decreasing $\phi$.
Best Answer
If for each partition $P$ of $[a,b]$ of size $n$ you can find a partition $Q$ on $[-b,-a]$ of size $m$ so that $$\sum_{i=1}^{n-1} (\sup_{x\in[x_i,x_{i+1}]}f(x))(x_{i+1}-x_i) = \sum_{j=1}^{m-1} (\sup_{x\in[y_j,y_{j+1}]}g(x))(y_{j+1}-y_j) $$ Then $\overline{\int_a^b} f = \overline{\int_{-b}^{-a}}g$. The same argument will show that $\underline{\int_a^b} f = \underline{\int_{-b}^{-a}}g$.