Say we have two continuous functions, $f, g : \mathbb{R} \rightarrow \mathbb{R}$. Moreover, let's say that $f$ and $g$ are "bijective" over two intervals $R_a = [a_1, a_2]$ and $R_b = [b_1, b_2]$, in the sense that for all $t_a \in [a_1, a_2]$ there exists $t_b \in [b_1, b_2]$ such that $f(t_a) = g(t_b)$, and vice versa. In other words, there is an isomorphism, $\phi : R_a \rightarrow R_b$, such that $f(t_a) = g(\phi(t_a)), \forall t_a \in R_a$ and $g(t_b) = f(\phi^{-1}(t_b)), \forall t_b \in R_b$.
Is then the following statement true?
$\int_{a_1}^{a_2}f(t) dt = \int_{b_1}^{b_2}g(t) dt$
Let $f, g:[0,1]\rightarrow\{0, 1\}$ defined by $f(x)=1$ on irrational on $f(x)=0$ on rationals. And $g(x)=0$ on points of the type $\frac{1}{n}$ for $n\in\mathbb{N}$ and $g(x)=1$ otherwise. Then we can easily define $\phi$ using any bijection between rationals and the set of all numbers of the type $\frac{1}{n}$ for $n\in\mathbb{N}$. Note that $g$ has Reimann integral $1$ whereas $f$ is not Reimann integrable.