I'm trying to prove the given equation, and have only managed to derive the following so far:
$$ \int_{4}^8 f(x)dx = \int_{4 - 2}^{8 - 2} f(x + 2)dx = \int_{2}^{6} f(x + 2)dx. $$
However, I am stuck at this point and do not know how to proceed on with the question.
First, we have that $\int_{6}^{8}f(x) \ dx = \int_{6}^{8}f(x-2) \ dx$ and with the sub $x-2 \to x$, we get that this is equal to $\int_{4}^{6}f(x) \ dx$. Then $$\int_{4}^{8}f(x) \ dx = \int_{4}^{6}f(x) \ dx + \int_{6}^{8}f(x) \ dx = \int_{4}^{6}f(x) \ dx + \int_{4}^{6}f(x) \ dx = 2\int_{4}^{6}f(x) \ dx$$
It suffices to show that $\int_4^6f(x)dx=\int_6^8f(x)dx$. We prove a more general statement:
Let $t>0$ be a fixed real number, $f: \mathbb{R}\to\mathbb{R}$ be a function subject to $f(x+t)=f(x)$ $\forall x\in\mathbb{R}$. If $f$ is Riemann-integrable on $[0, t]$ then $\forall a\in\mathbb{R}$, $f$ is integrable on $[a, a+t]$ and $$\int_0^t f(x)dx=\int_a^{a+t}f(x)dx$$ (and thus, $\forall a,b\in\mathbb{R}$, $\int_a^{a+t}f(x)dx=\int_b^{b+t}f(x)dx$, since they are both equal to $\int_0^t f(x)dx$).
A way to prove it is the change of variables. But we use a more inspiring method: directly applying the definition. Intuitively speaking,we prove that translation shall not change the area of a set. proof: Assume $k\in\mathbb{Z}$, $a\le kt\le a+t$. We only prove the case s.t. $a<kt< a+t$ strictly. (If one of the "$=$" holds, the proof is quite similar.) We prove the equality $\int_0^t f(x)dx=\int_a^{a+t}f(x)dx$ by establishing (1) $\int_0^{a+t-kt}f(x)dx=\int_{kt}^{a+t}f(x)dx$ and (2) $\int_{a+t-kt}^tf(x)dx=\int_a^{kt}f(x)dx$. We shall only prove (1) since (2) can be done with a same argument.
To prove (1), let $$kt=x_0<x_1<\cdots<x_n=a+t$$ be a partition of $[kt, a+t]$. Then $$0=x_0<x_1-kt<\cdots<x_n-kt=a+t-kt$$ is a partition of $[0, a+t-kt]$. We observe that the two partitions has the same length of their small intervals with maximal length (we abbreviate it simply as "length" from now on), since the latter is a translation of the former.
Now select $\eta_1, \eta_2,\cdots, \eta_n$ s.t. $\eta_i\in [x_{i-1}, x_i]$ and construct a Riemann sum of $\int_{kt}^{a+t}f(x)dx$: $$\sum_{i=1}^n f(\eta_i)(x_i-x_{i-1})$$ (This need not $\int_{kt}^{a+t}f(x)dx$ to exist, but we can always construct the Riemann sum.)
Since $\eta_i-kt\in [x_{i-1}-kt, x_i-kt]$, there is a corresponding Riemann sum of $\int_0^{a+t-kt}f(x)dx$: $$\sum_{i=1}^n f(\eta_i-kt)((x_i-kt)-(x_{i-1}-kt))$$
Since $f(\eta_i-kt)=f(\eta_i)$ and $(x_i-kt)-(x_{i-1}-kt)=x_i-x_{i-1}$, we conclude the two Riemann sums are equal: $$\sum_{i=1}^n f(\eta_i)(x_i-x_{i-1})=\sum_{i=1}^n f(\eta_i-kt)((x_i-kt)-(x_{i-1}-kt))$$.
Recall the two partitions has the same length, so we conclude:
For any Riemann sum of $\int_{kt}^{a+t}f(x)dx$, exists a Riemann sum of $\int_0^{a+t-kt}f(x)dx$ with the same length of its underlying partition.
As the last step, for any $\epsilon>0$, $\exists\delta>0$ s.t. for any Riemann sum $S$ of $\int_0^{a+t-kt}f(x)dx$ with length less than $\delta$, $|\int_0^{a+t-kt}f(x)dx-S|<\epsilon$. For any Riemann sum $S'$ of $\int_{kt}^{a+t}f(x)dx$ with length less than $\delta$, $S'=S$ where $S$ is a Riemann sum of $\int_0^{a+t-kt}f(x)dx$ with length less than $\delta$. $\implies |S'-\int_0^{a+t-kt}f(x)dx|<\epsilon$. By the definition of Riemann integral we conclude $f$ is integrable on $[kt, a+t]$ and $$\int_0^{a+t-kt} f(x)dx=\int_{kt}^{a+t}f(x)dx$$.
