I've come across a proof of the classic limit definition of $e = \lim_\limits{n \to \infty}(1 + \frac{1}{n})^n$ that starts with letting $t$ be any (real, I'm assuming) number in the interval $[1, 1 + \frac{1}{n}]$. The proof goes:
$$\frac{1}{1 + \frac{1}{n}} \leq \frac{1}{t} \leq 1\Rightarrow \int_1^{1 + \frac{1}{n}}\frac{1}{1 + \frac{1}{n}}dt \leq \int_1^{1 + \frac{1}{n}}\frac{1}{t}dt\leq \int_1^{1 + \frac{1}{n}}1dt.$$
I fail to see how the first inequality implies the second.
Assume that $$\fbox{ \( g(x)\ge f(x)\) }$$ Then, $$g(x)-f(x)\ge0$$
Let's define $h(x):=g(x)-f(x)$ Note that because of the above inequality, $h(x)\ge0$.
Using the definition of a limit, we obtain that the integral of $h(x)$ can be defined as$$\int_a^bh(x)dx = \lim_{n\to \infty}\frac{b-a}{n}\sum_{k=0}^{n-1}h\left(a+\frac{(b-a)k}{n} \right)$$ Note that this is the pure definition of a limit. We're cutting up the interval from $a$ to $b$ (so therefore it will have a size of $b-a$ into $n$ rectangles where the number of rectangles goes to infinity. The $\frac{b-a}{n}$ represents the individual width of each rectangle, and the summation on the right represents the corresponding height of the function $h(x)$ at the $x$ value of $k$.
Note that $\frac{b-a}{n}\sum_{k=0}^{n-1}h\left(a+\frac{(b-a)k}{n} \right)\ge0$ is true for $n$ rectangles and because $h(x)$ is always positive. Therefore, taking the limit as $n$ goes to infinity should still hold true that $\lim_{n\to \infty}\frac{b-a}{n}\sum_{k=0}^{n-1}h\left(a+\frac{(b-a)k}{n} \right)\ge0$.
If $$\lim_{n\to \infty}\frac{b-a}{n}\sum_{k=0}^{n-1}h\left(a+\frac{(b-a)k}{n} \right)\ge0$$ Then $$\int_a^bh(x)dx\ge0$$ $$\int_a^b(g(x)-f(x))dx\ge0$$ $$\fbox{ \( \int_a^b g(x)dx \ge \int_a^bf(x)dx\) }$$ Yippee!
This answers your question because it shows that the inequality's truth is preserved if you integrate both sides with the same bounds. I enjoy this proof (even if it proves something that I take for granted) because it uses the OG definition of an integral.