What does it mean to say a function is "defined" on/over $[a, b]$, or "continuous" over $[a, b]$, has an "average value" on $[a, b]$, etc.?

Question

I'm currently taking intro calculus, and all the time I see problems or proofs from the teacher with language like the following (not in any particular order):

1. Let $f$ be defined on $[a, b]$.

2. $f$ is continuous on the interval $[a, b]$.

3. Find the expected/average value of $f$ over the interval $[a, b]$.

4. Find the average rate of change of $f$ over $[a, b]$.

From what I gather, we are always supposed to assume that $[a, b]$ (whatever $a$ and $b$ are) is an interval of the function's domain. Am I correct in that assumption? And if so, why does this principle always seem to go unstated? Why, when we talk about a function over an interval, are we always talking about "over an interval of the domain ($x$-values)," such that $[a, b]$ is (apparently) the same as $a\leq x\leq b$?

Answer 1

1. The domain of $f$ is $[a, b]$. So it is the same as $f:[a, b] \to \mathbb{R}$.
   
   

2. $f:[a, b] \to \mathbb{R}$ is continuous at every point $x \in [a, b]$.
   
   

3. $\displaystyle \frac{1}{b-a} \int_a^b f(x) \, dx$. I assume you are familiar with integrals and "Mean Value Theorem for Integrals". This is related to average value of $f$ at node points $x_0, x_1, \dots , x_n \in [a, b]$ which is $\displaystyle \frac{\sum_{k=0}^n f(x_k)}{n} $. If you choose those points equally distrubuted, that is $x_k - x_{k-1} = \dfrac{b-a}{n}$ then you can easily see $\displaystyle \frac{\sum_{k=0}^n f(x_k)}{n} = \sum_{k=0}^n \frac{f(x_k)}{n} = \sum_{k=0}^n \frac{f(x_k)}{\frac{b-a}{x_k - x_{k-1}}} = \frac{1}{b-a} \sum_{k=0}^n f(x_k) (x_k - x_{k-1}) \to \frac{1}{b-a} \int_a^b f(x) \, dx$.
   
   

4. $\displaystyle \frac{f(b) - f(a)}{b - a} = \frac{\text{Change of $f(x)$ as $x$ changes from $a$ to $b$}}{\text{Change of $x$ from $a$ to $b$} } $. You might guess this is related to derivatives and "Mean Value Theorem for Derivatives" since $\displaystyle \frac{f(b) - f(a)}{b - a}$ is the slope of the line passing $(a,f(a))$ and $(b,f(b))$.

As @David stated in the comment, all of the above can go wrong if $[a, b] \subseteq \text{Domain of } f$ is not satisfied. For example, you cannot talk about continuity of $f$ at a point which is not in the domain. Thus, you are correct in "your assumption" and "that principle will always go unstated" as it is a "convention". This is because you are (or the problem or the proof is) concerned about "some properties of $f$" just in the inverval $[a, b]$.