The definition here says that the function is continuously differentiable on $[a, b]$.
Is it possible to compute the integral when the function is continuously differentiable on $(a, b)$?
I believe it is possible.
Let us take a function $y$ that is continuous on $[a, b]$ and differentiable on $(a, b) $.
We are to evaluate the integral, $$S=\int_{a}^{b} 2\pi y\sqrt{1+(y')^2} dx$$
Here, the domain of the integrand is $x \in (a, b) $. This tells us that the integrand function is discontinuous at points $a$ and $b$. Since we are computing the Riemann Integral, continuity at end points will not matter.
Even if in another case the integrand was defined at the end points, the areas of the two end rectangles would become insignificant. It would not bother us if the two rectangles were there or not( in the case of discontinuities ). So, the Riemann Integral would give us the same result as it would if the function were undefined at the end points.
In summary, the Riemann integral of end discontinuous functions( not infinite discontinuous ) can be computed.
Best Answer
The fact the domain of $f$ is $(a, b)$ rather than $[a, b]$ does not "[guarantee] $f$ is discontinuous at $a$ and $b$," since continuity/discontinuity at a point is only defined for points within the domain of a function, and $a$ and $b$ are not in the domain of $f$. So it just doesn't make sense to talk about continuity/discontinuity at $a$ or $b$.
The Riemann integral, defined in terms of partitions of intervals, makes literal sense only on a closed, bounded interval. To handle an open interval, one has to use a limiting process such as $$ \int_{(a,b)} f(x)\, dx = \lim_{a' \to a^{+}} \lim_{b' \to b^{-}} \int_{a'}^{b'} f(x)\, dx. $$
If $f$ is continuous on $(a, b)$, it does not follow that $f$ is bounded, i.e., that there exists a real number $M$ such that $|f(x)| \leq M$ for all $x$ in $(a, b)$, i.e., that the graph of $f$ lies between two horizontal lines $y = \pm M$.
On a closed, bounded interval $[a, b]$, boundedness is automatic from the extreme value theorem.
Boundedness of $f$ on $(a, b)$ needs to be added as a hypothesis for integrability. For example, $f(x) = 1/x$ is continuously differentiable on $(0, 1)$, but is not (even improperly) integrable on $(0, 1)$. (By contrast, $f(x) = 1/\sqrt{x}$ is not Riemann integrable on $(0, 1)$, but is improperly integrable on $(0, 1)$. This shows boundedness is not necessary for improper integrability, though boundedness is sufficient.)
If $f$ is bounded, and is continuous on $(a, b)$, then no matter how $f(a)$ and $f(b)$ are defined (as real numbers), $f$ is integrable on $[a, b]$, and the integral does not depend on the endpoint values.
This isn't entirely obvious, but your argument sketch furnishes the main idea.