enter image description here

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.

  • $\begingroup$ I think the continuity of $f'$ is sufficient but no way necessary. All that matters is that the function $y\sqrt{1+y'^{2}}$ should be Riemann integrable on $[a, b]$. The analysis of surface area of solids of revolution is similar to that of length of an arc of a curve. $\endgroup$
    – Paramanand Singh
    Jul 5, 2017 at 13:40
  • $\begingroup$ If $y$, as per our definition, is continuous on $[a, b]$, $y'$ is continuous on $(a, b) $. Since $y'$ isn't continuous at the end points, $y\sqrt{1+y'^2}$ is also not continuous at $a$ and $b$. It is not Riemann integrable. It is an improper integral in precise terms, isn't it? A function is only Riemann integrable if it is bounded and the interval in which it is bounded is closed. As I have learnt, the value of the function( when bounded in the open interval ) at the end points don't matter while computing the integral, an improper integral. Am I missing any point here? $\endgroup$
    – R004
    Jul 5, 2017 at 15:16
  • $\begingroup$ In the theory of Riemann integral, the value of a function at any finite number of points does not matter. So endpoints are anyway not relevant. What is relevant is that the function must be bounded and it should not be too much discontinuous. The rigorous meaning of too much discontinuous will be a bit difficult to explain here. Also improper integrals come into picture only when we know for sure that the function is unbounded. Otherwise the usual Riemann integral is sufficient. Your hypotheses do not specifically mention about the function being unbounded. $\endgroup$
    – Paramanand Singh
    Jul 5, 2017 at 16:30
  • $\begingroup$ An example of too much discontinuous is a function that takes value $1$ for all rational numbers and $0$ for all irrational numbers. Since there are infinitely many irrational numbers between rational numbers and vice versa, the function is infinitely discontinuous. I'm that case, the function is bounded but the limits of the Riemann sums do not converge. $\endgroup$
    – R004
    Jul 5, 2017 at 16:52
  • $\begingroup$ Here, they considered the integral of discontinuous functions to be improper. The author of the chosen answer also tells us the same. Could we discuss this? $\endgroup$
    – R004
    Jul 5, 2017 at 16:58

2 Answers 2

  1. 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 $b$.

  2. 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 $$ \int_{(a,b)} f(x)\, dx = \lim_{a' \to a^{+}} \lim_{b' \to b^{-}} \int_{a'}^{b'} f(x)\, dx. $$

  3. 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.)

  4. 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.

  • $\begingroup$ If I understand well, statement 1 follows from our choice of domain, does it not? $\endgroup$
    – R004
    Jul 5, 2017 at 12:54
  • 2
    $\begingroup$ Yes, I see what you mean. Here's a more specific wording of what I meant: Saying $f$ is continuous on $(a, b)$ does not say anything about whether or not $f$ extends continuously to $[a, b]$. Particularly, $f$ might be the restriction of a continuous function on $[a, b]$, though it also might not be. $\endgroup$ Jul 5, 2017 at 13:01
  • $\begingroup$ This clears it. Perfect! $\endgroup$
    – R004
    Jul 5, 2017 at 13:09
  • 1
    $\begingroup$ Your second point seems to use the idea of improper Riemann integrals unnecessarily. If your function is defined and bounded on some open interval $(a, b) $ then it is a trivial matter to extend the function to closed interval $[a, b] $ by defining it in any manner at the end points $a, b$ and then apply definition of Riemann integral for this function on closed interval $[a, b] $. Improper integrals are used only when the interval of integration is unbounded or the function is unbounded. For bounded functions and bounded intervals the usual definition of Riemann integral suffices. $\endgroup$
    – Paramanand Singh
    Jul 6, 2017 at 3:27
  • 1
    $\begingroup$ @ParamanandSingh: One way or another, one has to show "integrability" is sensible over an open interval. Your approach (my point 4) is lower-tech if one is coming from mathematical definitions, but seemed likely to be more cryptic for a student whose background is calculus (hence the use of limits in 2). Agreed, I should not have written "has to" in 2, since limits are not the only way to proceed. $\endgroup$ Jul 6, 2017 at 10:53

There is no question here, but I'll try to answer anyway.

This is not a matter of beliefs; it is a matter of definitions and of proofs. First of all, the concept of Riemann integral is defined only for bounded functions defined on a closed and bounded interval. So, if your function is not bounded, then it is trivially true that it is not Riemann integrable.

Note that when I wrote that “the concept of Riemann integral is defined only for bounded functions defined on a closed and bounded interval” I was by no means implying that all such function are Riemann integrable. I was only saying that if a function doesn't satisfy these conditions, then it is automatic that it is not is Riemann integrable.

Now let's see a concrete example which is simpler than the one of the area of a surface. The length of the graph of a function $f\colon[a,b]\longrightarrow\mathbb R$ is $\int_a^b\sqrt{1+f'(t)^2}\,dt$. Now consider the function$$\begin{array}{rccc}f\colon&[-1,1]&\longrightarrow&\mathbb R\\&x&\mapsto&\sqrt{1-x^2}\end{array}.$$Since its graph is a semicircle with radius $1$, the length of the graph should be $\pi$. Applying the previous formula, we get that the length of the graph is$$\int_{-1}^1\sqrt{1+f'(t)^2}\,dt=\int_{-1}^1\frac{dt}{\sqrt{1-t^2}}.$$But now we have a problem: the function whose integral we are trying to comput isn't even defined at $\pm1$. Furthermore, it is unbounded. So, it is not Riemann integrable.

What we can do is to work with improper integrals: the previous integrals should be interpreted as$$\left(\lim_{x\to1}\int_0^x\frac{dt}{\sqrt{1-t^2}}\right)+\left(\lim_{x\to-1}\int_x^0\frac{dt}{\sqrt{1-t^2}}\right)\text,$$if both limits exist. And they do! They are both equal to $\frac\pi2$ and so we do get that the length of the graph is equal to $\pi$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .