3
$\begingroup$

This is a question from my test: Find the area enclosed by the graph of $y=x^3-6x^2+11x-6$ and $y=0$.

This is actually very simple but the way I look at it, I can see two different answers to this question, depending on the answer to my title.

The first is where I assume area can be infinite and use a boundary from $-\infty$ to $\infty$. This is what I ended up writing as my answer in the test ($x=1;2;3$; are points of intersection): $$-\lim_{a\to-\infty}\int_{a}^{1}y \ dx + \int_{1}^{2}y \ dx - \int_{2}^{3}y \ dx+\lim_{b\to\infty}\int_{1}^{b}y \ dx = \infty$$

The other one is what my teacher told me the answer to the test question is, which is just: $$\int_{1}^{2}y \ dx - \int_{2}^{3}y \ dx = \frac{1}{2}$$ Now my question is, can the area enclosed actually be infinite? I personally think an infinite area should be possible, which is why we normally use boundaries in integrals to limit the area from becoming infinite.

From Wikipedia, it says something like $\int_{a}^{b}f(x) \ dx = \infty$ means that $f(x)$ doesn't bound a finite area between $a$ and $b$, while $\int_{-\infty}^{\infty}f(x) \ dx = \infty$ means the area under $f(x)$ is infinite.

So what do you think about this? I'd really appreciate your opinions or even facts on this matter. Can an area enclosed be infinite? And thus, would an answer of $\infty$ be a legitimate or false answer to the test question? (I don't plan on complaining for marks, this is just for my own curiosity and self-learning).

Sorry for the long question, and thanks in advance!

$\endgroup$
  • 1
    $\begingroup$ Perhaps the question would have been better stated as "find the area of the region bounded by" the two graphs. A bounded region in the plane is a region which can be enclosed within a circle. $\endgroup$ – John Wayland Bales Nov 28 '16 at 4:25
  • $\begingroup$ @JohnWaylandBales Since you got here first, feel free to expand that into an answer. $\endgroup$ – Mark S. Nov 29 '16 at 1:37
  • $\begingroup$ OK, I have submitted an answer. $\endgroup$ – John Wayland Bales Nov 29 '16 at 2:46
2
$\begingroup$

It is a reasonable question. Although usually if asked to find the area of a region or regions bounded by two graphs what is meant by "bounded" is that the regions all lie within the interior of some circle.

This is analogous to a bounded set on the number line being contained in some interval $[a,b]$. It is completely circumscribed.

However it is possible for to graphs to enclose a finite, yet unbounded region.

There are many examples, but one is as follows.

Find the area of the region "bounded" by the graphs of $y=0$ and $y=\dfrac{x}{x^4+1}$

Here is the graph of the region.

bounded finite region

This region is not bounded in the sense stated above. It cannot be contained in the interior of a circle. Yet it has a finite area.

\begin{equation} \int_{-\infty}^\infty\dfrac{|x|}{x^4+1}\,dx=\int_{0}^\infty\dfrac{2x}{x^4+1}\,dx\\ \end{equation}

Make the substitution $u=x^2$, $du=2x\,dx$ and this becomes

\begin{eqnarray} \int_{0}^\infty\dfrac{1}{u^2+1}\,du&=&\frac{1}{2}\arctan(u){\Large\vert}_{0}^\infty\\ &=&\left(\dfrac{\pi}{2}-0\right)\\ &=&\frac{\pi}{2} \end{eqnarray}

Therefore it is acceptable to say that, in a sense, an unbounded region is "bounded" by two graphs so long as the area enclosed is finite.

$\endgroup$
  • $\begingroup$ How about when the area enclosed is infinite? I think an unbounded region with a finite area would be considered as a convergent integral, so we can calculate the finite area. But what about an unbounded region which is divergent? When asked to calculate the area, would the area be infinite, or should we only consider the finite area regions? $\endgroup$ – Gyakenji Nov 29 '16 at 2:57
  • $\begingroup$ If the area approaches infinity then we can say that the region has infinite area. In your original problem, however, it was a question of what the teacher meant by saying the region "bounded" by the function. I do not know the policy of your teacher, but when I was teaching I was always happy to further explain the meaning of a question if the student thought the question was unclear. $\endgroup$ – John Wayland Bales Nov 29 '16 at 3:02
  • $\begingroup$ Actually I notice that your professor said "enclosed" not bounded. That is actually not a mathematical term so you could certainly ask what he meant by "enclosed." $\endgroup$ – John Wayland Bales Nov 29 '16 at 3:07
  • $\begingroup$ My professor doesn't speak English as a first language so perhaps it was just a mistake, but on the lectures, he is usually asking for the area between two curves (so the area bounded by the two functions?). I will ask him further about it. Sorry if I'm repeating, but assuming he meant the area bounded by the two functions, then would it be mathematically correct to say the area is infinite, given the integral will be divergent and area will approach infinite? Or would the area bounded by the functions be the finite areas only? Thanks so much in advance. $\endgroup$ – Gyakenji Nov 29 '16 at 3:17
  • $\begingroup$ The term "enclosed" is widely used in textbooks when talking about regions bounded by two curves, but I have never seen a textbook give an explicit definition of the term. There is a common mathematical concept of a "closed curve" which is a curve in the plane which begins and ends at the same point. So one could define an enclosed region as a region whose boundary consisted of one or more closed curves. With that definition, even the example I gave in my answer is not enclosed. $\endgroup$ – John Wayland Bales Nov 29 '16 at 3:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.