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!
Best Answer
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.
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.