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For context, this is the first math class in Bachelor's electrical engineering.

Our teacher has just given us a definition for the area under a curve as the following:

If $f(x)$ is an integrable and non-negative function on a closed interval $[a, b]$, then the area under the curve $y = f(x)$ in that range is the integral of $f(x)$ evaluated from $a$ to $b$.

I asked him why the non-negative condition is given, and we're a little stuck trying to find an answer.

In a real-life context, I understand that areas are negative. But when applied to stuff like physics, we'll have to account for the sign as it often denotes ideas like direction. Moreover, imo pure mathematics should allow for negative areas.

Does anyone have any answers?

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    $\begingroup$ Areas are not negative, and likewise, integrals are not areas. $\endgroup$
    – K.defaoite
    Oct 3 at 11:43

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Areas are not ever negative, no matter how pure the mathematics.

The statement in the question defines what the area under a curve is for the case of a non-negative and integrable function. It doesn't say anything about what the area under a curve might be defined to be if the function is negative or not integrable.

The point here is that the area under the curve is the area of the geometric figure defined by the $x$ axis and two parallel vertical lines $x=a$ and $x=b$ and the function value. If you want to define "area under the curve" for negative functions, you'll have to think of another definition for that, presumably the area between the function and the $x$ axis. But it won't ever be negative.

Regarding negative areas, let's draw an analogy with a map. Suppose you draw a north-south line through Hamburg, and then you say "areas west of this line are negative", so areas in Berlin are negative, and areas in Paris are positive. Then you can take the area of the Brandenburg gate and add it to the area of the Arc De Triomphe and get a region with zero area. That is clearly ridiculous, but it's no less ridiculous to claim that the region between the x axis and a negative function has a "negative area". The area of a bounded region is the amount of two-dimensional space it occupies, and that cannot ever become negative.

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  • $\begingroup$ Couldn't you take the area of a graph under the x-axis, which is always positive, but then put a negative sign in front of it to indicate the orientation of area? So then you have "negative" area, wherein the negative sign indicates the orientation of the area in the plane? $\endgroup$ Oct 3 at 13:51
  • $\begingroup$ Area is not ever negative. Suppose on a map that you draw a line and then you say "areas right of this line are negative", so areas in Berlin are negative and areas in Paris are positive. That is clearly ridiculous, but it's no less ridiculous to claim that the area between the x axis and a negative function has a "negative area". $\endgroup$ Oct 4 at 0:37
  • $\begingroup$ I understand that the area itself is never negative, but you can put a negative sign in front of area in order to denote orientation. Isn't that what "signed area" is all about? The same thing is done with distance. Am I missing something obvious? $\endgroup$ Oct 4 at 8:43
  • $\begingroup$ If you understand that the area itself is never negative could you please stop commenting under this answer? $\endgroup$ Oct 4 at 9:15
  • $\begingroup$ Sure. Just making sure I'm reasoning through it correctly. Thanks anyway. $\endgroup$ Oct 4 at 13:21
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Usually, a geometrical area is always between two curves or more. Let us consider the example of two curves $y=f(x)$ and $y=g(x)$. Further, if $f(x)> g(x)$ or $f(x) < g(x)$ in $(a,c)$; at $x=a,c$ the two may be equal. Then the geometrical area is $$A=\int_{a}^{c} |f(x)-g(x)|~ dx.$$

But if these two curves cross each other at a single point $x=b$, then the geometrical area is $$A=\int_{a}^{b} |f(x)-g(x)|~dx+\int_{b}^{c} |f(x)-g(x)|~dx,$$ without even knowing which one is more or less in $(a,b)$ or $(b,c)$. In case there are more crossing points $b_1,b_2,b_3,...b_n$, the integral will be broken into $n+1$ integrals.

Most often it is asked to find area area projected on the $x$-axis by a curve $y=(x)$ from $x=a$ to $x=c$. and the curve does not cross $x$-axis in the interval $(a,c).$ Then the geometrical area is $$A=\int_{a}^{c} |f(x)-0|~dx.$$

If the curve crosses x-axis at $x=b$, the the geometrical area will be $$A=\int_{a}^{b} |f(x)| dx+\int_{b}^{c} |f(x)| dx$$

There could also be a concept of vector area which could be positive or negative. So the integrals $\int_{a}^{c}f(x) dx$ and $\int_{a}^{c} (f(x)-g(x))~dx$ represent the vector area, which may not be the same as the geometrical area. For instance $$\int_{0}^{2\pi} \sin x dx=0$$ is the vector but the geometrical area made by $\sin x$ on $x-$axis from $x=0$ to $x=2\pi$ is $$\int_{0}^{2\pi} |\sin x|~dx=2 ~\text{sq. units}$$

I hope that this discussion may be helpful.

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  • $\begingroup$ If you include the absolute value then there is no mathematical need to split the integration. The splitting is only needed when you turn around and remove the absolute values. $\endgroup$
    – Ian
    Oct 3 at 10:47
  • $\begingroup$ Yes, but this is harmless and more cautious $\endgroup$
    – Z Ahmed
    Oct 3 at 11:10
  • $\begingroup$ I'm not so sure it's pedagogically harmless; you're drawing a contrast between the first indented expression and the second one even though they're exactly the same. $\endgroup$
    – Ian
    Oct 3 at 11:11
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This is a purely pedagogical gambit. The definition is phrased like that precisely to stop people like you asking awkward questions! Later, presumably, your course will introduce the idea that the area under the x-axis must be counted as negative. But for now, they are keeping it as simple as possible for people who may be unfamiliar with mathematical terminology.

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  • $\begingroup$ I find this response a bit rude. Not only does it not answer the OP's question, but they are just asking why a non-negative condition must be given in the definition. I don't find his question awkward considering he's taking a first-year math class for electrical engineering. $\endgroup$ Oct 4 at 4:22
  • $\begingroup$ @Accelerator: You have completely misread the tone of my answer. I am giving the OP credit, not blame, for asking an awkward question. $\endgroup$
    – TonyK
    Oct 4 at 8:49
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When the function goes negative, you can still define the area under the curve -- but in this case the "area" evaluates as negative.

The teacher is simplifying the scenario so as not to confuse the students.

Finding "the area under the curve" is a basic application of the technique of integral calculus, and this is the usual way of introducing the concept to students.

However, there is a danger that unless this is followed up with a proper exposition of what calculus actually is, then students may believe that "the area under the curve" is all there ever is to integral calculus.

What is of deeper concern is that the lecturer was unable to explain the above.

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  • $\begingroup$ Can the person who voted this down explain what I did wrong in the comments? $\endgroup$ Oct 3 at 18:39

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