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I am high school student with an interest in mathematics. My school allowed me to self-study Pre-calculus over the summer so that I can move into AP Calculus AB next school year. I'm using a course on edX called Discovery Precalculus and things are ok for the most part, but I've recently hit a rough patch. I've come across a question that's asking me to find the algebraic measure of the space between a curve and the x-axis. From what I know, this would imply using integrals, but how does this make sense if this is meant to be a precalculus course? I've never worked with integrals or derivatives before.

The Question

I am given the function $f(t) = \frac{1}{t}$. It says our activities are restricted to the interval $t\in[1,3]$, which I'm guessing, in the language of calculus, would be the limits. The next things it tells me to do is plot the function on a coordinate plane with a scale of 0.1. The $f(t)$ axis goes from $0$ to $1.1$ and the $t$ axis ranges from $0$ to $3.1$. Now, from what I know so far about functions, $f(t)$ should be representing the y-axis and $t$ should be representing the x-axis. Finally, I'm told to plot the function along the domain interval $[1,3]$. So after I've done all of this, it states: The function that represents the accumulated area under $f(t)$ on the interval $[1,x]$ where $x\in[1,3]$ will be called $L(x) $. What is the value of $L(1)$?

The issue

I'm pretty sure it's asking me to find the area between the point $[1,1]$ and $[1,3]$, but how do I do this? It can't possibly be asking me to use integrals when the concept has never been brought up within in the scope of the course. I tried to sort it out and the farthest I got was $\int_1^3$, which I'm not even sure is right. Is there some sort of intuitive thing that I'm missing?

edit - Here is the entire problem as presented in the course (This just includes the first part, where it asks for $L(1)$): enter image description here

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  • $\begingroup$ Perhaps the instructions mean for you to plot points with $x$-values incrementing by $0.1$. This would let you make fairly narrow rectangles under or over the curve, to get a lower and upper bound for the area. In other words, use rectangles of width $\Delta x=0.1$. $\endgroup$ – MPW Jul 3 '18 at 13:04
  • $\begingroup$ mathwords.com/t/trapezoid_rule.htm $\endgroup$ – Phil H Jul 3 '18 at 13:07
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    $\begingroup$ It's asking for $L(1)$, not $L(3)$? That would make things much easier, but to the point of silly. $\endgroup$ – aschepler Jul 3 '18 at 13:14
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    $\begingroup$ I think @aschepler is onto it, and I don't think it's silly either. You need calculus to compute $L(x)$ for $x>1$, but you can discover lots of information about it from the graph of $f$. That's a pretty good skill to have, especially in a precalculus course. $\endgroup$ – Matthew Leingang Jul 3 '18 at 13:19
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    $\begingroup$ My advice is to forget about integrals. There is no mention of integrals in the question. Assume you know nothing of integrals and literally answer the questions as they come. For example, L(1) = 0 because the area under f(x) from x = 1 to x = 1 is 0. Since you are being asked to carefully graph with a lot of detail, the later questions are probably asking about approximate areas. You are not going to get the exact value of L(2) (which, by the way, is the natural log of 2). $\endgroup$ – Leonard Blackburn Jul 3 '18 at 14:40
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Answer -

$ L(1) = 0 $


Explanation -

What does the question mean ?

This is an excellent question, which gives you hints about several important results in mathematics, including a very important property of definite integrals. In a nutshell, it is asking you this -

We are defining a function $ f(t) = {1 \over t } $ on the interval; $ [1,3] \subset \mathbb{R} $. If we define $L(x) = \int_{1}^{x} f(t) \ dt $, find $ L(1) $.

This basically means -

We agree that $ f(t) $ represents the reciprocal, $ {1 \over t} $ of the real number $ t $. $ t $ can be between 1 and 3, both inclusive. Now we also agree that $ L(x) $ represents the area of the graph of $ { 1 \over t } $ from $ t = 1 $ to $ t= x $. Then find $ L(1) $ ?

This simplifies to -

Find the area under the graph of $ { 1 \over t } $, from $ t = 1 $ to $ t = 1 $.

So, how do we get the answer from this ?

Now, this should be pretty obvious after the rewording. Basically, you are asked to find the area under the graph for the point $ t= 1 $.Now, the area under the point is just the area of the line segment under the point $ (1 , 1) $ till the $x$-axis, i.e. the area of the line segment joining $(1, 1)$ and $(1 , 0)$. And guess what ? that area is $0$. LOL :).













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Great, but you said something about important results in math ...

Well, to start of with, saying that the area under the point is $0$, is equivalent to saying (well, not absolutely equivalent, but thats another story)-

$ \int_{a}^{a}f(x)\ dx = 0 $

Now, that is quite obvious and simple, but is quite important just the same.

Also, it talks about drawing the graph in (pretty narrow), intervals of $ 0.1 units $ ($=1mm$, i guess), which can mean two things -

  1. Suggesting the geometrical interpretation of definite integration (See, the animation in the link). This is typically the first intutive notion of definite integration in calculus courses.

  2. Suggesting an intuitive notion of approximation ( numerical integration, see this and this ) techniques, such as the trapezoidal rule.

To me the first seems more likely than the second possibility (or maybe its a bit of both ).

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  • $\begingroup$ The definition of $L$ should be $L(x) = \int_1^x f(t)dt$. You mixed the integrating variable and the variable of $L$. $\endgroup$ – red_trumpet Aug 25 '18 at 10:15
  • $\begingroup$ I see, i will edit for clarity's sake. (It doesn't of course matter what the variable of integration is) ;-) $\endgroup$ – Devashish Kaushik Aug 25 '18 at 10:18
  • $\begingroup$ @red_trumpet See edited answer. $\endgroup$ – Devashish Kaushik Aug 25 '18 at 10:21
  • $\begingroup$ @DevashishKaushik Thank for taking the time to add such a detailed explanation! Since asking this, I was able to get my hands on an actual calculus textbook and have come many more problems like this. I understand now that it teaching about finding an approximation for the area under the curve in an attempt to provide an intuition for using integrals as well as provide an application for the use of $e $. $\endgroup$ – CaptainAmerica16 Aug 26 '18 at 11:12
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This comment under the question hit the nail mostly on the head, I think:

My advice is to forget about integrals. There is no mention of integrals in the question. Assume you know nothing of integrals and literally answer the questions as they come. For example, L(1) = 0 because the area under f(x) from x = 1 to x = 1 is 0. Since you are being asked to carefully graph with a lot of detail, the later questions are probably asking about approximate areas. You are not going to get the exact value of L(2) (which, by the way, is the natural log of 2)
- Leonard Blackburn Jul 3 at 14:40

Of course it has something to do with integrals. In particular it looks like the course authors are trying to introduce students to the kind of question that will later receive an answer in form of integrals. Perhaps the idea is that by having seen such an approximation done before they see limits and derivatives, they will later have an easier time keeping the big picture in mind instead of being bogged down in meaningless symbolic manipulations when they actually learn integrals.

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