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 $L(1)$):

Answer -
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 ?
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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 -
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.
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 ).