Which units describe the area under a curve? A car traveling in a straight line from its starting position at a speed of $65 $ miles per hour for $3$ hours is represented by the function $v(t)$ where $t$ is the number of hours traveled. What units describe the area under the curve?

*

*Miles per hour

*Miles per hour per hour

*Hours

*Miles

I just don't know what to do at all. I completely do not understand the problem. You don't even have to give me an answer I just want an explanation and I can't find any similar problems on google. My guess is that the answer is miles per hour per hour because each square would be a mile per hour and then all together...nevermind I don't understand. Please help.
 A: Think of drawing a rectangle under the curve, where one of the top corners touches the graph of $y=v(t)$.  Then the height of the rectangle has the same units as $v(t)$, i.e. miles per hour, and the width of the rectangle has the same units as $t$, i.e. hours.  So, to calculate the area of the rectangle, your units would match what you get when you multiply height by width, that is $$\frac{\text{miles}}{\text{hour}} \times \text{hours} = \text{miles}.$$  Since the rectangle will be part of the total area under the curve, the units will be the same as the total area.  See the image below:

A: In physics, every number is associated with a dimension, and the space of all possible dimensions is the (multiplicative) free abelian group $(U, \times)$ ( https://en.wikipedia.org/wiki/Free_abelian_group ) over the set $S = \{m, s, kg, K, A, cd, mol\}$ ( https://en.wikipedia.org/wiki/SI_base_unit ). Elements of $U$ are of the form $m^a s^b kg^c K^d A^e cd^f mol^g$; with $a, b, c, d, e, f, g \in \Bbb Z$
You can only add two numbers when they share the same dimensions. You can multiply any two numbers, and their product will have as dimension the product of the dimensions of the operands, as per the free group's multiplication operator (which behaves as you'd expect it to: $m \times m = m^2$, etc).
This dimensional behavior applies to all factors, including things like $dx$ in an integral.
So in your case, you have a function $t \to v(t)$, where $t$ is in seconds, and $v(t)$ is in meters per second.
The area under the curve is calculated with $A = \int v(t)dt$, with $dt$ an infinitesimal increment of the input, ie, a value in seconds. An integral is a sum, so does not change the dimension.
Consequently, $[A] = [speed][time] = m s^{-1} s = m$: the area under the curve is in meters.
