In the physics chapter we developed an intuitive understanding of integrals since we used this concept in order to calculate the position and the velocity of an object given the knowledge of its acceleration. It is now time to study the techniques of calculus more closely and with more mathematical rigour. Real math is not so much about calculating things, but about seeing patterns and relationships between concepts.

The only prerequisite knowledge for calculus is functions. You see, most problems that you will find on a calculus exam involve some function $f(t)$ and ask you to calculate some property of the function, hence the name calculus. Let us now look at two real-world examples where calculus ideas are used.

Suppose you are downloading a large file to your computer. At $t=0$ you clicked “Save as” in your browser and started the download. Let $f(t)$ represent the size of the downloaded data. The number $f(t)$ is what you would see if, at time $t$, you clicked on the partially-downloaded file and checked how much space it takes on your disk. If you are downloading a $700$[Mb] file, then the download progress bar at time $t$ will correspond to the fraction $\frac{f(t)}{700 \text{[Mb]}}$.

The derivative function $f^\prime(t)$ is a description of how the function $f(t)$ changes over time. In our example $f^\prime(t)$ is the download speed. Indeed, if you are downloading at 100 [kb/s], then the function $f(t)$ must increase by 100[kb] each second. If you maintain this download speed the file size will grow uniformly: $f(0)=0$[kb], $f(1)=100$[kb], $f(2)=200$[kb], $\ldots$, f(100)=10[Mb]. The “estimated time remaining” is calculated by dividing the size of the part remaining to be downloaded by the current download speed: \[ \text{time remaining } = \frac{ 700 - f(t) }{ f^\prime(t) } [s]. \] The bigger the derivative is, the faster the download will finish.

Let us now consider this problem from the point of view of the router which connects you to the Internet. The router knows what download rate $f^\prime(t)$[kb/s] you had at all times during the download because it was sending you the data. What is the download speed for you is the upload speed for the router.

The router does not have access to your computer and therefore does not know the actual file size $f(t)$ on your computer. Nevertheless, the router can infer the information about the size of the file $f(t)$ from the transmission rate $f^\prime(t)$. The sum (integral) of the download rate between $t=0$ and $t=\tau$ corresponds to the total new downloaded data that appeared on your computer. During this period, the change in the file size was \[ \Delta f = f(\tau)-f(0) = \int_0^\tau f'(t)\; dt. \] Assuming that the file size starts from zero $f(0)=0$[kb] at $t=0$, the router can use the integration procedure to find $f(\tau)$, the file size on your computer at $t=\tau$: \[ f(\tau) = \int_0^\tau f^\prime(t)\; dt. \]

This example illustrates two very important ideas. The first is the notion of an integral $\int_a^b\cdot dt$ which is the calculation of the total of a function during the time period from $t=a$ until $t=b$. The second idea illustrated is the inverse relationship between the integral and the derivative operations. If you know the slope of a function $f^\prime(t)$, you can find the value of the function $f(t)$ by integrating.