This quiz introduces integrals, which "undo" derivatives (and vice versa) much the same way logarithms "undo" exponentials.
Integrals form a major pillar of calculus, and have many practical applications such as computing the area bounded by a curve, and finding the distance traveled by an object directly from its velocity
We'll unpack, prove, and apply these ideas fully in the integrals chapter of this course. This quiz introducing integrals just scratches the surface.
So what is an integral? That's a deep question that'll take time to unpack, but here's a way to get us started on the path to an answer:
The graph below shows the region bounded by the graph of the -axis, and the vertical lines and
What happens to the area of this region, as you change
So the area bounded by the curve and the lines and depends on In other words, is a function.
We normally use either "" or "" to represent an independent variable in calculus; since "" is already 's variable, let's use "" as 's.
We call the integral of between and
There's much more to say about integrals as we'll see later, but this definition is a good start.
Let's find in an example that'll help us understand the connection between 's integral and the shape of its graph in the next problem:
Use geometry or the plot below to find the integral of
Note: For now, does not represent time.
Now, we want to see how the derivative "undoes" the integral, like the intro page promised. We'll use a little bit of algebra to do this.
's rate of change between and is roughly given by where is small. For example, in the last problem has Since 's rate of change at is about which is
This is true in general: the rate of change of at is In other words, the derivative of the integral of is itself.
We'll show why this is true using some geometry later in the course. For now, let's use this result to answer the following question:
If what is 's rate of change at
To summarize last problem's result, we take the integral of to get but if we take 's derivative, we get back. But there's more!
We saw in the last quiz that taking the derivative of a position function at gives us the velocity To keep things simple, assume
It turns out that the integral of gives us So, in summary,
integrals "undo" derivatives, and derivatives "undo" integrals.
Since this is an intro quiz, we're sweeping a lot under the rug. But this Fundamental Theorem of Calculus sits at the heart of calculus, so you can be sure we'll do a much deeper dive later in the course.
Like derivatives, integrals are used in kinematics, the study of motion. Let's take a quick look at an example we'll cover in greater depth later.
A sample velocity graph, for a rocket is shown below; the area under the graph between and is shaded.
Ignoring physical units for now, how far does the rocket travel between the launch time and
Hint: What does the Fundamental Theorem of Calculus mentioned on the last page tell you? You can assume that the rocket launches from the origin; i.e. where is the rocket's position function.
Finding a derivative is the same as finding a tangent line, which is a geometric problem. Integrals also solve geometric problems!
A particularly beautiful example involves Gabriel's Horn. We make the horn by revolving the graph of over the -axis and into the third dimension, like the 3D touch-interactive graph below shows.
The horn itself extends forever, so the interactive plot below shows just the piece between and whose surface area (SA) obeys We'll show where this result comes from after we develop some skill with integrals.
Estimate the total surface area of the full horn using the plot above. Does this match your intuition?
Now, suppose we want to compute the volume of Gabriel's Horn instead of its surface area. This is something we'll do at the end of the chapter on integrals.
Like in the last problem, the plot below shows the volume (V) of the part of the horn between and which is given by the formula
What value does the volume seem to approach as the length increases?
We defined integrals in terms of areas, but we haven't said how these areas are actually found. To wrap things up, let's address this gap.
Think of the area bounded by the graph between and and the -axis. This is shaded blue below.
We can approximate this shape with some rectangles, much the same way that pixels can be used to represent an image on a screen.
Adding up the rectangular areas gives us an estimate for the true area.
Many thin rectangles give a better estimate than a handful of broad ones. The plot below shows how this works for the function, we chose above.
is the number of rectangles, and the plot displays their total area.
Use this to estimate the blue area, and therefore the integral of between 0 and 1.
To summarize the last problem, we can always estimate an integral with (finite) sums of rectangular areas if we choose the heights just right.
The more rectangles we use, the thinner they must be, which means a finer resolution. It follows that if we could somehow use an "infinite" number of rectangles we would know the integral precisely.
But how do we make sense of infinite sums? Like derivatives and integrals, infinite sums are a central pillar of calculus, so we have a whole chapter set aside just for them.
The final quiz in this intro chapter, however, introduces us to infinite sums through the Tower of Lire puzzle.