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 $v(t).$

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 $y = f(x),$ the $x$ -axis, and the vertical lines $x =0$ and $x = b.$

What happens to the area of this region, $A,$ as you change $b \ ?$

Integrals in a Nutshell

So the area bounded by the curve $y=f(x)$ and the lines $x = 0, x = b,$ and $y = 0$ depends on $b.$ In other words, $A$ is a function.

We normally use either " $x$ " or " $t$ " to represent an independent variable in calculus; since " $x$ " is already $f$ 's variable, let's use " $t$ " as $A$ 's. $^{\dagger}$

We call $A(t)$ the integral of $f$ between $x= 0$ and $x = t.$

There's much more to say about integrals as we'll see later, but this definition is a good start.

Let's find $A(t)$ in an example that'll help us understand the connection between $f$ 's integral and the shape of its graph in the next problem:

Use geometry or the plot below to find the integral of $f(x) = 2 x.$

$^\dagger$ Note: For now, $t$ does not represent time.

Integrals in a Nutshell

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.

$A$ 's rate of change between $t=b$ and $t= b +h$ is roughly given by $\frac{\text{change in } A}{\text{change in } t} = \frac{A(b+h) - A(b)}{h} ,$ where $h > 0$ is small. For example, $A(t) = t^2$ in the last problem has $\begin{aligned} \frac{A(b+h) - A(b)}{h} & = \frac{(b+h)^2-b^2}{h} \\ & = \frac{(b^2 + 2h b +h^2) - b^2}{h} \\ & = \frac{2 bh + h^2}{h} \\ & = 2 b + h. \end{aligned}$ Since $h \approx 0,$ $A$ 's rate of change at $t= b$ is about $2b,$ which is $f(b).$

This is true in general: the rate of change of $A$ at $t$ is $f(t).$ In other words, the derivative of the integral of $f$ is $f$ 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 $f(x) = 4 x(1-x),$ what is $A$ 's rate of change at $t = \frac{1}{2} \ ?$

Integrals in a Nutshell

To summarize last problem's result, we take the integral of $f$ to get $A,$ but if we take $A$ 's derivative, we get $f$ back. But there's more!

We saw in the last quiz that taking the derivative of a position function $h$ at $t$ gives us the velocity $v(t).$ To keep things simple, assume $h(0) =0.$

It turns out that the integral of $v(t)$ gives us $h(t).$ 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, $v(t),$ for a rocket is shown below; the area under the graph between $t= 0$ and $t=a$ is shaded.

Ignoring physical units for now, how far does the rocket travel between the launch time $t= 0$ and $t= 9.5 \ ?$

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. $h(0) = 0$ where $h$ is the rocket's position function.

Integrals in a Nutshell

It doesn't move at all. About

4

distance units. About

8

distance units. About

12

distance units. About

24

distance units.

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 $y = 1/x, \, x \geq 1$ over the $x$ -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 $x = 1$ and $x = l+1,$ whose surface area (SA) obeys $\text{SA}(l) > 2\pi \ln(l + 1).$ 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?

Integrals in a Nutshell

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 $x = 1$ and $x = l+1,$ which is given by the formula $V(l)=\pi \left( 1 - \frac{1}{l+1} \right).$

What value does the volume seem to approach as the length increases?

Integrals in a Nutshell

1

e = 2.71828\dots

2 \sqrt{2} = 2.8284\dots

\pi = 3.14159 \dots

\infty

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 $y = f(x) = x(1-x)$ between $x=0$ and $x = 1$ and the $x$ -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, $f(x),$ we chose above.

$n$ is the number of rectangles, and the plot displays their total area.

Use this to estimate the blue area, and therefore the integral of $f$ between 0 and 1.

Integrals in a Nutshell

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.

Calculus in a Nutshell

Integrals in a Nutshell

Integrals in a Nutshell

Integrals in a Nutshell

Integrals in a Nutshell

Integrals in a Nutshell

Integrals in a Nutshell

Integrals in a Nutshell

Integrals in a Nutshell

Integrals in a Nutshell

Integrals in a Nutshell