In this post, we briefly review some precalculus and calculus that will be important for the course.

Introduction

Hi, I’m your TA. My name is James Marshall Reber. I have two last names, no hyphen. You can just call me James.

The idea behind this blog series is to collect things I’ve talked about in recitation in a nice format that you can reference. I’ll try writing these posts ahead of time and then coming back to update them with anything we talked about. This should help you guys with taking notes/studying for the exam. This is something new I’m trying, so if you like or hate this let me know.

Resources for the class can be found here or on Carmen. We will be following the Ximera textbook. Note that you’ll need to open Ximera through Carmen if you want to get credit for an assignment.

While it’s expected that you have taken Calculus before, I understand math might not be a priority in your life. If at any time you feel overwhelmed or completely lost, please let me know. Even if you don’t have a direct question, it helps me know where you’re at and lets me try to help you build a plan to get caught back up.

Today we will be covering review material. This is not indicative of what future material will be like. Please let me know if you are struggling with anything.

With this out of the way, let’s dive into some review. Note that we will not get through all of the review (and even this will not be comprehensive), but this hopefully gets you back into the mindset of doing math. All figures come from Ximera and are not my own.

Precalculus

This is a brief introduction to some of the key concepts from precalculus. This is not exhaustive, meaning this doesn’t cover everything needed. The goal of this is to jog your memory.

Functions

The most important thing here is the concept of a function. A function is an assignment of elements (or objects) in such a way that one object doesn’t get assigned to two things. We will (generally) not work with this level of abstraction in this class, and functions for us will be assignments of numbers to numbers (whatever numbers are). The collection of elements that we can plug into the function is called the domain, and the collection of objects that the function achieves (or gets mapped onto) is called the range. We write \(f : D \rightarrow R\) to indicate that \(f\) is a function which takes objects from \(D\) as inputs and outputs things in \(R\).

Example: An easy example is the function \(f : \mathbb{R} \rightarrow \mathbb{R}\) given by \(f(x) = x^2\). This takes a number \(x\) and outputs that number squared. Notice that

• the range of the function is \([0,\infty)\); we never get a negative number as the output,
• the domain of the function is all real numbers.

(Here, \(\mathbb{R}\) is a fancy way of writing real numbers. The definition of a real number is hard to get into, but you can look here if you want to see more.)

If you have the graph of a relationship, you can use something called the vertical line test to see whether or not it defines a function.

Vertical line test: Given a graph of a relationship of the real numbers, if for every vertical line \(x = a\) we have that it intersects the graph at most once, then the relationship defines a function.

Example: Consider the relationship \(x^2 + y^2 = 1\). If you plot this, you will get a circle. Taking the vertical line \(x=0\), you get something that crosses the graph twice, so this does not define a function. You can also see this algebraically. Notice that if you try to solve for \(y\), you get

\[y^2 = 1 - x^2 \implies y = \pm \sqrt{1 - x^2}.\]

There are two options for what \(y\) could be, so you cannot define a function \(f(x) = y\).

Given two functions \(f : D \rightarrow R\) and \(g : R \rightarrow H\), we can define their composition by

\[g \circ f(x) := g(f(x)).\]

This says to take \(x\) in \(D\) and apply \(f\), then take that output and plug it into \(g\). Notice this gives us a new function \(g \circ f : D \rightarrow H\).

A function is one-to-one if for any two inputs \(x\) and \(y\) with \(x \neq y\) we have \(f(x) \neq f(y)\). Another way of saying this is if \(f(x) = f(y)\), then we must have \(f(x) = f(y)\).

Example: Consider the functions \(f,g : \mathbb{R} \rightarrow \mathbb{R}\) given by \(f(x) = x^2\) and \(g(x) = x\).

• The function \(f\) is not one-to-one, because

\[f(-1) = (-1)^2 = 1 = 1^2 = f(1).\]

• The function \(g\) is one-to-one, because

\[g(x) = x = y = g(y),\]

so \(g(x) = g(y)\) implies \(x = y\).

Let \(f : D \rightarrow R\) be a function. A function \(g : R \rightarrow D\) is said to be the inverse of \(f\) if \(g \circ f(x) = x = g \circ f(x)\). We denote an inverse function by \(f^{-1}(x)\) – pay attention though, this is not the same thing as \((f(x))^{-1}\). An important property of one-to-one functions is the following:

Fact: If a function is one-to-one, then we can define an inverse function on its range.

Try to think about why this is true.

Example: Consider the function \(f(x) = 2x + 1\) on the real numbers.

• If \(f(x) = f(y)\), then \(2x + 1 = 2y + 1\), and we can use algebra to see that \(x = y\). The function is one-to-one.
• Let \(y\) be any real number. We claim there is a real number \(x\) so that \(f(x) = y\); in other words, the range of \(f(x)\) is all real numbers. To find the number \(x\), we have to solve

\[f(x) = 2x+1 = y.\]

Rearranging, we find

\[x = \frac{y-1}{2}.\]

So \(f \left( \frac{y-1}{2}\right) = y\) (you should test this yourself).

• Since the function is one-to-one, we can use the above fact to get that there is an inverse function. Notice in the previous bulletpoint we have constructed the inverse function. It will be

\[f^{-1}(x) = \frac{x-1}{2}.\]

Example: Consider the function \(f(x) = x^2\). As we pointed out earlier, this function is not one-to-one on \(\mathbb{R}\) since \(f(-1) = f(1)\). However, it is one-to-one on \([0,\infty)\). If we restrict the domain, so we consider

\[f : [0,\infty) \rightarrow \mathbb{R}, \ \ f(x) = x^2,\]

then we can define an inverse function. This will be the square root, which we write by

\[f^{-1}(x) = x^{1/2} = \sqrt{x}.\]

If you have the graph of a function, you can use the horizontal line test to check whether a function is one-to-one or not. In words, if you can draw a horizontal line which intersects (or crosses) the graph of the function more than once, then the function cannot be one-to-one. This makes it even easier to see why \(f(x) = x^2\) is not one-to-one; if you draw the horizontal line \(y=1\), you see that it intersects the graph twice. If you restrict \(f(x) = x^2\) to the domain \([0,\infty)\), then you can see that every horizontal line intersects the graph at most one time, so the function is one-to-one. This is not a very rigorous way of checking things, but it is useful if you need a quick answer.

There are many more important things one should talk about when it comes to functions, but we’ll leave it here for now.

Polynomials and Rational Functions

Our favorite example of a function is a polynomial. A function \(f : \mathbb{R} \rightarrow \mathbb{R}\) is a polynomial if we can find real numbers \(a_0, a_1, \ldots, a_n\) (with \(a_n \neq 0\)) so that for each real number \(x\) we have

\[f(x) = a_0 + a_1 x + \cdots + a_n x^n = \sum_{i=0}^n a_i x^i.\]

(Here, \(\Sigma\) is the Greek letter “sigma.” If you don’t remember what this notation is, we’ll come back to it later.)

If we can write \(f\) in the above form, then we say that \(f(x)\) has degree \(n\).

We say that a function \(f : \mathbb{R} \rightarrow \mathbb{R}\) is a rational function if there are two polynomials \(p(x)\) and \(q(x)\) so that \(f(x) = p(x)/q(x)\). Note that this is a little bit of a lie – the way we wrote it implies that \(f(x)\) has domain \(\mathbb{R}\), but it is not defined when \(q(x) = 0\).

Trigonometric Functions

Consider a right triangle as follows.

One way of viewing trigonometric functions is as a way of measuring the relationship between an angle and the ratio of a triangle’s sides. The two important ones are

\[\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \text{ and } \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}.\]

There is a third one that we can form from these two:

\[\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\text{opposite}}{\text{adjacent}}.\]

Finally, there are three more we can consider by “flipping” all of the functions:

\[\sec(\theta) = \frac{1}{\cos(\theta)}, \ \ \csc(\theta) = \frac{1}{\sin(\theta)}, \ \ \cot(\theta) = \frac{1}{\tan(\theta)}.\]

If we take a unit circle and draw a right triangle on the inside, we can relate the lengths of the sides of this triangle to the trigonometric functions above. See the following picture.

This is a helpful interpretation for deriving some trigonometric identities. For example, using this picture and the Pythagorean theorem, we have

\[\sin(\theta)^2 + \cos(\theta)^2 = 1.\]

You can play around with this identity to get a bunch of other identities. For example, dividing both sides of the equation by \(\cos(\theta)^2\) gives you

\[\tan(\theta)^2 + 1 = \sec(\theta)^2.\]

Knowing how to rearrange this identity will be extremely useful throughout the course.

Another useful one when it comes to integration is the double angle formula. This can be written in two different ways:

\[\cos(2\theta) = 2 \cos(\theta)^2 - 1 \text{ or } \cos(\theta)^2 = \frac{\cos(2\theta) + 1}{2}.\]

Again, knowing how to rearrange this one will be useful in the future.

Before finishing this section out, we should talk about inverse trig functions. Thinking about sine and cosine just in terms of a right triangle is limited, and a better way of thinking about them is with the unit circle. This let’s us think of them as functions with the domain \([0,2\pi)\). In real life, you’ll often times want to rotate more than just \(2\pi\), so it’s nice to extend these functions to all of the real numbers by making them periodic. This means that \(\sin(\theta + 2\pi) = \sin(\theta)\) for any \(\theta \in \mathbb{R}\) (and same with cosine).

Once we think about these functions as periodic functions, we have that they are functions taking real numbers and spitting our real numbers. As a result, we can plot them. For example, here is \(\sin(\theta)\):

Using the definition, it is immediately clear that \(\sin(\theta)\) is not one-to-one. It repeats itself, so it is impossible for it to be one-to-one. By the fact, this means it is impossible for there to be an inverse function for sine. However, just like when we thought about \(f(x) = x^2\), we still might want to “undo” sine. As a result, we restrict to a domain where \(\sin(\theta)\) is one-to-one. This is the darker region shown above, this happens for \(-\pi/2 \leq \theta \leq \pi/2\). The function arcsine is the function \(\arcsin : [-1,1] \rightarrow [-\pi/2, \pi/2]\) such that \(\arcsin(x) = \theta\) implies \(\sin(\theta) = x\). We can do the same thing for cosine and tangent as well, but the domains will be different. For reference:

• The range of arcsine is \([-\pi/2, \pi/2]\) and the domain is \([-1,1]\).
• The range of arccosine is \([0,\pi]\) and the domain is \([-1,1]\).
• The range of arctangent is \((-\pi/2,\pi/2)\) and the domain is \((-\infty, \infty)\) or \(\mathbb{R}\).

Warning: It is always true that \(\sin(\arcsin(x)) = x\), but it is not always true that \(\arcsin(\sin(\theta)) = \theta.\) This also holds for the other inverse trigonometric functions.

Exponential and Logarithmic Functions

We say that a function of the real numbers \(f(x)\) is an exponential function if there exists a real number \(b \neq 1\) so that

\[f(x) = b^x.\]

By using the horizontal line test, you can check that an exponential function is one-to-one, so we can define an inverse. The inverse function to an exponential function is a logarithmic function. We write \(\log_b(x)\) to denote the inverse of \(b^x\). Notice that this means

\[\log_b(x) = y \iff b^y = x.\]

You should take a second to plot some exponential functions and logarithmic functions.

Before moving on, let’s also review some common algebra when it comes to exponentials and logarithms. For any real number \(b \neq 1\) and any real numbers \(x\) and \(y\), we have

• \[b^{x+y} = b^x b^y,\]
• \[(b^x)^y = b^{xy},\]
• \[b^{-x} = \frac{1}{b^x},\]
• \[b^0 = 1.\]

Similarly,

• \[\log_b(x y) = \log_b(x) + \log_b(y),\]
• \[\log_b(x^y) = y \log_b(x),\]
• \[\log_b(x^{-1}) = - \log_b(x),\]
• \[\log_b(1) = 0.\]

Finally, an important identity is the change of base formula. Given \(a\) and \(b\) not equal to \(1\), we can write

\[\log_a(x) = \frac{\log_b(x)}{\log_b(a)}.\]

There’s actually an easy way to see why this is true. Notice that

\[x = b^{\log_b(x)}.\]

On the other hand,

\[b^{\log_b(x)} = x = a^{\log_a(x)} \text{ and } a = b^{\log_b(a)}.\]

If we replace \(a\) with \(b^{\log_b(a)}\), we get

\[b^{\log_b(x)} = a^{\log_a(x)} = (b^{\log_b(a)})^{\log_a(x)}.\]

Now we can use our algebra rule to get

\[b^{\log_b(x)} = b^{\log_b(a) \cdot \log_a(x)}.\]

Since \(b^x\) is one-to-one, we can deduce that

\[\log_b(x) = \log_b(a) \log_a(x),\]

and rearranging gives us

\[\log_a(x) = \frac{\log_b(x)}{\log_b(a)}.\]

The change of base formula is extremely useful, as it lets us work with just one base instead of having to think about all possible bases. The nicest base to work with in calculus is \(e\), where \(e\) is [Euler’s constant](https://en.wikipedia.org/wiki/E_(mathematical_constant). In fact, this base is so nice that we even gave its logarithm a unique symbol – we write \(\ln(x)\) for \(\log_e(x)\).

Calculus

Now we’ll do a crash course in calculus.

Lines

The core of calculus is the study of lines. At this point, you should be very confident when it comes to a line. Remember that a function of real numbers \(f(x)\) is a line if we can write it as

\[f(x) = mx + b.\]

In other words, a line is a polynomial of degree at most \(1\).

As we’ve seen in the above crash course, functions can get really complicated. What we’d like to say is that even though functions may get complicated, if we zoom in really close to a point we can still “pretend” that they are a line. Of course, this is not true for even the most basic non-line functions. Take the function \(f(x) = x^2\) and plot it in your favorite graphing calculator. As you zoom in on \(x=0\), you’ll see you always have this curve, so it will never look like a line. But, if you draw the line \(y=0\) on top and zoom in, you’ll notice that as you get closer the difference between the function and the line gets smaller. With this perspective, we predict that maybe a line can model the behavior of a function with some error that gets smaller as we zoom in more.

Take your favorite function with a jump, though, and you’ll see that this prediction fails. We can’t apply this philosophy with every function, but it seems like we can apply it to polynomials, which covers a lot of ground already. It also seems to work with trigonometric functions and exponential functions. What makes these functions so special compared to the function with a jump? Also, it seems like I just plucked the line \(y=0\) in the previous example out of thin air – how do you figure out the best line to model the function?

Let’s pretend we are Newton (or Leibniz depending on who you ask, see this) and let’s try to figure out how we would do this if we ignored all mathematical rigor. Remember that given two points \((x_0, y_0)\) and \((x_1, y_1)\), we can figure out the (unique) line connecting them. This line will have slope

\[m = \frac{y_1 - y_0}{x_1 - x_0},\]

and we can solve for what \(b\) should be:

\[y = mx + b = \left(\frac{y_1 - y_0}{x_1 - x_0} \right)x + b,\]

and if we use the fact that we want \(y = y_0\) when \(x = x_0\) then we can solve for \(b\) to get

\[b = y_0 - \left(\frac{y_1 - y_0}{x_1 - x_0}\right) x_0 = \frac{x_0 y_1 - x_1 y_0}{x_1 - x_0}.\]

This is a bit scary as is, but it’s sometimes useful to have this formula. A nicer way to write this is to just leave the slope as \(m\). In this case, we get

\[y_0 = mx_0 + b \implies b = y_0 - mx_0.\]

This gives rise to the point-slope formula: given a point on the line \((x_0, y_0)\) and the slope of a line \(m\), we can find out the equation for the line by solving

\[y - y_0 = m(x- x_0) + b.\]

Now let’s pretend these two points lie on the graph of a function \(y = f(x)\). So one point will be \((x_0, f(x_0))\) and the other point will be \((x_1, f(x_1))\). Using the above formula, we have that the line connecting these two points is given by

\[y = \frac{f(x_1) - f(x_0)}{x_1 - x_0} x + \frac{x_0 f(x_1) - x_1 f(x_0)}{x_1 - x_0}.\]

Notice that the quantity

\[\frac{f(x_1) - f(x_0)}{x_1 - x_0}\]

is measuring the rate of change of the function as we go from \(x_0\) to \(x_1\). We’d like to know what the rate of change is when \(x_1 = x_0\), or in other words exactly at a point. Plugging this in gives us non-sense though, so what we’d like to do instead is to “zoom in” as much as we can. This is what a limit represents. As we zoom in more and more, we hope that this gets closer and closer to some value, which we’ll call the derivative (or the instantaneous rate of change of the function at the point \(x_0\)). We denote the derivative by \(f'(x_0)\). The other quantity seems like its hard to work with, so instead of trying to do some similar trick there is note that we want to construct a line through the point \((x_0, f(x_0))\) with slope \(f'(x_0)\). Assuming this all works, we can use the point slope formula to get that the line best modeling our function near \(x_0\) is given by

\[y = f'(x_0) (x - x_0) + f(x_0).\]

Of course, there’s a big if here: assuming this all works. The next few sections are going to go into detail of how you actually do this.

Limits

Intuitively, a limit is a best approximation of the value of a function given surrounding information. We say that \(\lim_{x \rightarrow a} f(x) = L\) if \(f(x)\) gets close to \(L\) as \(x\) gets close to \(a\). Notice even this definition is vague (what does close even mean), but the idea should be clear – the limit is not supposed to detect what the function is at a point, but what the function should be if I kept following the pattern coming from the surrounding information.

Example: Consider the function

\[f(x) = \frac{x^2 - 2x}{x-2}.\]

Notice that if I try to plug in \(x=2\), I would get

\[f(2) = \frac{2^2 - 2(2)}{2-2} = \frac{0}{0},\]

and this is not defined. So \(x=2\) is not even in the domain of this function. The domain of the function is actually \((-\infty, 2) \cup (2,\infty)\). Suppose \(a > 2\). Notice

\[f(a) = \frac{a^2 - 2a}{a -2} = \frac{a(a-2)}{a-2}.\]

Since \(a-2 \neq 0\), I can cancel \(a-2\) from the top and bottom to get

\[f(a) = a.\]

So for every number bigger than \(2\), I get that the function is equal to \(x\). This means that

\[\lim_{x \rightarrow 2^+} f(x) = 2.\]

The \(+\) here means I’m moving from the right, so I’m taking values bigger than \(2\) (but not equal to \(2\)) and getting closer to \(2\). Also notice that the same argument tells me

\[\lim_{x \rightarrow 2^-} f(x) = 2.\]

Thus we can say

\[\lim_{x \rightarrow 2} f(x) = 2.\]

This reminds us of the following fact:

Fact: For the limit \(\lim_{x \rightarrow a} f(x)\) to exist and be equal to \(L\), we need

• \(\lim_{x \rightarrow a^+} f(x) = L\) and
• \(\lim_{x \rightarrow a^-} f(x) = L\).

This means for a limit problem, there’s three things we have to check:

• the limit from the right exists,
• the limit from the left exists,
• the limit from the right is equal to the limit from the left.

I recommend taking some time to brush up on limits. Though they will not be directly useful like in Calculus 1, conceptually they will keep coming back.

Continuity

A function \(f(x)\) is continuous if \(\lim_{x \rightarrow a}f(x) = f(a)\) for every point \(a\) in the domain. Note that

• polynomials and rational functions are continuous on their domains,
• exponential and logarithmic functions are continuous,
• trigonometric functions are continuous on their domains.

Derivatives

The derivative of a function \(f(x)\) is given in two different ways:

\[f'(x) = \lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a} = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}.\]

We also might write

\[f'(x) = \frac{df}{dx}(x).\]

The symbol \(\frac{d}{dx}\) will mean “take the derivative of the function with respect to \(x\).”

Warning: Not every function is differentiable! There are continuous functions which are not differentiable.

Notice that these are saying the same thing at the end of the day. While the definition of a derivative always works, it would be tedious to do a limit every time we need to take a derivative. Luckily there are some rules that make things easy.

• We have \(\frac{d}{dx}(f \pm g)(x) = f'(x) \pm g'(x).\)
• We have \(\frac{d}{dx} (fg)(x) = f'(x)g(x) + f(x)g'(x).\)
• We have \(\frac{d}{dx} \left(\frac{f}{g}\right)(x) = \frac{g(x)f'(x) - f(x)g'(x)}{g(x)^2}.\)
• We have \(\frac{d}{dx} (f \circ g)(x) =\frac{d}{dx} f(g(x)) = f'(g(x)) g'(x).\)
• If \(k\) is a constant, then \(\frac{d}{dx} k = 0.\)
• If \(n \neq 0\), then \(\frac{d}{dx} x^n = n x^{n-1}\).
• We have \(\frac{d}{dx} e^x = e^x\). In general, \(\frac{d}{dx} b^x = b^x \ln(b).\)
• We have \(\frac{d}{dx} \ln(x) = \frac{1}{x}.\) In general, \(\frac{d}{dx} \log_a(x) = \frac{1}{\ln(a) x}.\)
• We have \(\frac{d}{dx} \sin(x) = \cos(x)\).
• We have \(\frac{d}{dx} \cos(x) = -\sin(x)\).
• We have \(\frac{d}{dx} \tan(x) = \sec(x)^2.\)
• We have \(\frac{d}{dx} \sec(x) = \sec(x) \tan(x).\)
• We have \(\frac{d}{dx} \csc(x) = -\csc(x) \cot(x).\)
• We have $$\frac{d}{dx} \cot(x) = -\csc(x)^2.
• We have \(\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1-x^2}}.\)
• We have \(\frac{d}{dx} \arccos(x) = -\frac{1}{\sqrt{1-x^2}}.\)
• We have \(\frac{d}{dx} \arctan(x) = \frac{1}{1+x^2}.\)

As we mentioned before, we can also use the derivative to find the tangent line, or the best linear approximation of the function at a point. The formula for a tangent line of the function \(f(x)\) at a point \(x=a\) is given by

\[L_a(x) = f'(a) (x-a) + f(a).\]

Warning: I didn’t talk about implicit differentiation, but I recommend you look that up and review it.

Integrals

We say that a function \(F(x)\) is an antiderivative of a function \(f(x)\) if \(F'(x) = f(x)\).

Example: If \(f(x) = x^2\), then an antiderivative of it would be \(F(x) = \frac{1}{3}x^3.\)

Warning: Notice I said “an” and not “the.” Antiderivatives are not unique. We also have that \(F(x) = \frac{1}{3} x^3 + 4\) is an antiderivative of \(f(x)\). That being said though, we have the following fact.

Fact: Let \(f(x)\) be a function and \(F(x), G(x)\) two antiderivatives. There is a constant \(C\) so that \(F(x) = G(x) + C\).

So all possible antiderivatives of \(x^2\) are given by \(\frac{1}{3}x^3 + C\).

The indefinite integral is the collection of all possible antiderivatives of a function. We write this as

\[\int f(x) dx = F(x) + C, \text{ where } F'(x) = f(x).\]

Warning: The notes linked above have a lie in them. The indefinite integral is not a function, but rather it is a collection of functions. The antiderivative F(x) is a function.

Let me point out that finding an indefinite integral is much harder than taking a derivative. That being said, we still have a list of rules to help us out.

• If \(k\) is a constant, then \(\int k dx = kx + C.\)
• We have \(\int [f(x) \pm g(x)]dx = \int f(x)dx \pm \int g(x)dx\).
• If \(k\) is a constant, then \(\int k f(x)dx = k \int f(x)dx.\)
• We have \(\int x^{-1} dx = \ln(\mid x \mid) + C.\)
• If \(n \neq -1\), then \(\int x^n dx = \frac{x^{n+1}}{n+1} + C.\)
• We have \(\int e^x dx = e^x + C\).
• We have \(\int \cos(x) dx = \sin(x) + C\).
• We have \(\int \sin(x) dx = - \cos(x) + C\).

As you can tell at this point, you can work backwards from the derivative rules to get the antiderivative rules. I’ll leave the rest to you.

A definite integral is (at least ahead of time) completely different from an indefinite integral. This represents the (signed) area underneath the graph of a function. We write \(\int_a^b f(x)dx\) to mean the signed area underneath the graph of the function on the domain \(a \leq x \leq b\).

Example: Take the function \(f(x) = x\). Then \(\int_0^1 f(x) dx\) will just be the area of a right triangle with height one and base one. We get

\[\int_0^1 x dx = \frac{1}{2}.\]

The magic of integrals is that the indefinite integral and the definite integral are related by the fundamental theorem of calculus.

Fact: If \(f(x)\) is a continuous function on an interval \([a,b]\) and \(F(x)\) is any antiderivative of \(f(x)\), then

\[\int_a^b f(x) dx = F(b) - F(a).\]

You should take the time to practice using this theorem; Ximera has some examples.

This should hopefully give you a base to work with. Next time, we’ll pick up with more integration review.