# Your Integral Calculus Toolbox

Updated: Jan 12

My favorite metaphor for a math and science education is the toolbox metaphor: every technique you learn is like a tool, which can be used to solve a number of problems in the future. Some tools are very niche, and are used only in rare circumstances; others are so frequently applicable that you don’t even think about them anymore.

You can think of multiplication as your trusty hammer, while derivatives are more like a… um… a pickaxe. Don’t think too hard about it.

*When I depict derivatives with this specific pickaxe, suddenly they look so much more useful, don’t they? Buckle up though, because knowing how to differentiate stuff will come in handy today. *

*Like I said, don’t think too hard about it.*

*Don’t think about hitting your homework with a pickaxe.*

Jokes aside, there’s a whole bunch of stuff to learn about integral calculus, and I hope that by the end of this article you’ll have a better understanding of the toolset you’re being asked to use in this class. **The tools described below should help you master roughly the first half of MAT 17B (applies to 16B and 21B as well), depending on your professor. If you don't go to UC Davis, this will help you in your Calculus I or II class (depending on when you learn integrals).**

### Tool 0: What is an Integral?

At some point, someone told you that an integral is “the area under the curve”. And, at least technically, that person is not wrong. But “the area under the curve” is a bad definition to have in your head about an integral. In reality, integration is the act of adding an infinite number of tiny things together—normally getting a finite solution. Pretty wild, right? This definition will help you with the more visual problems in integral calculus, because when you get stuck, you can always ask yourself: “Which tiny things am I adding together in this problem?” Once you can answer that question, you should know what to integrate.

### Tool 1: The Basic Antiderivative

I assume you already know this one: an integral is the opposite of a derivative. (Did the name give it away?) That means that if the derivative of (x^2) is (2x), then the integral of (2x) must be (x^2). You should be able to do more versions of this antiderivative in your head quickly:

$$int 99 x^{100} d x=frac{99}{101} x^{101}$$

If you’re having trouble convincing yourself that the above equation is true, then spend some time with more problems like it. There is no shortcut around using this tool—it is the one you are most likely to continue using after this class is over.

### Tool 2: Basic Properties of Integrals

There are a few main points to discuss here. First of all, if you see that an integral consists of two terms *added* together, you can break the terms apart into two separate integrals. In other words,

$$int cos (x)+sin (x) d x=int cos (x) d x+int sin (x) d x$$

Notice that you cannot do the same thing with multiplication.

$$int x^{2} d x neq int x d x * int x d x$$

If you don’t believe me, just evaluate both sides of the equation above according to the basic antiderivative. They’re not equal. However, this does not mean that all hope is lost for multiplication in integral calculus. If your integral is multiplied by some constant (which is not dependent on your variable of integration), then you can pull it out into the front of your integral.

$$int k * sin (x) d x=k * int sin (x) d x$$

The final thing to mention here is about even and odd functions. As you may know, even functions have symmetry about the y-axis, which means that anything happening at (x>0) is also happening at (x <0 ). Meanwhile, odd functions have symmetry about the origin, which means that anything happening at (x > 0) is flipped over the x-axis for (x < 0). From these definitions of even and odd functions, you can likely convince yourself of the following two truths:

(int_{-a}^{a} y(x) d x=2 * int_{0}^{a} y(x) d x) For an even function

(int_{-a}^{a} y(x) d x=int_{0}^{a} y(x) d x-int_{0}^{a} y(x) d x=0) For an odd function

### Tool 3: The u-Substitution

At the risk of sounding pompous, I want to say that the u-substitution is an “easy to learn, but hard to master” kind of tool. I can teach you *how* to use it, but you have to figure out *when* to use it on your own.

The basic protocol of a u-substitution is this: if you can’t solve an integral with the given variable (normally x), write your own variable u, which is some function of x. If done properly, the substitution will simplify the integral into an immediately solvable function of u. There are a couple of rules to sort out before we jump in:

When you change x into u, you must also change dx into some function of du. This is the easiest mistake to make when doing a u-substitution.

If you have limits of integration in your problem, you may also need to adjust them before solving the integral. If you’re like me, and don’t enjoy doing this, simply convert all the u terms back into x terms after integration, and before evaluating the final answer.

Never leave your final answer in terms of u. The question was asked in terms of x! Don’t make rookie mistakes!

Like I said, I can’t teach you every possible use of the u-substitution, so here are a few examples of problems involving u-substitutions:

(int frac{x^{3}}{1+x^{4}} d x) In this classic example, we use (u=1+x^{4}); (d u=4 x^{3} d x).Denominators of annoying fractions are good targets for a u-substitution. In this case, the simplified solution is (frac{1}{4} int frac{1}{u} d u=frac{1}{4} ln (u)=frac{1}{4} ln left(1+x^{4}right)).

(int(x-2)(x-1)^{9} d x) I don’t think anyone wants to FOIL this guy. This time around, (u=x-1); (d u=d x). We end up with (int(u-1) u^{9} d u), which we expand to (int u^{10}-u^{9} d u=int u^{10} d u-int u^{9} d u) I’m sure you can do the rest on your own.

(int tan ^{2}(x) sec ^{2}(x) d x) In this sneaky integral, the goal is to realize that the derivative of (tan (x)) is (sec ^{2}(x)). So, we say: (u=tan (x)); (d u=sec ^{2}(x) d x). From there, the simplified integral is (int u^{2} d u), which you can solve on your own.

### Tool 4: Partial Fraction Decomposition

This topic is a bit too complicated to describe here—I think it requires an article of its own. You may have learned how to do this in high school, or perhaps you’re learning it for the first time now. Either way, this tool will help you to solve problems such as (int frac{x-1}{x^{2}+3 x} d x). If your professor wants you to solve problems like these using this tool, they will likely go out of their way to make sure you know how to use it.

### Tool 5: Volumes of Solids

This topic is one of the more interesting ones, in my opinion. The question typically looks like this: “I have some graph (y(x)), that I want to rotate around the x-axis. What is the volume of the shape I create with this rotation?”

The answer is typically the same—if the graph is rotated around an axis in a circular fashion, then it stands to reason that adding an infinite number of circles (cylinders with infinitesimal height) should give you the correct volume (remember how we defined an integral earlier!).

For example, consider the following problem:

What is the volume of the solid created by rotating the graph (y=sqrt{x}) around the (x)-axis, evaluated from the points (x = 2) to (x = 4)?

And infinite number of tiny cylinders with thickness (dx) should do the trick here. The solution takes the form: (int_{x_{min }}^{x_{max }} A(x) d x), where (A) represents the cross-sectional area of a given cylinder. We solve this problem by evaluating the integral: (int_{2}^{4} pi r^{2} d x=int_{2}^{4} pi(sqrt{x})^{2} d x). Note that the radius of the cylinder’s area is conveniently equal to (y). Of course, not all problems of this type are as simple as the example above, but they all have the same general idea to them.

### Tool 6: Finding Function Averages

Let’s say I have a function (y(x)), and I want to find the average (y) value of this function between (y = 10) and (y = 20). We can use a clever property of integrals to solve this problem. Remember how I said that an integral is a sum of tiny things? Well, the first step in finding an average is adding a bunch of things together. So, to get your average (y) value, simply add up all possible (y) values with an integral: (int y(x) d x), and then divide by the number of things you added!

…Uh oh. Didn’t I say that an integral involves adding an *infinite* number of small things together? That’s true, but never fear! There’s still a way to find that average, and it requires using our limits of integration. Simply divide by size of the range you integrated across. Your final answer will look like this:

$$frac{1}{left(x_{max }-x_{min }right)} int_{x_{min }}^{x_{max }} y(x) d x$$

### Tool 7: Arc Length

Arc length is one of those topics that conceptually rather difficult to understand, but mathematically isn’t that bad. The formula for arc length is shown below:

$$A R C(y)left(x_{min } rightarrow x_{max }right)=int_{x_{min }}^{x_{max }} sqrt{1+left(y^{prime}right)^{2}} d x$$

Remember that (y^{prime}) represents the derivative of the function (y(x)). I won’t go into the reasoning of why the above equation is true, but now that you have it, most arc length problems should be fairly simple for you to solve.

### Tool 8: Integration by Parts

You might be surprised, but integration by parts is actually one of the more likely techniques for you to find in another class down the line. This technique deserves its own article independent from this one, but here is the general strategy:

When confronted with an integral such as: (int x * e^{x} d x), you can metaphorically split the integral into two halves: (u) and (dv). Generally, label the easier half of your integral (u) (in this case (u = x)), and the scarier half of the integral (dv) (in this case (d v=e^{x} d x)). Find (du), the derivative of (u), and also find (v), the antiderivative of (dv). At the end of all these shenanigans, your final answer is:

$$left(u * vright)-int v * d u$$

Don’t ask me why—that’s a question for another time. The most interesting thing about these types of problems is that sometimes you need to perform integration by parts multiple times. If the integral (int v * d u) is still unsolvable, you can integrate that expression by parts as well, sometimes ending up in an infinite loop of integration. The stuff of nightmares—or of interesting math puzzles, depending on how you look at it.

### That’s All, Folks

There you have it! Those are all the general tools you need for solving integral calculus problems. As I said, this list will likely only take you through the first half of MAT 17B, depending on your professor. Still, it’s fairly comprehensive, and useful for other classes as well. (This is typically the same material that shows up in MAT 21B.) Go forth with your new knowledge, and win all of those tests!