Back To CourseMath 104: Calculus
13 chapters | 104 video lessons
Erin has taught math and science from grade school up to the post-graduate level. She holds a Ph.D. in Chemical Engineering.
Let's take a minute to review some techniques for integration. Here we're going to integrate the indefinite integral - that means it has no limits - of f(x)dx. We know that the integral of f(x)dx equals the anti-derivative as a function of x plus a constant of integration. If you take the derivative of the anti-derivative, you get back your original function. So what are some of the ways we know to find integrals?
First, we can use a table. This may be a table in a book, online or what you have memorized. For example, the integral of x^2dx = (1/3)x^3 + C. You know this because you know how to integrate polynomials. The integral of sin(x)dx = -cos(x) + C. The integral of e^(x)dx = e^(x) + C. For each of these cases, if you take the derivative of the right side, you end up with the integrand. This is true of all integrals; it's how you calculate an integral.
The second way we know to calculate integrals is by substitution. In this case, we're going to take an integral that depends on x, and we're going to make a substitution where u equals some new function of x. By plugging in u, we hope to end up with a simpler integral that we can integrate with respect to u. For example, we have the integral sin(2x)dx. I want to substitute u for 2x, so u=2x and du=2dx. So I can plug those in, both for 2x and for dx, and my integral becomes 1/2 sin(u)du. I can use a table to solve this, because the integral of sin(u) is -cos(u). I get -1/2 cos(u) + C. Now I want to plug in 2x where I have u - that's my original substitution - so I get x back in my final answer. I get -1/2 cos(2x) + C. If I take the derivative of this, I end up with sin(2x). That's solving by substitution, and that is by far what you are going to use the most when solving integrals by hand, but there are a couple of other methods that you should be aware of.
One is integration by parts. Here you have the integral of udv = uv minus the integral of vdu. This is just a rearrangement of the product rule. An example would be the integral of xe^(x)dx. Here I'm going to set x equal to a new variable, u so that du=dx. I'm going to set e^(x)dx equal to dv, so v has to be equal to e^x because the derivative of e^x is e^(x)dx. If I plug u, v, du and dv into the right side of my equation for integration by parts, I end up with xe^x (that's uv) minus the integral of e^(x)dx (that's vdu). At this point, I can use a table in my head because I have memorized this integral. The integral of e^(x)dx is e^x, so if I plug that in, my integral becomes xe^x - e^x + C. If I take the derivative, I end up with xe^x.
The last way you can solve an integral is by Riemann Sums. This is not an analytical way to solve it; that is, you aren't going to have numbers left. You're going to solve it numerically, on a computer or calculator - you're going to plug in actual numbers.
Let's do an example. Let's say you're given the integral of x^2(sin(x))dx. Your first step would be to see if you can remember this integral or look it up in a table. Most likely, any integral you come up with is either in a table or it isn't solvable. There are whole books on writing out integrals like this, all the possible integrals you can think of that actually have solutions. But let's say you don't have that book handy, so the first thing you try is substitution: u=sin(x) and du=cos(x)dx. This doesn't make much sense, because if you plug in sin(x) you get u but x^2 becomes arcsin^2(u). You just made life a little more complicated, so maybe that's not the best way to do it. Because substitution is still the first thing you want to go to, what about using u=x^2 and du=2xdx. That's good, but you still have sin(x), which would become sin(square root of u) and that again sounds really complicated. So maybe substitution isn't the method you want to look at.
Since you have one function multiplied by another function, maybe you could do integration by parts. In integration by parts, I'm going to set this first function, x^2, equal to u. That leaves sin(x)dx as dv, because remember, the integral of udv = uv minus the integral of vdu. So my integral has to be u times dv: if x^2=u, then sin(x)dx=dv. If u=x^2, then du=2xdx. If dv=sin(x)dx, then v= -cos(x), because if I take the derivative of -cos(x), I end up with sin(x)dx. So I have u, v, du and dv. If I plug all of these into my equation for integration by parts, I get -x^2(cos(x)) + the integral of cos(x)2xdx. (The '+' comes from the minus sign of v combined with the minus sign in our equation.) I can rewrite this as -x^2(cos(x)) + 2 times the integral of x(cos(x))dx. I don't actually know the integral of x(cos(x))dx off the top of my head. It looks a little simpler than x^2(sin(x)), but it's still not something I just know.Continue reading... Create an account to read entire transcript...
Let's take a look at this whole equation: the integral of x^2(sin(x))dx= -x^2(cos(x)) + 2 times the integral of x(cos(x))dx. Let's make the first term a different color, so we know it's from the original integration by parts. Let's take the second term, 2 times the integral of x(cos(x))dx, and integrate that by parts. So I'm going to do the same thing all over again. In my new integration by parts, I'm going to set x equal to u and cos(x)dx equal to dv. So du=dx and v=sin(x). If I plug all of those into integration by parts, I end up with x(sin(x)), that's uv, minus the integral of sin(x)dx, that's vdu. I'm almost there. Now I can take this term, split it up and multiply the 2 out, so I get 2x(sin(x)) - 2 times the integral of sin(x)dx. Finally I have an integral, sin(x)dx, that I know off the top of my head. It's just -cos(x) + C.
We just did a lot of work to find this, so let's make sure we did it correctly and take the derivative of the right-hand side. We have x^2(sin(x))dx= -x^2(cos(x)) + 2x(sin(x)) + 2cos(x) + C. Let's look at one term at a time. The derivative of -x^2(cos(x)), by the product rule, is the first times the derivative of the second plus the second times the derivative of the first: -x^2(-sin(x)) - 2x(cos(x). For my second term, 2x(sin(x), to find the derivative I'm again going to have to use the product rule. I get 2x(cos(x) + 2sin(x). It's getting longer and longer here! What about the third term, 2cos(x) + C? The derivative of that is just -2sin(x) + 0, because the derivative of a constant is zero.
Let's write this all out. I have x^2(sin(x)) - 2x(cos(x)) + 2x(cos(x)) + 2sin(x) - 2sin(x). This is fantastic. The 2x(cos(x)) terms cancel with one another, and the 2sin(x) terms cancel with one another. So the entire derivative simplifies down to x^2(sin(x)), which was my original integrand, so this indeed is the integral of x^2(sin(x))dx.
Let's review. At the end of the day, you have a few tools at your command to solve problems of integration. You can solve problems just by memorization of certain integrals. These are the really common integrals that you want to be able to pull off the top of your head, like sin(x), 1/x and x^2. You can also find integrals in a table, online or in a book.
But not all of these integrals can be found in one of these places, either in your head or some reference. So you need to use a couple of other tools, like for example substitution. Here you're taking an integral that depends on x, turning it into an integral that depends on u and hoping that new integral is easier to solve. You can also solve it by parts. This is like using the product rule in reverse. You can also use algebra and trigonometry to simplify your integrand, f(x), into something that is more manageable. When you use trigonometry, you might want to use a different kind of substitution based on the rules of geometry and right triangles. Finally, if all of these fail, you can solve a definite integral numerically. That is, you can calculate the value of your integral using Riemann Sums.
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Back To CourseMath 104: Calculus
13 chapters | 104 video lessons