# Using Integration By Parts

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Lesson Transcript
Instructor: Kelly Sjol
Your mother may have warned you not to bite off more than you can chew. The same thing is true with integration. In this lesson, learn how integration by parts can help you split a big interval into bite-sized pieces!

## Integration by Parts

I absolutely love doing jigsaw puzzles. I think they are the best pastime in the world. I can sit for hours and do a 1,000-, 2,000- or 5,000-piece jigsaw puzzle. One of the tricks to doing jigsaw puzzles that I've found is to take the entire puzzle and break it up into smaller, more manageable chunks. So instead of looking at all 5,000 pieces at once, you're only looking at 200 pieces that are all roughly the same color and trying to put those together first. Once you put those together, you can put that small chunk back into your big puzzle. It makes the process a little less daunting if you can do this for each part of the puzzle; the puzzle is done before you know it.

So what does all of this have to do with math? It is similar to a concept we call integration by parts. In integration by parts, you have an integral like udv. v is your variable of integration and u is your integrand, so it's going to be a function as well. You can write the integral udv as being equal to uv minus the integral of vdu. What you've done is taken out part of the gigantic interval, udv, and made it look a bit easier. You took out uv, and instead of integrating udv, you're left with vdu. At first this doesn't look exactly like the puzzle analogy. You have a small integral equal to two separate things (that is like the puzzle), but the two separate things aren't necessarily easier to put together. I mean, if vdu is no easier to integrate than udv, you haven't saved yourself any work by pulling those pieces out of your puzzle. We'll talk about that. First, where does this come from, and, more importantly, how can you remember it?

It is the same thing as the product rule. Imagine you have a function uv, and u and v are each a function, like f(x) and g(x). If you take the derivative of uv, you end up with u times the derivative of v plus v times the derivative of u: (uv)` = uv`+vu`. We remember this as the first times the derivative of the second plus the second times the derivative of the first. Do you see how that relates to integration by parts? Think about it this way. If you integrate both sides of the equation, you end up with uv= the integral of udv + the integral of vdu. This is your integration by parts, except we've moved vdu to the other side of the equation. So if you ever forget the equation for integration by parts, see if you can remember the product rule for differentiation. Let's do an example.

## First Example of Integration by Parts

Let's say we have the integral of (xe^x)dx. This isn't an integral I know off the top of my head. This is instead one function multiplied by another function. Maybe I can solve this by substitution. No, I don't think a u substitution would work here. Since I've got two functions multiplied by one another, maybe I can do integration by parts. For integration by parts, my integral, (xe^x)dx, needs to look like the integral udv. So what do I want to call u and what do I want to call dv? Let's substitute u for x. What's left, (e^x)dx, has to be equal to dv, because (xe^x)dx has to equal udv. If I make that substitution, where u=x and dv=(e^x)dx, then I can find out what du is and what v is. What do I mean by that? Well, I've got u=x; let's take the derivative of that: du=dx. Okay, so I have u and du, and dv is (e^x)dx. What is v? To find v, I have to integrate (e^x)dx, because I want to find the anti-derivative of this equation. This is like undoing the derivative, right? Well, I know that if I take the derivative of e^x I get e^x, and if I take the integral of e^x I get e^x. So if dv=(e^x)dx, then v=e^x.

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