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Switching bounds of definite integral

FUN‑6 (EU)
FUN‑6.A (LO)
FUN‑6.A.1 (EK)
FUN‑6.A.2 (EK)

Video transcript

- [Voiceover] We've already seen one definition of the definite integral, and many of them are closely related to this definition that we've already seen is the definite integral from a to b of f of x d of x is this area shaded in blue, and we can approximate it by splitting it into n rectangles. So let's say that's the first rectangle, one. That's the second rectangle, two, and you're going to go all the way to the nth rectangle, so this would be the n minus oneth rectangle. For the sake of this argument I'm going to make in this video, we're going to assume that they're all the same width. So this is the nth rectangle. They all have the same width, and we see they're definitions of integration where you don't have to have the same width here, but let's say that each of those widths are delta x, and the way that we calculate delta x is we take b minus a and we divide it by n, which is common sense, or this is what you learned in division. We're just taking this length and dividing it by n to get n equals spacings of delta x. So if you do this, you'll say, okay, well we see this multiple times, you can approximate it. You can approximate this area using these rectangles as the sum from i equals one to n. So you're summing n of these rectangle's areas where the height of each of these rectangles are going to be f of x sub i, where x sub i is the point at which you're taking the function value to find out its height. So that could be x of one, x of two, x of three, so on and so forth, and you're multiplying that times your delta x. So you take x sub two, f of x sub two is that height right there. You multiply it times delta x. You get the area. We saw that when we looked at Riemann sums and using that to approximate. We said, hey the one definition of the definite integral is that since this is the area this is going to be the limit as n approaches infinity of this where delta x is defined as that. So let me just copy and paste that. So that's one way to think about it. Now, given this definition, what do you think this, or maybe another way to think about it, how do you think this expression that I'm writing right over here based on this definition should relate to this expression? So notice, all I've done is I've segued from a to b. I'm now going from b to a. How do you think these two things should relate? I encourage you to look at all of this to come to that conclusion, and pause the video to do so. Well let's just think about what's going to happen. This is going to be, if I were to literally just take this and copy and paste it, which is exactly what I'm going to do, if I just took this, by definition, since I swapped these two bounds, I'm going to want to swap these two. Instead of b minus a it's going to be a minus b now. It's going to be a minus b. This value right over here. Let me make these color-coded maybe. So this orange delta x is going to be the negative of this green delta x. This is the negative of that right over there, and everything else is the same. So what am I going to end up doing? Well I'm essentially going to end up having the negative value of this. So this is going to be equal to the negative of the integral from a to b of f of x dx. So this is the result we get, which is another really important integration property, that if you swap the bounds of integration, and it really just comes from this idea, instead of delta x being b minus a, if you swap the bounds of integration, it's going to be a minus b. We're going to get the negative delta x, or the negative of your original delta x, which is going to give you the negative of this original value right over here. Once again, this is a really, really useful integration property where you're trying to make sense of some integrals and even sometimes solve some of them.