Class 12 math (India)
- Integrating scaled version of function
- Integrating sums of functions
- Definite integral over a single point
- Definite integrals on adjacent intervals
- Definite integral of shifted function
- Switching bounds of definite integral
- Worked examples: Finding definite integrals using algebraic properties
- Using multiple properties of definite integrals
- Definite integral properties (no graph): function combination
- Worked examples: Definite integral properties 2
- Definite integral properties (no graph): breaking interval
- Warmup: Definite integral properties (no graph)
- Finding definite integrals using algebraic properties
- Examples leveraging integration properties
- Definite integrals properties review
Examples leveraging integration properties
Watch how some integral properties can actually be used.
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- at4:39, Sal says that e^(x-2) is a shifted function of e^x. In the "definite integral of shifted function" video, we learned that f(x-c) is a shifted function. I understand that (e^x)-2 is a shifted function of e^x, but I don't understand how both (e^x)-2 and e^(x-2) can result in the same shifted function. (e^x)-2 and e^(x-2) have very different values. Is the answer that the function of x is as a power of e so that f(x-2) is e^(x-2)? If so, what would the area under (e^x)-2 equal? the sum of the areas under e^x and 2 between a and b, which would be an addition of 2 and not a shift?(6 votes)
- The graph of e^(x-2) is the graph of e^x shifted right by 2. The graph of (e^x) -2 is the graph of e^x shifted down by 2. You are correct that they are totally different. I don't think he talks about the second one in this video.
So if you shift a graph to the right, the integral will be the same if you shift the limits of integration to the right. Thus in this example, the integral of e^(x-2) from x=2 to x=3 was the same as the integral of e^x from x=0 to x=1.
If you shift a graph up or down, the area under the curve will change by the area of a rectangle with width equal to the width of the interval of integration, and height equal to the height of the vertical shift. The integral of e^x from x=0 to x=1 was e - 1, so the integral of (e^x) -2 from x=0 to x=1 is (e - 1) + (1 * -2) = e - 3.(12 votes)
- Negative areas were touched on in the previous video dealing with switching the bounds. In the second function, Sal sketched the constant portion of the function above the x axis. Am I right in thinking the rest of the function would be the original graph flipped over under the x axis?(5 votes)
- You are right in thinking that the graph of -e^x would be e^x flipped over the x axis. 3-e^x would then be that flipped function moved up 3 units at each and every location.(5 votes)
- at 5.03 the graph shifts to the right from e^x to e^X-2, if you move to the right should it not be e^x + 2.
Are you not taking any/all values of x and increasing by 2 ?(3 votes)
- Think about it like this: If the graph is shifted by "c" units right, the x value is "c units" greater than the "y value" --> x = c + y. However, the general format of most equations you'll see is f(x) = something. Since f(x) represents the y value, then if x = c + y, y = x - c. This means you solve for f(x - c) when the graph is shifted right.(4 votes)
- could anyone explain how did sal do the 3rd example? why is e^x-2 the shifted version of e^x(1 vote)
- [Voiceover] Over here, we have the graph of the function Y is equal to E to the X. Actually, let me make that clear. That right over there is Y is equal to E to the X. This is, of course, our X axis. This is our Y axis. And so, this blue area right over here, this blue area, we can denote that as the definite integral from zero to one of E to the X, DX and let's say we know in the future, we'll learn how to come up with this number but let's just say we know that that area is equal to E minus one. Given that, I have these other three definite integrals and what I encourage you to do is try to pause this video and try to evaluate what this three other definite integrals are using only this information and what we already know about definite integral properties. I am assuming that you have given a go at it. You might first say, okay, let's see E to the X plus two. You might be tempted to decompose this into definite integral of E to the X, DX plus the definite integral from one to-- So that you could decompose this from one to one at E to X, DX plus one to one, plus the definite integral from one to one of two DX but you have to remember if you're taking the definite integral where your bounds are the same. This is the definite integral from one to one. This is going to be zero. This is going to be equal to zero. Once again, you can imagine that you're trying to find the area of something that has no width. You're not moving at all. You have no width along the X axis here. That was pretty straightforward. From one to one, your area is going to be zero. Now, let's think through this one over here. Here my bounds are different. They are the same as these bounds right over here. I guess that gives me a pretty good clue that maybe if I can break this up a little bit, especially if one of these is in the form of E to the X, I might be able to evaluate this definite integral. Let's do that. The first thing I could do is I could break this up as the definite integral from zero to one of 3DX and then I could say Plus the definite integral from zero to one of negative E to the X, DX and then I can take the negative. That negative is really just a negative one. We saw that you could take that scaling factor out of the integral so let's do that. This is the same thing as the integral from zero to one. 3DX minus the integral from zero to one, E to the X, DX. Well, lucky for us, we already know what, let me do this in that same color. We already know what this, I have trouble changing colors. We already know what this right over here is. That's going to be equal to E minus one. If you subtract an E minus one, you're going to get, actually, let me just write it out so we don�t get confused. We're going to subtract an E minus one, what is this going to be equal to? The definite integral from zero to one of 3DX. We just have to graph that to imagine it. We're going from zero to one and the function is that one, two, three. I'm not doing it at the same scale. The function looks like that. This right over here is just this area. Under the function F of X is equal to three. from zero to one So, what's this area? Well, this is just our base. Our width is one. Our height is three. This is just going to evaluate to three. That's just going to be equal to three and so, this is going to be equal to three and you're going to have minus E, just distributing the negative side and then plus one. And then of course, three plus one is going to be equal to four. This is going to be equal to four minus four, minus E and we're done. We're able to solve or I guess we say, evaluate or find the value of this definite integral using only integration properties in the value of this definite integral. Now, what about this one over here? This one is interesting way. I have different bounds here. Definitely go from two to three. Having E to X minus two. And when you see something like this, especially, if you're trying to evaluate this only with integration properties, the big thing that may or may not jump out of you is that E to the X minus two is shifted version of E to the X. It's shifted by two to the right. Let me draw that. Instead of this point being there, this point is going to be there and so, the curve is going to look something like let me draw my best attempt. My best attempt to do it. It's going to be like there. It's going to be like there. It's going to look something. I'm trying my best to just shift it over. So each of these, this is going to be shifted to right over here. There you go. E to the X minus two. We're going from two to three and so, what are the bounds now? The bounds are from two to three. We're going from this point right over here to three, which is going to be over here. Just going to be over here. Take this point, you shift one and two. You're going to get right over here and what do we just do? We're thinking about this area right over here. We shifted the curve over to the right by two. This is the curve of E to the X minus two and we shifted the interval event that we're trying to find the area under. We shifted that by two. This is going to have the exact same area. I just shifted everything to the right by two. This area is also going to be E minus one. In my mind, this one's a little bit tricky. You just have to realize that this is just a shifted version of that and that we also shifted the two bounds to the right and we already saw that in one of the integration property videos.