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Current time:0:00Total duration:6:22

Examples leveraging integration properties

Video transcript

so over here we have the graph of the function y is equal to e to the X actually let me make make that clear so that right over there is y is equal to e to the X this of course is 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 0 to 1 of e to the X DX and let's say we know in the future we'll learn how to actually come up with this number but let's just say we know that that area is equal to e minus 1 so 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 these three other definite integrals are using only this information and what we already know about definite integral properties so I am assuming that you have given a go at it so you might first okay let's see e to the X plus 2 you might be tempted to kind of decompose this into the definite integral of e to the X DX plus the definite integral from 1 to water so that you could compose this from 1 to 1 of e to the X DX plus 1 to 1 plus the definite integral from 1 to 1 of 2 DX but do you have to remember if you're taking the definite integral where your bounds are the same this is a definite go from 1 to 1 this is going to be 0 this is going to be equal to 0 once again you could imagine that you're trying to find an area of something that has no width you're not moving it all along you have no width along the x-axis here so that was that was pretty straightforward from 1 to 1 your area is going to be 0 now let's think through this one over here so here my bounds are different they are the same as this as these bounds right over here so 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 informed in the form of e to the X I might be able to evaluate this definite integral so let's do that so the first thing I could do is I could break this up as the definite integral from 0 to 1 of 3 DX and then I could say plus the definite integral from 0 to 1 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 so this is the same thing as the integral from 0 to 1 3 DX minus the integral from 0 to 1 e to the X DX well lucky for us we already know we already know what let me just 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 1 so if you subtract an e minus 1 you're going to get actually let me just write it out so we don't get confused we're going to subtract and e minus 1 now what is this going to be equal to the definite integral from 0 to 1 of 3 DX so that we just have to graph that to imagine it so we are going from 0 to 1 and the function is at 1 2 3 I'm not doing it at the same scale so the function looks like that so this right over here is just this area it's under under the function f of X is equal to 3 from 0 to 1 so what's this area well this is just our base our width is 1 our height is 3 so this is just going to evaluate to 3 that's just going to be equal to 3 and so we get this is going to be equal to 3 and you're going to have minus e just distributing the negative sign and then plus 1 plus 1 and then of course 3 plus 1 is going to be equal to 3 plus 1 it's going to be equal to 4 so this is going to be equal to 4 minus 4 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 and the value of this definite integral now what about this one over here this one is interesting way I might have different bounds here definitely go from 2 to 3 ezx have an e to the X minus two and when you see something like this if you're trying to evaluate this.only with integration properties the the big thing that may or may not jump out at you is it e to the X minus 2 is shifted version of e to the X it's shifted by 2 to the right so let me let me let me draw that so instead of this point being there this point is going to be there and so the curve is going to look the curve is going to look something like I'm gonna draw my best attempt my best attempt to do it so it's going to be like there it's going to be like there so it's going to look it's going to look something 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 so there you go that's e to the X minus 2 so we're going from 2 to 3 and so what are the bounds now the bounds are from 2 to 3 so we're going from this point right over here to 3 which is going to be over here just going to be over here so you take this point you shift 1 and 2 you're going to get right over here and what did we just do we're thinking about this area right over here well we shifted the curve over to the right by 2 this is the curve of e to the X minus 2 and we shifted the the interval xi that we're trying to find the area under we shifted that by 2 so this is going to have the exact same area I just shifted everything to the right by 2 so this area is also going to be e minus 1 in my mind this one's a little bit tricky you just have to realize that this 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