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Current time:0:00Total duration:4:26

Video transcript

let's say that we know that this area under the curve y is equal to f of X let me label it y is equal to f of X so under this curve above the x-axis between a and B which we denote as the definite integral from A to B of f of X D of X let's say we know what this is let's say it's equal to I don't know let's say this area is equal to 5 so given that can you figure out what the definite integral what the definite integral from the definite integral from a plus some constant C let me do this in a different color so what is going to be the definite integral of f of X minus C DX from a plus C to B plus C so this might look a little daunting but I encourage you to kind of try to visualize what's going to happen you try to pick a C in your brain and try to graph them and pause the video try to think about what this is going to be given what we know about this so I'm assuming you've had a go at it so what is f of X minus C well that's essentially the function f of X shifted to the right by C so let's do that so that's going to look like so if we take that function we shift it to the right let's say that let's say that this distance right over here is C so if you shift it to the right by C it's going to look something like that so I just copied and pasted my original one it's going to look something like that and I can even color code it let me give it a so this thing right over here this is the graph of this is the graph of y is equal to Y is equal to f of X minus C and so all I did it really just shifts everything over by C it just shifted everything shifted everything over shift it everything over by by C and you wrote you in one way to Ramin this is something you probably learned in precalculus class or an algebra class and the key thing to realize is okay when X is e to see you're essentially inputting so when X is equal to C when X is equal to C you're inputting 0 because you're gonna get C minus C you're going to get input 0 into F so you're gonna get the same value here when F so when X is equal to C for X minus C you're inputting 0 into the function you get the same value there as when you just took the function and you just input it 0 into it so that's some of the logic why when you take X minus C you're shifting to the right by C now let's think about the bounds here this is a plus C so a plus let's shift so a plus C is going to get us right over there roughly and so this is a plus C and B plus C is going to get us is going to get us right over here so this point right over here is B plus C so our new bounds our new bounds look like this our new bounds we're going to go from A plus C to B plus C and so this is the area actually let me do it in that yellow color since that's what I made the definite integral in so we care about this area now we care about this area and I think it might starting to jump out at you what this is going to be equal to this right over here is going to be this exact thing we just shifted everything we shifted the function to the right we shifted the balance to the right and so this is going to be the same thing as the integral from A to B of f of X DX which in this case and I just kind of made that up in this case it's going to be equal to 5 but the important thing to realize this is a tricky one you'll see this sometimes sometimes in math competitions or in kind of a difficult test but sometimes it can actually help you solve an integral once again if you're tackling a really well as we'll see in the future we'll tackle some really interesting problems we're identifying this can be valuable you might say hey wait this is some bizarre thing how do I figure this out and just realize well this is just shifting this this is just shifting this area over to the right by C so it's going to have the exact same the exact same value you