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# Definite integral of shifted function

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 is 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, let's
say this area is equal to five. So given that, can you
figure out what the definite integral, what 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 here. Try to pick a c in your
brain and try to graph them. And pause the video and 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 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 by original one. It's going to look something like that. And I can even color code it. 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 over by c. And this is something you probably learned in precalculus class or in algebra class. And the key thing to realize is, okay, when x is equal to c, you're
essentially inputting. So when x is equal to c,
you're inputting zero, 'cause you're gonna get c minus c. Your gonna input zero 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 zero into the function. You're going to get the same value there as when you just took the function and you just inputted zero 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 gonna get us
right over there, roughly. And so, this is a plus c and
b plus c is gonna get us, is gonna 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 gonna
go from a plus c, to b plus c. And so this is the area, actually
let me do it 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, it's
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 bounds 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 kinda made that up, in this case is going to be equal to five. But the important thing to realize, and 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, as we'll see in the future, we'll tackle some really
interesting problems, where identifying this can be valuable. You might say, "Hey, wait,
this is some bizarre thing. "How do I figure this out?" And you just realize 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 value.