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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 4

Lesson 6: Definite integral properties- 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: Definite integral properties 2
- Worked examples: Finding definite integrals using algebraic properties
- Definite integral properties (no graph): function combination
- 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

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# Definite integral of shifted function

When you shift a function, what happens to its integral?

## Want to join the conversation?

- I am sorry I dont get this straight ahead, but why do we shift it to the right when we have -c and not to the left? Sal doesnt get very deep on the explanation of this...(40 votes)
- I'll try to explain but I don't claim to be a master at math nor teaching.

so, the question is, since it is MINUS c why does it shift to the RIGHT. Shouldn't it shift to the left since the x-axis is a number line and negative numbers go towards the left?

That is an extremely easy mistake to make because it does kinda make sense. However, try to think of it like this: When you plug an x-value into a function, the x value never changes only the y-value does. In other words, the input to a function is always going to be whatever you input, no matter what.

For example, take y=x^2 this is a function. You put in 2 and you get out 4. The 2 you put in(x-value) is always 2. The y-value is the one that got changed. The x-value was squared to change it into a y-value.

Here is where this will make sense(hopefully). lets take that function and shift it. Now lets look at y=(x-2)^2 we think it should shift to the left because of x-2. We think this because we mistakenly believe that when we put an x value into that function then that x value will move 2 to the left. As stated before, only y values move. So, plugging any number(aka x) into this will give us the y value that would appear at x minus 2. Looking at our function, y=(x-2)^2 if we plug in 0 it will spit out 4. 4 is the y value we would have gotten at x=-2 if this function was not shifted.

So essentially, the -2 shifts it to the right because it takes x, then subtracts 2 from it, then goes through the rest of the function(in this case that means squaring it), then puts that y value at the original x value, which is 2. Since it does this for every x value that means the whole entire function is shifted to the right.

To condense it more and talk about Sal's function, f(x-c), pretend that when you take any x, the negative c shifts it to the left by c then takes whatever y value is at x-c and puts that y value at x.(30 votes)

- Can someone please explain to me why the function is shifted to the right if it is a negative sign? I know this as a general rule but I am stuck on why this happens(2 votes)
- Let's say we have a function such as y = x²

So, one of the most notable features of this parabola is its vertex, in this case its lowest point, which happens to be at (0, 0).

Let's compare this function to y = (x- 2)² which is also a parabola. Where is its vertex? The lowest possible point of the parabola will occur when x - 2 equals zero. That happens when x is 2 with this equation. All other values will be greater than zero, as we can see below.

From the equation, which is in the "vertex form" y = (x-`2`

)²

x | y = (x- 2)²`-1 9`

`0 4`

`1 1`

`2 0`

Lowest value of the function: the vertex`3 1`

`4 4`

The values show that the lowest point of this parabola is at x = 2, with coordinates of (2, 0)

So, even though the equation shows a negative, the feature we are watching has moved to the right.(17 votes)

- Why is there a dx at the end of each function?(3 votes)
- Did you finish the section just prior to this on Riemann Sums?

Recall we were calculating the area under the curve via subdividing the region into rectangles whose height was f(x) and whose width was Δx. We found that the accuracy of the estimate of the area got better and better as the number of subdivisions increases, in which case the width, Δx, is obviously decreasing.

It made intuitive sense, then, that to get better accuracy, we need to keep taking narrower and narrower slices, that is, taking the limit of sum of the rectangles as the width, Δx, approaches zero. Once we say that, then we are dealing with Δx which is infinitesimal (cannot be measured). We denote the infinitesimal Δx as dx.

When we were summing via Riemann approximations we used the symbol ∑ to indicate the summing of each of the slices. The integral symbol ∫ is actually an elongated S which stands for sum and is used when summing infinitesimals. So the only difference between the Riemann sum form ∑f(x_i)Δx_i and the integral form ∫f(x)dx, is that the latter is summing the areas in infinitesimally thin slices, where as the former is made up of slices whose widths can be measured, either way, the slice area has a height component, f(x) and a width component, dx for infinitesimals, or Δx if not.(11 votes)

- is f(x-2) same as y=x-2?(3 votes)
- No. f(x-2) means, when you are given an x, first subtract 2, and then put that value into your function. Say your function is f(x) = x^2, then, f(x-2) = (x-2)^2; if your function were f(x) = x + 5, then f(x-2) = x - 2 + 5 = x + 3.(8 votes)

- Is their a video containing information regarding shifting of graphs if their is can someone can navigate me to specific topic.(2 votes)
- A fun way to think of the idea of "shifting everything": if the original function is to be shifted to the right by c units, but the upper and lower bounds both get to new values that are their respective origins and a common addend (for example, a common number other than c). Will the definite integral of the new bounds equal to the original integral?(2 votes)
- That's an interesting question. I'd say that just about every function would have a different integral if you shift the bounds, but there would be some exceptions. Obviously, if it was just a straight line, such as y = 5, the integral would be the same for any 2 sets of bounds, as long as the upper bound was the same distance from the lower bound in both cases, giving you an equal-sized rectangle each time. Also there would be the case of a sine curve, or something similar, where the graph would be exactly the same from pi/2 to pi as it is from 5*pi/2 to 3*pi, and so on.

Those are the only ones I can think of, but I'm sure there are others(1 vote)

- Where do we put shifts on the y-axis?(2 votes)
- If the f(x) shifts right after you subtract C, then why do points A and B shift right when you ADD C?(1 vote)
- By doing (x-2) you're changing the input x, not f(x). Basically by changing the input x that goes into the equation negatively, you're shifting it all to the right. A good example that helps me is to imagine y=x^2 as f(x). If you then do f(x-2) you get the equation y=(x-2)^2 instead of y=x^2-2.(2 votes)

- I'm seeing lots of questions about why is c negative when we are shifting right? I found a really simply answer on another vid. When we move something "c" units right, then our x value is c units greater than our y value. For example, if (0,0) is shifted to (2,0) which is 2 units right, the x value is two units greater than the y value. This is the same thing as saying, x = 2 + y, or in more general format, x = c + y. However, whenever we use f(x) = something, we are finding what the y value is in terms of x, not how the x value is in terms of y. If x = c + y, then y = x - c. Thus for a horizontal shift, we always think "opposite sign".

This is extra, but if we wanted to know what to do for a vertical shift instead, then it's just how you'd expect it. If we move c units up, we get: y = x + c. If we move c units down, we get y = x - c. The reason its simple is because vertical shift has to do with the y-axis which refers to f(x). The only reason horizontal shift is different is because when we think "horizontal", our mind tends to think "x" and not f(x).(1 vote) - So then, if the integral of f(x) is equal to 5, the integral of f(x-c) would also be equal to five? Right?(1 vote)

## 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.