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# Definite integral of shifted function

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

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• 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... •  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.
• 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 • 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`` ` we can directly read that the vertex is at (2, 0) or we can plot the curve by making a value table:
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.
• Why is there a dx at the end of each function? • 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.
• is f(x-2) same as y=x-2? • Is their a video containing information regarding shifting of graphs if their is can someone can navigate me to specific topic. • 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? • 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? • 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) • 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) 