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## Calculus, all content (2017 edition)

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

Lesson 2: Indefinite integrals intro- Reverse power rule
- Reverse power rule
- Reverse power rule: negative and fractional powers
- Rewriting before integrating: challenge problem
- Reverse power rule: sums & multiples
- Visually determining antiderivative
- Graphs of indefinite integrals
- Reverse power rule review

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# Graphs of indefinite integrals

More practice identifying the graph of the antiderivative of a function.

## Want to join the conversation?

- Has Sal done videos on transformations of graph? I mean when do we shift the graph, when do we scale up/down the graph etc?

Can anyone, please provide the link to understand this concept of scaling/shifting the graph?(9 votes)- Ajmal, right here: https://www.khanacademy.org/math/algebra2/manipulating-functions/shifting-functions/v/shifting-and-reflecting-functions

There are a ton of videos and pretty cool practice exercises that give students a more intuitive sense of how algebraic changes affect the position and shape of the subsequent graph.(15 votes)

- why there is dx in the expression, and also what is it for.?(4 votes)
- In the case of integrals, the
`dx`

tells us what our variable of integration will be; in this case, we're integrating with respect to x. More generally, the`dx`

is a "differential element of x", which represents an infinitesimal, near-0 change in the value of x.

Right now, you've probably only seen integrals with one variable, but as you get further into calculus, you'll come across things like u-substitution and multiple integrals where it's important to define which variable will be used for the integration.

Lastly, if you're curious: although this isn't rigorous, we can think about differentials this way. Say you have a function`f`

and its derivative`df/dx`

(that's the Leibniz notation for derivatives, which looks like a fraction and can be useful for visualizing operations like this "algebraically"). If we want to find the antiderivative of`df/dx`

, we need to sum up all of the infinitesimal changes in`f`

-- in other words, we need to find the sum of all of the`df`

s. So we say that`df/dx = df/dx`

(that's just an identity), then we multiply through by dx to get`df = (df/dx) dx`

. Then by integrating both sides, we get`int()df = f = int(df/dx) dx`

.

I hope you find some of that helpful.(9 votes)

- Hello, sorry but I do not understand yet. Why did you not chose D?(2 votes)
- I think that he didn't choose the answer "D" because it is supposed to be connected to the "y" part of the graph and that one is not centered in the graph. So it would not be directly on top of the line of "y"(2 votes)

- why doesn't sal do anything with the dx(1 vote)
- The
`dx`

in integral notation represents an infinitesimal (a number that is approaching zero, but will never equal zero). You may want to review the basic integral calculus videos.

https://www.khanacademy.org/math/integral-calculus/indefinite-definite-integrals(3 votes)

- can you use substitution to solve indefinite integral,if yes how?(1 vote)
- If you know how to do substitution for any definite integral, then you know how to do it for any indefinite integrals and visa versa.

Is there a specific indefinite integral example you have in mind?(3 votes)

## Video transcript

- [Instructor] Find the
general indefinite integral. And so we have the integral of two x dx. Which of the graphs shown below, which of the graphs below shows several members of the family? So if we're talking about, so if we're taking the
integral of two x dx, we're talking about the
antiderivative of two x. And what's that going to be? Well, it's going to be two x to the second power. 'Cause this was two x to the first power, so we increment the exponent to two. And then we divide by that
newly incremented exponent, so this is going to be x squared. And you might have done that
on your own, you said, okay, I know that the derivative
of x squared is two x, so the antiderivative
of two x is x squared. But we aren't quite done yet. Because remember, this isn't the only
antiderivative of this. We could add any constant here. If we add some constant here and we take the derivative
of it, we still get two x. Because the derivative of a
constant with respect to x, it's not changing with respect to x, so its derivative is zero. So the antiderivatives, I
guess you could say here, take this form, take the form of x squared plus C. Now what does that mean visually? So, let me draw, I can draw
a neater version of that. So slightly better. So if that's my y-axis and this is my x-axis, we know what y equals
x squared looks like. Y equals x squared looks like, I'll just draw the general shape. So y equals x squared looks like this. Now, what happens if I add a C? Let's say if I add, let's say y is equal
to x squared plus two. And two is a valid C, so we could say, so let me write this down. This right over here is
y is equal to x squared. But remember, and I
guess you could say that, in this case, our C is zero. But what if our C was some positive value? So let's say it is y is equal to x squared plus, I don't know, y is
equal to x squared plus five. Well, then we're going to have
a y-intercept here at five, so essentially we're just
gonna shift up the graph by our constant right over
here, which is positive five. So we shift up by positive five, and we will get something
that looks like this. We just shifted it up. Now, you might be saying, okay, well, that kind of looks like
this choice right over here. But this choice also, also has some choices that start down here, that
we're adding a constant. But you remember, this
constant can be any constant. It could be a negative value. So in this case, C is five. In this case, C is zero. But C could also be negative five. So C could also be negative five. So if we wanted to do y is equal to x squared plus negative five, which is really x squared minus five, then the graph would look like this. It would shift x squared down, down by five. So this one is shifted up by five. This one is shifted down by five. So you would shift by the constant. If it's a positive
constant, you're going up. If it's a negative constant,
you are going down. So B is definitely the class of solutions to
this indefinite integral. You take any, any of the functions that are represented by these graphs, you take their derivative,
you're going to get two x. Or another way to think about it, the antiderivative of
this or the integral, the indefinite integral of two x dx is gonna be x squared plus C, which would be represented
by things that look like, so essentially things, essentially y equals x
squared shifted up or down. So I could keep drawing
over and over again.