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### Course: Integral Calculus (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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# Visually determining antiderivative

Given the graph of a function, can you identify the graph of its antiderivative? Created by Sal Khan.

## Want to join the conversation?

- Is e^x (and I suppose n*e^x as well) the only type of function that results in itself when you take the (anti-)derivative? Or are there others?(15 votes)
- On differentiation:

a*e^(x + b) is the only function that has a "differentiating period" of 1, so to say.

e^(-x + b) has a period of two: its first and second derivatives are -e^(-x + b) and e^(-x + b). But e^(x + b) also has a period of two, so the 'general' period-2 function is a*e^(x + b) + c*e^(-x + d).

What about a period of four? sin and cos have that, so the 'general' period-4 function is a*sin (x + b) + c*cos (x + d).

What about other periods? Here you'll need complex numbers: more specifically roots of unity. A n:th root of unity is a number u such that u^n = 1, and there are always n of them. The first root of unity is 1; the two 2nd roots of unity are 1 and -1; the three 3rd roots of unity are 1, (1 + i sqrt 3)/2 and (1 - i sqrt 3)/2; the four 4th roots of unity are 1, i, -1 and -i; etc.

A function with a "differentiating period" of n satisfies the following differential equation:

y^(n) = y, where ^(n) means the n:th derivative.

Once you know how to deal with differential equations, it's fairly straightforward to show that the solution to that differential equation is:

y = ∑{k = 1 to n} a_n * e^(u_n * x + b_n)

where a_n and b_n are arbitrary parameters and u_n are the n n:th roots of unity.

For antiderivatives, there is no such function, because of the constants of integration.

The first antiderivative of e^x is e^x + C; the second, e^x + Cx + D; the third, e^x + Cx^2 + Dx + E; etc. They start building up a polynomial tail.(16 votes)

- At2:20, how is the slope of the first graph close to 1?(3 votes)
- If you draw the tangent line (as Sal did), it appears to be inclined at about a 45 degree angle. Since the arctangent of 45 degrees is 1, the slope is about 1. The rise and run between any two points are roughly equal, so the slope is about 1. Choosing two points and calculating the slope between them gives about 1. Choose your method for getting the slope, but using any method you like, the slope at the given point is close to 1.(12 votes)

- Even though I saw all of the videos completely, that come before the next exercise, I could get, like, 1 right in 5 in the exercise. I guess sometimes on Khan Academy the videos maybe have just 1 example, OR the exercises are lifted half a level higher than the videos. I guess that's just not cool. maybe u guys should at least, like, describe all the concepts that come in the exercise, before the exercise.

I guess I was not clear in my question, but I don't know.(2 votes)- Don't expect that you can just watch one video and then get 100% of the questions correct all the time. The exercises are helping to connect the video you just watched with previous concepts that you still need to know. I would guess that most of the concepts are described before the exercise, but it is including every concept from all the years of content before Integral Calculus.

As @TripleB491 said, make sure you closely go over the hints for the exercises if you get them wrong. Even go over the hints when you get it right, just to make sure you're following the same or similar steps.(6 votes)

- at1:30

Why the function needs to have a positive-slope if the Derivative of it F'(x) has a positive slope?(2 votes)- If your function has a positive slope, then the derivative will have a positive
but not necessarily will the derivative a positive slope as well.**value**

Conversely: if your derivative has a positive(regardless of the derivatives own slope), then the function has a positive slope.**value**(3 votes)

- If the antiderivative of e^x with respect to x is equal to e^x, then the antiderivative of
*e^(-1/x)*with respect to x should be e^x?(1 vote)- Er... No. I'm not sure what path you took to get there; if you'd like, I can check to see what happened. But, interestingly, the function
`e^(-1/x)`

has**no**elementary antiderivative. We actually have to express it in terms of an "exponential integral function", which (you guessed it) is defined based on the integral of a function similar to the one you've written. More about that can be found here: http://functions.wolfram.com/GammaBetaErf/ExpIntegralEi/ .

One interesting phenomenon in integral calculus is that some functions that are deceptively simple to write end up having no antiderivative that can be expressed in terms of elementary functions (which are pretty much the only things covered up to this point in most math curricula, or at least the one I went through -- polynomials, trigonometry, logarithms, exponentials, rationals, complex numbers). Another example is the set of "Fresnel integrals," the integrals of`cos(x^2)`

and`sin(x^2)`

-- those functions don't have elementary antiderivatives, either.(3 votes)

- Why sal starts to compare these four graphs? Do we have the right to force all those graphs to have the same X and Y values?(2 votes)
- He was not comparing the four graphs. He was just eliminating the wrong answers by evaluating them. In this case, the only graph that has no flaws upon evaluation is the correct graph as an answer to the question. No one has the right to force such graphs, which is the reason why he used the technique I described above so that decisions are backed by reasoning.(1 vote)

- i didn't understand the following:

1) why is the slope positive?

2) why the third graph has negative slope?(1 vote)- 1) The slope, or the instantaneous rate of change, is positive because the function increases as x also increases. The slope is the change in the value of the function, divided by the change in x. (+)/(+)=(+)

2) Remember that the value of the derivative at a given value of x is the slope of the function. derivative value = slope. The derivative is negative, so the slope is negative.(2 votes)

- what is the difference between capital and lowercase?(1 vote)
- f(x) is the function of the graph on the left, it is a derivative of F(x) which is another function. You can also say that F(x) is the antiderivative of f(x). Sal is trying to find the graph of F(x) which is one of the four graphs on the right.(2 votes)

- isnt the slope of the first graph on the left corner negative..and how at2:17, does SAL say that the slope is close to 1?.?(1 vote)
- it's positive because as x increases y increases, too(2 votes)

- The domain of "e" is x>=0, how come we have value when x<0 ? what's the function of it?(1 vote)
- e is just a natural number, equal to 2.71828 18284 59045 23536...

It is irrational, meaning it cannot be expressed as a ratio of two integers, and it is non-repeating.

e is not a function; just a number.

As such, it does not make sense to speak of e having a domain or range like a function.

f(x) = e^x has the domain -∞ < x < +∞ and the range 0 < f(x) < +∞(2 votes)

## Video transcript

Let's say that this right over here is the
graph of lowercase f of x. That's lowercase f of x there. And let's say that we have some other
function capital F of x. And if you were to take its derivative,
so, capital F prime of x, that's equal to lowercase f of x, lower
case f of x. So given that, which of these, which of
these, could be the graph of capital, of capital F of
x? And I encourage you to pause this video,
and try and think about it on your own before we work
through it. Well if, if this curve is going to be the
derivative of one of them. That means that any, for any x value it's describing what the instantaneous
rate of change. Or what the slope of the tangent line is,
of which ever one of these is the possible capital
F of x. So, let's just look at a couple of things
right here so what, what do we know about lower case
f of x? What do we know of which is the derivative
of one of these? Well, one thing we know is it's always
positive. It, it has as we go to negative infinity,
it asymptotes towards 0. But it's always positive. So since this is describing the slope of
one of these. That means that the slope of one of these
always, or out of the candidates, has to always be
positive. And, if we look at this, the slope of the tangent line here is, indeed,
always positive. The slope of the tangent line here does
look like it's positive. Every time we increase an x, we're
increasing by y. Here it's positive. But here it's negative. When we increase by x, we decrease by y. So, we can rule, we can rule this one out. Now, what else, what else do we know? Well, this is the derivative. This is telling us the slope of the
tangent line. So, for example, when x is equal to, when
x is equal to negative 4, f of, f of negative 4 is pretty close to
0. It's pretty close to 0. So, it's slightly, slightly more than 0. So, that tells us that the slope of
tangent line of capital F of x has to be pretty close to 0, when x
equals negative 4. So, let's see, when x equals negative 4,
the slope of tangent line, here, isn't close to 0, this
actually looks closer to 1. So, we could rule this one out. Over here, when x is equal to negative 4,
the slope of the tangent line, yeah, that actually does
look pretty close to 0. So I won't rule that one out. And over here, the slope of the tangent
line, when x is equal to negative 4, that also looks
pretty close to 0. So these are still both in the running. So let's see how we can think of it
different. So let's just pick another point. When x is equal to, when x is equal to 0,
f of 0 looks like it's pretty close to 1. I don't know if it's exactly to 1. Actually it looks almost exactly. Almost exactly equal to 1. So when capital F of, so at capital F of
0, the slope of the tangent line needs to be
pretty close to 1. So over here the slope of the tangent
line, when x is equal to 0, that looks smaller than
1. So this slope is definitely not 1. While over here, when x is equal to 0. The slope of the tangent line does look,
the slope of the tangent line does look pretty, pretty close,
pretty close to 1. So this, right over here, looks like the best candidate for capital, for capital F
of x. So that one right over there. Lemme, that is capital F of x. And you might say hey these look similar
to each other. And fact they look almost or actually they
do look almost identical. And you might remember from what you knew
about differentiation, that these actually look
like the basic exponential function. Were, were I didn't ask you to find out
what the actual function was just the possible
anti-derivative of this function would be. This is the derivative, lower case f is
the, is the derivative of capital f, or you could say that capital f is an
anti derivative of lower case f. And when you just inspect this, this looks
like this, the, the function, both of these functions is,
are e to the x. Because the derivative of e to the x is e
to the x.