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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 6

Lesson 8: Finding antiderivatives and indefinite integrals: basic rules and notation: reverse power rule- Reverse power rule
- Reverse power rule
- Reverse power rule: negative and fractional powers
- Indefinite integrals: sums & multiples
- Reverse power rule: sums & multiples
- Rewriting before integrating
- Reverse power rule: rewriting before integrating
- Reverse power rule review

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# Reverse power rule

Can you find a function whose derivative is x^n? Created by Sal Khan.

## Want to join the conversation?

- Why didn't he use quotient rule here?(45 votes)
- The reason is that n should NOT be viewed as a variable. even though we do not know what n is, we treat n as a constant. He is simply deriving the formula for integrating x raised to some exponent. The idea is that the formula works for any exponent, but when you actually do a specific problem n would be an actual number. So the quotient rule is not used because x is actually being divided by a constant, we just do not know what the constant is because the idea is that it could be any number and the formula would still work.(62 votes)

- Here's a question, at4:47, is it truly necessary for you to factor out the five before continuing the problem? Or can you simply take the anti derivative of the expression with the 5 in the expression?(17 votes)
- Since the 5 is a constant, you can factor it out or leave it inside the integral. It won't change the answer in the end. However, it is common practice to pull constants out of the integral because it (usually) makes taking the integral easier to read and you can find any mistakes easily if you made them.

tl;dr version, it's completely optionally(49 votes)

- What is the significance of the dx at the end of ∫(f(x))dx. I understand that dx means an infinitesimally small change in x, but what role does it play (or maybe I should ask, what does it represent) in the integration? Thanks.(5 votes)
- To add to what redthumb.liberty said, which is quite true, there is something that beginning integral calculus students often get wrong regarding the dx.

The`dx`

is not just some notation that you can tack on. You may interpret the`dx`

as the derivative of the variable`x`

. This gets important when you start learning standard integral forms.

For example, here is a standard integral form:`∫ cos (u) du = sin (u) + C`

So, some students will incorrectly see:`∫ cos (x²) dx`

and say its integral must be`sin (x²) + C`

. But this is wrong. Since you are treating x² as the u, you must have the derivative of x² as your du. So, you would need`2xdx = du`

. Thus, it is`∫ (2x)cos (x²) dx = sin (x²) + C`

Whereas,`∫ cos (x²) dx`

because it doesn't have the derivative of`x²`

, is an advanced level integral that couldn't possibly be solved by any method taught in an introductory integral calculus course.(16 votes)

- Is there a name for the symbol used for representing the anti derivative?(4 votes)
- Well, there are two ways of representing the antiderivative.

Generally, the lowercase letter is used for functions; e.g.: f(x). The letters here should also be italicized, but Khan Academy does weird things with the formatting if I were to try to do that.

Anyways, the antiderivative of f(x) is often written as F(x). Thus, F'(x) = f(x). This really cannot be used for anything other than indeﬁnite integrals (which is what antiderivatives are).

The integral sign, ∫ has a bit more of a story for it. This symbol is more speciﬁc to integrals. When this symbol is used for an indeﬁnite integral, one could almost say that an antiderivative is being written in the form of an integral.

A deﬁnite integral basically ﬁnds the sum of an inﬁnite number of parts (learn about that in the next section). Because of this, it would make sense to use an S as the symbol for integration--- S for*sum*, just like Σ (a Greek S) is used for summation.

Back in the day (over a century ago), English (and other languages) used two forms of a lowercase S. There was a long S, written as ſ (basically an f without the cross-stroke), and there was a terminal S, written as s. The long S was used at the beginning and the middle of a word. To illuſtrate this, I will ſtart uſing it. The terminal S was uſed at the ends of words. Now, perhaps you’ll notice that an italicized f looks kind of like an integral ſign with a croſs-ſtroke. Well, when you italicize an ſ, you baſically get an integral ſign. This is actually where it comes from. In fact, the integral ſign is really the only place where the long S is even uſed anymore.

The integral sign is an italicized long S---the S stands for sum.

Juſt a kind of ſide-bar, if you are familiar with German, you’ve probably ſeen that they have an S symbol that looks kind of like a B. Well, look at the long S and terminal S next to each other: ſs. Now, imagine them being merged into one character: ß. There you go!(16 votes)

- At1:03: why don't you differentiate n+1?(3 votes)
- Because n isn't changing with respect to x.(18 votes)

- If n cannot be -1, how will you integrate x^(-1), aka 1/x?(5 votes)
- The actual antiderivative of 1/x is the absolute value of ln|x|, commonly mistaken with ln(x). This is due to the end behavior of the graph of the natural log of x and the domain and range of the graph. According to the AP course, usually ln|x| is accepted. I hope this helps, and please do feel free to ask further questions as needed.(8 votes)

- This has no relation to the video,but it was on my mind What is the inverse of

x^x(x powered x)

Plz show some steps to proceed(3 votes)- That is an advanced-level problem that requires the Lambert W-function (also known as the product log) to solve

Let us define a function of x, W(x) such that

W(xe^(x)) = x

(This is known as the Lambert-W function or product log and there are a variety of equivalent ways it can be defined.)

y = x^x

Invert: x = y^y

log x = y log y

note that y = e^(log y)

log x = log y (e^(log y))

W(log x) = W{log y (e^(log y))}

W(log x) = log y

e^W(log x) = y

Note: This can be rewritten as y = log (x) / W(log x)

Writing it this way requires using a common property of the W function:

e^[W(u)] = u / W(u)(8 votes)

- What happens if you have an original function f(x) = x^1. The derivative of this would be just x^0, which is the same as f'(x) = 1. So how do you find the integral of f'(x)? Do you just say f'(x) * dx, and then put (x^1)/1 in the function?(4 votes)
- Yep, just remember that if it is an indefinite integral, you would also have the + C at the end of it.
`f(x) = x`

f'(x) = 1 dx

integral of f'(x) dx = x + C(5 votes)

- Do we always need to write down the constant 'c', right after the equation? If this is so, then why can't we write down the answer as "-5x^-1" instead of "-5x^-1 + c"?(3 votes)
- The c acknowledges that the solution is a general one. If in a particular instance you determined that a particular solution had the value c=0, then you could write -5x^-1 and it would be interpreted and a particular solution. The differences between them will become clearer as you study Differential Equations, for which the solutions are not values, but functions.

For instance:

https://www.khanacademy.org/math/differential-equations/first-order-differential-equations/separable-equations/v/particular-solution-to-differential-equation-example

https://www.khanacademy.org/math/differential-equations/second-order-differential-equations/linear-homogeneous-2nd-order/v/2nd-order-linear-homogeneous-differential-equations-2(5 votes)

- So what is the antiderivative of x^n, when n is -1?(1 vote)
- ln x + c.

The derivative of ln x is x^-1, which you can find through implicit differentiation.

Start with y = ln x.

This is the same equation as x = e^y.

Differentiate both sides with respect to x, and obtain 1 = (e^y)*y'

Solve for y': y' = 1 / e^y

Since y = ln x: y' = 1 / e^(ln x)

and since e^(ln x) = x: y' = 1 / x.

So 1 / x is the derivative of ln x,

which means ln x + c is the (most general) antiderivative of 1 / x.(5 votes)

## Video transcript

Let's take the
derivative with respect to x of x to the n
plus 1-th power over n plus 1 plus some constant c. And we're going to
assume here, because we want this expression
to be defined, we're going to assume that
n does not equal negative 1. If it equaled negative
1, we'd be dividing by 0, and we haven't defined
what that means. So let's take the
derivative here. So this is going to be equal
to-- well, the derivative of x to the n plus 1
over n plus 1, we can just use the
power rule over here. So our exponent is n plus 1. We can bring it out front. So it's going to be n
plus 1 times x to the-- I want to use that same color. Colors are the hard part--
times x to the-- instead of n plus 1, we subtract
1 from the exponent. This is just the power rule. So n plus 1 minus
1 is going to be n. And then we can't forget that we
were dividing by this n plus 1. So we have divided by n plus 1. And then we have plus c. The derivative of a constant
with respect to x-- a constant does not change as x
changes, so it is just going to be 0, so plus 0. And since n is not
equal to negative 1, we know that this is
going to be defined. This is just going to
be something divided by itself, which is
just going to be 1. And this whole thing
simplifies to x to the n. So the derivative of
this thing-- and this is a very general terms--
is equal to x to the n. So given that, what is
the antiderivative-- let me switch colors here. What is the antiderivative
of x to the n? And remember, this
is just the kind of strange-looking
notation we use. It'll make more sense when we
start doing definite integrals. But what is the
antiderivative of x to the n? And we could say the
antiderivative with respect to x, if we want to. And another way of calling this
is the indefinite integral. Well, we know this
is saying x to the n is the derivative of what? Well, we just figured it out. It's the derivative
of this thing, and we've written it
in very general terms. We're actually encapsulating
multiple constants here. We could have x to
the n plus 1 over n plus 1 plus 0, plus 1, plus
2, plus pi, plus a billion. So this is going to be equal
to x to the n plus 1 over n plus 1 plus c. So this is pretty powerful. You can kind of view this
as the reverse power rule. And it applies
for any n, as long as n does not equal negative 1. Let me make that very clear. n does not equal negative 1. Once again, this thing
would be undefined if n were equal to negative 1. So let's do a couple
of examples just to apply this-- you could
call it the reverse power rule if you want, or
the anti-power rule. So let's take the antiderivative
of x to the fifth power. What is the antiderivative
of x to the fifth? Well, all we have to
say is, well, look, the 5 is equal to the n. We just have to increment
the exponent by 1. So this is going to be equal
to x to the 5 plus 1 power. And then we divide
by that same value. Whatever the exponent was
when you increment it by 1, we divide by that same
value, divided by 5 plus 1. And of course, we
want to encapsulate all of the possible
antiderivatives, so you put the c
right over there. So this is going to be equal to
x to the sixth over 6 plus c. And you can verify. Take the derivative of
this using the power rule, you indeed get x to the fifth. Let's try another one. Let's try-- now
we'll do it in blue. Let's try the antiderivative
of-- let's make it interesting. Let's make it 5 times x to
the negative 2 power dx. So how would we evaluate this? Well, one simplification
you can do-- and I haven't rigorously
proven it to you just yet-- but we know that scalars
can go in and out of the derivative operator when
you're multiplying by a scalar. So this is, indeed, equal to
5 times the antiderivative of x to the negative
2 power, dx. And now we can just use,
I guess we could call it this anti-power rule,
so this is going to be equal to 5 times x to
the negative 2 power plus 1 over the negative 2 power plus
1 plus some constant right over here. And then we can rewrite this as
5 times negative 2 power plus 1 is x to the negative 1
over negative 2 plus 1 is negative 1,
plus some constant. And this is equal to 5 times
negative x to the negative 1 plus some constant. And then if we want, we
can distribute the 5. So this is equal to negative
5x to the negative 1. Now, we could write plus
5 times some constant, but this is just an
arbitrary constant. So this is still just
an arbitrary constant. So maybe we could
[INAUDIBLE] this. If you want it to show that
it's a different constant, you could say this
is c1, c1, c1. You multiply 5 times c1,
you get another constant. We could just call that c,
which is equal to 5 times c1. But there you have it. Negative 5x to the
negative 1 plus c. And once again, all of these,
try to evaluate the derivative, and you will see that you
get this business, right over there.