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### 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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# Indefinite integrals: sums & multiples

An indefinite integral of a sum is the same as the sum of the integrals of the component parts. Constants can be "taken out" of integrals.

## Want to join the conversation?

- Sal explained that the definite integral is the area under the curve from a to b (a is an under bound and b is an upper bound). However, there is no such thing as under bound or upper bound in the indefinite integral. Then what does indefinite integral refer to in the graph?(5 votes)
- Suppose we have a function, f(x). The indefinite integral of the function will be another function, F(x), such that F(c) is equal to the area under the curve generated by f(x) between x=0 and x=c.(5 votes)

- When we are dealing with integrals and using the symbol ∫, how am I going to know if it is a definite integral or an indefnite integral? Is there some way to distinguish the two?(1 vote)
- Definite integrals have bounds of integration written on the top and bottom of the integral symbol while indefinite integrals do not.(11 votes)

- If we have a function like
`4x+7`

and then we take the anti derivative, shouldn't this mean the 4x becomes`2x^2 + c`

and the 7 becomes`7x+c`

causing the final equation to be`2x^2 + 7x +2c`

?(2 votes)- c is an arbitrary constant, so multiplying it by another constant does not matter and we can remove the factor of 2.(7 votes)

- Definite integrals have the same properties as indefinite integrals, don't they ?(4 votes)
- If you're algebraically doing integration, these properties will work with either type.(2 votes)

- can someone explain to me how the reverse power rule works for constant numbers like 8 or 9?(2 votes)
- Imagine 8 as 8x^0. Integrating this using the reverse power rule, we get [8x^(0+1)]/(0+1) = 8x. You can verify that 8x is the antiderivative of 8 by differentiating 8x.(6 votes)

- Is this really a proof of the two properties? I can't see how.(4 votes)
- at1:38, if it were definite integral, then it makes sense. so i assume this property applies to indefinite integral too(2 votes)
- Indeed! Most properties of integrals are common between a definite and indefinite integral. It's just that in the former, you have to do an extra step of plugging in numbers, while that doesn't need to be done in the latter.(3 votes)

- why are we taking derivatives what's the point feels like waste of time(2 votes)
- We're just deriving different results. Plus, it's a good way to see the Fundamental Theorem in action!(2 votes)

- in4:07is it the the fundumental theorom that sal is applying?because there is a constant there and why are we allowed to treat constant like that?(2 votes)
- That is not the fundamental theorem of calculus. Sal showed us that property in this video, so you probably didn't know it beforehand.(1 vote)

- If I'm asked to find the antiderivative of 4x+7, is that sum rule? I'm not sure because its a variable plus a constant rather than adding two variables. But I don't know what else to do...(1 vote)
- This late reply is for others who may have the same question. It's a very good question. I am just a student here myself, but perhaps I can help. To me, it is useful to think of 7 as 7x^0.

x^0 = 1 and 1 * 7 = 7; therefore 7 = 7x^0.

You could then apply the sum rule and the reverse power rule.

int.(4x + 7) dx = (4x^(1+1))/(1+1) + C1 + (7x^(0+1))/(0+1) + C2

= (4x^2)/2+ C1 + (7x^1)/1 + C2 = 2x^2 + 7x + C(3 votes)

## Video transcript

- [Instructor] So we have listed here are two significant properties
of indefinite integrals. And we will see in the future that they are very, very powerful. All this is saying is
the indefinite integral of the sum of two different functions is equal to the sum of
the indefinite integral of each of those functions. This one right over here
says the indefinite integral of a constant, that's not
gonna be a function of x, of a constant times f
of x is the same thing as the constant times the
indefinite integral of f of x. So one way to think about it is we took the constant out of the integral, which we'll see in the future, both of these are very useful techniques. Now, if you're satisfied with
them as they are written, then that's fine, you can move on. If you want a little bit of a proof, What I'm going to do
here to give an argument for why this is true is use
the derivative properties. Take the derivative of both sides and see that the equality holds once we get rid of the integrals. So let's do that. Let's take the derivative
with respect to x of both sides of this. Derivative with respect to x. The left side here, well,
this will just become whatever's inside the indefinite integral. This will just become f of x plus g of x plus g of x. Now what would this become? Well, we could just go to
our derivative properties. The derivative of the sum of two things, that's just the same thing as
the sum of the derivatives. So this'll be a little bit lengthy. So this is going to be the
derivative with respect to x of this first part plus the
derivative with respect to x of this second part. So this first part is the
integral of f of x dx, we're gonna add it. And this is the integral of g of x dx. So let met write it down. This is f of x and then this is g of x. Now, what are these things? Well, these things, let me
just write this equal sign right over here. So then this is going to be equal to the derivative of this with respect to x is just going to be f of x. And then the derivative
with respect to here is just going to be g of x. And this is obviously true. So now let's tackle this. Well, let's just do the same thing. Let's take the derivative of both sides. So the derivative with
respect to x of that and the derivative with
respect to x of that. So the left-hand side will
clearly become c times f of x. The right-hand side is going to become, well, we know from our
derivative properties, the derivative of a
constant times something is the same thing as the
constant times the derivative of that something. So then we have the integral, indefinite integral of f of x dx. And then this thing is
just going to be f of x. So this is all going to be
equal to c times f of x. So once again, you can see that
the equality clearly holds. So hopefully this makes you feel good that those properties are true. But the more important thing
is you know when to use it. So, for example, if I were
to take the integral of, let's say, x squared plus cosine of x, the indefinite
integral of that, we now know it's going to
be useful in the future. Say, well, this is the same thing as the integral of x squared dx plus the integral of cosine of x dx. So this is the same
thing as that plus that. Then you can separately evaluate them. And this is helpful, because
we know that if we are trying to figure out the integral of, let's say, pi times sine of x dx, then we can take this constant out. Pi is in no way dependent on x, it's just going to stay be equal to pi. So we can take it out and
that is going to be equal to pi times the integral of sine of x. Two very useful properties, and hopefully you feel a lot
better about them both now.