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# Indefinite integrals: sums & multiples

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.C (LO)
,
FUN‑6.C.1 (EK)
,
FUN‑6.C.2 (EK)
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?
(4 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.
(4 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.
(8 votes)
• Is this really a proof of the two properties? I can't see how.
(4 votes)
• Definite integrals have the same properties as indefinite integrals, don't they ?
(3 votes)
• If you're algebraically doing integration, these properties will work with either type.
(2 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.
(2 votes)
• in is 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)
• 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)
• What's the difference between definite integrals and indefinite integrals ?
(1 vote)
• An indefinite integral results in a set of functions whose derivatives are equal to the integrand.
∫𝑓(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝐶
𝐹 '(𝑥) = 𝑓(𝑥)

A definite integral is when we evaluate 𝐹(𝑏) − 𝐹(𝑎), which gives us the area under 𝑓(𝑥) over the interval [𝑎, 𝑏].
∫[𝑎, 𝑏] 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎)
(2 votes)
• I don't quite understand how the derivative cancels out the indefinite integral. Sal didn't mention a property like that in previous videos.
(1 vote)
• According to the first part of the fundamental theorem of calculus, the definition of an indefinite integral of a function f is another function F such that F'(x) = f(x).
(2 votes)
• Starting at , how does Sal just write that d/dx[∫(f(x) + g(x))dx] is equal to (f(x) + g(x))? I feel like I'm missing something very obvious here?
(1 vote)
• d/dx(∫(f(x) + g(x))dx) = d/dx(∫f(x)dx + ∫g(x)dx) = d/dx(∫f(x)dx) + d/dx(∫g(x)dx) = f(x) + g(x)
(1 vote)

## 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.