Indefinite Integrals: Study with Video Lessons, Practice Problems & Examples

Introduction to Indefinite Integrals

In recent videos, we've spent some time talking about antiderivatives. And recall that this is just the reverse process for taking a derivative. Now what we're going to learn in this video is the concept of something known as the indefinite integral. Now this might sound like something that's totally scary and abstract and new, but don't worry about it. Because in this video, we're going to learn that the process for the indefinite integral is actually the exact same process as taking the antiderivative.

So most of this video will probably seem pretty straightforward. The only real difference is learning this new notation that we use for dealing with integrals. And since you're very commonly going to see this in this course in the future for problems that we see, let's go ahead and just jump right into things. So this is what an indefinite integral looks like. Notice how it's just this kind of elongated s symbol that we see.

This is the operation, which means we need to do this antiderivative process, which is known as the indefinite integral. Now the function that we're dealing with, that we're integrating, is what's known as the integrand. So this is the function that we see inside. And recall that if we have this function, little f of x, where we see this lowercase f here, then the antiderivative is big F of x. So what happens is if we integrate this function right here, we're going to get the antiderivative, big F of x, plus this constant c that we've seen before.

Now we actually have a name to this constant now, it's called the constant of integration. And that's because when we do this integral, we get this constant that pops out. Now another thing that's new is this piece right here. Notice we have this little dx. What exactly does that mean?

Well, this little dx is what's known as the variable of integration. And all the variable of integration tells us is what is what variable we're integrating. So that means this function is going to be dealing with x's, like 3x squared or 2x or something like that. If we solve something like a y here instead, that means we'd have like 3y squared or 2y or something like that. So this just tells us what we're integrating with respect to.

So this is what it looks like. And so when finding the indefinite integral of this function, you get the antiderivative and the constant. So, it's literally just the same process as taking the antiderivative. We just have this new elongated s symbol and this variable of integration. So let's actually jump into some examples to see what this indefinite integral looks like.

And I think you're going to find this is a very familiar process. So we'll start with this first example, a here, where we're asked to find the indefinite integral of 0. What exactly would this be? Well, this would be the integrand that we're dealing with right here. We've seen this integrand as this inside function, which means we need to find the antiderivative of the 0.

And recall that if we take the derivative of a constant, it will give us 0. And so that means that the integral of 0 would give us a constant. It's just the reverse process. So this right here would be our solution, and that would be the answer, for example, a. We just get some constant c that pops up.

We don't know what that constant is, but we're just going to get a constant. And this is what happens whenever you take the integral of 0. Whenever you take the integral, that's just going to give you a constant because the antiderivative of any of 0 is a constant because the derivative of any constant is 0. So it's just going backwards. Now we'll move to example b.

Example b asks us to find the indefinite integral of 3 with respect to x. Now notice here how we're integrating a constant. If we take the antiderivative of some constant number, all we need to do is slap an x onto that number. Because remember that taking the derivative of 3x would just give us 3. So taking the integral of 3 would give us 3x, and then you need to add the plus the constant of integration c.

You always want to add a plus c when you're doing these indefinite integrals or antiderivatives. So this is what happens whenever you take the integral of a constant number. If you take the integral of a constant, you're just going to take that constant and multiply it by x, and then, of course, add the plus c. And this will always be the case when you're dealing with the indefinite integral. Now for this last example here, example c, we're asked to find the indefinite integral of 3x squared.

Now how exactly could we deal with this? Well, this is a situation where we need to think about, well, could we have taken the derivative of to give us 3x squared? And looking at this, this kind of looks like a power rule was done. Because you see we have this number 3 in front, then we have the power which is reduced by 1. It's 3 reduced by 1, which is 2.

So what I'm going to do is say that my integral is x cubed. And if I think about it, the derivative of x cubed using the power rule would give us this 3x squared. So x cubed plus the constant of integration c would be the solution to this example. So this is how you can solve all three of these situations where we're dealing with integrals. And just like we've seen for antiderivatives, you can check your answer of the indefinite integral by taking the derivative of whatever the function it is you were dealing with.

So going back up here, we can just take the derivatives to see if we got these correctly. So what I can do is take the derivative of a constant c. The derivative of any constant is going to give me 0, which matches with the integrand up here, so that's correct. We can also take the derivative of this function, 3x+c. The derivative of 3x with respect to x is 3.

The derivative of any constant is 0, and 3 matches the integrand right there. And then lastly, we can take the derivative of this function over here. The derivative of x cubed plus c, that's just going to give me 3x squared, because that's being using the power rule here that's going to give me the 3x squared. The derivative of any constant is 0, which matches with what we had in the integrand up here. So this right here would be the solution, that's how you can check your answers and do indefinite integrals.

Hope you found this video helpful. Let's try getting a bit more practice.

Find the following indefinite integral.

$\int-300dx$

• A. $-300x+C$
• B. $-300x$
• C. $300x+C$
• D. $-300+C$

Find the following indefinite integral.

$\int4x^3dx$

• A. $x^4$
• B. $x^4+C$
• C. $\frac43x^4+C$
• D. $\frac43x^4$

Find the following indefinite integral.

$\int100x^{99}dx$

• A. $\frac{100}{99}x^{100}$
• B. $x^{100}$
• C. $\frac{100}{99}x^{100}+C$
• D. $x^{100}+C$

Power Rule for Indefinite Integrals

In recent videos, we've been spending some time talking about indefinite integrals and how to, in essence, reverse the process of taking a derivative. Now one thing that I always had a hard time with as a student was the kind of outside-of-the-box thinking required. We need to think about what was the derivative before that led us to this function. That was a struggle for me. But in this video, we're going to learn there actually is a straightforward way of taking these common integrals that we see. Just like how we have a power rule for derivatives, we actually have a power rule for integrals as well, which you're going to learn in this video.

So let's just get right into things. Now, to think about the rules when dealing with integration, what we need to do is think about first the rules for derivatives. Because we know the integral is just going to be the reverse process of a derivative. Now one of the common rules that we know is the power rule when we have a function like this, where we have x raised to some power of \( n \). And we know that when dealing with this, for derivatives, what you want to do is multiply the front of this term by the exponent that you have and then decrease the exponent by 1.

So the rule looks something like this. You bring this to the front and then decrease the exponent by 1. So how exactly would an integral power rule work? Well, what we need to do is go the opposite of the logic we had for the derivative power rule. So if we wanted an integral power rule, what I'm first going to do is increase the exponent by 1 rather than decrease it.

So we'll take this exponent here that we have and we'll increase it by 1. And then what you want to do is divide, rather than multiply, by the new exponent. So you're going to take this whole exponent here and divide it. So, this would actually be the rule when it comes to integrals. As you can see, we're just doing opposite operations in the opposite order.

For the derivative power rule, we multiply first and then decrease. For the integral power rule, we increase first and then divide. So it's just the opposite order here. And don't forget, you need to add a plus \( c \) whenever you do an indefinite integral. Now one other thing I'll mention about this power rule here is that \( n \) cannot be equal to negative one when you're doing the integral power rule.

But the only reason for this is that we added negative one down here for \( n \). Notice we get negative one plus 1, which is 0, and we cannot divide by 0. So that's the reason this rule is here. And we actually are going to see in the future how to deal with situations where \( n \) is equal to negative one. But for now, we're not going to worry about that.

Just focus on using this rule right here. Now, to apply this to an example, let's say we have the integral of \( x \) to the 6th power. Now to do this, we could use our power rule. We could also think about what the derivative would be in the first place, but that's going to be some pretty crazy outside-of-the-box thinking with something like this. So instead, what I can do is use this power rule where I take the exponent that we have and add 1 to it and then divide by that exponent, and then, of course, you're going to have the plus \( c \).

So what this is going to end up being is \( x \) to the 6 plus 1, which is 7, divided by that new exponent of 7 plus \( c \). And this here would be the answer for our indefinite integral. Now to really make sure we have this down, let's try one more example down here where we're asked to find the indefinite integral of this function, \( g(t) \). And we can see \( g(t) \) is equal to \( t \) to the 4th power. So setting this up, we're going to have the indefinite integral of \( t \) to the 4th power, and we're going to do this with respect to \( t \).

Now what I can do here is the same power rule that we have up there. I just take our exponent and I add 1 to it, then I divide by that new exponent of 4+1, and then we have the plus constant of integration \( c \). Now \( t \) to the 4+1 is the same thing as \( t \) to the 5th power, and all of that is going to be divided by 5, and then we have plus \( c \). So this right here would be the solution to our integral. That is how we can solve this example.

As you can see, it's a very straightforward, quick way of solving these types of problems. And again, the process for integration is the same process as just reversing or taking the antiderivative. So if you wanted to check your solution here, all you need to do is take the derivative of this whole function here with respect to \( t \). Now I'm not going to go through that stuff in this video. We've seen this a bunch of times before.

But if you do the power rule for derivatives and you try it on this function right here, you would get back to the original function you had, \( t \) to the 4th power. So this is how you can solve situations where you have this kind of \( x \) to the \( n \), or you have some kind of power function and you need to integrate it with respect to that variable. Hope you found this video helpful. Let's get some more practice.

Find $h\left(x\right)$ by evaluating the following indefinite integral.

$h\left(x\right)=\int x^8dx$

• A. $h\left(x\right)=x^9+C$
• B. $h\left(x\right)=\frac{x^9}{9}+C$
• C. $h\left(x\right)=\frac{x^9}{9}$
• D. $h\left(x\right)=x^9$

Find $h\left(x\right)$ by evaluating the following indefinite integral.

$h\left(x\right)=x^{100}dx$

• A. $h\left(x\right)=\frac{x^{101}}{101}+C$
• B. $h\left(x\right)=\frac{x^{101}}{100}+C$
• C. $h\left(x\right)=\frac{x^{101}}{101}$
• D. $h\left(x\right)=\frac{x^{101}}{100}$

Find $h\left(x\right)$ by evaluating the following indefinite integral.

$h\left(x\right)=25x^4dx$

• A. $h\left(x\right)=25x^5$
• B. $h\left(x\right)=5x^5$
• C. $h\left(x\right)=5x^5+C$
• D. $h\left(x\right)=25x^5+C$

Additional Rules for Indefinite Integrals

In recent videos, we've been spending a lot of time talking about indefinite integrals and the various rules that you can use for solving integrals. So we should know at this point how to integrate something like \( x^2 \), how we could do an integral of this. We should also know how to deal with something like, say, \( 3x \) or even something like a constant, like, say, 6. These are things we should be able to individually integrate. But what would happen if we had some sort of combination and we needed to integrate all of these at once?

Well, right at first, this might look really confusing, but in this video, we're going to be going over some rules that we'll learn about to integrate a function like this, and I think you'll find it's pretty straightforward. So let's just go ahead and get right into things. Now you may recall that when learning about derivatives, we talked about the sum and difference rules or the constant multiple rules, and it turns out these same rules also apply for integrals. So let's say, for example, we're dealing with the sum and difference, where we have 2 functions being added or subtracted, and we need to integrate the entire thing. Well, if we wanted to do this, we could treat these integrals as being 2 separate integrals.

So you could take this integral and basically distribute it into each function and integrate them separately. So if 2 functions are being added or subtracted, you can integrate the functions individually and then add or subtract the results. So let's say, for example, we we're dealing with this integral, the integral of \( x + 6 \). Well, since these two functions are being added together, I could also say this is the same thing as the integral of \( x \) plus the integral of \( 6 \). Now if we integrate \( x \), we know how to use the power rule to solve this integral.

So doing this, we're going to get \( \frac{x^2}{2} \) for this integral. We're going to get \( 6x \) for this constant here. We just need to latch an x on to whatever constant we have, then we need plus the constant of integration \( c \). And this right here would be the solution for this integral. So as you can see, it's very straightforward.

Just use the sum and difference rule, just split it into multiple integrals, and solve the problem. Now something that students will often be curious about is why we get this plus \( c \) that occurs here rather than getting 2 plus \( c \)'s for each integral that we did. Well, the reason for that is because \( c \) is just a constant. So if we had one constant pop out here and another constant pop out there, so say constant 1 and constant 2, these 2 would add to just give you another constant. So we don't need to write multiple constants for each integral we do.

Whenever we're integrating some function, we just need 1 plus \( c \) at the end to cap the whole thing off. So that is how you can use the sum and difference rules. Now another popular rule we learned for derivatives is the constant multiple rule. And this rule also applies for integrals. So let's say that we're integrating some constant multiplied by the function.

Well, you could take whatever this constant is right here and bring it to the outside of your function, and then integrate the function separately. Now I will mention that there is a situation where leaving the constant inside the integral could be helpful, and that's if you had the integral of some constant by itself. So if you were just integrating a constant, then it would probably be best to leave it inside the integral, because pulling it to the outside really isn't going to do much for you. But if you ever have a constant multiplied by the function, then it can be very helpful to take that constant and pull it on the outside. This is a rule that you can do.

So let's say, for example, we have the integral of \( 5x^3 \). Well, I could also write this as 5, taking that constant and pulling it to the outside times the integral of what we have inside the function, which is \( x^3 \). This is going to be with respect to \( x \). Now if I want to integrate \( x^3 \), well, I know that I can just use the power rule that we've already learned about. So doing this, I'm going to get \( x^4\ / 4\).

That's going to be what that integral comes out to. And keep in mind, all of this is going to be multiplied by the 5 that we have out in front here. So we're going to have this multiplied by 5, and that's going to be \( 5 \times \frac{x^4}{4} + c \), and this would be the solution to our problem. So these are the rules that you could apply to integrals, the sum and difference rule as well as the constant multiple rule. But let's see if we can really combine this idea altogether and solve a problem like this.

So here we have this problem down here, where we have the function \( x^2 - 3x + 6 \), and we want to find the indefinite integral of this function. Now this function down here is actually the function we talked about at the start. And we're now going to learn how we can apply multiple rules that we've learned about in combination to solve a problem like this. This can allow you to find integrals of more complicated functions, like polynomials that we have down here. So if I want to find the integral of this function with respect to \( x \), it's going to be the integral of this entire thing that we have here.

So it's going to be \( x^2 - 3x + 6 \), and we're integrating this entire thing with respect to \( x \). Now what I can do is first apply the sum and difference rule. Wherever the place I see a difference or a sum, I can just treat this as an individual integral for each term that I see. So doing this, we're going to have the integral of \( x^2 \). That's going to be minus the integral of \( 3x \).

That's going to be plus the integral of 6. And, of course, all of this is going to be with respect to \( x \). So we're going to integrate all of this with respect to \( x \). And what I can also do is now use the constant multiple rule for any place that I see constants multiplied by a function. I can see we have one constant there multiplied by a function.

So what I can do is treat this as the integral of \( x^2 \) with respect to \( x \). Is going to be minus, and I'll take that 3 there and pull it outside of the integral, and then we'll have the integral of \( x \) with respect to \( x \). As you can see, this made the integral here much more simple since we're only integrating \( x \). Then we're going to have plus the integral of 6. Now like I mentioned before, you could take the 6 and pull it outside the integral if you wanted to.

But since we have a constant being integrated by itself, I'm just going to go ahead and leave that 6 inside the integral, since I think that's going to make things a little more simple. So now that I've done this, let's go ahead and integrate. So we can use the power rule on each of these terms. So I can use the power rule on \( x^2 \), which is going to give me \( \frac{x^3}{3} \). Then I can go ahead and subtract this from 3 multiplied by the integral of \( x \), which using the power rule is going to give me \( \frac{x^2}{2} \).

Then we'll have plus the integral of 6, which will just be 6x, plus the constant of integration \( c \). So this right here would be the answer to the problem, and that is how you can integrate these longer polynomials. So as you can see, the process is pretty straightforward. It's the same rules that we've learned for derivatives. We can also apply these to integrals when it comes to sums and differences or constant multiples.

And a lot of times, you'll need to use a lot of these in combination when dealing with longer, more complicated functions or polynomials. So hope you found this video helpful, and let's go ahead and move on.

Find $f\left(x\right)$ by evaluating the following indefinite integral.

$f\left(x\right)=\int\left(100x^2-35x-\frac{13}{2}\right)dx$

• A. $f\left(x\right)=\frac{100}{3}x^3-\frac{35}{2}x^2-\frac{13}{2}x+C$
• B. $f\left(x\right)=50x^3-35x^2-\frac{13}{2}x+C$
• C. $f\left(x\right)=50x^3-35x^2+C$
• D. $f\left(x\right)=100x^3-35x^2-\frac{13}{2}x+C$

Find $f\left(x\right)$ by evaluating the following indefinite integral.

$f\left(x\right)=\int\left(8x^7+10x-20\right)dx$

• A. $f\left(x\right)=x^8+5x^2-20x+C$
• B. $f\left(x\right)=8x^8+10x^2-20x+C$
• C. $f\left(x\right)=x^8+5x^2+C$
• D. $f\left(x\right)=8x^8+10x^2+C$

Additional Rules for Indefinite Integrals Example 1

Let's try solving this problem. It is determined that the marginal profit \( p' (x) \) in hundreds of dollars per unit from selling a certain number of parachutes is given by the function below. Find the total profit function \( p(x) \), assuming that \( p(0) = 0 \). So this is the function that we're dealing with, and we're trying to find the total profit function \( p(x) \). Now we're given \( p' (x) \), and \( p(x) \) would be our profit function.

So how exactly could I find this? Well, notice that \( p(x) \) would just be the reverse of the derivative, because \( p' (x) \) is the derivative of \( p(x) \). So what we need to do is find some sort of way to, in essence, take the antiderivative of \( p' (x) \). Now we can do this by recognizing it's the same process as just integrating everything that we have on this right side here. Because doing this will give us our total profit function \( p(x) \).

So let's see what we can do here. Well, what I can do is use the rules for integrals since I recognize that we are subtracting 2 terms here. So since we're subtracting, both the 0.03x and the 1.5, well, what I'm gonna do here is I'm gonna split this integral so we'll have the integral of 0.03x with respect to x, minus the integral of 1.5 with respect to x. Now after doing this, what I can also do is pull this constant outside of the integral. So we're going to have that this function \( p(x) \) is equal to 0.03 times the integral of x with respect to x minus, and then I'll leave this 1.5 inside the integral just because the constant is by itself, so what would be integral of 1.5 with respect to x.

Now what I can do from here is use the power rule. And using the power rule in this term, we add 1 to the exponent and then divide by that exponent. So doing this will give us 0.03, this constant on the outside, multiplied by \( x^{1+1} \), which is \( x^2 \), all divided by 2. Then I'll have minus the integral of 1.5 with respect to x, which is just 1.5x. And then don't forget we have to add the constant of integration \( c \).

Now from here, I can actually simplify this fraction on this side, because 0.03 divided by 2 is 0.015. So we're going to get 0.015 \( x^2 \) minus 1.5x plus \( c \). And this right here is the general equation for our total profit function. Now this looks good at the moment here, but notice that we also have this condition right here, that \( p(0) = 0 \).

And so what that means is I need to actually solve for my constant \( c \) here by plugging 0 in for \( x \). Now if I plug 0 in for \( x \) here, notice that this term will entirely go away because 0 times anything is just 0, and so will this term. So all we're gonna be left with is \( c \). So that \( p(0) = c \), and we have here that \( p(0) = 0 \), so that means \( c = 0 \). So going back to this function over here, I can take that \( c \) and replace it with 0, which is the same thing as just removing this entire, like, last piece of the function here, leaving us with just this.

So this would be our total profit function \( p(x) \), and that is the solution to the problem. Let's move on.

Additional Rules for Indefinite Integrals Example 2

Let's try this problem. The marginal revenue from a ceramics company is given by the function \( r'(x) \) in dollars per unit from selling a certain number base. Find the total revenue function \( r(x) \), assuming that \( r(0) = 0 \). So notice we have the marginal revenue function right there, and we're trying to find the total revenue function. To do that, well, we're given \( r'(x) \), and we're trying to find \( r(x) \).

So what I need to do is take the antiderivative, or basically integrate \( r'(x) \). And in doing this, I can integrate everything on this right side here. Now to integrate this, I can apply the rules for integrals that I know. What I can see is that we have the sum between these terms that we have. So I can integrate each one of these terms separately.

So we have the integral of 1.68x squared. It's gonna be plus the integral of x with respect to x, and it's gonna be plus the integral of 2.5. Now from here, I can pull out these constants that I see. So I see that we have this constant 1.68. So pulling that out to the front row, \(1.68 \times \int x^2 dx\), that's going to be plus the integral of x, then we'll have plus the integral of 2.5.

Now I just need to apply the power rules, which I've already learned about. In doing that, what I'm basically doing is taking this exponent, adding 1 to it, and then dividing it by that new exponent. So I have \( x^{2+1} / (2+1) \). So going down here, we're gonna have 1.68, and that's going to be \( x^{3} \) all divided by 3. So we're gonna have 1.68x33, and then we're gonna have plus the integral of x, which using the power rule, we'll get x22. And then we're going to have plus the integral of 2.5, which is gonna be 2.5x.

Now, taking into account that \( r(0) = 0 \), we know that \(\int r'(x) \, dx = r(x) + C\) and setting \(x = 0\), \(r(0) = 0.56x^3 + 0.5x^2 + 2.5x + C\). Substituting \(x=0\) gives \(C = 0\).

Thus, the total revenue function \( r(x) \) is \( r(x) = 0.56x^3 + 0.5x^2 + 2.5x \). This is the revenue function and the solution to the problem. Hope you found this explanation helpful, and let's move on.