AP®︎/College Calculus BC
Introduction to improper integrals
Improper integrals are definite integrals where one or both of the boundaries is at infinity, or where the integrand has a vertical asymptote in the interval of integration. As crazy as it may sound, we can actually calculate some improper integrals using some clever methods that involve limits. Created by Sal Khan.
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- This still doesn't make sense to me. I see how the area could be approaching 1 but if it ever actually reaches 1 when moving infinitively then it would go over 1 extremely slightly. If its moving out to infinity, i don't see how it could have a set area. Can anyone explain this? Is my point valid?(51 votes)
- With any arbitrarily big value for n, you'd get a value arbitrarily close to 1 but never bigger than 1. Only at infinity is the area 1. And there isn't anything beyond infinity, so it doesn't go over 1. It's exactly 1.
Infinities can be really confusing.(134 votes)
- What if 0 is your lower boundary?(18 votes)
- Good question! When dealing with improper integrals we need to handle one "problem point" at a time. Infinity (plus or minus) is always a problem point, and we also have problem points wherever the function "blows up," as this one does at x = 0.
Because we can't deal with both problem points at the same time, we split the integral into two (or more) parts. The integral from 0 to ∞ is equal to the integral from 0 to a plus the integral from a to ∞, where a is an arbitrary positive constant. So if we're asked to analyze the integral of this function from 0 to ∞, we would choose a constant (we can use any constant, but it makes sense to choose one that makes calculations easy, such as 1) and evaluate two separate improper integrals: from 0 to 1, and from 1 to ∞. The integral from 0 to 1 would be evaluated as the limit as n approaches zero of the integral from n to 1, and the integral from 1 to ∞ would be evaluated as explained in this video.
Note that the overall integral converges only if both of these converge. When you break up an improper integral into multiple improper integrals, you know the overall integral diverges as soon as you establish that any one of the components diverges.(37 votes)
- I'm confused as to how the integral of 1/(x^2) became -(1/x) at1:52. Can someone please clarify?(7 votes)
- It may be easier to see if you think of it
as "how the integral of x^(-2) became -x^(-1)"
What is the derivative of -x^(-1) ?(12 votes)
- Do you not have to add +c to the end of the integrals he is taking?(2 votes)
- No. The + C is for indefinite integrals. These integrals, while improper, do have bounds and so there is no need of the +C.(14 votes)
- What is a good definition for "improper integrals"?(6 votes)
- I would say an improper integral is an integral with one or more of the following qualities:
1. Where at some point in the interval from the lower bound to the upper bound of the integration limits, there is a discontinuity in the actual function you are integrating. Take the integral from -1 to 1 of (1/x^2)*dx as an example, as the function is discontinuous at x=0.
2. One of the integration limits contains positive or negative infinity.
An example with both conditions would be the integral from 0 to infinity of 1/(x^2) *dx(7 votes)
- Is it EXACTLY equal to one? I think as 'n' approaches infiniti, the integral tends to 1.(5 votes)
- We see that the limit at2:39can be evaluated exactly, so the area is EXACTLY equal to one. It is not uncommon for people to misunderstand the concept of infinity. It takes some time to get used to it. For instance, some people would argue to death that 0.999... is less than 1, while in reality, these two are completely equal.(10 votes)
- What exactly is the definition of an improper integral?(3 votes)
- "An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration."
In other words: it's a definite integral (an integral with limits of integration) where either a, b, or both are infinity, or the function being integrated has unbounded behavior between the limits of integration.(2 votes)
- I know L'Hopital's rule may be useful here, is there a video abut improper integrals and L'Hopital's rule?(2 votes)
- L'Hopital's is only applicable when you get a value like infinity over infinity. Or Zero over Zero. When you get that, take the derivative of the highest power function like (x)/(x^2) as x approaches infinity is 1/2. It's a little confusing and difficult to explain but that's the jist of it(4 votes)
- Instead of having infinity as the upper bound, couldn't the upper bound be x? That way, the upper bound can be as large as you want it to be-- it will essentially be infinity.(1 vote)
- Well, infinity is sometimes easier to deal with than just plugging in a bunch of x values especially when you have it in the form 1/infinity or something similar because 1/infinity is basically just 0.(5 votes)
- Why can't we just evaluate it like a regular integral? We find the indefinite integral, and plug in the top and bottom values (even when it is infinity). Then we get the answer, or if we get + or - infinity, we know it diverges. I don't see how putting it in limit form can help us solve the problem. Can someone explain? Thanks in advance.(1 vote)
- Because infinity is not a real number, and so it cannot be plugged into any of our familiar functions. Writing out the limit makes it clear what we are actually doing when we write a '∞' as the bound.(4 votes)
What I want to figure out in this video is the area under the curve y equals 1 over x squared, with x equals 1 as our lower boundary and have no upper boundary, just keep on going forever and forever. It really is essentially as x approaches infinity. So I want to figure out what this entire area is. And one way that we can denote that is with an improper definite integral, or an improper integral. And we would denote it as 1 is our lower boundary, but we're just going to keep on going forever as our upper boundary. So our upper boundary is infinity. And we're taking the integral of 1 over x squared dx. And so let me be very clear. This right over here is an improper integral. Now how do we actually deal with this? Well, by definition this is the same thing as the limit as n approaches infinity of the integral from 1 to n of 1 over x squared dx. And this is nice, because we know how to evaluate this. This is just a definite integral where the upper boundary is n. And then we know how to take limits. We can figure out what the limit is as n approaches infinity. So let's figure out if we can actually evaluate this thing. So the second fundamental theorem of calculus, or the second part of the fundamental theorem of calculus, tells us that this piece right over here-- just let me write the limit part. So this part I'll just rewrite. The limit as n approaches infinity of-- and we're going to use the second fundamental theorem of calculus. We're going to evaluate the antiderivative of 1 over x squared or x to the negative 2. So the antiderivative of x to the negative 2 is negative x to the negative 1. So negative x to the negative 1 or negative 1 over x. So negative 1/x is the antiderivative. And we're going to evaluate at n and evaluate it at 1. So this is going to be equal to the limit as n approaches infinity. Let's see, if we evaluate this thing at n, we get negative 1 over n. And from that we're going to subtract this thing evaluated at 1. So it's negative 1 over 1, or it's negative 1. So this right over here is negative 1. And so we're going to find the limit as n approaches infinity of this business. This stuff right here is just the stuff right here. I haven't found the limit yet. So this is going to be equal to the limit as n approaches infinity of-- let's see, this is positive 1-- and we can even write that minus 1 over n-- of 1 minus 1 over n. And lucky for us, this limit actually exists. Limit as n approaches infinity, this term right over here is going to get closer and closer and closer to 0. 1 over infinity you can essentially view as 0. So this right over here is going to be equal to 1, which is pretty neat. We have this area that has no right boundary. It just keeps on going forever. But we still have a finite area, and the area is actually exactly equal to 1. So in this case we had an improper integral. And because we were actually able to evaluate it and come up with the number that this limit actually existed, we say that this improper integral right over here is convergent. If for whatever reason this was unbounded and we couldn't come up with some type of a finite number here, if the area was infinite, we would say that it is divergent. So right over here we figured out a kind of neat thing. This area is exactly 1.