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## Class 12 math (India)

### Course: Class 12 math (India)>Unit 10

Lesson 10: Improper integrals

# Introduction to improper integrals

AP.CALC:
LIM‑6 (EU)
,
LIM‑6.A (LO)
,
LIM‑6.A.1 (EK)
,
LIM‑6.A.2 (EK)
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?
• 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.
• What if 0 is your lower boundary?
• 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.
• I'm confused as to how the integral of 1/(x^2) became -(1/x) at . Can someone please clarify?
• 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) ?
• Do you not have to add +c to the end of the integrals he is taking?
• No. The + C is for indefinite integrals. These integrals, while improper, do have bounds and so there is no need of the +C.
• What is a good definition for "improper integrals"?
• 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.

or
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
• Is it EXACTLY equal to one? I think as 'n' approaches infiniti, the integral tends to 1.
• We see that the limit at can 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.
• What exactly is the definition of an improper integral?
• "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.
• I know L'Hopital's rule may be useful here, is there a video abut improper integrals and L'Hopital's rule?
• 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