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.
Improper integral with two infinite bounds
Improper integral with two infinite bounds
Improper integrals are used to calculate the solutions to definite integrals that may not necessarily exist because the functions are either discontinuous or have indefinite limits. The solution to an improper integral can be calculated as a limit, as long as the limit exists and can be calculated. In this case, the improper interval is convergent. If the limit doesn't exist, or if it converges to ±∞, then the improper integral is divergent. For a function f (x ) over the interval [a. ∞], . For the function f (x ) over the interval [−∞, b ], . In general, to solve an improper integral, replace the ±∞, limit with a variable (t ), and then take the limit of the integral as t approaches ±∞Need more help understanding improper integrals?
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3. The differential equation describing the shape of a wire of constant linear density w (a constant) hanging where is the horizontal tension in the wire at its lowest under its own weight is point (a constant). Usin.
Recall that the definition of an integral requires the function f (x ) to be bounded on the bounded interval [a ,b ] (where a and b are two real numbers). It is natural then to wonder what happens to this definition if 1 the function f (x ) becomes unbounded (we call this case Type I ); 2 the interval [a ,b ] becomes unbounded (that is or )(we call this case Type II ).
In both cases, we say that the integral is improper.
Case Type I: Consider the function f (x ) defined on the interval [a ,b ] (where a and b are real numbers). We have two cases f (x ) becomes unbounded around a or unbounded around b (see the images below)
For the sake of illustration, we considered a positive function. The integral represents the area of the region bounded by the graph of f (x ), the x-axis and the lines x =a and x =b. Assume f (x ) is unbounded at a. Then the trick behind evaluating the area is to compute the area of the region bounded by the graph of f (x ), the x-axis and the lines x =c and x =b. Then we let c get closer and closer to a (check the figure below)
Note that the integral is well defined. In other words, it is not an improper integral.
If the function is unbounded at b. then we will have
Remark. What happened if the function f (x ) is unbounded at more than one point on the interval [a ,b ]. Very easy, first you need to study f (x ) on [a ,b ] and find out where the function is unbounded. Let us say that f (x ) is unbounded at and for example, with . Then you must choose a number between and (that is ) and then write
Then you must evaluate every single integral to obtain the integral . Note that the single integrals do not present a bad behavior other than at the end points (and not for both of them).
Example. Consider the function defined on [0,1]. It is easy to see that f (x ) is unbounded at x = 0 and . Therefore, in order to study the integral
and then study every single integral alone.
Case Type II: Consider the function f (x ) defined on the interval or . In other words, the domain is unbounded not the function (see the figures below).
The same as for the Type I, we considered a positive function just for the sake of illustrating what we are doing. The following picture gives a clear idea about what we will do (using the area approach)
This is a problem in my calculus text book. Let R be the region to the right of x=1 that is bounded by the x-axis and the curve y=1/x. When this region is revolved about the x-axis it generates a solid whose surface is known as Gabriel's Horn. Show that the solid has a finite volume but its surface has an infinite area. [Note: It has been suggested that if one could saturate the interior of the solid with paint and allow it to seep through to the surface, then one could paint an infinite surface with a finite amount of paint! What do you think?]
You just need to do the integrals, and they will tell the story. Briefly, and without the details (you can fill those in) the volume elements are the cross-sectional areas and so vary as 1/x² while the area elements are the perimeters of the cross-sections and so vary as 1/x. The integral of 1/x² varies as 1/x, and so goes to zero as x goes to infinity, while the integral of 1/x varies as log(x) and so goes to infinity as x goes to infinity.
As to the paint -- sure, no problem. As long as you set no lower limit on the volume of paint needed to cover any given surface area. Look at the volume to surface area ratios.
In calculus. an improper integral is the limit of a definite integral. as an endpoint, or both endpoints, of the interval approaches either a specified real number or ∞ or −∞.
If the function f being integrated from a to c has a discontinuity at c. especially in the form of a vertical asymptote. or if c = ∞, then there may be no more convenient way to compute the integral
than by finding the limit
In some cases, the integral from a to c is not even defined, because the integrals of the positive and negative parts of f (x ) dx from a to c are both infinite, but nonetheless the limit may exist. Such cases are "properly improper" integrals, i.e. their values cannot be defined except as such limits.
can be interpreted as
From the point of view of mathematical analysis it is not necessary to interpret it that way, since it may be interpreted instead as a Lebesgue integral over the set (0, ∞). On the other hand, the use of the limit of definite integrals over finite ranges is clearly useful, if only as a way to calculate actual values.
cannot be interpreted as a Lebesgue integral, since
This is therefore a "properly" improper integral, whose value is given by
One can speak of the singularities of an improper integral, meaning those points of the extended real number line at which limits are used.
Such an integral is often written symbolically just like a standard definite integral, perhaps with infinity as a limit of integration. But that conceals the limiting process. By using the more advanced Lebesgue integral. rather than the Riemann integral. one can in some cases bypass this requirement, but if one simply wants to evaluate the limit to a definite answer, that technical fix may not necessarily help. It is more or less essential in the theoretical treatment for the Fourier transform. with pervasive use of integrals over the whole real line.