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Improper integral with two infinite bounds

A worked example of a challenging improper integral that involves two infinite bounds and an inverse trig substitution. Created by Sal Khan.

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  • mr pants teal style avatar for user Wiebke Janßen
    At : Why do you substitute 5tan*delta for x?
    (64 votes)
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  • mr pants teal style avatar for user Wiebke Janßen
    Could we also have said we compute the improper integral with one infinite bound and multiply the result by 2 if the graph is symmetric to the x - axis?
    (20 votes)
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    • male robot hal style avatar for user aoiqed
      In general, yes, as long as you are certain that your function is even, i.e. symmetric about the y axis. Proving a function is even is not the goal of this video, so Sal ignored that useful tidbit you noticed.
      In Sal's example, the function turns out to be even, and as the calculations in the last minute of the video show, computing one of the integrals and doubling it gives the correct answer.
      (12 votes)
  • blobby green style avatar for user hirkento
    at , how do you conclude that arctan of negative infinity is -pi/2?
    (13 votes)
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    • blobby green style avatar for user austin.saintvincent
      Remember that the tan or arctan of an angle on the unit circle represents its slope. A slope of negative infinity is a vertical line drawn downwards from the origin ( if using the center of the unit circle as the reference point). This angle is 270 degrees which is expressed as 3pi/2 or -pi/2 depending on which way you approach it. In this case, we approached it from the negative direction.
      (10 votes)
  • mr pants teal style avatar for user Diana Dai
    Just to make clear... an improper integral is basically just a definite integral with one or both boundries being -infinite or +infinite?
    (7 votes)
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    • piceratops ultimate style avatar for user jimstanley49
      Close. An improper integral can also have a singularity within the bounds. For example:
      ∫ 1/x from 0 to 1
      As x → 0, 1/x → ∞
      So an improper integral is either bounded by ±infinity, or approaches infinity at one or both of the bounds.
      (19 votes)
  • blobby green style avatar for user Paula Tyler
    Is trig substitution the only method of integration we could use here?
    (4 votes)
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    • piceratops ultimate style avatar for user Daniel Schneider
      No, you could also have used a u-substitution. Divide the numerator and denominator by 25, and you can write the integrand as 10/(1 + x^2 /25) dx, or 10/(1 + (x/5)^2) dx. If we substitute u = x/5, so that du = (1/5)dx, we see that the integrand becomes 50/(1 + u^2) du. Integrating this function is easy--its antiderivative is 50arctan(u) + C or 50arctan(x/5) + C, which is what Sal obtained on the video via his trig substitution. Hope that helps!
      (16 votes)
  • blobby green style avatar for user robb.demars
    Where did the tangents and arctangents come into play? Everything was making sense up until that point
    (8 votes)
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    • blobby green style avatar for user deadpanchain
      well, to understand the role of tangents and arctangents you need to have some basic
      knowledge about integration and calculus...if you do then observe sal using trig
      substitution
      to solve the integration of 250/25 + x^2. you can even use a direct formula ;
      integration 1/a^2 + x^2 where a=5 & x=x
      or you can use u substitution . Hope that helps! :)
      (0 votes)
  • leaf green style avatar for user Raphael David
    The integral in this video demonstrates an area under the curve of 50pi. But the very next video "Divergent Improper Integral" shows an area of infinity under the curve of 1/x. The curve on this page (250/(25+x^2)) looks like it should be at least twice as large as that under the curve of 1/x. It goes to infinity in two different directions, whereas 1/x only goes to infinity in one direction.
    How can the curve on this page yield an area of only 50pi, where as the curve of 1/x yields an area of infinity?
    (3 votes)
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    • mr pink red style avatar for user Georgi Nikolov
      1/x^2 becomes smaller faster than 1/x, and all the other numbers (multiplying by 250 and adding 25) are just "applied" to a finite number, thus the results can be only finite.

      You can compare that to the discrete case. The sum from 1 to infinity of 1/x is infinity (the sum is divergent), but the sum of 1/x^2 is finite (=pi^2/6).
      (6 votes)
  • piceratops ultimate style avatar for user Nikhil
    Couldn't the improper integral be written as the lim as n--> infinity of the integral from -n to n?
    (5 votes)
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  • mr pants teal style avatar for user Wiebke Janßen
    At : Why is dx equal to 5 sec^2*delta*d*delta?
    (4 votes)
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  • leaf blue style avatar for user Trent Kierstead
    It really blows my mind that a region with infinite bounds can have a finite area! How does that work? Does it just approach that value as its area? Sorry if my question doesn't make sense, I'm just trying to wrap my head around this.
    (2 votes)
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Video transcript

Right here we have the graph of y is equal to 250 over 25 plus x squared. And what I'm curious about in this video is the total area under this curve and above the x-axis. So I'm talking about everything that I'm shading in white here, including what we can't see, as we keep moving to the right and we keep moving to the left. So I'm talking about from x at negative infinity all the way to x at infinity. So first, how would we actually denote this? Well, it would be an improper integral. We would denote this area as the indefinite integral from x is equal to negative infinity to x is equal to infinity of our function, 250 over 25 plus x squared, dx. Now, we've already seen improper integrals where one of our boundaries was infinity. But how do you do it when you have one boundary at positive infinity and one boundary at negative infinity? You can't take a limit to two different things. And so the way that we're going to tackle this is to actually break up this area into two different improper integrals, one improper integral that describes this area right over here in blue from negative infinity to 0. So we'll say that this is equal to the improper integral that goes from negative infinity to 0 of 250 over 25 plus x squared dx, plus the improper integral that goes from 0 to positive infinity. So plus the improper, or the definite, integral from 0 to positive infinity of 250 over 25 plus x squared dx. And now we can start to make sense of this. So what we have in blue can be rewritten. This is equal to the limit as n approaches negative infinity of the definite integral from n to 0 of 250 over 25 plus x squared dx. Plus-- and I'm running out of real estate here-- the limit as-- since I already used n, let me use m now-- the limit as m approaches positive infinity of the definite integral from 0 to m of 250 over 25 plus x squared dx. So now all we have to do is evaluate these definite integrals. And to do that, we just have to figure out an antiderivative of 250 over 25 plus x squared. So let's try to figure out what that is. I'll do it over here on the left. So we need to figure out the antiderivative of 250 over 25 plus x squared. And it might already jump out at you that trig substitution might be a good thing to do. You see this pattern of a squared plus x squared, where in this case, a would be 5. So we can make the substitution that x is equal to a tangent theta, 5 tangent theta. And since we're going to have to reverse substitute later on, we can also put in the constraint-- well, we'd say x/5 is equal to tangent of theta, which is completely consistent with this first statement. And so if we wanted to have theta expressed as a function of x, we can put the constraint that theta is equal to arctangent of x/5. So once again, this is completely consistent with this over here. x can be 5 tangent of theta, and theta can be equal to arctangent of x/5. So now let's do the substitution. Actually, before that, we also have to figure out what dx is equal to. So dx is equal to-- I'll do it right over here-- well, the derivative of this with respect to theta is 5 secant squared theta d theta. So now we're ready to substitute back in. So all of this business is going to be equal to 250 times dx. Well, 250 times dx is 250 times 5 secant squared theta d theta. So that's 250 dx is this business up here. All of that over 25 plus x squared. Well, x squared is going to be 25 tangent squared of theta. And now we can try to simplify all of this business. This is equal to 250 times 5 secant squared theta over 25 times 1 plus tangent squared theta d theta. So 25 goes into 250 10 times. And then 1 plus tangent squared theta is secant squared theta. You can prove it for yourself if you write tan squared theta as sine squared theta over cosine squared theta. Add that to 1, which is the same thing as cosine squared theta over cosine squared theta. And then you can use some basic trig identities to realize that this is equal to secant squared theta, which simplifies our expression a good bit. You get secant squared theta over secant squared theta, which is just 1. And so you're just left with the 10 times the 5, 50 d theta. So this is equal to 50-- I'll take the 50 outside the integral-- 50 d theta, which is equal to 50 theta. We can write the plus C, just to show that these are all of the antiderivatives. But we only need the most basic antiderivative to evaluate these definite integrals. But we right now only have it in terms of theta. Let's write this in terms of x. We set the constraint that theta is equal to arctangent of x/5. So this is equal to 50 arctangent of x/5 plus C. These are all the antiderivatives. We can pick C is equal to 0 to find an antiderivative of these things to evaluate the definite integrals. So let's do that. So what we have in blue, we can rewrite as the limit as n approaches negative infinity of the antiderivative of this, or an antiderivative of this, which we could say is 50 arctangent of x/5. And we're going to evaluate it at 0 and n. And to this, we're going to add the limit as m approaches positive infinity of just the antiderivative of this, 50 arctangent-- not "actangent," arctangent-- arctangent of x/5 evaluated from 0 to m. And let me put a little parentheses right around this x/5. So what's this going to be? This is going to be 50-- so let me write this. This is going to be the limit as n approaches negative infinity of 50 arctangent of 0/5 / minus 50 arctangent of n/5. And then to that, we're going to add-- let me give myself even a little bit more space-- the limit as m approaches infinity of 50 arctan of m/5 minus 50 arctan-- I think you see where all this is going-- minus 50 arctan of 0/5. So now we can evaluate these things. And to help us evaluate them, let's think about a unit circle. So let's think about a unit circle so we can really visualize the arctangent function. So one way to think about the tangent is the slope of the line that is on the angle or that helps define the terminal side of an angle. So for example, if we have an angle that looks like this-- this is an angle between the positive x-axis and this line right over here. If we have an angle just like that, the tangent of that angle is going to be the slope of this line. And so one way to think about it is, OK, if I want the arctangent of 0, it's essentially saying, OK, let's get an angle where the terminal side has slope 0. Well, an angle where the terminal side has slope 0 is an angle of 0. So arctangent of 0 is 0 radians. So it's 50 times 0. That's just going to be 0. And we could also write that over here. This is just going to be 0. So what we're left with is the limit as n approaches negative infinity of negative 50 arctangent of n/5 plus the limit as m approaches positive infinity of 50 arctangent of m/5. So let's think about what these are going to be. So it's the limit as n approaches negative infinity. One way to view that is the limit as the slope of this terminal side approaches negative infinity. So that's getting more and more negative, more and more negative. It's at negative infinity, or it's approaching negative infinity when our angle right over here is negative pi/2 radians. Or you could say the limit of arctan of n/5 as n approaches negative infinity, this part right over here, is going to approach, as n approaches negative infinity, it's going to be negative pi/2. And then we're just going to multiply that times the negative 50. So this is going to be negative 50 times negative pi/2, which is going to be positive 25 pi. So this is going to be 25 pi. And similarly, arctan of m/5 as m approaches infinity-- well, as m approaches infinity, the slope of the terminal side of the angle approaches infinity. So the slope is getting higher, higher. It approaches infinity as it gets closer and closer to vertical. So the arctan of m/5 as m approaches infinity is going to be positive pi/2. This is an angle of positive pi/2. So what we see in orange right over here is going to be positive 50 times pi/2, which is going to be 25 pi. So the area that we have in blue, going back to our original problem, is 25 pi. The area that we have in orange is 25 pi. And so if we wanted to answer our original question, what's the total area under this curve-- which is kind of a cool question to try to answer-- we get it's 25 pi plus 25 pi, which is 50 pi. And we're done.