Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Improper integrals. Integrals with infinite limits of integration are improper integrals of type 1. So if we have F of X being continuous on A bounded on one end but Infinity on the other, so on A to Infinity, then we can see that the integration of A to Infinity F of XDX would really equal the limit as B goes to Infinity of A to BF of XDX. Because basically when we evaluate our integral, we can't stick in Infinity. So we substitute that idea with some variable and then say what happens is that variable goes to Infinity. This is true if it's negative Infinity instead of positive Infinity. So we're just going to put the variable on the bottom bound instead of the top bound. Now the third case, if it's negative Infinity and positive Infinity, what we're going to do is we're just going to look for some real number C within negative Infinity to Infinity, split it up into two different integrals, and then use the Part 1 and the Part 2. And each case, if the limit is finite, we say that the improper integral converges and that the limit is the value of the improper integral. If the limit fails to exist, the improper integral diverges. Frequently the limit failing to exist means it goes to Infinity and negative Infinity, but it doesn't have to. So kind of the idea general is if we can get a number, then it converges. If we can't get a single number then it doesn't. So we're going to have some tests for convergence. These are just a few of our tests. Direct comparison test says let F&G be continuous on the interval A to Infinity with 0 less than or equal to F of X less than or equal to G of X for all X greater than or equal to a. Then the interval A to Infinity of F of XDX converges if A to Infinity of G of XDX converges. So basically if the bigger one has some sort of a limit, if the bigger one goes to a number, then the smaller function would also have to go to a number. We don't know what the number is, we just know that it has to go to a number. The same thing occurs if we can convince that the smaller one diverges, then the bigger one has to diverge. So if a goes to Infinity of G of XDX diverges. If a going to Infinity of F of XDX diverges. So if I can say the smaller one diverges, then the bigger one has to diverge. If I can say the biggest one converges, then the smaller one has to converge. That's direct comparison. The limit comparison test if the positive functions F&G are continuous on A to Infinity, and if the limit of the ratio of those two functions equals L where L is some number, some positive number, then the integration of A to Infinity of F of XDX and the integration of A to Infinity of G of XDX both converge or both diverge. Now, if the limit doesn't go to some L, we it's inconclusive. It doesn't tell us anything. So if we look at one to Infinity of DX over XP, first of all, remember that that could really be thought of as X to the negative PDX. We're going to look at a couple cases first. If P doesn't equal 1, well limit B goes to Infinity of one to B of DX over XP. The limit as B goes to Infinity. Add 1 to the exponent and divide by the new exponent B to 1. So I'm going to put in my upper bound of B minus my lower bound of one and I'm going to simplify. So I'm going to get the limit as B goes to Infinity of 1 / 1 -, P times the quantity B to the 1 -, P -, 1, because one to any power is 1. So what's going to happen is if P is greater than one, if P is greater than 125700, we'd get one over some constant, but we'd get the B to 1 -, 101 -, 1001 -, 1,000,000. So this is going to get really, really, really small. So we get 1 / 1 -, P times the quantity 0 -, 1 or -1 times the 1 / 1 -. P gives us 1 / P - 1. Now, if P is less than one, say -100 negative 1000, negative a million, this B to the 1 -, P is going to get really big because it's going to be B to the 101, B to the 1001. So this is going to be really, really big. Infinity -1 is still going to give us Infinity even if we have it divided by some constant. So basically what this says is if P is greater than one, it's going to converge, and if P is less than one, it's going to diverge. Now let's look at what happens when P = 1. So if we have limit as B goes to Infinity of the integral 1 to BDX over X, that's just the limit as B goes to Infinity of the natural log of the absolute value of X. Putting in the upper bound minus the lower bound. The natural log of Infinity is something really really big. So Infinity -0 is going to give us Infinity. So to conclude this, the integral of one to Infinity DX over XP is going to converge if P is greater than one and it's going to converge to the number 1 / P -, 1. It's going to diverge if P is less than or equal to one. We're going to use that a lot. Let's look at some other things. If we have one to Infinity of DX over X 1.001 as its power. So the limit as B goes to Infinity one over BDX over X to the 1.001 power. So taking our integration, we get negative X to the negative .001 / .001 from B to 1. And that's just the power rule. Add 1 to the exponent divided by the new exponent. When we stick in our B, we get our B and our one we get negative B to the negative .001 / .001 + 1 to some power over .001. That's really just 1000 at the end. And that B to the negative exponent can come down South B to something or Infinity to something small when it's in the denominator. So something big in the denominator is really going to make it go to something small. So we're going to get the -1 / 1.001 -, 1. Oh, so we know that this is going to converge because it goes to 1000 here. If we use the rule we just developed, P was 1.001, which is greater than one. Hence it converges to one over your P -, 1. One over .001 is 1000. So we found that by actually doing the math. But we could also use the rule we just developed to show that also we have a type 2 case. Also integrals of functions that become infinite at a point within the interval of integration are improper integrals of type 2. So F of X is continuous on AB and is discontinuous at A. Then we would look at A to B of F of X DX as the limit C goes to A from the right side of C to BF of XDX. If F of X is continuous on AB and is discontinuous at B, then we'd have the integral ABF of XDX equaling limit as C goes to B from the left side of acf of XDX. If F of X is discontinuous at some C in the middle, then we're going to split up that integral into acf of XDX plus CBF of XDX and use parts one and two. If the limit exists and is finite, then it converges. If the limit does not exist, it diverges. So if we look here, we have zero to four. When we put in the upper bound, we can see that that would not exist. So we're going to say the limit as a goes to four from the left hand side of 0 to a of DX over sqrt 4 -, X. We're going to use the power rule, add 1 to the exponent, and divide by the new exponent. We're also going to use the chain rule by taking the derivative of the inside, which is -1. So we get the limit as a goes to four from the left hand side of -2 square root 4 -, X from zero to a putting in our upper bound, putting in our lower bound. Then we can see that 4 -, 4 is really 0. So that first term is going to cancel 2 times sqrt 4 is going to give us 4. If we look at another example, this one, we can't have zero. We have negative Infinity to one, but at zero it would be undefined. So we're going to have limit as B goes to 0 from the left -8 to B of DX over X to the 1/3, plus the limit of C goes to 0 from the right of C to the one DX over X to the one third. Using the power rule, putting in the upper bound minus the lower bound and evaluating, we can see that we're just going to end up with -9 halves because when we stick 0 in, once we've integrated, it's then defined. But back here, it wasn't defined. So we have to be careful when we're putting the 0IN. If we have zero to two of X + 1 / sqrt 4 -, X ^2 DX, we can see we can't put the two in there. So we're going to have limit as a goes to two from the left. And we're going to split this up into two pieces. We're going to have X / sqrt 4 -, X ^2, and then we can use AU sub there. And then we're going to have 1 / sqrt 4 -, X ^2. So if we do our U sub here, we get U equal 4 - X ^2 DU is -2 DX 1/2 DU equals XDX, right? There's my XDX. So we get -1 half the integral one over squared of Udu doing the power rule. That's the -1 half power. So add one, divide by the new exponent. So we're going to get the limit as a approaches 2 from the left of the negative square root, 4 -, X ^2 from zero to a. Now, if we look at this other piece, we don't know how to do that one the way it's written, but we could use our trigonometric substitution. If we do our triangle, we realize it's a minus. So the hypotenuse is going to be our two. If we come over here, we're going to have X so that we have sine Theta equaling X / 2. If we have sine Theta equaling X / 2, we could have two sine Theta equaling X SO2 cosine Theta D Theta would equal DX. So when we do this integration, our DX turns into two cosine Theta D Theta. We get the square root of four minus the X ^2, so 4 -, 4 sine squared Theta. If we factor out a four we get one minus sine squared Theta which is just cosine squared Theta. When we square root it, we end up with two cosine thetas which cancel. So we just really end up with the integral of D Theta which is Theta and then the Theta was just sine inverse of X / 2. So when we put that piece in, we have the sine inverse of X / 2 from zero to a putting in our upper bounds minus our lower bounds for the limit portion. And we can see that it ends up being 2 + π halves the next one if I have negative Infinity to 22 DX over X ^2 + 4. So limit as B goes to negative Infinity of beta 22 DX over X ^2 + 4. Once again, we're going to use our trigonometric substitution. We're going to draw our triangle. We're going to have X and two. SO tangent Theta is X / 2. SO 2 tangent Theta is the X or secant squared Theta D Theta is 1/2 DX. So if we have this 2 DX 2 two secant squared Theta D Theta because if we multiply by two over here, we get 2 secant squared Theta D Theta. The X ^2 is 2 tangent Theta quantity squared +4. That's really just 4 tangent squared Theta plus one. If we factor out of four, we can see the force cancel tangent squared is secant squared. Those cancel, we get just D Theta, which is Theta plus C or tangent inverse of X / 2 + C So if we come back to this problem, this is going to evaluate into tangent inverse of X / 2 and we're going to put in our lower bound and our upper bound. So we get this to be tangent inverse of 2 / 2 putting in our upper bound 2 / 2, tangent inverse of 2 / 2, tangent inverse of 1. So what angle gives us out a tangent value of 1? And that's going to be π force and then minus tangent inverse of negative Infinity over 2. So what angle gives us out negative Infinity for tangent? So now we want to think about what happens as B goes to a negative Infinity. So what's the tangent inverse for negative Infinity or what angle gives us out negative Infinity? If we think about looking at our tangent graph, the Y values are going to negative Infinity when the X is approaching or when the angle is approaching negative π halves. So we get π force minus negative π halves or three Pi force, which means it converges. Thank you and have a wonderful day.