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Which of the following integrals are improper integrals?1 [~ 6+3y & I /' G-x-"dx m [' z& Iv / e-3 r dxO1and IIb) Olonly0III only0I and IV...

Question

Which of the following integrals are improper integrals?1 [~ 6+3y & I /' G-x-"dx m [' z& Iv / e-3 r dxO1and IIb) Olonly0III only0I and IV

Which of the following integrals are improper integrals? 1 [~ 6+3y & I /' G-x-"dx m [' z& Iv / e-3 r dx O1and II b) Olonly 0III only 0I and IV



Answers

Which of the following integrals is improper? Explain your answer, but do not evaluate the integral.
$$
\begin{array}{ll}{\text { (a) } \int_{0}^{2} \frac{d x}{x^{1 / 3}}} & {\text { (b) } \int_{1}^{\infty} \frac{d x}{x^{0.2}} \quad {\text { (c) } \int_{-1}^{\infty} e^{-x} d x}} \\ {\text { (d) } \int_{0}^{1} e^{-x} d x} & {\text { (e) } \int_{0}^{\pi} \sec x d x} \quad {\text { (f) } \int_{0}^{\infty} \sin x d x} \\ {\text { (g) } \int_{0}^{1} \sin x d x} & {\text { (h) } \int_{0}^{1} \frac{d x}{\sqrt{3-x^{2}}}} \quad {\text { (i) } \int_{1}^{\infty} \ln x d x} \\ {\text { (j) } \int_{0}^{3} \ln x d x}\end{array}
$$

Okay, so we're given a bunch of integral is and were asked to find which of these integral XYZ improper And there are two ways and integral can be improper. One way is if the interval of integration is infinite and one or both of these end points can be infinite for this one to apply. And the other way is if the inte grant may tend to infinity and the inte grant is the function that you're integrating. So this one applies if the function is discontinuous, um, at any point, um, within the given interval Oh, for the integral. So the first integral we have his, um the integral from 0 to 2 of D x over X to the one third. Okay. And we could just rewrite this as integral from 0 to 2 of one over X to the one third the X, which just helps visualize the function better. So if you look at our function which is one over X to the one third you see that are variable is that is in the denominator and if you remember off fractions, denominator cannot be zero. But if you look at the interval were given which is from 0 to 2. If you would plug zero into this function So that would be like saying if we find f of zero, so that's gonna be 1/0 to the one third that's going to give us 1/0, which is undefined. So dysfunction is discontinuous within the interval given which would make this integral improper. The next integral we have is the integral from one to infinity of D X over X to the 0.2. And just by looking at the interval were given. We see that we have, uh, infinity far as part of our interval of integration. So we already know that this integral must be improper. So for the next integral, we have the integral from one to from negative one to infinity of e to the negative x d x. And for this one, we also have an infinity in our interval of integration. So this integral is also improper. So the next integral we have is the integral from zero toe one off e to the negative x d x in our function here is e to the negative X we can do a rough sketch what this graph would look like. Look, something like this where the graph approaches infinity at both ends and the interval were given is 0 to 1. And if we look at the graph of dysfunction, we see that it is continuous and defined within the interval were given and therefore the integral is proper. Yeah, so the next integral we have is the integral from zero to pi of C can't x d x and our function here is C can X. And if we do a quick sketch of this function within the interval given shit, it would look something similar to this Mhm. Yeah, and you can see that this function has an ass, um, tote at pi over two. And since dysfunction is undefined within the given interval of zero dupuy, this integral is improper. So the next integral we have is the integral from zero to infinity off sign of x d x and since our interval of integration has infinity as one of the endpoints, this integral is improper. So the next integral we have is the integral from 0 to 1 of sign of x d x. And here are function A sign of X. So we do a quick sketch of this function. It will look something like this where continues on in either direction And since dysfunction is continuous, it will be continuous and defined within our given interval of 01 And therefore this integral is proper. So the next integral we have is thean integral from zero toe one of D x over Uh huh square root of three minus x squared. I'm just going to rewrite this as the integral from 01 of one over square root of three minus x squared d x. So we're given the interval from 0 to 1. And if we evaluate the function at either end point So if we look at f of zero, it would be one over the square root of three minus zero squared that would equal one over the square root of three. If we evaluate the function at the other end point f of one, get one over the square root of three minus one squared, which would equal one over the square root of two. And since this function doesn't tend to infinity within the interval given to us, it is continuous and defined within this interval and therefore it is a proper integral. So the next integral were given is the integral from one to infinity of l N x d x. And since one of our endpoints is infinity, this integral is improper and the last integral we have have to squeeze it in right here were given the integral from 0 to 3 off L N X dx and here are function is Ln X, which is the natural log. And if you remember, Ln X is defined for values of X greater than zero. But since our interval is 0 to 3 Ln x is undefined at our lower end point. So our function is not continuous within this given interval and therefore this integral is improper.

So far, eh? Look at the graph ofthe ten and ice. First interval from zero to pie before is finite and there's a function of ten attacks has no infinite. It is continued on the interval from their tow pilot for so it is not improper. B also look at it A graph of ten an axe Wei will find Han and pile off too. Yes, infinite. So hand and eggs has infinite is continually on this angel from there to pie. So it is improper Integral. See what axe goes to make to Juan. Denominator of dysfunction. I like square minus X minus two goes to zero. This function got to infinite. What fax goes to get you one So it is improper for the this into a wall is given it so it is improper, integral.

So for a mm You realize that the function is a fraction. So they do know later shouldn't be able to service. So we are saying that two X. My next one should not be equal to observe. So this implies that are two eggs Should not be equal to one. Which implies that our eggs should be equal to one divided by soon. So since they find share is called genius is continuous on the intifada. And the whatever 12 this in place so it is in place that's the giving is in place. That this implies that the kim feel insignia is proper it's proper. Okay then for the second part B. We have DNC girl. So you have been sick girl from several one 0 to 1 one divided by two eggs. My next one the ace. So where where our f of eggs is giving us one divided by two Eggs -1. So they are saying this implies us two X. My next one should not be equal to observe. So this implies that our eggs should be equal to one divided by two. So so because because the fund shame the fraction is continuous. It's continuous on the interval zero. Well divided by soon dale. It's open and oh peanuts one if I did buy two includes that's when But this continues this continuous this current thing years. This continues at X. Equal to Mexico too one one divided by soon one divided by it soon the they seek uh and seeing good is improper is in girl power in proper thankful see of C. See we help the etc. Girl from negative infinity to infinity of one divided by two eggs. My next one the s. You can write this as being cinchona from negative infinity. So one divided by two. We have one divided by two x -1. The eggs plus the integra. one divided by its own infinity. one divided by two x -1. The eggs. So and this so these diverges diverges. So then this implies that the in secret is improper. His improper because their limits and limits of integration. Integration. Uh huh. Lots of mine. It's infinity. Then for D. With the mm. We have the n. c. a girl from 1 to 2 Lane of eggs. My last one the X. So we are seeing that eggs my next one should be greater than zero. So this in place that are eggs should be greater done wow. So then the being speaker you would think uh is improper is improper because because the function is continuous continues on the end of uh one. So but this quint seniors That's x. equal to one

In this problem were given a few into roles and were asked to determine whether these into girls are improper or proper. So for our first integral, we're just gonna look at our bounds and determine whether this integral is continuous at all points. So if we plug in Tangent of Pi over four, we get one just good. And then if we plug in tangent of zero, we get zero, which is good, and every point between zero and pie fourths works. Four are integral and therefore, since its continuous all points part A is a proper integral. So for part B, we do the same thing. Tangent of Pie is zero, which is good, and tangent of zero is zero, which is awesome. But if we look at all the points between zero and pie, we'll notice that tangent of Pie House is actually undefined. It's 1/0, and that means that we have a GIs continuity in our integral, which means that we have an improper into a girl for part B. So for Part C again looking at those bounds If we plug in one, we get d X over one minus one minus two which is D X over negative, too. And that's fine. But if we plug in negative one, we get D X over one plus one minus two. Which leaves us with 1/0 TX, which obviously means that we have an improper integral because of that disk continuity. And finally, if we look a d. Ah, we can easily identify that we have an improper integral in this case because one of our bounds is infinite. This is a clear and easy sign to say, or to show that we have an improper, integral


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