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Problem 1. (9 pts each) Determine whether each integral is convergent or divergent. dr Vz (9 + =)...

Question

Problem 1. (9 pts each) Determine whether each integral is convergent or divergent. dr Vz (9 + =)

Problem 1. (9 pts each) Determine whether each integral is convergent or divergent. dr Vz (9 + =)



Answers

Determine whether each integral is convergent or divergent. Evaluate those that are convergent.

$ \displaystyle \int_0^9 \frac{1}{\sqrt[3]{x - 1}}\ dx $

In this problem were given the improper integral one over X minus one to the 1/3 from 0 to 9 and were asked to determine whether this integral is convergent or divergent. So in this case, neither of our bounds are discontinuous. But this integral is discontinuous at X equals one. If we plug X equals one in, we get 1/0, which is undefined. So I'm going to split this integral up into two and a girls, one being from zero 21 of our integral plus one that goes from one to nine of our integral. So solve this since we have to, and girls that are the same besides the bounds. I'm gonna begin by just solving the indefinite integral one over the cube root of X minus one D x. So to solve this, we're going to use a use substitution where u equals X minus one and d you equals D X. And if we plugged that into our integral, we get the integral of one over the cube root of you. Do you? And if we saw this, we get three halves, huge, the 2/3 plus c. And if we put back in our value for you. We get three halves times X plus one to the 2/3 plus C. And finally, if you plug back in our bounds for both of our integral, we're going to get three halves X minus one to the 2/3 from 0 to 1. Plus three have times X minus one to the 2/3 from 1 to 9. And finally, I'm going to take the limit of this and plug in those bounds. So I'm going to start with this first integral. So if we just take the limit of that, we get the limits. As T approaches one from the left of three have Times team minus one. The 2/3 minus three halves times negative one. And if we put in one, we get three halves. Oops. We re write that we get three halves as our answer, and if we go ahead and do our other one, we will get the limit as T approaches one from the right of three halves times four minus three halves. Times T minus one to the 2/3. And again, if we plug in our value for T, we get six. So if we go ahead and plug this in, I accidentally added a minus. Sign here, so I'm gonna go hand race out so we actually get negative three halves here. My apologies. If we go ahead and add these together we get negative three halves plus six, which equals negative three halves plus 12/2, which equals nine halves in This means that our integral converges to nine halves.

In this video, we're gonna go through the answer to question number 19 jumped 5.8. What could be so for us to evaluate the integral between minus infinity, Infinity of X squared over nine. Course, extend six the ex. So this is, uh, or rather into ground is an even function. So I could write. This is two times the interval between zero in for an affinity. The next squared over nine was X for the six. The ex Now that he's made a substitution. So if we let you be able to execute Mmm. The u the ex. It was a three x square. The ex sequence of wall over three x squared. Yeah. This is gonna be equal to two. That the limits stay the same because they were acute. Zero in the limit is exposed. Infinity. You also goes to infinity. Cynical X glad of a nine clothes yeast bread times 1/3. What a wreck squared. Do you this x squared, miss, excluding a council backwards. 2/3. That's we know infinity. One of the nine. Plus, he's glad. Do you? Okay, so now we're looking at this integrated and we're thinking this looks something like, Ah, my function is gonna integrate to a narc tan. But before that, we need to make a second substitution. So if we let where were you equal to you over three. And I'll just write this in every way. Swiftboats Very that it's a kind of a factor off a nine. All right, this is you with three square. Therefore, to you is equal to three times it'll be a It was the next page. So therefore we got to over three times one of the nine in school, which means they were. And if it is again, the into girls at the limit of the integral is don't change e 1/1 minus, do you? So plus, Dewey squared plans by three Okey dokey at a three And this three, we're gonna cancel. It's about two of the nine times the integral of 1/1. Plus after you squared, that's gonna be, uh, 10 of w. This means a one infinity in the limit. W goes to infinity Arc Thanh goes to you high over too Onda zero our time. It's zero different Final answer. It's part of a nine and it's interval converge

In this problem were given the improper integral X squared over nine plus x to the sixth d X from negative infinity to infinity and were asked to determine whether this integral is convergent or divergent. So because we have two bounds that go to infinity, I'm going to rewrite this as two into girls, with one having the bounds zero to negative infinity of our function and one going from zero to infinity and adding these two together. So because we have two in a girl's now that are exactly the same besides the bounds, I'm going to begin by just solving the indefinite integral X squared over nine plus x to the sixth d X. So if I do this, I can rewrite this as X squared over nine plus x cubed squared, which same thing as Exodus six. And if we do this, I can see that we have a U substitution that we can use where you equals X cubed and d'you equals three x squared. And if we rewrite our function with this, you substitution, we get 1/3 time's the integral won over nine plus u squared d you. And now that we have this, we can see that this looks a lot like one over a squared plus x squared. Where is a number? And that is our formula for the integral of tangent. Inverse. So we can rewrite this if we take the integral as 1/9 times the tangent in verse of you over three. Because nine is the same thing as three squared. And now, if we plug back in our ex, we can do 1/9 tangent in verse of X cubed over three plus c. And now I'm going to plug back in our bounds for our first in a girl. I'm gonna have t approach negative infinity. And for the second, I'm going to have tea approach infinity so we can rewrite this as 1/9 tangent Inverse of X cubed over at over three from t 20 plus 1/9 tangent in verse of X cubed over three from zero to t. And now I'm going to take the limit of both of these. So for my 1st 1 I'm gonna have t approach negative infinity of this function. So if we put in our bounds, we would get negative 1/9 tangent in verse of tea cubed over three, plus the limit as T approaches infinity 1st 2nd expression of 1/9 tangent in verse of T squared over three. So if we take the limit of this will get negative 1/9 times negative high over to plus 1/9 times high over to. And this is because if we plug in tangent inverse of infinity or Tanja inverse of negative infinity and we look at a quick sketch of the tangent graph will notice that as tangent approaches infinity, it approaches high over two, and as it approaches negative infinity, it approaches negative pi over two. That's where we got those values. And finally, if we do some simplifying, we get pi over 18 plus pi over 18 which leaves us with high over nine. And because we got a number value, this means that are integral is convergent two pi over nine

Upper integral. We're gonna take the limit as a goes to infinity of the integral from zero to a of D X over X plus nine square. Uh, the anti derivative of one over X plus nine square is negative one over X plus night when they evaluate this from 0 to 8. So putting some values in when we put a into this expression will get negative one over eight last night. And when we put zero, then we're going to get a plus one overnight. Since a tends to infinity, the first term goes to zero and we're just left with one over.


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