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IuesLiuil 0Let X(t) (1 t)+itwith 0 < t < 1. The contour integral 5,14l" dzis equal to which one of the following values?...

Question

IuesLiuil 0Let X(t) (1 t)+itwith 0 < t < 1. The contour integral 5,14l" dzis equal to which one of the following values?

IuesLiuil 0 Let X(t) (1 t)+itwith 0 < t < 1. The contour integral 5,14l" dz is equal to which one of the following values?



Answers

Evaluate the following iterated integrals. $$\int_{0}^{1} \int_{0}^{1} t^{2} e^{x t} d s d t$$

So here we are given a double integral. So it's important to remember when we're doing a double integral or triple integral integration, we're basically looking for uh we're only or assuming that the non and the variable not being integrated, the parts of that are just going to be a constant. So since we're integrating for the variable T. We can bring out the coastline S cube out of the integral expression. Since this would just be a constant. So this would just be the integral of one would just be T in this case. And once we substitute for S squared this would be the integral from 0 to 1 of the square co sign of X cubed Diaz. So now we can use U substitution where we take s cubed as you and then we take d'you equivalent to three S squared. The uh T. S. And this would be one third would be S equals zero. S equals one of co sign of you to you. So that uh the integral, of course you would just be sign of you. So this would be one third sign of S cute. From zero to once a sign of zero is just zero Sign of a sign of one would be equivalent. So this would just be 1/3 sign of one. And this gives our final answer

They have to keep in indignities. Indignation off one minus d caused by D dd for solving it will write it as you'd want to because by Dee Dee minus genuine too. Dean Gauss. Bye d dd So this becomes signed by D A Bond by. I didn't do this suit. This becomes do you now for this? This is I so let Saudi for solving days we'll apply the rule, which is since costs by D isn't even funds. Um and D isn't hard funds in therefore hard into even fun son begins our friends in Therefore this becomes and you don't do is on offense and into cause variety which is even fence and Dede is equal to or defense. Um, Andi, this becomes a geo sins. In addition, off our fund, son is zero. Therefore, this far belongs jello. Hence we can say that went on Saudi We're on indigo, But it Jones to Gino. This is the answer to a given problem

So this iterated into girl, we're going to use a U substitution to solve this. So we're going to say you is equal to us, Rusty. So do you. There's going to be equal to d s, okay, within general, from zero to one. And then when we have Tio, when we use a u substitution, we have to change our integration bounds. So you re plugging s is equal to zero. We're going get use equal to t place. Looking s is equal to one that one plus t. So this is it. Aah! Into girl squared of you do your duty already so we know how to solve this were not a find. That internal of this is going to be two thirds You three over too. So I waited from Tito won over tea DT, pull out the two thirds in general from zero to one. Ah, Now use FTC here, so FTC will get one plus t three over too minus three minus T to three over too, dt. But to do that to over three and again, we're going to use another U substitution. I'm not going to write this one out. Actually, I will but I won't go into too much detail. Just kind of go through the answer, right? That's for the second part. The second interval here. And what will this integration give us? It will give us to fifth one plus T to the five over two. Ah, minus two fifths. T to the five over two, and this guy's gonna be evaluated from zero to one. Okay, so pull out the two fifths, and when we plug in one, we will get to to the five over two and one minus. When we plug in zero, we will just get one to the five over two and minus zero. So our answer not going to simplify here was for over fifteen to to the to the to the file, over to minus two, and we're done.

Hi there. And this problem were asked to compute this definite integral. Now it's the definite integral of vector valued function. So from the end of this chapter, let's remember our definition of the integral of a vector valued function, uh is we're gonna have three separate into girls. We can just add together. So it will take the definite integral of this first component which, let's rewrite that as Tito the 1/3 power. As long as we're at it, that will be our I component. Plus again definite integral from 0 to 1 of our function, one over t plus one that were J component and finally definitely a girl from 0 to 1 of e to the minus t. That'll be RK component. So we just to each of these separately, let's begin with this 1st 1 here. Let's find an anti derivative. In order to use the fundamental theorem of calculus of the anti derivative of tea to the 1/3 we bumped power up by one to get 4/3 and then divide by that in front. So we have 3/4 t of the 4/3 we'll plug in one and zero, and that will be our I coordinate next for the J Cordant uh, anti derivative one over T plus one. Now, since it's one over a function, we suspect the natural log of that bottom function. And there's no need to use U substitution here, since a derivative of T plus one is just one. So the anti derivative truly is a natural log of t plus one on again. We will plug in one end zero Jay here and finally for K component this last one here, anti derivative eat of the minus t is minus you. The minus T. We'll plug in zero on one. That will be our key component. Okay, so let's compute these now. So if you plug in 11 of the 4/3 power is just one. So plugging in one gives us just 3/4 times one, which is one plugging zero clearly gives us zero. So that's quick for the I component. Now let's move on to the J component. Ah, natural log of one plus one says natural. Log of to minus. I was plugging zero for tea. Natural log of zero plus one. This natural log of one, the natural log of one is zero on any log of one is just zero. So that's our K component. Is there a J component now for K? I was plugging one for so minus eat of the minus one and minus. That's a minus minus, so we end up with plus and eat. Of the zero is just one. There's RK component. So we're almost done. Just we'll clean this up a little bit. The end up with 3/4 I That's it for the I component plus natural lot of to Jay, that's RJ component. And finally for K. Let's see, we can rewrite this a little queen. Or maybe that's right. The one first and then eat of the minus one is the same as one over E. So that's RK component and we are done


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