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Evaluate the definite integral:8tdt 2 (1+ e 8) -8tdt = 2 (1+ e 8t) =(Type an exact answer in terms of e.)...

Question

Evaluate the definite integral:8tdt 2 (1+ e 8) -8tdt = 2 (1+ e 8t) =(Type an exact answer in terms of e.)

Evaluate the definite integral: 8t dt 2 (1+ e 8) - 8t dt = 2 (1+ e 8t) = (Type an exact answer in terms of e.)



Answers

Evaluate the given definite integral. $$\int_{1}^{2} e^{-2 t} d t$$

Yes, we have the integral of negative E to part two. And they're to be went over to G G G. Well, you never went overseas. It's all in a T evaluated at ***. Be another each emergency that gives me Ellen from the E minus. L in E you are too based on our property. We could bring that in. That's just gonna be over in a Don't be too hard to So the negative comm positives are easy. Cancel. Well, that just have Ellen won over E. That gives me Ellen of one minus own E. Excuse me. Photo Ellen of one cancels Ellen. He just wanted to get a new one.

This question we got to find the value of this integral. So Different integral. So let's segregate the terms. So we have here is to T. D. T. That should be integrated within the limits of minus two. And then we have here is two minus T which is again within the limits of minus two. The integration of areas to these areas to be within the limits of minus 2 to 2. And then we have integration of areas to minus T as minus of areas to minus T within the limits of minus 2 to 2, compressing the Parliament. We have e square minus eras tu minus students the lower limit and minus comes out and we have ages too minus two minus materials to minus of minus two which will become plus two. So this will become E square minus series two minus two minus series two minus two plus e square. So this can be simplified as there are like tons and these are two like terms and these are also to like terms, so e squared z squared to e squared and we have to time series two minus two. So too is taken out, we have e square minus series two minus two as the final. Um So thank you

In this question we have to make a definite integration and let I represent that integration by I And it's limited to 0- two. And the function we have done the earlier days it is to the body minus a raise to the power minus T. Duty. Okay, so I am integrating it without substitution. If I integrate years to the party, which integration is serious to the party and integration of it is to the par minus T. Is minus or minus it is to the par minus T. Okay, this is the integration and now we have to put the limits here. So first of all I'm solving the inner bracket it provides me plus are very close to the power minus city. Okay now we have to put your upper limit. So the Parliament I'm going to replace that is two. If we put here to then we get a raise to the power to last year. S to the power minus two minus. If I put the lower limit it is areas to the power zero plus it is to be power zero again. Okay. No. So that we have to so this so to so this you can see here that This term changes into areas to the ball, 2-plus one of the areas to the ball to minus the changes into one and one that provides you, too. So this is the final answer. Years. Okay, Thank you.

We want to evaluate the interval to t E to the negative T square with respect to teeth. So in this chapter, we are told that integrating into the X d X is equal to just evil X plus. So maybe we can make a substitution into this to make it look like this Eat and you might know is if you let you equal to negative t squared and then we take the differential or we take the derivative to find the differential. D'you is equal to negative to t t. And then I divide the negative to over so we could do you over Negative too. There's equal to t e t. I noticed that the t d t here. Er actually, I don't even want to divide by negative too. I just want to negative over to DT because you could notice that the two d t there But we have the two tea and the d t. And then that just becomes negative. D u on the e. So the negative T square just becomes e to the U. And now all we need to do is a great integrate this so I can pull that negative out front. You need to the you should just beat me to you. We add on our constant of integration and then we can replace you again, which we said Waas negative t squid. And so this is what we get when we evaluate this indefinite into


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