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Evaluate the integral Cyds, C: x = t^3, y = t^2, 0 ≤ t ≤1....

Question

Evaluate the integral Cyds, C: x = t^3, y = t^2, 0 ≤ t ≤1.

Evaluate the integral Cyds, C: x = t^3, y = t^2, 0 ≤ t ≤ 1.



Answers

Evaluate the integral. $$\int_{1}^{0}\left(t^{3}+t^{2}\right) d t$$

We want to evaluate you. Definite integral of one over tea from 1 to 1 over X one X is bigger than zero. So the driven, the anti derivative of one of her tea is just natural. Log T We get natural log of teeth evaluated on 1 to 1 over X. Then we can go ahead and plug in one over X and then plug in one extract so we get natural log. Actually, you know what this is So first plugging in one over X, we get natural log of one of Rex minus natural log of one well natural log of 10 And then we can rewrite the natural law of one of her ex as X to the negative first power. And then let's factor that negative one out. So we get negative natural log of X, and so this here would be our definite integral

Because we have polynomial divided by polynomial and they're both too The first power I would go ahead and do long division here, so go using the polynomial division to simplify the inside weaken right. This is for that's the kosher. Remainder is negative three and then we have one plus bt on the bottom. So So this is from using just doing the polynomial division and then you can see our new integral is easier than the original. So the internal afore we know that she's fourteen here. If this three in front of the T and this plus one is bothering you, you can go ahead and use the use up here, from which you can see that do you equals three dt So I won't write out the full details there. But if this may help you integrate, then we have minus three natural log one plus three tea and then we'LL also have to divide by three due to this in the use of our end point zero one and then cross off those threes. So plug in the one first for tea, we just get a four and then plug it into the natural line your natural log before. And then when you plug in zero for tea both. Herms Laura zero. So that's your final answer.

Okay, So pulling out the constant we can be right. The integral is four times the integral from 4 0 to 1 off, one over T squared, plus one dt applying our our contention proves we can set. This is equal to four times the arc tangent off T. And then we're gonna add value with this from 0 to 1. This is four times the arc tangent of one minus the arc tangent off zero, which is equal to four times pi over four minus zero. And this our final answer is high.

All right here. We're trying to evaluate the integral off from zero to a one off director, two T and 40 et for this. Uh, yeah, we can actually, um, just factor out the numbers Understood in the same integral once. So, uh oh, yeah. The other things equal to to this, which we realize that this integral here is it's the same integral Valerie twice. So let's just do that separately. Thereto one off t t on this isjust t squared over to evaluate at zero 21 So we on that worked a sequel to half off One square to my This is the old squared actress 1/2. All right, So that's that way. We can just plug it in two times. 1/2 on foreign one and that gives one and two. That's it.


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