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1. Find the line integralyd8,whereis the curve given by 7() = (1,t,4/2), 0 < t<1....

Question

1. Find the line integralyd8,whereis the curve given by 7() = (1,t,4/2), 0 < t<1.

1. Find the line integral yd8, where is the curve given by 7() = (1,t,4/2), 0 < t<1.



Answers

$1-16$ Evaluate the line integral, where $C$ is the given curve.
$$
\begin{array}{l}{\int_{C}(x+2 y) d x+x^{2} d y, \quad C \text { consists of line segments from }} \\ {(0,0) \text { to }(2,1) \text { and } \text { from }(2,1) \text { to }(3,0)}\end{array}
$$

Okay. So we're trying to find the length of the car lt equals equals sign of pity to T. And sign or to buy tea from 100 two 140 Okay. Mhm. Okay so the length of the cuff of the function of the vector function of T. From T equals A. To B. Is given by norm of our primal T. D. T. Uh And we just integrate that from A. To B. So we need to find the lower and upper limits. Okay now the given function given points. I'm just gonna call This point p. one at this point as p. two. And we're looking for T. Such that Albay is equal to p. one for the lower limit. And the easiest way to do this is to look at the 121 Function Y. O. T. So the at p. one we must have Y. Of A equals zero. So which means To Obey two times a. is equal to zero which means A. Equals. Yeah. And I'm sorry. Similar. I think with the upper limit our P. R. B. Must be peace up to. And then we're going to look at the Y. O. B. Is equal to this has before. So two times B is equal to four. Which implies that B has to be too. Okay so now we have the setup of the integral um for the lower limit and a plummet and we need to take the derivative of our respective T. And then we get negative pi times sine pity two and two pi course I of to pity where we take the norm, we get square it. Hi square Science square. Or pity plus four plus four pi square, coarsen square or to buy tea. Sorry that means that the length of the car. We found A. To B zero And B. two B 2. And then the nor mobile priority is this. We found this above plus four plus four pi square, Cool Science Square or to Pity Bt. And when we knew markedly integrate this using the calculator, we get the value to be around 10 points 33 11

The question says that we have to evaluate the line integral where C. Is the curve integral over the regency by D. S. Were see x equals two T square why is constituted? And the range is from 0 to 3. Now moving towards the solution recall that DS is given by under root of D X by DT square plus de vie by DT square duty. So here in this case the U. S. Will be called to root Toting Square plus two square DT which will be equal to two into root T square plus one. Did. So now we need to substitute the values of X. Y. And the years integral over the region. C. Y ideas will be quarto integration from 0 to 3 to T into two under root T square plus one dating. So it will be equal to integration from 0 to 3 40 under root t square less one DT. Now we will be solving this integral. So moving further, substitute your T square plus one, it calls to to and you will get to T DT equals. Today you now limits will change from 0-3 to one took 10. So now the integral will be uh integration from 1 to 10 to undergo to you D you. So it will be equal to two and 2 two x 3. U three x 2 Lawen going from 1 to 10. Putting the limit. You will get your uh value for integral as four by three and two. Uh 10 and two. Route 10 minus four by three into one. Into route one which will be called to 40. Route 10 minus four by three. And this will be our answer. Thank you.

So we're working with the curve y equals seven over X squared, and we're going to be evaluating the area of the region under this curve. But it's bound by the X axis, so X, so why has to be positive and is also bound by X equals one. So we're going to It's found on the left by X equals one. So let's just write that in that That's the left. So if we were to draw a picture of this would see that we have this sort of curve going like that that goes all the way to infinity getting closer and closer to some number. Um and so we're going to be cut off by the X axis and on the left by Y equals one. So we're cut off right about here or by X equals one. So we're going to go from X equals one all the way to infinity. So when we're writing an integral, we're going to have one to infinity of our function. Seven over X squared, evaluated with respect to X. Now we can't go all the way to infinity, so we're just going to have to approximately a limit that as T approaches infinity. So we're gonna try to get as close to infinity as possible without actually trying to reach infinity. So now we're going to be able to take the anti derivative. So I'm going to rewrite this as the limit as T approaches infinity of the integral from one to t of seven times X to the negative too. So the anti derivative is going to be equal to the limit as T approaches infinity of seven times X rays, the exponents by one to the negative one. And now multiply our divide by the new exponents. So that's times negative one evaluated from one to t. So now we're going to plug in tea and one for our upper and lower bounds so we'll have the limit as t approaches Infinity of negative seven over T minus negative 7/1. So for this first term, as T gets larger and larger and larger in the denominator, the larger the denominator a fraction of this fraction the closer, the fractions, getting to zero. So we're just going to end up with zero when t is equal to basically infinity, and that's going to be added to seven. So our solution for the area is going to be equal to seven. Under this curve from X equals 12 X equals of


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