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Integrate byline answer

this video contains the solutions to the line integrals practice problems for this first one we have to compute the line integral integral over C of FDS where C is the upper half unit circle parametrized by r of s equals cosine of s the sine of s for s between 0 and pi and f of XY equals x plus y so there's a lot of information being given to you here but the main thing to keep in mind is that the curve that we're talking about is this unit circle and since its parametrize with X equaling cosine and Y equaling sine that's our standard parametrization so we know that's going to be a counterclockwise parameterization and so we're really doing is integrating this function from this point 1 comma 0 to this point negative 1 comma 0 so the first thing we need to do is figure out whether this curve is parameterized by arc length and the way that we do that is by figuring out the magnitude of r prime of us in other words figuring out the speed of this parameterization so since R of s is cosine s sine s R prime of s is negative sine of s cosine of s what's the magnitude of that vector well it's the square root of negative sine of s squared plus the cosine of s squared that's sine squared plus cosine squared and we know that that's 1 and the spirit of 1 is 1 so that means that this curve is in fact parameterize by arc length so we can simply integrate from the starting value of s to the ending value of x from 0 to PI of our function X plus y but now we're going to take the X component of our parameterization and our y-coordinate component of our parameterization and substitute that into our expression that's what we really get is instead of X plus y we get cosine s plus sine s and then we can just integrate that with respect to s since our curve is parametrized by arc length so when we take our antiderivative antiderivative of cosine is sine so we get sine of s minus the antiderivative of sine the antiderivative to sine is negative cosine of s plug in PI plug in 0 and subtract so we get sine of Pi minus cosine pi minus the quantity sine of 0 minus cosine 0 that's minus minus 1 minus minus 1 which is 2 we have another line integral here this time the curve is a line segment given by this parameterization and our function is x squared plus y squared so again we need to figure out whether our curve is parametrized by arc length so we need to look at our prime of T which is going to be the vector 6 comma 8 and we can already probably see that the magnitude of R prime of T is not going...