Hello, everyone.Tom from everystepcalculus.com and everystepphysics.com.A problem on line integral, today from my menu.And this is off the internet from Patrick JMT. So, he’s my favorite. Um, let’s do it.Index 8 to get to my menu. And we gotta press second and the cursor down so you can go screen by screen to get to the L section.Which is Line Integral, here.And since they give us rt in the problem, you can see the problem on your screen. We choose RT from the menu.And that was the, um.Let’s go back and look at that menu, quick. Notice we have RT which is the original function XT plus YT plus ZT and then we have the magnitude of this of the derivative of that.But in other words, here’s exactly what it is.XT, YT, ZT which is the function there times the magnitude which is x prime of t squared, y prime of t squared, z prime of t squared.Write this all on your paper, of course.We’re going to enter our function.You have to press Alpha before you enter anything into these entry lines.You’re going to press Alpha 2 times x times y plus 3 times z.Again, I show you what you’ve entered.You can change it if you’ve made a mistake.Looks pretty good.Now we’re going to enter our lower point from the problem.So you’re going to press Alpha 0, 0, 0. I say it’s okay.Upper limit or point.Alpha 2, 5, 4. Now notice that the farthest point, really really we’re taking so it’s different from 0 0 0 in the beginning point, we would subtract the beginning from the end point here. And then when you integrate this with respect to T, that’s where the whole system starts.Then we get XT which is 2t, 5t, and 4t.Then we start putting it together here.Here’s the original function XT, YT, ZTand the magnitude square root of 45. And then we, in integration, we always take the constants out of the integral before you work the integral.So, we factored the inside, took the 4 out and the square root of 45 out and this is what is left here for integration.So we’re going to integrate that, you know add 1 to the 2 here to get 3 and divide by 3. And so then we have this here which we’re going to do over the limit of 1 to 0.Write all this down, this is exactly the way it’s done.T equals 1 then.Substitute 1 for all the t’s in the integral.And so here, I use quotation marks here because of the calculator but you’re going to use parentheses when you put this on your paper. So you look like you know what you’re doing.These calculations are exactly right.We come up with 38 times the square root of 5. And then at t = 0 and here’s the replacements of t with all the zeroes in it and that of course equals zero.We’re going to take the upper minus the lower which is 38 square root of 5 minus 0.So the answer is 38 square root of 5.

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