:: Transcript ::

Hello, Tom from Every Step Calculus Every Step Physics .com. This is a problem submitted by a student. How do you evaluate this line integral? A line integral the original formula is the integral of a function times d R which is the derivative of the RT function here and then C is the what the parameters are going through which in this case is a curve. Trying to find the integral of this curve here. index(8) to get to my menu. Scroll down here to line integrals because that’s what they are asking you to do in the problem in your test or whatever, homework. So we’re gonna go down here to line integrals, press Enter . And we’re gonna choose number six I and J because that’s what they give you here, we’re in I and J okay? So you have to get used to that and my programs a little bit. You have to think a little bit in college so I and J we’re gonna choose number six and then in calculus 3 this is an M this might be calculus 2 here because we’re only dealing in XY & IJ parameters not XYZ and ijk parameters. So if it’s calculus two then this is an M which which you know tells you that this whole function here is an M and this is an N function before the J ok? So enter the m function term before the I, here’s the term before the I okay? And we’re gonna put that in now you’re gonna press alpha, that’s the only thing you have to do in my program, just press alpha and then you can enter the two times x times y plus one here in the calculator here okay? And then, so we did that, and then the end function here and we’re going to enter that x-squared plus 1 by pressing alpha first and then putting that term into the calculator okay? Now I’ve already done that in the simulator to save time. So you can see that it matches what they’re talking about the calculator always puts the XY term before the consonants you know so switch them around a little bit but no big deal that makes sense to us. And we’re saying it’s okay because you can change it in case you made a mistake in entering it. Our line segments given no they’re not, a segment is XY point, to another XY point, or XYZ point to the number another XYZ point depending upon what they tell you in the problem but no line segments so we’re gonna say no. But they do mention I and J in the in the problem okay so here’s I and J where is it? Here’s I and J we’re gonna say yes to that and we’re gonna enter the term before the i hits which is T okay so we we’d enter alpha t normally alpha t. And then we’re gonna enter the term before the J well here it is over here now this is a tricky problem because sine of PI over 2 is really 1 okay so you can put that in here you can enter sine of PI over 2 times T squared just like it says here but the calculator well we’ll assume that you’re talking about sine of PI over 2 is really 1 so it’s 1 times T squared. So if you enter that it’s just gonna come up with T squared anyways okay so here’s the I and the J that they’ve entered here and now we’re going to put in the range range is zero less or equal than t and then less or equal than one okay? So the lower limit is zero you’d put alpha zero in here and then you put alpha one just in case they gave you something different here. But generally when you parameterize something in a problem like this it’s always 0 to 1 ok? So we say it’s okay so now we start to do the the derivatives etc here’s the X term which is T which is T right here and the derivative of that is one here’s Y as T squared the derivative is 2t DT here’s the original integral and we’re going to put the all these what we found for X&Y; into the problem here which we’ve done I have to use quotation mark so the calculator does that but you’re gonna put parentheses around here like for x put parentheses around for T squared okay and you’re in your paper. Then we of course we multiply it out and condense it and here’s the real which you’ve entered here. And we’re gonna integrate the problem which is here and there you could do that really easy couldn’t you? I couldn’t, but you could over the range of 0 & 1 so we integrated this problem now at t equals 1 we compute it through there and we get seven halves at t equals zero we computing its 1/2 upper – lower always in physics and calculus upper – low which we’ve determined. And the answer is 3 pretty neat huh? everystepcalculus.com go to my site buy my programs are only 40 dollars and you’ll pass calculus and you don’t have to study this stuff hours and hours and hours. I’m pretty good at line integrals right now because I’ve been doing it for three weeks 25 years actually but then cleaning it up here for three weeks for other problems and almost all the problems I can I can find so have a good one, I hope you pass calculus!

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