Split (1)

train · 116k rows

role stringclasses 2 values	content stringlengths 0 2.1k	session_id int64 10 21.7k	sequence_id int64 0 2.38k	annotations listlengths 0 8
student	ohh, sorry, off**	15,082	24	[]
volunteer	let's see, where did we leave me last time?	15,082	25	[]
volunteer	Right	15,082	26	[]
volunteer	So, last time we had glossed over.	15,082	27	[]
volunteer	the various	15,082	28	[]
volunteer	um forms of integration by fractions.	15,082	29	[]
volunteer	mainly this table.	15,082	30	[]
volunteer	which just covers like the form of the restroom, but you notice that they don't really go beyond um 3rd degree polynomials here.	15,082	31	[]
volunteer	Um, that is because	15,082	32	[]
volunteer	you can typically	15,082	33	[]
volunteer	uh	15,082	34	[]
volunteer	rewrite	15,082	35	[]
volunteer	the expression	15,082	36	[]
volunteer	um for a higher norm for a higher-order polynomial.	15,082	37	[]
volunteer	If you can, if you can't if you can factor it and then kind of rewrite it as um	15,082	38	[]
volunteer	a group of smaller polynomials.	15,082	39	[]
volunteer	or lesser order polynomials that preach you the best way to go.	15,082	40	[]
volunteer	But I think, and I also don't think they're gonna give you	15,082	41	[]
volunteer	they're gonna have you do problems that are	15,082	42	[]
volunteer	higher order than 3,	15,082	43	[]
student	right,	15,082	44	[]
volunteer	right, but I, I don't think we actually did any problems.	15,082	45	[]
volunteer	Um	15,082	46	[]
volunteer	let's see, what do we cover last time? We had finished doing	15,082	47	[]
volunteer	the particular	15,082	48	[]
volunteer	um the previous chapter covered	15,082	49	[]
volunteer	particular integrals	15,082	50	[]
volunteer	which involved	15,082	51	[]
volunteer	um	15,082	52	[]
volunteer	just in just trig properties.	15,082	53	[]
volunteer	and logarithmic properties, right? We had this table, which I really wished they would have gave this its own table.	15,082	54	[]
volunteer	rather than the way they have it written out.	15,082	55	[]
student	yeah, that would have been very helpful	15,082	56	[]
volunteer	Uh, but again, these are all just various um	15,082	57	[]
volunteer	various forms.	15,082	58	[]
volunteer	that I borrowed that are already known, they've been proven, we've seen them proven. Um, they just kind of help the integration process.	15,082	59	[]
volunteer	right? So that we, so that we can get whatever problems we have, we can just, if we can figure out how to write them, um, typically by substitution, if we can figure out how to write them in the forms that we know, basically, uh, um, these	15,082	60	[]
volunteer	uh	15,082	61	[]
volunteer	forms which are, uh, derivatives of either 2nd order or 1st order.	15,082	62	[]
volunteer	um, polynomials	15,082	63	[]
volunteer	Well, square rooted second-ordered polynomials.	15,082	64	[]
volunteer	Um.	15,082	65	[]
volunteer	then these already have known solutions.	15,082	66	[]
student	yes right	15,082	67	[]
volunteer	and then it just becomes a form of, you know, how do you apply it?	15,082	68	[]
volunteer	right? And then now we get to, well, what happens if we have	15,082	69	[]
volunteer	um rational	15,082	70	[]
volunteer	expressions with	15,082	71	[]
volunteer	polynomials and denominator, as always, partial fresh, where um	15,082	72	[]
volunteer	they do, they have a B term in the middle.	15,082	73	[]
volunteer	right, because the previous polynomials were always x-a minus x, right, which could be	15,082	74	[]
volunteer	factored into um X + A or X minus A or so on and so forth.	15,082	75	[]
volunteer	but never really, um,	15,082	76	[]
volunteer	the full expanded form didn't cover having an uh	15,082	77	[]
volunteer	an X term in the second-order polynomial. Now we're covering that other aspect where now, OK, we have	15,082	78	[]
volunteer	a second-order polynomials that can be factored into two different expressions, but have this um	15,082	79	[]
volunteer	what is this? A + B term?	15,082	80	[]
volunteer	for X	15,082	81	[]
volunteer	or even 3 or polynomials.	15,082	82	[]
volunteer	Um.	15,082	83	[]
volunteer	which I think actually	15,082	84	[]
volunteer	the last few polyno the last few expressions also covered that.	15,082	85	[]
volunteer	Where is it	15,082	86	[]
volunteer	Yes, here	15,082	87	[]
volunteer	Yeah	15,082	88	[]
volunteer	wild	15,082	89	[]
volunteer	Wild.	15,082	90	[]
volunteer	Anyway, so logically	15,082	91	[]
volunteer	you can kind of see um the progression. It's just typically as these things go, that, you know, integration is a skill.	15,082	92	[]
volunteer	Um.	15,082	93	[]
volunteer	and only we get better, if you get better at a scale is to just keep practicing.	15,082	94	[]
volunteer	So	15,082	95	[]
volunteer	that is	15,082	96	[]
volunteer	kind of what I've planned today.	15,082	97	[]
volunteer	um, unless you've already practiced, done problems. We can go over the example problems	15,082	98	[]
volunteer	Um,	15,082	99	[]
volunteer	if not, then we can move on to	15,082	100	[]
volunteer	if you have done them, then we can move on to um	15,082	101	[]
student	I did 7.4	15,082	102	[]
volunteer	integration by parts.	15,082	103	[]
volunteer	You did 7.4. OK. So you haven't done 7.5.	15,082	104	[]
volunteer	So we can go through the example problems and then move on to actual problems. And then, um, touch on, actually, you'll probably have time to do integration by parts depending on how	15,082	105	[]
volunteer	many practice problems we do.	15,082	106	[]
volunteer	Yeah	15,082	107	[]
volunteer	Yeah, and I need to practice. I was trying to do 7.5.	15,082	108	[]
volunteer	and kind of hit a wall right away.	15,082	109	[]
volunteer	That's why I said I'm	15,082	110	[]
volunteer	I need to practice.	15,082	111	[]
volunteer	Most of this stuff should be obvious, but	15,082	112	[]
volunteer	OK, so	15,082	113	[]
volunteer	can you, um, do I need to zoom in or anything? I don't think I can zoom in comfortably.	15,082	114	[]
volunteer	and very tight	15,082	115	[]
volunteer	There we go	15,082	116	[]
volunteer	All right. So do we need any	15,082	117	[]
volunteer	well, let's just go through the theory.	15,082	118	[]
volunteer	It always helps	15,082	119	[]
student	yes	15,082	120	[]
volunteer	Hopefully it won't confuse us	15,082	121	[]
volunteer	So recall that a rational fraction is function as a fraction, function is defined as a ratio of 2 polynomials.	15,082	122	[]
volunteer	in the form of PPX divided by Q x.	15,082	123	[]

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