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0:00
you're refining some of these probably that I understand we ran out of the key because I I assigned those problems last year deal so when I'm the Moodle have downloaded the notebook I saw that they were not there then I went to last year's in download of that and it was there those exercise solutions were there I don't understand because I thought a coffee last year solution apologize anyway long story short is I re uploaded an updated chapter 5 solution
0:33
key that got all the missing problems in which actually also included a bunch of problems with accent inception 524 I will still have to check for chapter 6 7 8 to make sure all the answers are there for those too but I think I will go over some of those problems every that's part of our lecture today before we do that let's just review what we've learned in section in chapter 5 overall except for well except for the very beginning some of the definition
1:03
everyone livability let's review the theorems some of which are not on Mathematica here
1:17
I'll just labeled as properties of integrals and they're going assumption every single one of these properties is that all functions that I write down will be riemann integrable alter of the given intervals assume as a background assumption as soon as the background assumption that all functions that I mentioned are remaining negative all over the relevant of your intervals unless that otherwise so I'm not really
2:10
going to state these properties as theorems I'm not going to state them as equations and inequalities one property is linearity well you could write down as a bunch of sub properties you could say the integral of a constant times a function is a constant that constant times the integral careful with our notation here F is a function C is a constant
2:50
you pull the constant on the front that's a linear in property you can also set up an integral on the sum of two functions it is the stumbling corresponding integrals the same interval you can combine these you could write a linear combination of F and G in fact you can write a linear combination of any finite group of functions and in fact properties see that I'm left to write down for a the integral of a finite linear combination of functions really includes properties AMD
3:36
right I could just write down property see here I'm writing it separate but it really includes properties a and B
4:06
okay that really includes ad right property B would follow from this with n equal to 2 and C 1 and C 1 C 2 both equal to 1 property a would follow if all the C's accept the first 1 over 0 so we only had one function pull the constant that property really includes a and B by the way you should not rush i way from writing that in summation notation it says the sum K goes from 1 to N well ok the integral of a sum which I'll put in parentheses here though
4:56
it's not traditional to put parentheses is the corresponding sum of the integrals in effect it's kind of like
5:18
it's kind of like you can distribute the summation sign through the parentheses here all well the C photo can be distributed into the integral effectively and the summation sign in the sentence can be distributed into the integral it's kind of more to imagine that okay don't they don't take that people just notice how about way of writing it is equivalent to this way so you should be comfortable seeing it like that one of the properties sweet AB
5:53
if f of X is always less than or the G of X for all X in some interval and again assuming the F Fujio room on the bigger bowl then the integral of F is less than urgently being an old Jeep in particular that also means maybe I should list I've is to to some some properties here that also means of G of X or a constant function call it m the output was always in
6:44
then the integral of F with less than or equal to capital M times B minus a which is the same as the integral of the constant function M over the interval from A to B the area of the rectangle that's the geometric interpretation at least when the graph about is above the axis the area under the graph of F is less than for that area and M is positive but actually this stuff works even if F goes below the axis and even if if a more negative this
7:19
still works in some quality so it works that same direction but the area interpretation
7:30
what other properties going on well the next section was more big theorems which I'm not going to write on the board of God in the Mathematica here section 5.2 or those theorems whose proofs are pretty tough on Moncada functions are not real on the niggerball the book states that has if f is riemann integrable then it's bounded on the interval contrapositive of that that would say if you've got a unknown function like it's got a vertical asymptote on the
8:08
given closed interval then it would not be real integral that's the contrapositive of a statement though once more I will remind you that does not now that fact is not contradicted by the existence of some improper integrals where your functions got a vertical asymptote because we're not talking about improper integrals here there's an equivalent condition freedom and the ability in terms of the oscillation I'll just say that you can look it up if
8:43
you want to look like more importantly ultimately for us we want to know that continuous functions are even different monotone functions are riemann integrable there are certainly plenty of functions that are neither continuous nor monotone that are not very difficult but these conditions are sufficient if you know where your functions continuous or you know it's long time even if it's not continuous and your function
9:16
we I knew the area on your graph makes sense according Fremont what other properties of integrals some that I haven't mentioned beyond passing here I'll talk about a little bit more here today um what I'm about to write down was actually more of a definition you can call it a property but it's not something you prove it's a definition if a happens to be bigger than B then the integral from A to B well it's this lower limit is bigger than upper limit
10:08
by definition this is the office of the integral from B a it's a definition there it's not something we prove we define that to be the case we also by definition say that the integral from a constant to itself of any function is zero actually I guess you could say that's any function defined on this degenerate interval it's just one point because any function is going to be continuous on the domain that consists of one point
10:52
roll the degenerate case these are definitions really do they make good sense I couldn't they do we could probably try to motivate them and went to physics actually emotion kind of situation saying that the function you're integrating is your velocity this is just a sign never here you don't have to write this down let me even emphasize the velocity by writing the empty let's the velocity in meters per second say what's the integral of the velocity represent in physics
11:47
well let's integral to speed is distance traveled but velocity is allowed to be negative its displacement changing positions what that equals displacement change in position which can be positive or negative let's say position with zero is right at the edge of this table here positive position starts on the right and negative positions are to the left if I move back and forth I move this way with a positive velocity and then turn around and
12:27
move back this way I have a negative velocity but maybe I stopped right here the integral of my velocities might change in position it's this positive quantity i won here - here over that a long time I could have kept going past the initial position of zero into negative positions underlying then my change in position would have been negative this could be interpreted that way whether a has less than B or vice versa typically we think of a as being less than me and so I'm
13:07
moving as time goes by and I change my position maybe my change of position is negative here but worth I switched it around the integral should be the opposite of the original integral what does that correspond to it corresponds to physically moving backwards in time right letting it's likely we videotaped myself like we're doing right now and watched it in Reverse might change in position would then be positive going from there to there as time goes backwards that makes sense
13:47
my original motion for overtime ended up a negative change in position but then if I run time backwards then they go from there there and that's a positive change in position the opposite of the negative transition and that we're going to be justified is saying you know if no time elapse system there's no motion so it's the civil senator properties but this is really a definition the following equation that comes from breaking up an interval of two pieces
14:31
is true certainly it can be proved or should be proved as a theorem when a is between B's sorry when C is between a and B and in fact the book does prove it as a theorem in that case however once you've got these definitions here this is true actually even if C is not between a me though it's harder to understand meaning so this fact is true actually no matter what a B and C Oh
15:15
sort of think about one case think about over here let's pretend laws pretend seize over here well I wanna verify this equation works even when C is over there C is not between a and B however in this picture a is between C me so I could certainly write down now the integral from C to B would equal the integral from C to a plus the integral from A to B which makes sense if you think about areas imagine the graph is about the axis
16:12
you can make sense of this kind of thing by thinking about areas the tall area is the sum of the two smaller areas that doesn't quite look like this can I make it look like that
16:40
it's not clear that you definitely in let's see if you solve this equation for the integral of F from A to B by subtracting this integral to both sides
17:04
everybody performance subtract that indigo for both sides cancels on the right side on the left side you get this minus this equals the integral i'm going to switch around the order the integral of F from A to B is the integral of that from C to B minus the integral of F from CA doesn't look like this yet I tell you to use this fact the negative of the integral from C a is the same as plus the integral from A to C
18:02
that now matches this
18:08
okay so this has to be proved using the definition of remodeling ability or in fact the equivalent form in the case where C is between a and B but once you've got this definition and this one as well C could equal a or B then this property can be seen to hold no matter what okay so these are fundamental properties of integrals that you should know then there's the fundamental theorem of calculus we use the word fundamental twice but they are
18:44
fundamental there are all of basic importance let me state the fundamental theorem of calculus the way the book does and this time I will stated it in a stronger form the books takes it in suppose you've got a function a little app that's riemann integrable on some closed interval and okay in Part one Part A we define this new function capital F by integrating a little well over intervals of this form or maybe acts to find our different values of capital f of X
19:52
technically you need something that I didn't mention if a function is riemann integrable on some interval some closed interval it's also remembered memorable on any closed sub interval of that the interval from a to x is a sub interval of the interval from A to B I'm assuming X is less than or equal to B here you can say for sure that capital F is continuous one thing you can say for sure it as long as little up as remonda maybe bull
20:32
little F could be discontinuous moreover and more importantly if little F is continuous at a number I think the book goes ahead and calls an X I'll call something else I'll call the C at some point C in the interval then capital F is differentiable at C and capital F prime of C equals Willems so I think the book uses an X again I'm using different letter here's my X I use there is my variable parabola but you can you
21:34
know variables they can always be thought of as being constants if you want them to be you want to say I'm imagining this to be a fixed number you can go ahead and do that
21:44
this flu
22:00
if little up happens to be continuous over the entire normal then capital F would be differentiable over that entire interval and would be an antiderivative of little F over that entire interval that's why this is sometimes called the antiderivative construction form of the fundamental theorem of calculus you're constructing it in root of using an integral areas can be used to finding derivatives of the idea Part B is
22:32
the flip side of that and if your root lives can be used to fight areas your capital G is an antiderivative a little left on the interval then you can evaluate the integral a little out by doing capital G of B minus capital G of a that's the one you're more used to part a says areas or more precisely sine areas where you count areas when the graph is below the axis as negative contributions to the integral areas can be used to define the antiderivatives Part B
23:33
is really the flip side anti derivatives can use to find sine areas you can prove Part B actually pretty easily from Part A that's actually number fourteen let's go ahead and do that
24:02
be from a so we're going to assume we've got an anti derivative of the given function on the interval G prime of x equals f of X for all X in the interval flying new york from here it doesn't necessarily mean f is continuous
24:52
derivatives do satisfy the intermediate value property we've seen but are not necessarily convinced
25:12
yo okay number 14 they go ahead and they add the extra assumption that every little a business it can be proved without that assumption part b can be proved without the assumption the little F is continuous and in fact that the point of problem 12 essentially
25:51
12 and 13 combined that the point of 12 and 13 combined I'm not gonna talk about the problems but you can look at them is that you can prove Part B of the fundamental theorem of calculus here without assuming a little F is continuous it's still true even if a little F is not continuous is the point because derivatives don't have to be continuous G can be an antiderivative of little F without a little update details
26:22
so we're going to use part a to prove that I could need the continuity of little F because the second part in Part A assumes little F is continuous at the point here I should say on a higher level
26:52
so I can use part a is that is the idea I'm trying to use part a from Part B well I'm going to use Part A I'm gonna need a little F to be continuous now this is what the book says to do number 14 is a supers so what do you do okay so I'm assuming capital G is an antiderivative of little F I want to show that this equation right there is true how do you use part a go ahead and define a capital
27:24
look and sense a little F is continuous on the closed interval from A to B half a lock prime of x equals little f of X for all X in that closed interval in other words capital up is an antiderivative of little F that's what Part A says because that's where little F is always continuous time your interval okay well okay what are you honey related to G go back to chapter four for both capital F and G are anti derivatives of the same function we know from Chapter four as a
28:21
consequence the mean value theorem that means they differ by a constant on the interval they're anti derivatives of the same function in the same interval therefore they differ by a constant therefore essentially by the mean value theorem though it's more direct application of the concept of function theorem I'll go ahead and say five a constant function theorem the theorem that says if your derivative is always zero over function capital f of X equals capital G of X plus C
29:12
for some constant C and for all X I should say there exists the see you know R such that this is true for all X in the interval and that's the connection between capital F and G that we need
29:36
this means capital G of B minus capital G of a can be related to capital S with being capital F of a big issue the ruthless around with capital G there and capital F there that would have been okay well even like this though therefore capital G of B minus capital G of X equals capital F of B even when your C minus capital F of a minus C and that can be related to the integral by the definition of what capital it is this is the integral of
30:23
little F from A to B the C's cancel later minus the integral a little F from A to A that's zero
30:49
so what scans of entry
30:55
take some creativity but but but at the same time it kind of is the only thing you can do you want to prove this statement here assuming capital G is a derivative blew up show that that equation is true you gotta use part a I guess we have to create a function capital F and say well this formula is a capital of sin a derivative and with enough thought this is kind of the only thing you can imagine doing and fortunately it pops out question would uh
31:31
theorem statements for my chapter for you know 10 C possibly not that an exam 3 test but in the final possibility but you should still know things from chapter 4 to help you do things in chapter 5 and I might get your proof on exam 3 like this perhaps where I'd say you've got to use the constant function theorem or the fact that the antiderivative of the same function on the same here multiplied by a constant when I say different by constant
32:03
by the way that means you know you just track G of X for both sides here f of X and minus G of x equals C that's what it means that differ by a constant but you can write that like this too
32:28
let's go ahead and look at a few other problems light number 11 but I didn't have a solution for it initially and seven covert today but the solution is there now this file actually number 11 is pretty quick to prove if you allow yourself to use number 10 so number 10 is a more important issue
33:10
let's go ahead and do it that way let's let's look at number eleven first we use number ten to give a pretty quick proof of number eleven and then we'll spend time thinking about number ten so number eleven says suppose little air and little GE are continuous on some closed interval from A to B and that their integrals are the same the conclusion you want to prove is then there exists a C and the interval from A to B sub C
34:05
number 10 says partay at least for 10 per day says I have you got a function that's continuous and it's integral of zero then there must be some C where an F of C equals zero say that again if F is continuous and it's integral to zero then the function itself has to be zero somewhere the integral is zero the function can either be a horizontal line as there are two or would have to be partially above the axis and person below I think by the intermediate value theorem is got
34:54
across somewhere that fact can be used to prove this with a little truck what H be defined by the difference f of X minus G of X why am i doing that trying to use number ten that says if the integral equals zero then the function must be zero somewhere I'm assuming these two integrals are the same which would mean the difference of one with the other would be zero you have to define a function can you just use kind of the backwards of what we are one of the properties
35:55
that we already defined this morning and say the integral from A to B of F - interval A to B of G is equal to zero and I use that property backwards to say the integral of A to B is f minus G and we know that those are equal so therefore that would be the interval A to B to 0 which was I lost you there but I think you're I think you're on the same track as me yeah I'm just being a more official mother okay I lost you there in those words okay which I which I know what happens you
36:28
know it's reversed - I know I talk too fast - I try to say it sometimes what's the integral of H H is out minus G and yeah my linearity property one of the properties we talked about which is what the integral about - the equal interval of Evan is integral of G but by assumption that must be zero
37:13
so the integral of H is zero therefore my number ten a there must be a see where H of C is zero and therefore of that C is the same C where F of C equals juice
37:34
and a implies there just a C or maybe such that H or C is zero and applying F of C minus G of C equals zero and applying F of Z equals C so we can use use 10a then fairly easy though slightly tricky things but you start thinking used to these tricks the idea with these tricks they aren't educated tricks they are they do come about by thinking about natural things to do look a try so the more important issue is what about 10a
38:38
10a says suppose you've got one function it's integral it is zero show there exists a C in the interval so that F of C is zero the quickest proof of this actually uses the fundamental theorem of calculus for a applying capital F domain in interval from A to B D by the equation capital f of X the integral from a to X of little F by the fundamental theorem of calculus capital F prime equals the left what
39:33
we're being higher interval because a little F is continuous over the entire interval we're assuming the integral a little out from A to B is zero that means absolute capital F of B must be 0
40:02
also note that capital F of a is zero because you'll be integrating little F from A to A and capital F prime of X equals little f of X for all X in the interval this doesn't finish it I need something else my goal is to show there's some C so we'll of C is 0 what other thing do you think I need to get to that final conclusion this is all preamble rolle's theorem or the mean value theorem you can say apply to capital F rolle's theorem
40:50
which is really a special case of the mean value theorem so you can also call the mean value theorem implies there exists a C in the open interval actually maybe such that capital F prime of C is 0 but capital F prime of the fun will through the calculus is Lilith
41:39
Part B I guess I technically assigns but didn't give you a key for as well 10 be the key estar now Part B says as little F is continuous and non-negative over the interval non-negative means greater than zero then the fact that the integral is zero implies the function itself must be always zero if you're assuming it's not negative and you're assuming it's continuous over the interval and this integral
42:17
is zero then in fact the function must be the constant function always equal to zero that proof actually requires something from Chapter three I'm not going to go through the full proof but let's talk about the idea you could do it by contradiction assume to the contrary that that are some C call it where f of C is nonzero we're already assuming F is non-negative so then in fact f of C must be positive
42:54
the graph must be up there and you remember from chapter three continuity problem that was just the situation but assume we've got a function that's continuous at C and positive at C and you have to prove what were the letters there was a UN of e
43:20
and an amma such that the function is above that horizontal line for all x between UV this horizontal line is not part of the graph of the function the graph that the function has to be above this horizontal line the continuity at c is necessary to prove that or you only did have to prove continuity at just one point where it's hearing a continuity over the entire interval your contradiction here is going to be because of this because the function has to be above that
43:57
horizontal line that's going to mean the integral can't be zero it would have to be positive the horizontal line at him where m is positive over this interval the integral over that interval would have to be positive the integral over the entire interval can't be negative and the negative else were ultimately a contradiction would be that the integral would be positive not zero
44:31
I'm just trying to say you can't cancel out a positive contribution to the integral from this subinterval by saying there's a negative contribution somewhere else because the function is non-negative that would be the idea of identity without going into a ton of details so you should study that
44:59
um another area of application of the stuff is two differential equations some differential equations are easy to solve there's a differential equation dy/dx equals x cubed minus x is the differential equation y of 0 is 2 this is what I call cure a derivative problem this kind of problem actually comes up very rarely in differential equations because it's just - it's just an ant derivative what's what's an antiderivative of X cubed minus 4x I'm looking for a function whose
45:48
derivative is that and also it was graph goes to the point 0 2 there just be the integral X to the 4th over 4 minus would be 2x squared and then here with that plus C choose the C to be 2 in this case to make sure it goes through that point no problem with finding the solution there and here's the graphical interpretation of it this is something called a slope field these two graphs that you see are both anti derivatives of this function executes
46:27
minus 4x the slope field those hash marks that you see are is created by taking each point in the plane taking its x-coordinate and plugging that x-coordinate into this expression here to find the slope of the solution curve dy/dx represents the slope solution
46:54
for example when X is 1 there's one down there see the slope field has negative slopes along that vertical line when you plug in x equals 1/2 here you get a negative 3 1 minus 4 along the vertical line at x equals 1 all these little marks have slope negative 3 and solution curves have a slope of negative 3 that's what the differential equation is saying if I get function at every point your derivative whose slope at that value of x equals this thing
47:31
that's why those the graphs of solutions have to follow this slope and to also satisfy this initial condition y 0 is 2 if the constant of integration would be to where you can see it's the red graph that goes to the point x equals 0 Y is 2 but some functions Oh like this example
47:59
can't be solved in an ordinary kind of way if you try to integrate sine of x squared in an ordinary kind of white you will not be successful it has no elementary into derivative it does have a name and a derivative involving Fresnel s not kidney of X sorry Fresnel got to this function first somebody named president 50 years ago quite sure Gazza this function first evidently it's useful amount that's we give it a name friends
48:37
the less why yes was that principal is the middle initial or something no Fresnel asks because it's related to the sine function there's a presence C as well related to the
49:01
presidency
49:04
but it isn't us to business a derivative of sine of x squared and if I choose my integration constant to make this happen I believe Fresnel s of 0 is 0 so this repeat really this thing plus 2 no it doesn't plot super best batting role plus 2 would be the unique solution of the initial value problem by the fundamental of the compass here this is an application of different calculus Part A the true doesn't affect the derivative right into root 2 is 0 I'm creating a function
49:54
with an integral where the variables in the upper limit of the integral it is an antiderivative of sine of x squared so this red graph is the graph of this function prostitute
50:21
um I'm not an expert but I've heard Fresnel functions are useful for optics have you ever seen frozen elections okay okay there we going that bad that's true all right from the from the mouth of the expert about it Fresno functions are useful in optics we're going to finish class today by verifying that's true in fact did you know that did you know that's true actually when my son started saying the world because Tom's and I'm like what's wrong and he's
51:18
like it's Elmo I'm like well what is that lots of love anyone getting lots of laughter usually when you do substitution it's supposed to help you do an inner door and actually in this case it wouldn't help you the integral the left because the integral on the right I look at that interval on the right and say that's more durable than the integral the left looks like it is though I'm going to do the integral on the right I wouldn't have to use integration by parts three times
52:01
it wouldn't be true and they could do it in governments but you know the equality between these integrals means you could do this integral then by doing this integral but yes not obvious that these are equal how would you prove they're equal it's a substitution relate them essentially start with the integral on the left what should your substitute well or should I call my new variable maybe Z you might hope Z equals natural log of W would
52:55
be the thing to do I mean it seems like the only thing to try to get an C cubed involved at least
53:05
GZ is going to be one over wuw you can also say that w is e to the Z nursing community and that also means DW is wdz it's also easy the Z Z Z so bunch of running around with different equations here but this doesn't work what happens now when you do the substitution replace natural log of W with Z I do get a Z cubed replace DW with either Z Z so every go the integrand is right now what about the limits of integration you do need to change them W goes from 1 to e use this
54:06
equation to figure out what the Z does when W is 1 Z is natural log of 1 which is 0 and when W is e Z is natural log V which is 1 there you go these two integrals are the same in value now if you actually had to do this one do it by parts three times or use Mathematica or use a table
54:32
so sometimes substitution does lead you to an integral it's easier to do but it doesn't look what she's easy necessarily wanting to do a little bit from section 524 but I think we've been long enough today
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