|Start||--from 0 to pi over 2. Okay, whats the sin of pi over 2? 1.
and the sin of 0 is 0, which equals 1.
Get the idea?
These aren't so bad.
This is just like when you do derivatives; you say, great why did I do all that weird stuff? I say, well, now you understand it.
|0:33||So I showed you how to do a bunch of these
indefinite integrals before, but let's practice a couple.
I mean, these definite-- sorry-- so this is called a definite integral.
So an indefinite integral is when you do the antiderivative with the +C and a definite integral is when you plug in the numbers.
|1:11||Write a little dx there when you do integrals.
dx has a purpose.
For now, just sort of remember that it's there.
Kind of like when you took a derivative, you have dy, dx, sort of appears.
when you do the integral you have the dx.
|1:31||Which stands for the fact there's a derivative inside,
and when you integrate it, it disappears.
So this is called the indefinite integral.
So this is just the x cubed over 3, minus 3 x squared over 2, plus C.
Notice the dx is now gone.
In the process of integration you get rid of that.
You're going to need that dx for stuff, which we'll see... maybe next week? Certainly the following.
|2:01||And if I wanted a definite integral,
let's say I was going from 1 to 2.
Why is it called indefinite? Well, cause there's an infinite number of possibilities.
It just depends what you pick for C.
So far, so good?
So the definite integral is the same antiderivative,
|2:32||but now I'm going to evaluate it from 1 to
So I'm going to plug in 2, I'm going to plug in 1, and I'm going to subtract.
So this would be 2 cubed over 3, minus 3 times 2 squared over 2, minus-- uh oh-- 1 cubed over 3, minus 3 times 1 squared over 2.
|3:07||And you can figure out what that is.
I don't really care.
But you can figure out what that is.
I'll leave the fractions for you guys.
You're all going to be doctors, you'll be excellent at this stuff.
|3:33||Let's practice a couple, make sure everybody's good at this.
|4:21||Okay, let's see if you can do that one.
|4:36||Alright, so we're going to find the integral from 0 to 3.
You always say from the bottom to the top.
of x cubed minus 6 x dx.
So this is x to the 4th over 4, minus 6 x squared over 2, which is 3 x squared from 0 to 3.
|5:01||if you plug that in you get 81 over 4, minus 27,
minus, that part's just gonna come out zero.
And if you do this you should get minus 27 over 4, or minus six and three-quarters So how can that be negative?
So this is giving the area under the curve.
What happened is,
|5:41||the curve's doing something like that, okay?
That's not 6, I'm sorry.
That's the square root of 6, and 3 is somewhere here.
So this area underneath is bigger than that area above.
So this is another thing I should tell you about, this is what's called "signed area".
|6:01||So the area above the x-axis but under the curve comes out positive.
And the area below the x axis, so kind of above the curve, comes out negative.
And the negative part turned out to be bigger than the positive part on this curve, okay, on this area.
So, for example, suppose we look at our favorite function sine x.
|6:44||What do you think we would get if we did the area from 0 to 2pi?
We should get zero.
Because the part above the x axis should cancel with the part below the x axis.
So let's do that integral.
The integral from 0 to 2pi of sin(x) dx
|7:03||we get minus cosine x from 0 to 2pi.
The cosine of 2pi is the same thing as the cosine of 0.
So you get minus 1, minus, minus 1, which comes out 0.
|7:34||That's what we'd expect.
So by the way, sometimes, on these kind of calculus problems, if you know what the graph looks like, you can just get right to the answer.
you don't even have to do the evaluation.
So sometimes by symmetry, or by knowing the geometry of what's going on you'll be able to get to an answer very quickly.
Let's do another example of that, so you see what I mean.
|8:26||So I wanna do the area under y = x, f(x) = x, from 0 to 6.
|8:31||But what does that look like?
That's just a triangle, right?
y = x is a straight line, goes from the origin, it'll go to point (6,6).
The area would be one-half base times height, you get 18.
Now let's make sure that our calculus agrees.
|9:10||So if I did the integral from 0 to 6 of x dx,
that's x squared over 2, from 0 to 6,
which is 6 squared over 2, minus 0, that's 18.
That's exactly what we thought we were going to get.
So sometimes you should be able to look at this and figure out exactly what the area is.
|9:32||Other times it wont be so easy.
Let's just do a couple more, make sure we understand everything.
|10:32||So if we do the integral from 1 to 9 of 2 t squared, plus t squared radical t, minus 1, all over t squared
Now of course, t, x, the letter doesn't matter, right?
lets see if you can do that one.
|11:25||It's the integral from 1 to 9.
The top is 2 t squared, plus t squared time the square root of t, minus 1, all over t squared.
|11:40||So how do you do this?
First of all, you're gonna want to break this up -- the integrand, as it's called.
You do the integral from 1 to 9 of 2 t squared over t squared, plus t squared radical t over t squared, minus 1 over t squared, dt
|12:05||Which then simplifies to the integral from 1 to 9 of 2, plus the square root of t, which is t to the one-half,
minus 1 over t squared, which is t to the minus 2, dt.
Did you get to this stage?
Yes? [ question inaudible ] You wanted to bring the t squared in the denominator up and make it t to the minus 2?
|12:30||But you'd be distributing and end up in the same spot.
So it's unnecessary work, but sure, feel free.
But you know, it's sort of doing it the hard way.
But that doesn't stop lots of you guys.
You generally want to do this as efficiently as possible, but you don't have to.
When you see multiple things on top and one thing on the bottom, usually we do that to confuse you and see if you're clever enough to break them into the individual things over the denominator.
|13:00||So we do it the other way on purpose.
Alright, what's the integral of 2? It's going to be 2 t.
The integral of t to the one-half is going to be t to the three-halves divided by three-halves.
And the integral of t to the minus 2 is gonna be t to the minus 1 over minus 1, all evaluated from 1 to 9.
|13:31||Or if we want to simplify that
it would be 2 t plus two-thirds t to the three-halves plus 1 over 1 from 1 to 9.
So far so good? Loving this?
You know, I got my microphone back.
I got an email Monday night from the audiovisual people
|14:00||I guess somebody actually comes around at the end of the day
and closes things up and said, the microphone's missing from your classroom,
one of you people is bad, so I emailed back and said, well at 4 o'clock it wasn't here.
So I got a very nice email saying, oh thank you, now we've narrowed down the suspects.
I guess they found the person.
They probably killed him or her. [ laughter ] The head is probably mounted outside ESS as a lesson for those of you in the future. [ laughter ]
|14:31||Just a certain example for others.
Alright. We plug in 2 we get 18 plus square root of 9 is 3, 3 cubed is 27, two-thirds of 27 is 18, plus 1 over 9.
Minus: 2, plus two-thirds, plus 1.
So that's 36 and one-ninth, right?
|15:02||36 and one-ninth would be 325 over 9,
and that's 3 and two-thirds which we could turn to ninths and that would be 33 ninths
and that gives you 292 over 9.
Am I close?
32.44. How'd I do? Do I get to go to college?
That's why I stand in the front.
So I'll screw those up though and give you a chance to laugh at me, I promise.
[ To student ] This is your third time around?
You should know by now, right?
Yeah you. You've seen me do this before. [ laughter ]
|16:02||Okay, let's give you one more of these.
Let's give you a really entertaining one.
These are really fun, aren't they?
Much less painful than the other stuff. Good.
I'm gonna make it painful, now.
|16:40||Okay, who needs more time?
You should be able to do that in your head.
Got it? What's the answer?
Yeah, should be able to do it in your head.
|17:11||Who needs more time? Everyone?
What's the integral of e to the x? e to the x.
That's the easy part. No plus C.
|17:36||Okay, so this is e to the x so we can get you with the logs so you gotta be careful.
that's e to the log 5 minus e to the log 3.
What's e to the log 5? 5.
e to log 3? 3. There you go.
|18:01||That wasn't hard. Should be able to do that one in your head, as I said.
e to the log of 5 is 5, e to the log of 3 is 3.
Make sure you know that, e to the log x is x.
"ln" is pronounced "log", or "lnnnnnnn".
Let's give you a more serious one.
|18:30||I'm looking for ones in the book.
Ooh, this is painful.
Then i'll go back and do some other stuff.
I think I'll give you guys about three of them for practice right now.
|19:02||Good, I want you guys to be good at these.
Here's three to work on.
This one says the integral from 0 to 2 of y minus 1 times 2 y plus 1 dy.
This one says the integral from 0 to pi over 4, of secant theta tangent theta d theta, if you can read that one.
|19:33||This says the integral from 1 to 2 of x cubed plus 3 x to the 6th all over x to the 4th.
Can you all read those?
Let's do these.
So how do we do the first one?
Well what I would do first is I would FOIL this out.
|20:00||So this is 2 y squared -- minus 2 y plus y -- is minus y, minus 1, dy.
You all multiplied that out correctly?
That becomes 2 y cubed over 3 minus y squared over 2, minus y, from 0 to 2.
which is 16 over 3 minus 2 minus 2, minus 0, which comes out four-thirds.
|20:40||[ heavy Long Island accent ] "Faw Toids". As they say on the South Shore, "faw" thirds.
Alright, that was easy.
What is the integral of secant theta tangent theta?
It's secant theta.
Because the derivative of secant theta is secant theta tan theta.
It's going backwards.
So this is secant theta from 0 to pi over 4.
|21:07||Now comes the fun part, what's the secant of pi over 4?
If you don't know, at least on your exam write secant of pi over 4.
Get yourself some partial credit.
Might even get full credit depending on the professor.
The cosine of pi over 4 is radical 2 over 2.
You turn it upside down you just get the square root of 2.
|21:34||Now that we've got the square root of 2, minus 1, if you do it on your calculator it's about 0.414.
Why is the secant of pi over 4 radical 2?
You take radical 2 over 2 and you turn it upside down, you get this, equal to that.
If you're not sure, cross-multiply.
We know what the cosine of pi over 4 was.
It's 1 over the square root of 2.
Good, what do you get when you turn it upside down?
The square root of 2.
Well, it's also that -- it's both!
It's 1 over the square root of 2, or square root of 2 over 2, depending on who your teacher was in high school Use that triangle, okay?
|22:31||As long as you get the right answer, that's what matters.
Some of you memorized the unit circle, some of you still sort of do it in your head.
I could put that back up on my insta -- that was on the insta already, yeah, it's in there.
That stuff's on there from last semester.
I could delete it and put it back so you could re-like it.
I'm the old guy who puts up math equations and gets a hundred likes.
I don't know what that says about my life. [ laughter ] But what gets the most likes?
|23:01||Bonnie! Hello, the puppy pictures.
She's a little older than that now so it's kinda sad.
Okay, so we could break this up, we could make this its the integral from 1 to 2 of x cubed over x to the 4th, which is 1 over x, plus 3 x to the 6th over x to the 4th, which is 3 x squared, dx.
|23:30||That equals natural log of the absolute value of x, plus x cubed, from 1 to 2.
We don't really need the absolute value, because the absolute value of 2 is just 2.
But you could write it if you want.
And you get log 2 plus 8, minus log 1 plus 1.
What's the natural log of 1? 0, because the log of 1 is always 0.
|24:05||So you get log 2 plus 7, or about 7.7.
Got the idea?
So one of the fun things about these antiderivatives is they stand for the area under a curve.
But integrals also have other meanings.
An integral generally is just a sum.
|24:31||So, how many of you have taken physics?
Good, so you should start using integrals in physics.
It can get really hairy when you get into Physics 132.
132? No? Physics 126? 127 -- yeah, you'll see it in there.
Who do you have for 127? What professor?
Oh, Linwood Lee! He's been here a long time.
|25:01||But I've been here longer.
He came here in 1965.
But I came here in 1961.
Of course, I was a baby, but... Professor Lee has been teaching a long time.
He used to be the tennis champion at my tennis club.
I used to play him when I was your age, and when he was my age -- the age I am now.
How's he doing? He's fine. Okay. Good answer.
|25:31||Don't want to get yourself in trouble -- you're worried Ill call him up and tell him.
I would never do that.
Well this is a secret, no one even knows what we're saying right now!
No! It's not like we're being recorded or anything.
[ laughter ] If you're doing physics you learn that there's displacement and there's distance.
So displacement is the changing of position, where were you when you started and where were you when you finished?
So in physics they're allowed to ask questions like, "you get up in the morning, you drive to school and stop and get a bagel,
|26:04||you go out to lunch, you come back to school for a little while,
you go to your job, and come back home.
What's your net displacement?
Your net displacement is zero, cause you're back where you started.
But the distance is that whole route, okay?
So integrals can be used to find either displacement or distance.
For displacement, it's easy. For distance you have to factor in
|26:32||that you might be going different directions
at different times.
So suppose we have an object that moves along a line, and we're given the velocity
|27:02||at any time t
Find the displacement...
That says 15, if you can't read it.
Find the displacement in the first 6 seconds.
|27:30||In other words, what's the total displacement from time t=0 to time t=6?
So all you would do would be to integrate from 0 to 6 t squared minus 8 t plus 15.
If you integrate velocity, you get position, because if you take the derivative of position So if you integrate that, you do the antiderivative, you can figure out how far this thing has gone.
|28:02||So that comes out t cubed over 3 minus 8 t squared over 2 plus 15 t.
from 0 to 6.
That is equal to t cubed over 3 minus 4 t squared plus 15 t
|28:39||from 0 to 6.
That is equal to 6 cubed over 3, minus 4 time 6 squared, plus 15 times 6.
That's the first one.
The second one, when you plug in 0, you're just gonna get 0.
You can ignore it. Minus 0.
So let's see, that's 162 minus 144, it's 18.
|29:08||So that's the net displacement of this object from time t = 0 to t = 6.
Here's the fun thing, since lots of you haven't taken physics.
Velocity is a vector. What that means... to put it in simple terms is --
|29:30||If you go in one direction, it's positive.
If you go in the opposite direction, it's negative.
You could say, for example, to the right is positive, to the left is negative.
Up is positive, down is negative -- doesn't really matter.
But there's a difference between going this way and this way.
So this object has velocity t squared minus 8 t plus 15.
If you factored that...
|30:07||and you put that on a sign number line --
-- remember this from last semester? --
and you found the value of velocity
you'll find it goes like this.
Which means the object is kinda doing something like that.
|30:30||Until you get t = 3 it's going to the right.
Then it turns around and goes to the left until t = 5 and it finishes somewhere around there at t = 6.
So notice part of the time it's going to the right, part of the time it's going to the left, and then we found the displacement is this.
It's where you started versus where you finish.
This is actually on the x-axis,
|31:00||so the displacement, it says 18
So you're 18 units away from where you started.
But the total distance traveled is not going to be the same.
To get the distance, you have to break up the integral.
Because when it's moving to the left, you get negative values, you don't want to subtract those off when you're doing distance.
So if we were doing the distance instead of the displacement, you'd need three integrals.
We would want to go from 0 to 3 of t squared minus 8 t plus 15.
|31:36||And then, because we're going to the left,
we want to subtract the answer we would get from 3 to 5,
and then you would add from 5 to 6.
|32:01||Does that make sense?
This one's a little more tedious, obviously.
>>Student: Why are you subtracting the second one?
I'm subtracting the second one because velocity is negative, I would get a negative answer and i want to make it positive. Correct.
That make sense?
So the simple rule you say to yourself is, wherever velocity is negative, I just subtract the integral.
|32:30||I know most of you are not taking physics, but
you get to do this in 126 anyway.
So you want to break up the velocity, find out when it's positive, find out when it's negative, so let's just do one.
Something nice and easy.
|33:19||Here we are. Find the distance traveled in 5 seconds|
|33:31||Do you love math?
So, first thing you'll want to do is you're going to want to factor that incredibly difficult quadratic t squared minus 4 t plus 3.
You factor that, you get t minus 3, t minus 1 equals 0.
So far, so good?
|34:00||You have a number line and you should get that.
Now if you want to find the distance this thing has traveled, well, there're gonna be three integrals.
You do the integral from 0 to 1 of t squared minus 4 t plus 3 dt minus the integral from 1 to 3 t squared minus 4 t plus 3 dt
|34:30||plus the integral from 3 to 4 of t squared minus 4 t plus 3 dt.
How'd we do on this part?
Oh, I said 5 seconds. Sorry, 3 to 5.
If you break it up like this, that should be worth some partial credit.
|35:03||Alright, let's do the antiderivative.
So we have, t cubed over 3 -- t cubed over 3 minus 2 t squared plus 3 t from 0 to 1.
Minus t cubed over 3 minus 2 t squared plus 3 t
|35:30||from 1 to 3.
Plus t cubed over 3 minus 2 t squared plus 3 t from 3 to 5.
So far, so good?
We already get to this point?
So they're all the same antiderivative.
So, you could probably come up with more efficient ways to find the actual answer.
|36:02||So if you plug in you get
one-third minus 2 plus 3, minus 0
That's four-thirds. So for the first second
you travel a distance of four-thirds units
Now here you plug in and you get...
9 minus 18, plus 9.
|36:35||Minus: one-third, minus 2, plus 3.
so minus, minus four-thirds.
That gives you plus four-thirds.
The third part is annoying.
125 over 3 minus 50, plus 15
|37:03||minus: 9 minus 18 plus 9.
That's 20 over 3.
You should get a total distance 28 over 3 units.
Feet, meters, whatever.
|37:32||Alright, that's enough.