| Start | so we're trying to find the area under a curve and you find it by adding up a bunch of rectangles |
| 0:34 | okay umm jackie are you ready?
can you start? jackie can you start? okay,we're trying to find the area underneath a curve so |
| 1:00 | lets uh lets just do something where its not infinite and delta x and all that stuff lets do the straightforward one first suppose i want to the find the area under the curve x^2+5 |
| 1:37 | its going to go from 1 to 13 you're going to use 6 rectangles first of all x^2+5 is a parabola it sort of looks like that-doesnt really matter this is x=1 |
| 2:04 | that is x=13 we'd love to find exactly what the area is but thats hard to do so we're just going to do an approximation and we're going to say well what if i just take a bunch of rectangles underneath that i showed you last time but if you have rectangles that should get pretty close we're going to cut this into 6ths equal size rectangles of course they dont have to be equal but |
| 2:32 | we make them equal size and to get the height of the rectangle i take the interval and i start at the left side of each interval and i go up to the curve and that gives you the height of the rectangles like that now i know this height is going to be too small remember this curve is sort of blown up |
| 3:02 | and thats going to get you a guess
so these are what we call left hand rectangles
so how wide is each rectangle?
each rectangles width of 2 from 1 to 3 from 3 to 5 from 5 to 7 so there all 2 wide so when im doing my fancy notation thats going to be my delta x |
| 3:30 | thats going to be how wide they are
ok?
so the first rectangle has a width of 2 and a height defined by evaluating the equation at x=1 next one still has a width of 2 and you get the height by going to f(3) f(3).. go up to the curve, then across next one f(5) and so on |
| 4:09 | ok?
and i could certainly factor 2 out so i could also say this is 2 times f(1)+f(3)+f(5)+f(7)+f(9)+f(11) |
| 4:32 | ok?
then i can start evaluating f(1), f(3), f(5) i can find by plugging into the equation right? f(1) is 1^2+5 f(3) is 3^2+5 and so on do we understand this? called a left hand rectangle left hand because you use the left hand side of each interval some folks call it left hand point some just call it left ok? and i know im going to be underestimating the area of the curve |
| 5:03 | because each of these rectangles when i draw them comes under the graph
so far so good?
now im not going to actually calculate this because i have a life now what if i want to do right hand rectangles? now |
| 5:31 | when i go to 3
and go across i end up above the curve
i go to 7..
9..
11..
13
okay you all see how that developed?
the piece thats under and the pieces that are above are not exactly equal because the graph is curved on a diagonal not a straight line but its close so if we want to do right end point rectangles |
| 6:01 | right hand rectangles
i take each of these intervals and i use the right side
to calculate the area
so for right hand rectangles
each one has a width of 2
and now
im going to start at 3
because this first interval is 1 to 3 and i use the right side
right side
and so on
whos confused on that?
feel free to raise your hand no? no confusion? |
| 6:30 | confused, why are you confused>?
student asks question sure each one of these rectangles has an interval, right? 1 to 3 for example so the left hand side is f(1) this is x=1 the right side is x=3 but when i use left end point im going to use f(1) and when i do right end point f(3) and then you see the height of the rectangle starting at 3 i go up to the curve and i go across next we have from 3 to 5 |
| 7:01 | and the right side i go up to 5 where it hits the curve
and go across
ok?
so this will be f(5) f(7) f(9) f(11) and f(13) ok? now i could be fancy |
| 7:31 | say.. i dont have to use the left or the right side i could use the middle i could do that sort of like that |
| 8:00 | thats using 6 rectangles by the way of course you can use more rectangles so if we want to use the midpoint notice what'll happen is you can get a better understanding if i use a bigger number for example at x=10 the left side will overestimate the right side will underestimate |
| 8:31 | so i use the midpoint to get up to the curve some of its under the curve some of its above the curve so those 2 pieces might cancel eachother out its another way of estimating what the area would be so for midpoints you still have these to do but now you use the midpoint of each interval so f(2) f(4) f(6) |
| 9:00 | f(8)
f(10)
f(12)
ok?
so i could use the left side the left side is f(1) right side is f(3) midpoint is f(2) that make sense? so each interval i could start from the left i could start from the right i could start from the middle okay and for each you get different answers and this one would be approximately halfway inbetween these two |
| 9:33 | on its own the midpoint is a very nice estimate
or i could take 1 and 3 and add them together
if you really want to be fancy u use trapezoids
trapezoids are very nice but we dont learn that
that gives you a very nice estimate
k? then theres something called Simpson's Rule.
which is precise gives you the exact answer for polynomials (of degree 3 or less) but for now we just do these and of course we learn how to do it the real way so all this is good math its kind of like when you took 125 |
| 10:02 | and you learn the definition of derivative
then you learn all the shortcuts
and you say why'd we do it the other way?
because now you understand it better do people understand it better? no but we gave it a shot ok so thats how you do the area with a finite number of rectangles suppose i dont use a finite number suppose i want to do an infinite number or a big number |
| 10:45 | so imagine that theres lots of rectangles
im going from 1..
eventually i end up at 13 but its going to take awhile
ok?
so how wide do i want to make each rectangle? well i want to have some end rectangles the whole interval |
| 11:00 | has the width of 13
so the width of each one
would be 12/n
ok?
you take the endpoints you subtract-thats how wide the interval is then divide by n so if this goes from 1 to 100 which would be 99/n that make sense? ok? if this is 1 |
| 11:32 | whats the coordinate, whats the x coordinate there?
so i take the one and im going to add 12/n thats that point i moved 12/n to the right and n could be any number i want it to be so when i move 6 (12/6)+2 and 1+2 is 3 ok? but if n is 1,000 the first point would be 1 the second point would be 1+ 12 one thousandths and then the next point would be |
| 12:00 | 1+12 one thousandths plus another 12 one thousandths
that make sense?
i can have n=a million its be 1+ 12/1,000,000 then the next one would be plus another 12/1,000,000 + another 12/1,000,000 k? so in general point i could have in the first coordinate is 1 then ill have 1+ (12/n) then ill have 1 +... 2(12/n) |
| 12:30 | 1+...
3(12/n)
and so on and thats
going to be the next coordinate to be those points
and i have to add all those up
and where do i stop?
well.. i started at 1 plus this is actually 0 12/n's so the last one would be n-1 12/n's |
| 13:08 | and thats kind of annoying that n-1 thing so another way i could do this is instead of starting on the left side i could start on the right side and then the last one would be 1+n(12/n)'s so this last point |
| 13:30 | 1 ive added n
n rectangles, of course
this gives me 13th's
okay but you have to do this
you have to use the abstract notation
okay so what does that look like?
well i'm gonna have area ok so what did i do i backed out the width so the width is (b-a)/n right? so take 12-1 |
| 14:00 | sorry (13-1)/n and ive got an f at.. 1+(12/n) f at.. 2+(12/n) no, sorry sorry ignore that 1+2(12/n) 1+3(12/n) |
| 14:40 | that last one was f(13)
confused?
yeah, id be confused if i was you guys we'll clear it up we'll do some concrete examples in a minute ok? we're gonna do these the same thing i did before all these rectangles have the same width i find the width by taking the whole interval |
| 15:00 | and saying im going to have n slices of it
ok?
so im going to take the distance from 1 to 13 which is 12 you and cut it n times so the width of each rectangle n'ths is 12/n and to get the height of each rectangle i take f's and you're gonna start at the right end point of each interval so the first interval is 1 and i move 12/n to the right the second one is 1 and i move 2 12/n's to the right and then i move 3 12/n's to the right and i keep going until i move n |
| 15:32 | 12/n's
ok?
we dont really want to do that thats messy but thats the formula that you want to work with so now lets put this in math language thats the fancy way of writing it so what do i need? the width of this-whatever curve we've got |
| 16:01 | this is a plug in b this whole thing is b-a wide and each of these rectangles has a width and i get the width of each rectangle |
| 16:30 | by taking the whole b-a and dividing it by n
okay, so i call that delta x
its just more compact
delta being the difference between the x's
ok?
and i start at a and i add 1 delta x 2 delta x, 3 delta x i keep going until i get n delta x's of course i could just continue to go to infinity if i wanted so if i want to use the fancy sigma notation you guys love sigma notation right? half of you dont know sigma? i think i did that last time |
| 17:01 | a few of you
the way that they do it in the book they sort of leave that out
so lets do an example
lets actually do a problem where we can figure this out
okay so we're going to find the area under the curve y=x^2, f(x)=x^2, and go from 0 to 1
nice..simple. right?
thats the curve |
| 17:32 | f(1) is just a parabola
k?
so i need 2 things i need to know how wide is each rectangle? well how far is it from 0 to 1? 1 ok so delta x will equal (1-0)/n which is 1/n ok? |
| 18:08 | so how far-how can i come up with a nice way of writing all the other stuff?
well lets see i could start f at 1/n and have f at really sorry f at 0+1/n plus f at 0+2(1/n) |
| 18:31 | plus f at 0+3(1/n) okay, go all the way up to f at 0+n(1/n) the width of each of these is 1/n thats what we're working with |
| 19:00 | i can simplify that i'll wait till you get this copied so this is f(1/n) f(2/n) f(3/n) |
| 19:39 | f(n/n) also known as 1 if i wanted to use the fancy sigma notation that would look like this |
| 20:21 | so what does that sigma mean?
sigma means sum you're gonna add things up i start at i=1 |
| 20:31 | and i go up to n
if any of you have done computer science
this is a "do-loop", okay?
this is the beginning of the increment and you step by 1 each time until you get to the end of the do-loop. k, if you dont know computer science its the same concept you start i=1, then add 1 each time until you get to n then add them all up and thats whats going on ok? i start with i=1 then i=2 then i=3 all the way to i=n then i add them all up |
| 21:03 | ok?
i dont know if we're going to actually test you on anything with sigma notation im not sure but im not in charge of tests okay this is what we're looking for so for example on the homework one of the questions thats what theyre looking for on part 1 now theres one more step of course whats f(x)? f(x) is x^2 so we would say the area |
| 21:30 | equals..
and i can pull the 1/n out of that sigma
why can i take the 1/n out?
well it doesnt have any i's in it so since it doesnt have any i's in it when i plug in i its not affecting this at all ok? and whats f(x)? f(x) is just x^2 so 1/n^2 ok? |
| 22:03 | so if i had sin of x
now itd be sin of 1/n
ok?
if i was finding the area under e^x thatd be e^1/n now suppose i want to do an infinite number of things well then lets take the limit |
| 22:35 | heres the fun part
how do you evaluate that limit?
anyone have any idea how to evaluate that limit? well, we'll figure it out |
| 23:08 | so lets look again at what we've got we've got.. i'm just going to rewrite it over here okay, thats the same thing as.. |
| 23:38 | and i can just distribute that
square it, right?
now remember i said you dont care about this n because im just doing im only plugging in for i so i can pull out the n^2 and get 1/n^3 |
| 24:07 | i^2
and theres a formula for that last step
what is
the sum from 1 to n over i of i^2
1^2..
+ 2^2..
+ 3^2..
ok?
thats really what that sigma means you may not remember this from high school math or you may never have learned it |
| 24:39 | its equal to that its okay you werent supposed to memorize that you werent supposed to know that-we would tell you so in other words, i could take this and replace it with that and you get 1/n^3 times n(n+1)(2n+1)/6 |
| 25:13 | you hate your lives yet?
yes you do? question? well most people dont understand either so where dont you understand *student: inaudible* |
| 25:30 | how did i replace that?
so this is a formula that im telling you k? if you want to find the sum from i=1 to n of i^2 remember you want to add up the sum of squares the formula is n(n+1)(2n+1)/6 lets test it 1^2+2^2+3^2+4^2 should be 4 times 4+1 is 5 times 8+1 is 9 |
| 26:00 | over 6
and i believe thats 30
1+4 is 5
plus 9 is 14
plus 16 is 30
it works
i tested it for 1 but i think it works for all of them right?
thats not really math you prove this-you can prove this by abduction so this is a formula that i just give you so by telling you this you can now find that limit |
| 26:37 | ok?
and now we're going to do the limit of this thing this is what im really trying to figure out |
| 27:00 | any idea what that limit is?
you can do it in your heads multiply out the top ah where should i put it guess i can do it over here that becomes the limit n goes to infinity |
| 27:31 | 2n^2+3n+1
over 6n^2
remember what you do?
the n^2's are the same that gets you to 2/6 or 1/3 that horizontal asymptote rule remember you did this for like a minute back in 125 you dont remember every minute of 125? really? sorry |
| 28:01 | anyway
so thats how you find the sum
so all the answer is the area under the curve
y=x^2
from 0 to 1
is exactly 1/3
we're going to prove in a couple minutes
ok?
so i am totally lost no idea what im doing lets do another one of these maybe after 2 you'll be a little less lost ok? suppose... im going to erase this i'll give you one more minute to write it down |
| 28:32 | alright lets do another one of these
this is exactly the same way you would do problem .2 a and b
if thats of any help
now you have 2 options
you could try this on your own first
or we could do it as a team
who wants to team?
really? theres no i in team |
| 29:01 | we're going to do this all together right?
you know the response to someone who says there no i in team me, theres a me in team its got an m its got an e that got me 10 laps when i pointed it out to my coach alright so how will we do this? well its kind of like the other one i havent really changed very much so lets figure out what i need to do i need to find (b-a)/n |
| 29:32 | same as before
if this was from x=1 to x=5
then i would do 4/n
ok?
if this was from x=1 to x=10 that would be 9/n ok? just take the difference between the x's, over n so i'm going to find.. im going to start at 1 |
| 30:01 | and im going to have f(1+1/n then 1+2(1/n) plus f(1+3(1/n)) just like before i start at 1 and then just add a bunch of 1/n's |
| 30:32 | the width of each of these intervals
so far so good?
now what is f(x)? f(x) is x^2+1 ok? |
| 31:03 | thats my first one |
| 31:31 | thats really what im doing
ok yes?
the function is f(x)=x^2+1 the function is f(x)=x^2+1 so i take this i have to plug into x^2+1 so my first one is (1+1/n)^2+1 plus (2/n)^2+1 + (3/n)^2+1 |
| 32:00 | up to (1+n/n)^2+1
because thats my function
its right hand
you use right hand rectangles
yeah, why do we use right hand?
you can do left hand, midpoint, its just theres lots of other math stuff which wont be needed in notation later if you use right hand, it just works better ok? im amazed i thought attendance would be dropping now that i started videoing but i guess not |
| 32:34 | its very impressive we'll see how long you guys hang in there before you start cutting class
i mean..chemistry
whens the first chem exam?
2 weeks? okay so we wont see you guys that class right? yeah i usually get about 1/3 of the class that day thats when i give away all the answers alright, so now thats a curve you can answer |
| 33:01 | if i wanted to make this more compact
i would put this in sigma form
do i have room to write it down there?
im trying to write very large i get that so where i have the lowest numbers |
| 33:31 | i now replace it with a letter
so i use i, i could use any letter i wanted
k?
and i start i at 1 and i go up to n do you guys all see that? is that visible? for those of you who cant see past that desk? no so i'll write it again where should i put it guess i'll put it here |
| 34:10 | ok? thats what that says thats a 1 alright now we come up with a formula so if i wanted to find the actual area |
| 34:32 | all i have to do is let n go to infinity
why does letting n go to infinity work?
its not about an infinite number of rectangles their all infinite tests that theyve been so its going to be little strips when i add them up im getting exactly the length of that-the area under that curve okay? you saw this with first semester calculus you always do the limit to infinity and then we taught you the short cut and you hated us short cut is coming |
| 35:00 | ok?
so lets evaluate this now before i evaluate that youre going to need a few formulas actually 3 |
| 35:49 | while im at it |
| 36:11 | i'll explain why you need those in a minute |
| 36:30 | i'll post it later
for those of you who have been following me so far
its mostly been pictures of bobby but we're working on the other stuff
bobby will be back soon
alright
right jackie?
*unrelated to math* umm ok these are going to be very helpful when you want to evaluate that ok so im going to do 1/n |
| 37:02 | times sigma 1+ i/n there...plus 1 alright, lets multiply it out |
| 37:53 | now all i do is i multiply this out foil that out im ignoring the limit for now |
| 38:16 | ok heres stuff you dont need to know about sigmas and things you could break this up i could do the first one and the second one and the third one in other words i could make this |
| 38:32 | 1/n
i=1
n 2
plus 1/n
i=1 to n
of
2i/n
plus 1/n
i=1 to n
sub i squared over n squared
ok?
just going to distribute that sigma-yes? |
| 39:03 | that step to this step?
the square.. yes i foiled this out i took 1+i/n and i multiplied it by 1+i/n 1+i/n times 1+i/n 1x1 i/n i/n |
| 39:31 | i^2/n^2
ok?
i foiled anyway, alright so now that we did the sigmas so thats when these formulas are going to show up ok? so what does this really mean? this means do the sum from i= 1 to n to number 2 then we take 2 and add another 2 and add another 2 and add another 2 n times thats what this formula says |
| 40:01 | you take a you keep adding it n times you get n times n so im here im just going to get 1/n times 2n |
| 40:32 | alright now im going to find this sum well i could pull 2/n out i=1 to n of i so i just took the 2/n and i pulled it out and the sum of i= 1 to n of i is going to be this |
| 41:05 | and its very easy to get lost i dont like these either why is it 2? well i have this 2 over n and i pulled it out and i get 2/n^2 the sum of i=1 to n of 2 is n times a constant |
| 41:36 | just doin some algebra folks you can even pull an n^2 out and get 1/n^3 |
| 42:04 | so that now becomes ill read that 1/n(2n) plus 2/n^2 and then that formula is n(n+1)/2 you should have learned that back when you did series when you just added up 1 2 3 4 all the way up to a number its n times n number plus 1 over 2 |
| 42:32 | and the last one
1/n^3
this formula i showed you a few minutes ago
ok?
if i had a cubic term id need the bottom formula but we dont have a cubic term here though i might have one on homework problem # 22b-yes? |
| 43:05 | we're finding the area we're just going to keep going this is a bunch of algebra alright so what did i do what i did was i wrote out the equation for the area and now im really just simplifying it over and over with all of this sigma stuff ok |
| 43:30 | now that i got the area i'm gonna do sigma stuff because its in each one of these sigmas i can now replace with one of those 4 formulas this is now 2 this one the 2's cancel one of the n's cancels and you get (n+1)/n and this one, one of the n's cancels and you get n+1(2n+1) over 6 |
| 44:09 | yes this is n^2 sorry
ok?
and now im just going to take the limit as n goes to infinity well the limit as n goes to infinity of this |
| 44:31 | is uh..
well (n+1)/n is just 1
the other one we figured out a couple minutes ago is 1/3
this equals 3 and a third
or 10/3
so what i'll do later is ill write all of this up
put it up in my documents for you guys, okay?
it also has this stuff in my book so you may not be aware of this i wrote the princeton review AP calculus book some people know this a very small number of you |
| 45:00 | so if you dont know my biography im one of the people that helped princeton review many years ago its not khan academy but i have my own stuff so we have a couple other kahns in the class umm anyway, so back when i worked for princeton review i wrote up the AP calculus book which we've now sold a zillion copies you might have used it in high school so i will take pages from that that are relevant to the course and i'll put them up on blackboard with the part that says documents from david |
| 45:32 | somehow the documents section got eliminated
from blackboard
but i can stick it back in ok?
so ill put that stuff up there in the next couple days i dont do the infinite stuff because thats not on the AP but ill write up a couple pages of notes for everybody ok? just takes a couple days you have to be patient all the other stuff however will be in there the finite number of rectangles the left and the right and all that practice k? |
| 46:02 | alright so
how was that for fun? yes?
okay so this is the same as the previous problem where you get 1/3 here notice that n+1 and an n so the power is n to the 1 and n to the 1, same power coefficients are 1 for each so that just becomes 1 this is a 2 you get 2+1+1/3 |
| 46:30 | add it up you get 3 and a third alright?
well you really dont remember how to do limits? okay what happens when n gets really large well these 2 numbers really approach each other so there the same so you get 1 |
| 47:00 | another way i think of this is its n/n +1/n
1+(1/n)
1/n goes to infinity
that goes to 0
its 1
ok?
gotta remember your limits no we have plenty of time left this class goes to 12:20 ill stay forever doing calculus |
| 47:46 | im going to have you practice one of these instead
alright so lets practice one more of these
cause we love these right?
|
| 48:20 | find the area under the parameter f(x)=x^2-2 from x=3 to x=5 |
| 48:34 | see how you do okay so you should have that You should have gotten to that step and then your step from there |
| 49:00 | thats what you're going to be trying to evaluate
ok?
where did that come from take the 3+2i/n sticking it in for x alright a couple more minutes foil out 3+i(2/n) you get that you just have to foil 3+2i/n times 3+2i/n |
| 49:32 | i know this is hard like i said
this is supposed to be the most abstract stuff that we do
you've never really done this before
why does i=1?
well i is counting..this is a counter it helps me keep track of how many 2/n's i use so each time i go from 3 i move 2/n to the right each time |
| 50:01 | ok?
so i start once and then i just keep going so thats why i have 1 2/n 2 2/n's 3 2/n's etc. yes? well like i said i have 2/n then i move 2/n again 2/n again so i have i(2/n) which is just 2i/n ok? |
| 50:31 | i have i of these
2/n's
so at i=1
i just keep adding 1 to it each time
think about a number line
this point is 3
and i move 2/n to the right
and i move 2/n to the right again
so i did it 2 times
so i move 2/n to the right again
3 times
ok?
|
| 51:00 | so each time if i blow up the number line
im moving 2/n to the right
this i helps me keep track of how many times i do that-yes?
why 2/n? 5-3=2 okay and then divide that by n intervals k? now youre going to be here |
| 51:32 | i want to simplify that ok lets simplify it some more |
| 52:01 | remember this just comes from foiling out 3+(2i/n) k? subtract 2 now i can make that into 3 different sigmas |
| 52:44 | okay so now i can break that up-yes?
i didnt get 7 that 9 plus 12i/n+ 41^2/n^2 -2 9-2 is 7 |
| 53:02 | as i keep saying i know this is hard dont think im not sympathetic i didnt test this last semester but we're covering it so i have to teach it okay 2/n now i can use my formula this is just a constant 7 so when i multiply that out i get the constant times n and thats 7n |
| 53:32 | next one i have 12/n and the sum of pi equals 1 to n of i is just.. [n(n+1)]/2 third one |
| 54:04 | 4/n^2
[n(n+1)(2n+1)]/6
thats...
14+...
did i lose an n somewhere?
|
| 54:36 | i lost an an oh ive got 2/n sorry sorry about that forgot to distribute 2/n thats a mistake you wont make on the exam |
| 55:15 | ok?
simplify it a bit youre now going to take the limit as n goes to infinity im going to get 14+12+ well...8 times a third |
| 55:42 | ok?
and i'll write up some stuff about this-yes? well, because this is a third |