Big Picture: Integrals

Big Picture: Integrals

PROFESSOR: Hi. Well, if you’re ready,
this will be the other big side of calculus. We still have two functions,
as before. Let me call them the height and
the slope: y of x and the slope, s of x. Function one and function two. That’s what calculus is about. And earlier, we figured out how
to go from function one if we knew that, how to
find the slope. Of course, it was easy when
function one was just a straight line. Then the slope was just
up divided by across. But when function
one was curving, we had to do something. We had to take just a small
distance across, a small distance up and divide,
and then let small get smaller and smaller. The result of that was function
two, the derivative. Today, we’re going
the other way. We know the slope. We know s of x, and we want
to find the height, y of x, of the graph. We know how the graph is sloping
at every point, and we have to put all that information
together to find its height. OK. Let me say the easiest way to
do it, when it works, is to recognize if we are given a
formula for the slope, to recognize maybe we know a height
that goes with it. Let me take an example. So the one, and most important
case of a height is x to some power. x to n. That’s one that everybody
learns. When y is x to the nth, the
result of this process that produces function two going in
that direction is dy/dx is that n comes down and we
have one lower power. So if we happen to have that
as our slope, that would be our height. Nothing more to do. But many other slopes
are possible. Let’s just stay with this
another minute. What if the slope was
x to the nth itself? What if we started with function
two as x to the nth. Suppose dy/dx is x to the nth
power, like x squared. Where does x squared
come from? Well, look at this rule. That rule said that for these
power functions, the power drops by one. So if I want to end up with x
to the nth, going backwards, function one had better be,
involve an x to the n plus 1. But that’s not perfect, of
course, because when I take the slope of that one, this
factor, this power, this exponent, n plus n
will come down. Just as n came down to here, n
plus 1 will come down, and therefore I’d better divide by
n plus 1 so that when I do take the slope, when I do go to
number two, the n plus 1’s will cancel. The power will drop by one, and
I’ll have x to the nth. OK, there’s an example– quite a useful one– of going from step two to
step one just by kind of recognizing what
you want there. So this is one set that
everybody learns. Another is sines and cosines,
if you can fit it into that. Another is either
the x or logx. That’s pretty much the list.
And then you learn, in the future, ways to change things
around to fit into one of those forms. But then,
of course, there are many, many cases– many, many functions, too– that
you don’t fit, can’t fit into a form where you can
recognize from some list which you either have learned or you
find on the web or you find in the calculator. A lot of lists have been made
to help you go from two to one, but today we have
to understand what is the actual process. What is the, what is the the
reverse process to this one? And of course, this one involved
a limit as delta x went to 0, because always I have
to remember that things can be curving, things can be
changing, I can’t assume that they’re saying the same. And then, the reverse
direction. Maybe I just tell you first
what the symbol is. If I have this s of x and I want
to get back to y of x– so this is from two to one– one will be y of x. The symbol for y
of x will be– it’s, I’m just really wanting to
draw that integral symbol. The integral, I would say the
integral of s of x dx. And you’ll see why– this is, that’s an s– you’ll see why that’s a
reasonable way to write it. But of course, first, we need
the idea behind it. OK, so how am I going
to proceed? Step one, I’ll take steps, I
won’t try to get immediately to the case of continuous
change. I’ll take single, individual
steps. Let me do that. And then, I’m going to
take smaller steps. And then in the limit, I’m
taking continuous steps. OK, so first, big steps. So let me put down, for example,
suppose I have y’s. Suppose the y’s stepped up like
0, 1, 4, 9, 16, whatever. So those are heights, and, of a
graph that’s sort of pieces of straight lines, only
changing a few times. What would be the slopes? Going now, this is
one going to two. The slope is s. Well, the slope, if the step
size is one and I go up by one, the slope will be 1. Here I go up by 3. Here I go up by 5. Here I go up by 7. So if I, to go from there to
there, I’m taking differences. I’m taking delta y’s. These s’s are delta y’s. How do I go backwards? Suppose I gave you 1, 3, 5, 7,
and I started y off at 0. How could you recover
the rest of the y’s? Well, that I would, that 1 plus
3 is the 4, 1 plus 3 plus 5 is the 9, 1 plus
3 adds 7 more. You’re up to 16 and onwards. So if going this way is a
subtraction at each step, going this way is a kind
of running addition. I add up all the slopes to
see how far I’ve climbed. Let me do a second example,
just to make that point. Let me start you this time,
let me start you with the slopes, because that’s
today’s job. Suppose the slopes are
4, 3, 2, 1, 0. So if these were speeds,
I would say, OK, I’m slowing down. I’m slowing down, but I’m
still moving forward. Positive speeds, but putting
the brakes on. What would be the distances? If the trip meter starts, start
the trip meter at 0. OK. Then in the first second or
hour, the first delta x we would go 4. And what goes there? 7, right? I’m doing, I’m accumulating,
adding up the distances. Here I was at 4. I went another three
to 7, to 9, 10. Here’s, I didn’t move. No speed, zero slope, stays
flat, hit the top at 10. OK, I could do this with any
bunch of numbers and I can do it with letters. So now I move from arithmetic
to algebra. Algebra just means I’ll do it
with any letters, but I’m not yet doing it continuously,
which is what calculus will do. So with letters, I have here y0,
y1, y2, y3, y4, let’s say. So those are the y’s. Now what are the slopes between
them if the step across is 1? So what are the steps upward,
what are the delta y’s? Well, y1 minus y0, y2 minus y1,
y3 minus y2, y4 minus y3. And then my question is these
are the s’s, these are the delta y’s, you could say. These are the y’s. What happens if I add
all those delta y’s? Do you see what happens? What happens if I add those
four changes to get the total change? Well, when I add those, do you
see that y1 will cancel minus y1, y2 will cancel minus y2,
y3 will cancel minus y3. So the sum– and I use just a sigma symbol,
but I’ll just say sum and you know what I mean– of these delta y’s is what? What happened after all
those cancellations? Did everything cancel? No way. y4 is still there, minus
y0 is still there. So it was y4 minus y0. The last y minus the first y. I’ll just write y last minus
y 0, y first. y end minus y start is the sum of
the delta y’s. Simple algebra. Reminding us again and again and
again that the opposite, the inverse to go the other way
from two to one, we add pieces to get back to the y’s. Now I’m coming closer to what I
want, but I’m moving toward calculus now. So calculus, I got there by
delta y’s over delta x’s. So in moving toward calculus,
what am I doing? I’m thinking of the changes
delta y over smaller steps delta x. So I just want to
take this step. I want to divide by delta x
and multiply by delta x. Why do I do that? That’s because it’s this delta
y over delta x that is– it’s those ratios, whatever
the size of delta x is. And it’s going to get
smaller and smaller. I’m going to look at the change
over very short steps. Then it’s that ratio
that make sense. Delta y over delta x is
a reasonable number. It’s close to the s. It’s close to dy dx, but it
hasn’t got there yet. I’m multiplying by the delta x,
the small step that’s going to 0 but hasn’t got there yet. And I’m adding and I get the
last one minus the first one. Now here comes the
limiting step. So the limiting step will be
the limit of this left hand side, this sum. So in the limit, I’ll have more
and more and more things. As delta x gets smaller, if
I’m thinking of some fixed total change in x, I’m chopping
that up into smaller and smaller pieces, more
and more pieces. So more and more pieces of the
slopes at different points along times the size of the
piece give this answer. So now can I jump to the
way that I would write this in the limit? So now let delta x go to 0. And I ask the right hand
side, y last minus y first is not changing. y at the end– I’ll write something different,
y end minus y start, just to make that
same point again. But it’s this that’s changing. As delta x goes to 0,
this becomes dy dx. The little delta x’s
are going to 0. Here’s the way I write it. So in that limit, I can’t
legally write that sigma, so this integral symbol is kind
of copied from that sigma. But it’s telling you that
a limit has happened. And in that limit, this
is dy dx and this, the notation, is dx. I’ve got what I predicted
here, with s of x there. So what I hope this discussion
has, by starting with numbers, by going to algebra, by looking
at the sum of those things, which was simple, and
then by going to the limit, which was not simple. So a whole lot of limit has been
not fully explained, and I think the right thing to do
now is to do an example. So let me move to an example. So I’ll take a particular
function and follow this process, this limiting process
and see what it gives. And it will give us function one
and, as a bonus, it will give us a new meaning
for function one. Let’s do it. So now I’m going to take
a particular s of x. So here’s x. Let me take s of x
to be 2 minus 2x. I didn’t want to take one
totally simple that I already had started the lecture with,
but it’s not difficult either. We’ll be able to see what’s
happening here. OK, so let me graph it, because
I want to do this now with the graph. So at x equal 1, s of x has
dropped to 0, where when x was 0, it started at 2. So it started somewhere here. Here was 1, halfway down. It’s going to come down
in a straight line. And let me stop there. It could continue, but let me
call y end is going to be 0 and y start is going to be– well, we’ll see about
that, sorry. s at the end is 0. S at the end is 0. I don’t yet know what
y at the end is. It’s not 0. So what’s my idea? Well, not mine. Newton and Leibniz and a lot
of people had these ideas. It’s kind of interesting. So Archimedes. He goes way, way back, before
Newton or Leibniz or anybody conceived of them. Archimedes figured out
how to deal with a curve with a parabola. Archimedes got from a parabola,
he got from x squared, the parabola, back to
a height by special ideas. He was one of the great
mathematicians of all time. But even Archimedes didn’t see
what you now see, this connection between function
one and function two. If he had seen that, he would
have gone further. All right, now let’s see it. Let me take a delta
x equal to 1/4. So this is delta x here, and
this one is two delta x’s, and this one is three delta x’s, and
the one is four delta x’s in my original delta x, which is
1/4, which is small but not really small. So now what do I do? Look at this first period. The slope, the s function,
function two– see, over here is going
to be function one. This is going to be the y,
the integral of that. But I don’t know what
it is yet, so it’s pretty open to question. OK, so now let’s get there. So the point is that
over this interval, the slope is changing. It’s changing a significant
amount. Not too much, but
it’s changing. And I don’t know, from the
algebra, I don’t know how to deal with that. I’m just going to take a value
within this sum value and stay with it within that interval,
and I’ll take the starting value. So over this first delta x, I’m
going to pretend that the slope stays at 2. So I’m pretending that this
is my slope function. Then over my next delta x, I’m
going to pretend that it stays at probably 1 1/2. And then I’m going to pretend
that it stays at 1, and I’m going to pretend that
it stays at 1/2. If you allow me to go back to
distance and speed, I’m chopping up the full time, the
day, let’s say, into four pieces and in each piece, the
speed is changing, which I’m not ready to deal with,
which algebra isn’t ready to deal with. So the best I could do was say
OK, so suppose the speed is constant at what it was at the
start of that short time. So those would be delta t’s
rather than delta x’s. The s would be representing
speed, but no difference in the picture. So now let me do these things. Now I’m going to do this
addition, which won’t give me exactly the right y because
those rectangles are not exactly right. But I’ll get them better by
taking smaller delta x’s. Let me see, what
do I have here? Over this first time, I
have my slope, which I’m taking to b 2. So that’s the delta y over delta
x, that’s the s, and then times the delta x. And what is that? We might as well just face it,
that that 2 times that delta x, we can think of that
as the area in that tall, thin rectangle. Well, I’ve introduced the word
area for the first time. It never showed up on
the previous board. It’s the extra insight
that’s coming today. Now over the second short
period, I’m going to keep fix my speed at 1 1/2. 1 1/2 times delta x, because my
speed I’m setting at 1 1/2, and this is how long I go so
this is a distance or a change in height, a change in y. And you see what’s coming. The next one will be a 1 times
the delta x, and the last will be 1/2 times the delta x. This is adding the way
I did in the algebra. And what do I get? Well, this, again, is the
area of that piece. This one is the area of that
piece, this one is the area of that piece. I get an overestimate because
the true slopes dropped a little within each piece. I get some quantity which I can
figure out, but it’s not the right answer. It’s not the final answer. And what is now the main
step to get there? Chop delta x into half,
you could say. Why not cut it in half? Now I’ll have a different
picture. Can you see what this
picture is doing? Now over the first little half
of the old step I am up here, but then I drop to here. Can I do this with an eraser? A little bit got chopped away, a
little bit got chopped away. Where has it gone? I’m going to have
this zig zag. That wasn’t too bad. I’m replacing that with a sum of
eight pieces, because delta x is now down to 1/8. This is what we said
about that sum. That sum has got more and more
terms because it has a term for every little delta x, and
the size of that term is about like delta x multiplied
by an s. So what I getting in the limit
is a kind of running sum, a running counter, a mileage
meter, a trip mileage, that’s adding up distance
based on speed. Do you see what I’m getting
in the graph picture? What happens to the shaded part
as delta x gets smaller and smaller? This shaded part is going
to be the curve. These little long pieces are
going to get reduced, reduced, reduced, and in the end
the total height at 1 is going to be– ta-da, this is the moment– the area under the
slope curve. This y turns out to be
the area under the s of x curve, or y. So at x equal 1, what is it? What’s the area out to 1? Well, we’ve got a
triangle there. Its base is 1, its
height is 2. The area of a triangle is 1/2
the base times the height, so I have 1/2 times 1 times 2. I’ve got 1. The area at the end is 1. But– well, I shouldn’t say but. I should let you applaud first.
What if I only went that far, halfway? What if that was s end? What if I want to know what
is y at x equal 1/2? Then it’ll be of course, just
the area up to that point. Can I remove this part of the
picture for a moment? I’m always looking at area. And the area of that, do we know
what that area would be? It’s not a triangle anymore,
it’s some kind of a trapezoid. As delta x goes to 0, I’m going
to get the correct area, which will be what? Let’s see, I have 1/2 as the
base and the average height is about 1 1/2. Can I do that little
calculation? The base is 1/2 and the average
height is 1 1/2. I think I get 3/4. So halfway along, it’s
got up to 3/4. Where is 3/4? So this is 1, this is 3/4,
this is 1/2, this is 1/4. So at 1, it’s at 3/4. Halfway along, its at 3/4. I would like to know
that graph now. I’m ready to jump
to the limit. Let me do it the way I said at
the very start of the lecture. Let me take this and
try to guess. So, I’m taking a shortcut. Because do we go through this
horrible process every time we want to do an integral? Of course we don’t. The best way is, can we find a
y function, a function, one, that has that derivative? Let’s just try it. I’m allowed to take it in
two pieces, that’s a very valuable fact. So what has the derivative 2? If the slope is 2, what’s
the function? If the speed is 2, what’s
the distance? It’s constant speed, 2, times
the total distance. The slope of the 2x line
is 2, clearly. What about the 2x? Which function has
the slope 2x? Well, we saw it over here. The function that has the slope
2x is x squared, because when I take the slope of x
squared, the 2 comes down. The 2 shows up, I have
one smaller power, x to the first power. This is the correct y, and I
hope that my graph gets those points right. At x equal to 1, this is
2 minus 1, this is the correct height, 1. At x equal to 1/2, all right,
here is the moment of truth. Now set x equal to 1/2, and what
do you get for this y? You get 2 times 1/2–
that’s 1– minus 1/2 squared, 1/4. Hey, miracle. 3/4. This area I figured
to be 3/4 and this approach also gave 3/4. Either way, multiplying
those is 3/4, subtracting those is 3/4. What does my graph look like? What does the graph
of that look like? What’s the slope at the start? The slope at the start
is s at the start. And s at the start, when
x is 0, the slope is 2. So it starts out with
a slope of 2. But it’s slowing down, it’s a
little bit like this one where the car was slowing down, we’re
not picking up distance so quickly, we’re not picking
up height so quickly. But we’re still going
forward, we’re still picking up some height. So it starts with a slope of 2,
bends around to there, and I guess maybe that is– yes. That picture is almost
good, but not great. So the slope is 2 and there. And what is the slope
at this point? You can’t tell from my picture,
which isn’t perfect. The slope, I’m told
what it is. When x is 1, the slope
is 2 minus 2. Slope 0. The slope is 0. We’re not picking up any more
height, any more area. And of course, that’s right. At this point, we’re not
picking up more area. If I continue beyond here,
we’re losing area because below the axis, I’ll count as
negative area just because if it was speed, I’d be
going backwards. That’s what will happen here. I’ll start down. If that continued, this
would still be the correct thing to graph. If I do graph it, that’s
actually the top at x equal 1, and then it starts down and
probably by, I don’t know where, x equal something,
maybe by x equal 2. Oh yeah, you can see. By x equal to 2, it’s
got down to 0 again. When x is 2, this is now 0 and
you can see that when x is 2, we’ll have the bad area– the car going backwards– will be identical to
the forward area. The total area is 0, and I’ll be
at this point when x is 2. Let me just recap a moment. Today was about going
from function two back to function one. The quickest way to do it is
to find a function one that gives that function two
and then you’re in. But if you can’t do that or if
you want to understand what the real, behind it, limiting
process is, it’s like the algebra but it’s this expression
here that’s concealing so much
mathematics. Delta x going to 0, these ratios
going to the actual function, and the delta x I
replaced by the symbol dx, indicating an infinitesimal. We’ll see it more. Thank you. NARRATOR: This has been
a production of MIT OpenCourseWare and
Gilbert Strang. Funding for this video was
provided by the Lord Foundation. To help OCW continue to provide
free and open access to MIT courses, please
make a donation at

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  1. @themanisheab180237ce I went to a catholic high school in the W.I. When I took my G.C.E's I only did o-level math. I guess I should have taken a-level math instead: since the first time I encountered intergral calculus was at college.

  2. @gerardrbain1972 gerardrbain the question is which high school did you go to lol, this is high school math, this is basic calculus!

  3. here in Brazil, we learn it in college.

    high school stops at derivades, and, most part of schools dont reach neither limits.

  4. Good lecture and terrific idea on giving a big picture on the topic. We are sometimes sink too deep in the bolts and nuts of the areas and lost the overview and forget why we are there…

    Hope there will be more lectures…

  5. i love this kind of teaching suddenly things are crystal claer and make sense from the beginning…. great math
    this is how the real father of math , thought about math.
    other make it too complex and in the end it doesnt make sense so the student have to memorize and not really understand.

  6. EXACTLY how it should be, the method is a million times more important than the result, and some results such as simple equations and basic math become memory recall anyway

  7. exposition is outstanding! this is how math ought to be taught. Thank you to all who played part in the production of educational such as this one.

  8. This series has been great for reteaching myself the Calculus I have forgotten before the next college term starts. In my past classes, the professor could never explain what dy/dx meant, or why dx was found after the function behind the integration symbol. I finally fully understand the notation. Thank you.

  9. Why he did not explain limits in first lectures and then derivatives, then integrals? Definition of derivative includes limit so it would make sense to introduce limits first. I appreciate the effort of making this videos but it is by far not the best explanations.

  10. A good source for complete instructional videos for Calculus I and Calculus II college level courses:

  11. Very good explanation professor! I have studied this integral part of calculus in a different way… Now I got a good picture of where this all came from. Thanks professor and all MIT people

  12. Thank you Professor Strang and MIT. My summer college calculus taecher is giving us definitions, equations, and then practice problems. I had difficulty seeing the entirety of the integral idea.

  13. Love his lectures, just read that he was a Rhodes scholar, very impressive. I hope to be 1/2 as good as whatever it is I do as he is with teaching and mathematics. 

  14. this was the most beautiful and simple lecture about this topic i ever saw, he clearly understands it to be able to explain it so well

  15. I actually understand why integrals work now thanks to this guy! if you want to understand why integration finds the area under a curve, watch this video.

  16. he really knows his stuff but he absolutely CAN NOT explain it to anyone who doesn't already have a working understanding of integrals (and differentials).

  17. The nice thing is that once you really understand the intimate relationship between integral and differential calculus, this will confirm what you start to discover on your own. 

  18. I think I would have saved a lot of time and effort in my research, if Dr Strang was my lecturer in university.

  19. IF ONLY I had seen this video in 1988, I might have gotten > a C in Calculus 101! Beautiful. The whole connection with area has always mystified me.

  20. I've taken calc courses at UMD, even gotten good grades. This explanation sheds light on the ONE thing I never got – WHY calc works. Wonderful lecture!

  21. Holy shit. So basically, delta X is the base, and the height is the slope. The sum of these areas gives you the area in the aggregate.

  22. I knew the formulas and how to solve questions in Calculus. But for the first time I am able to visualize it and truly understand it. Thank you, Prof. Strang.

  23. I got A's in both calculus 1 and calc 2 and yet I did not know the basic concepts behind the math until now. Public Texas Universities for you….

  24. I wish my teacher taught like this… My teacher uses power points and it's horrible. I have to see how the problem is solved without steps being skipped.

  25. Bah…get to 33:32 and I still don't see how to actually perform the limit. We jump right to guessing the derivative..

  26. even i have learnt all of these things that prof delivers in this lecture, it's still an interesting lecture 🙂 thanks sir

  27. if i have a problem set of two functions & i have to determine one from the other how do i know which is function 1 & which is function 2 which will be the derivative & which is the integral is there kind of rule to know that?

  28. Some check space pictureing area detopic defind communiative space minus integrate equal Big Picture.
    Try comment.

  29. Prof. Strang's lectures have brought me to a culminating point in my efforts to understand Calculus. I shan't rest on my laurels now.

  30. Decided to watch these as a "refresher" since it has been years since I used any calculus…He does a great job explaining it, and he has an enthusiastic personality, which made it seem a lot less dry. I wish I had instructors like this when I was an undergrad…

  31. This is the most awful and useless approach. Hes teaching math as if it was a complex form of arithmetics, he dont teach folks to think and reason. Waste of time

  32. A much better approach is to show the ideas behind derivatives and integrals as separate concepts, THEN prove they are inverses of each other. Faster, and most straightforward, AND lays the ground work for other things like numerical methods.

  33. Your big picture of the integrals are very insightful but your mathematical graphs cannot be implicitly right in special case of space time relativity cause the function sin dx is interrelative different in special case of space volume geometry pertaining to the inerant shape shifting of the slopes. You définetely know the function of a graph is co dependant of the speed and the height and the slope but in overall timing spécial relativity of gravitation has his hear say in the definition of those numbers. You can measure a phenomena by a graph but the intricacies of a special gravitation graph often implied us into errors by the contingencies of special intertwined relativity. Your mapping of integrals is right by compute graphing but when special relativity is involve, the definition of the slopes are the stringent variables to be define cause it's not a thumb rule.

  34. Thank you internet! This is the best time to live if you are yearning for knowledge. Thank you Professor Strang and MIT.

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