You probably recall the name, Johannes Kepler, from when your high school science class covered astronomy. He was the first to figure out that the paths that the planets take around the sun are ellipses, which is Kepler’s first law of planetary motion. He came up with two other laws as well. You can learn more about Kepler by clicking here

But Kepler’s primary skill was not astronomy but mathematics. Legend has it that at Kepler’s own wedding reception he sipped his wine and pondered. But rather than pondering the nuptial joys that awaited him that very evening, he showed himself to be a true mathematician to the core. He occupied his mind with the problem of finding the volume of wine in the barrels from which his drink had been poured. He noticed the caterers periodically checking each barrel with a dipstick. If a barrel were a perfect cylinder, Kepler reasoned, then it would be easy to determine the volume of wine remaining from the mark it made on the dipstick. But barrels aren’t perfect cylinders. They are wider in the middle and narrower at the top and bottom. And in between the sides make a graceful curve that appears to be the arc of a circle. So what could possibly be the formula for the volume of such a shape, especially when it is only partially full? It was this mystery that stood that night between Johannes Kepler and marital bliss.

Since I cannot reproduce Kepler’s exact line of thinking, it is here that my tale must
turn fanciful. It seems that Johannes had a second cousin named Artie, whose hobby it
was to collect manhole covers from cities all over the world. Now Artie had told Johannes
all about his collection at the bachelor party the night before. And he had mentioned the
peculiar fact that despite covers from Vienna, London, Moscow, Constantinopal, and
dozens of other cities all differing in diameter, every manhole cover from everywhere
Artie had collected all had *exactly* the same thickness.

This got Johannes to thinking. The volume of a manhole cover is easy because it is
nothing more than a short cylinder. If its thickness is `h` and its radius
is `r` then its volume is given by

V = prwhich is the formula for the volume of a cylinder, where thickness of the manhole cover is the same as the cylinder's height.^{2}h eq. 10.1-1

Let's suppose that all the manhole covers are one inch thick. Here is my version of Johannes Kepler's line of thought: If he went to Artie's collection, he could pick a manhole cover that was the same radius as the base of the barrel. He could find another that was the same radius as the barrel was 1 inch above its base. And he could find another that was the same radius as the barrel was 2 inches above its base, and so on, always matching the radius of the next manhole cover to the radius of the barrel at the height where the cover is to be stacked. If he were to stack up these manhole covers in order until they were as high as the barrel, he would have a pretty good approximation of the shape of the barrel. And indeed if he were to add up the volumes of all the covers he had stacked, he would have a pretty good approximation of the volume of the barrel.

Furthermore, if he only stacked the manhole covers as high as the wine remaining in the barrel, he could add up those volumes and have a pretty good idea of the volume of wine it took to fill the barrel to that height.

But the volume of wine computed this way would not be exact. It would be off by a glass or two. Why? Because the stack of manhole covers does not exactly trace the shape of the sides of the barrel. Take the bottom manhole cover, for example. Its radius is the same as the base of the barrel. But the top of that manhole cover is already an inch high, and it still has the same radius. Yet at one inch up on the barrel, its radius is already greater than that of the bottom manhole cover. So you can see that there is some small measure of inaccuracy in computing the volume this way.

Just then one of the caterers removed an empty barrel from the table. It had left a wet ring on the butcher-paper that covered the table, and when the caterer lifted the barrel, it tore away the disk of paper inside the wet ring. Kepler watched as the caterer hauled out the barrel with the disk of paper stuck to its bottom.

If I used paper disks instead of manhole covers, he thought, I could stack them into a much more accurate approximation of the barrel. Of course it would take a lot more disks to do it that way than it would take manhole covers to do it with cousin Artie's collection. That's because paper is so much thinner than a manhole cover. But that is also why stacking paper disks, each one the same radius as the barrel is at the height it is to be stacked, is so much more accurate. Take the bottom paper disk in such a stack. Instead of having a thickness of one inch, it has thickness of about 0.003 inches. At a height of 0.003 inches, the barrel is nearly the same radius as it is at zero inches height. The barrel's radius simply can't change very much in the span of just 0.003 inches.

The principle of adding up the volumes of the disks of paper works in just
the same way as stacking the manhole covers. Each paper disk is really just
a *very* short cylinder (`h = 0.003` inches

It's still not perfect though. The barrel's radius *does* change just
a tiny amount in 0.003 inches, and that will lead to a tiny error in the
measurement of the barrel's volume. But you can imagine that if you used
thinner paper and more sheets, you could get an even more accurate approximation
of the barrel's volume. Still there'd be a miniscule error. The error would
be smaller than you'd get using thicker paper, but it would still exist.
Indeed no matter
how thin the paper you used, there would still be some error. But imagine taking the
limit as the thickness of the paper goes to zero and the number of paper disks
goes to infinity. The limit of that sum, if it exists, would seem intuitively to match
the volume of the barrel exactly. And that limit is what integrals are all about.

We shall leave solving the wine barrel problem for a later section. Instead what we shall do now is use the same method suggested above to solve a much simpler problem -- one for which we already know the answer in advance. We know that the graph of

f(x) = mx eq. 10.1-2is a straight line that passes through the origin. The slope of the line is given by the value of

What we are going to do is use the method Kepler used on the wine barrels to
approximate the area of this triangle. That is, we will slice it up and approximate
each slice with a rectangle. Then we will add up the areas of the rectangles to
approximate the area of the triangle. Since this will only approximate the
area of the triangle, we expect that the approximate area arrived at by this
method will not be exactly equal to the area we got using the tried and true
formula for the area of a triangle. But if we try using thinner slices and more
of them, we should see the resulting approximation draw closer to the actual
area of `(1/2)ma ^{2}`. And if we take the limit as the
thickness of the slices goes to zero and the number of slices goes to infinity,
then it would certainly be satisfying if, in the limit, the approximation turned
out to be

Figure 10-2a shows one way we might slice the tringle into four slices.
Each slice has a thickness of `a/4`. Observe that the left-hand
edges of the four slices are at
` x = 0`` x = a/4`` x = 2a/4`` x = 3a/4`

hThat is we have used the value of the function,_{1}= f(0) = 0 h_{2}= f(a/4) = ma/4 h_{3}= f(2a/4) = 2ma/4 h_{4}= f(3a/4) = 3ma/4

area approximation = 0 + maWe know the exact area of the triangle is^{2}/16 + 2ma^{2}/16 + 3ma^{2}/16 = 6ma^{2}/16 = 3ma^{2}/8 eq. 10.1-3

But what if we make more slices? Let's try eight of them. You can see how
this looks in figure 10-2b. Again we have used the value of the function,
`f(x)`, at the left-hand edge of each slice as the height of each rectangle.
But this time the thickness of each rectangle is `a/8`, or half of what
it was last time (since there are twice as many rectangles this time). Again we
find the area of each rectangle as its thickness times its height. And the heights
are:

hTaking these heights times the thickness (which is the same for all of the rectangles) and summing it all up, we get:_{1}= 0 h_{2}= a/8 h_{3}= 2a/8 h_{4}= 3a/8 h_{5}= 4a/8 h_{6}= 5a/8 h_{7}= 6a/8 h_{8}= 7a/8

area approximation = 0 + maThis time we are only short of the magic area of^{2}/64 + 2ma^{2}/64 + 3ma^{2}/64 + 4ma^{2}/64 + 5ma^{2}/64 + 6ma^{2}/64 + 7ma^{2}/64 = 28ma^{2}/64 = 7ma^{2}/16 eq. 10.1-4

If you divide the base of the triangle, which is length `a`, into `n` sections,
then each will have length of `a/n`.
The left-hand edge of the first section will be at
`x = 0a/n``x = a/n``x = 2a/n``k`th section will be at
`x = (k-1)/n`

That is for each integer, `k`, from `1` to `n`, we have
a rectangular slice. Again we will take each rectangle's height to be
`f(x)` where `x` is the left-hand edge of the rectangle
(that is `x = (k-1)a/n``f(x) = f((k-1)/n) = h _{k} = (k-1)ma/n`

ATo find the total area of the approximation, add up the_{k}= (a/n)((k-1)ma/n) = (k-1)ma^{2}/n^{2}eq. 10.1-5

Now substitute the expression we got for the area of the^{ }n A_{total}= å A_{k}eq. 10.1-6a_{ }k=1

Notice that the factor,^{ }n A_{total}= å (k-1)ma^{2}/n^{2}eq. 10.1-6b_{ }k=1

This equation is saying simply, take the factor,^{ }n A_{total}= (ma^{2}/n^{2}) å k-1 eq. 10.1-6c_{ }k=1

S = 0 + 1 + 2 + ... + n-2 + n-2 + n-1 eq. 10.1-7aBut because addition is commutative, you can also write that same sum backward and get the same thing:

S = n-1 + n-2 + n-3 + ... + 2 + 1 + 0 eq. 10.1-7bNow add these two equations together column by column and you get

S = 0 + 1 + 2 + ... + n-3 + n-2 + n-1 S = n-1 + n-2 + n-3 + ... + 2 + 1 + 0On the bottom line, how many~~2S = n-1 + n-1 + n-1 + ... + n-1 + n-1 + n-1 eq. 10.1-7c~~

2S = n(n-1) eq. 10.1-7dor equivalently

S = n(n-1)/2 = nWe're near the end of this now. Remember that^{2}/2 - n/2 eq. 10.1-7e

APerhaps you had surmised a while back that the pattern relating the number of rectangles to how good the approximation was is this: If you use_{total}= (ma^{2}/n^{2}) (n^{2}/2 - n/2) = ma^{2}/2 - ma^{2}/(2n) eq. 10.1-8

Now look at what happens when you take the limit as `n` goes to infinity.
Notice that the `ma ^{2}/2` term is unaffected by
the value of

AIn other words, in the limit as you keep slicing the triangle thinner and thinner into more and more rectangular slices to approximate its area,_{total}= ma^{2}/2 = (1/2)ma^{2}eq. 10.1-9

It is important that you understand this concept of slicing a shape into more and more, thinner and thinner, slices and taking the limit, as we did here. Take some time to mull it over in your mind and to review the material on this page until the concept feels comfortable to you.

**1)** Here's one almost the same as the one above. Let

f(x) = 3x + 4We would like to find the area enclosed by the

A = (1/2)(hSo 14 is the answer that we know ahead of time. But the way we want to do this problem is by slicing the trapezoid into_{left}+ h_{right})w A = (1/2)(4 + 10) × 2 = 14

**Step 1: If you divide the width of the trapezoid into **`n`**
pieces, how long will each piece be?** Remember that the trapezoid's width
is 2. Click here when you think you know.

**Step 2: If there are **`n`** sections, and each has the length
you determined in step 1, then at what **`x`** does the left-hand edge
of the **`k`**th slice happen?** If you have trouble with this, think about
where the left-hand edge of the first one is, then where the left-hand edge
of the second one is, and so on. Try to discern a pattern. When you think
you have it, click here.

**Step 3:** The height of the `k`th rectangle will be `f(x)`, where
`x` is where the left-hand edge of the `k`th rectangle is. You
determined that `x` in step 2. **Now use
** `f(x) = 3x + 4`** to determine the height
of the **`k`**th rectangle.**
Click here when you think you've got it.

**Step 4:** The widths of all the rectangles are the same. You found
that width back in step 1. In step 3, you found the height of the
`k`th rectangle. **So what is the area of the** `k`**th rectangle?**
Click here when you've worked it out.

**Step 5: Now write the summation expression for the sum of the
areas of the rectangles.** Use sigma notation for this. When you've written
your sigma-expression, click here to check
it.

**Step 6:** Note that the expression in the summation is itself the
sum of two things. **So write them as separate sums.** That is use the sigma
notation to write it as the sum over `k` of the first expression
plus the sum over `k` of the second expression. Doing this will
make the final steps easier.
Click here when you've got it.

**Step 7: From each sigma expression, factor out the stuff that
is constant with respect to your index variable,** `k`. Those factors
should each come out to the left of the sigma-expression they are extracted
from. Take a crack at it, and then click here
to see if you did it right.

**Step 8: Now take the sums of each of the sigma expressions.** The first
one is identical to the one we did back in equation 10.1-7a.
The second one is especially easy. It's just asking you to take the sum of
`1` added to itself `n` times. Click
here when you think you've got it.

**Step 9: Take the limit of the result you got in step 8 as** `n`** goes to
infinity.** This should be easy for you. But just in case you get confused,
you can click here to see how it's done.

**2)** In the last two examples, the one in the text and the step-by-step coached
one, you might be saying, "Big deal. I've just learned how to take the area
of a triangle or trapezoid the hard way. Why not just use the formulas and
do it the easy way?"
Indeed as a practical matter, you should use the formulas when you encounter
having to find the area of a triangle or trapezoid. But what do you do when
you encounter having to find the area of a shape that has a curvy edge?

The figure to the right shows a shaded area that is bounded by the
function, `f(x) = x ^{2}`

Remember that in both the triangle and the trapezoid problems you
had, at one point, to find a way of summing the integers from zero
to `n-1`. As you will see, it will turn out that in this one
you will have to have a way of summing the squares of the integers
from zero to `n-1`. Without going into the details of how
to derive this formula, here it is for your reference later in the
problem:

n-1 n n(n-1)(2n-1) å kIf you are interested in knowing a way to derive this, click here, but it is optional material and is not especially important to where we will eventually go with this slicing concept.^{2}= å (k-1)^{2}=~~k=0 k=1 6~~

**Step 1: If you slice the shaded area into **`n`** vertical slices, what is
the width of each slice?** Remember that the width of the base of this
shape is `a`. When you get it, click here.

**Step 2: If there are **`n`** slices, and each has the width
you determined in step 1, then at what **`x`** does the left-hand edge
of the **`k`**th slice happen?** Proceed with this the same way
you did in the first problem. Where does the first slice begin? Where does
the second begin? And the third? See the pattern? When you do, write it down
and then click here.

**Step 3: Plug the result from step 2 into **`f(x)`** to find the
height of the **`k`**th rectangle**. When you've done it,
click here.

**Step 4: Determine the area of the **`k`**th slice.** You know the
width and you know the height. Just put them together, and then
click here.

**Step 5: Write the summation expression for the sum of the areas of the rectangles.**
Do this the same way you did with the first problem. The only difference is that
you have a new expression (that you got from step 4) to put in the summation.
When you're done, click here.

**Step 6: Factor out that stuff that is constant with respect to** `k`.
Move that stuff to the left of the summation sign. Then check it by
clicking here.

**Step 7: Use the ****formula given to add up
the sum.** Do you see that your summation expression is the same as
something in the formula? Just substitute. Then click here.

**Step 8: Multiply it out.** This is just an algebra exercise. When you're
done, click here.

**Step 9: Take the limit as **`n`** goes to infinity.** Any term
with no `n` in it stays. See if you can figure out what happens to the other
terms. Then click here.

**3)**
Just one more of these slice-and-take-the-limit problems before we move on to
discuss an insight that makes finding the area under the curve a lot easier.
The figure shows the function, `f(x) = 2 ^{x}`

As in the past examples, you will reach a point where you will have to take the sum of an expression. In this case you will have to find

n S = å uwhere^{k-1}k=1

n (u - 1)S = (u - 1) å uSince^{k-1}k=1

n (u - 1)S = å (u - 1) uNow multiply the^{k-1}k=1

n (u - 1)S = å uThink about what the right-hand side of this equation means. For^{k}- u^{k-1}k=1

Since `u ^{0} = 1`

(u - 1)S = uAnd we solve for the sum,^{n}- 1

uSo now you are ready to attack the problem of finding the area under the^{n}- 1 n S =~~= å u~~^{k-1}u - 1 k=1

**Step 1: Same as last time. What is the width of each slice?**
You have `n` slices and you can see the total width of the area
you are slicing. You figure it out, then click here.

**Step 2: At what **`x`** is the left-hand edge of the
**`k`**th rectangle?** Again if you have trouble with this,
figure out where the left-hand edge of the first one is, then the second,
and so on. Then try to discern the pattern. When you've got it,
click here.

**Step 3: What is the height of the **`k`**th rectangle?**
Remember that the height of each rectangle is `f(x)`, where `x`
is where the left-hand edge of that rectangle is.
Click here when you are ready.

**Step 4: What is the area of the **`k`**th rectangle?**
Just multiply the base (which you got in step 1) by the height (which
you got in step 3). Then click here.

**Step 5: Write the summation formula for the sum of the areas of
all the rectangles.** When you've got it, click here.

**Step 6: Is there any factor in the summation that constant with
respect to the index variable, **`k`**?** If so, factor it to the outside
of the summation. Then click here.

**Step 7: What exponential identity will serve you well when applied to
the exponential function that remains in the summation?** Observe that
the exponent is the product of two things. One of them is constant with
respect to the index variable, `k`, and the other is not. Is there
a way of separating these two expressions in a way that will simplify
things? I know this step is tricky, but try to tough it out. Then
click here.

**Step 8: Apply the ****summation formula for
exponentials to what you got in step 7.** It's easier if you keep the
substituted variable suggested in the step 7 solution for now. When
you've got it, click here.

**Step 9: Take the limit as **`n`** goes to infinity.**
Hint: You have the recurring expression, `a/n`, in
this limit. Try substituting `h = a/n``n` goes to infinity, `h` will have to go
to zero. See if you can see a familiar limit hiding in the resulting
equation. When you are ready, click here.

The sums we have been taking, where we slice an area under a curve into
rectangular slices and sum them up -- these are called *Riemann Sums*.
Click here to learn more about Georg Friedrich Bernhard Riemann. Riemann was not the first
to think up this kind of sum; Kepler might have been the first, and it is even possible that
some of the ancient Greek mathematicians had thought along these lines. But Riemann was the first to show that if you use
a *continuous function* to bound the area you are interested in, then
when you take the limit as the number of slices goes to infinity, the limit
will *always exist*, and it will be equal to the area that you have bounded.

You probably noticed that in every example above, we always used the left-hand
edge of each rectangle to define its height. It turns out that you can use the
right-hand edge and get the exact same result when you take the limit. The
two types of sums are called *left Riemann sum* and *right Riemann sum*
respectively. The figure shows a right Riemann sum of
`f(x) = x ^{2}`

But notice that `f(x)` is an increasing function where we took the
sum. If it had been a decreasing function, then it would have been vice-versa.
The left Riemann sum would have been the one in excess of the area under the curve and
the right Riemann sum would have been the one that was deficient. Try to visualize it
to see why.

Let's set up the right Riemann sum of `f(x) = x ^{2}`

hThe area,_{k}= (ka/n)^{2}= k^{2}a^{2}/n^{2}eq. 10.1-10

ASumming these_{k}= (a/n)h_{k}= (a/n)k^{2}a^{2}/n^{2}= k^{2}a^{3}/n^{3}eq. 10.1-11

I will end this section by telling you that^{ }n^{ }n A_{total}= å k^{2}a^{3}/n^{3}= (a^{3}/n^{3}) å k^{2}eq. 10.1-12_{ }k=1_{ }k=1

n n(2n+1)(n+1) å kand let you prove for yourself that in this case, you get the same thing for the limit of this right Riemann sum as you got for the limit of the left Riemann sum. And remember that Riemann demonstrated that you get the same limit whether you use the left or right sum in^{2}=~~k=1 6~~

We have looked at the cases where you take the height of the rectangle
by taking `f(x)` of the left-hand edge and at the right-hand edge.
I have assured you (and later you will see proof) that in the limit, these
two sums end up the same. What do you suppose would happen if you took the
height of the rectangle as `f(x)` halfway between the left and right
edges? or anywhere else in between? Remember that `f(x)` is assumed
to be continuous. How does `f(x)` at an in between point compare
to `f(x)` at either edge as the rectangles get narrower and narrower?
So will you still end up with the same sum in the limit? (By the way, when
you use `f(x)` at the halfway point between left and right edges,
that sum is called the *center Riemann sum*)

How about if you used the mean of `f(x)`
of the left-hand edge and `f(x)` of the right-hand edge? Do you think
you would still end up with the same sum in the limit?

Section 10.2: Living Backward -- The Fundamental Theorem of Calculus