1. A history of genius
There is quite a big debate as to who invented calculus. The most popular answer
is Sir Isaac Newton. However, their is a less-famous mathematitian, Gottfried
Leibniz, who also is credited with the invention. Newton was the first to patent
the idea, but who actually was the first to invent it was up to debate. (Travel tip
: When in Germany, Leibniz is the inventor of calculus, period. Seriously, they’re
a little touchy on the subject.)
Around 1675, Leibniz had a breakthrough in his study of mathematics. A
colleague posed a problem to him. Place a pocket watch on a table with the chain
pointing straight down. Now hold the other end of the chain and pull it in a straight
line to the right. Watch the path of the watch as it is pull by the chain. Consider
figure 5.19 in your text for help visualizing it. The path us shaped somewhat like
the graph of the function f (x) = x1 for x > 0. The problem is, find the area under
the curve created by the watch (and bounded below by the horizontal line created
by the other end of the chain).
Leibniz noticed something novel about this problem. Consider the figure again.
Suppose you freeze time at a certain point in the process. The chain will be
diagonal and straight, and the watch will be at a certain point on its path of
motion. Look at how the chain interacts with the curve at the point where the
watch is. Do you see it? The straight chain is tangent to the path of the watch.
Also note that by “freezing” the process repeatedly, will create triangles with one
side the length of the chain and a one side a certain horizontal length taken by the
end of the chain. The sum of these triangles appoximates the area under the curve.
If we imagine “freezing” more and more often, the approximation gets better. If
we consider the limit of infinitessimally small stages, we triangles will sum up to
the area.
This is very hard to visualize and explain only in words. Once again, the figure
should help visualize what we are talking about. The important thing here is
that Leibniz noticed something that will turn out to be fundamental to the study
of calculus (pun mostly intended). Leibniz noticed that the tangent lines to the
curve and the area under the curve are closely related. This should be somewhat
surprising. Area is a measurment of size and a function from the plane onto R,
while derivatives are rates of change. Those are not obviously closely connected.
2. Brief notation
Before we move on, we need a piece of notation that Leibniz created. Consider
a nice, smooth function f (x). For simplicity, say f (x) > 0, and let’s consider
1
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it on an interval [a, b]. Recall, that we can approximate the area under a curve
with rectangles. We do this because the areas of rectangles are very easy to
compute. Suppose we break up the interval [a, b] in small pieces, each with length
say dx. Create rectangles using these horizontal segments of length dx as the
horizontal sides. The vertical lengths will be the value determined by f (x) at
the left endpoint. So the area of each rectangle is y ⋅ dx, where y is the value of
f (x) at the left endpoint of the segment. The sum of the areas of these rectangles
approximate the area under f (x). Consider the figure below for visualization.
Recall that as we use more and more rectangles, our dx gets smaller and smaller.
So the rectangles get thinner. Also recall that the sum of the areas of these
rectangles more closely approximates the area under f (x).
If we let ∑ y ⋅ dx is the sum of the area of all the rectangles and A is the area
dx
under f (x), we have that A = lim ∑ y ⋅ dx. Since this is a pain to write, we use
dx→0
dx
some notation.
A=∫
b
a
y ⋅ dx.
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This notation just means the area under the curve on the interval [a, b]. This
elongated S is stand for the sum or summa. This notation is simply a shorter way
to write the limit of the sums of the rectangles considered above.
3. Leibniz’s method
We know go through a process very similar to how Leibniz found this strong
connection between area and tangents. In the next section, we will give a name to
this connection and see all the wonderful ramifications of it.
Consider a curve A that is strictly above the x-axis and strictly increasing on
an interval [a, b]. (It doesn’t have to be, but this case makes visualizing the
problem easier. We can remove the restriction and consider area under the x-axis
as negative area.) We do require that the curve is differentiable. “ Differentiable”
means that the curve has a derivative everywhere. x1 is not differentiable because
f ′ (0) is not defined. Now consider a second curve C. Unlike A, C will not be
arbitrary; it will be very dependent on A. We will also consider C on the same
dz
is the slope
interval [a, b]. Let P = (x, z) is a point on the curve A and say
dx
of the tangent at P . We let Q = (x, y) be the corresponding point on C, where
dz
. That is, the y values on the curve C are defined to be exactly the slopes
y=
dx
of the curve A. Notice that P and Q have the same x values. Again to reinforce,
the y values of C are determined by the slopes of the corresponding tangents on
A.
For example, if A was the curve f (x) = 21 x2 , the corresponding curve C would
be y = x. This is because at a certain point (x, z) on 21 x2 , the slope at that point is
given by f ′ (x) = x. So the corresponding point Q will be (x, y) where y = f ′ (x) = x.
Consider figure 5.21 in your text for help visualizing this.
Notice that by straight multiplications, dz = y ⋅ dx. dz is called the differential
of z and dx is the differential of x. Let’s concentrate on the curve A for now.
Consider the figure again. At the point P , imagine the tangent line intersecting
dz
the curve at P . Recall that it’s slope is dx
. Take a small piece of this diagonal
line and form a right triangle where this line fragment is the diagonal. Notice that
no matter the size of the diagonal we choose, the opposite and adjacent have the
dz
same ratio, dx
. Now let dz be the vertical value of the opposite, and let dx be the
horizontal value of the adjacent.
dz
Remember that by definition at the point Q = (x, y) on C, y = dx
, which implies
that dz = y ⋅ dx. Now consider the rectangle determined by the width, dx and
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height y under the curve C. That is, it has the same width as the triangle we drew
dz
under A, and it’s height is determined by the value y = dx
. What is the area of
this rectangle? Well, it is just base times height, so the area is y ⋅ dx. Again, recall
that y ⋅ dx = dz, where dz is the vertical value of the triangle. The important part
is that the area of this rectangle is exactly equal to the vertical value dz. So we
already see that the area under the curve C is somewhat related to the derivative
of the curve A.
Of course, we will not be satisfied with just one rectangle and triangle. Let’s pick
another point P ′ on A such that P ′ = (x + dx, dz ′ ). It will have a corresponding Q
′
on C with Q = (x + dx, dz
dx ). We go through the process again, drawing the triangle
and finding the area of the corresponding rectangle under C and again see that
the area of this second rectangle will be equal to the vertical value of the opposite
of this second triangle. We can do this for as many points of P as we wish.
Notice that the rectangles that we determine approximate the area under the
curve C. If we let the dx values of the triangles under A be arbitrarily small, the
resulting rectangles under C will give better approximations for the area under C.
Now for the punchline. If f (x) is the curve A, let c = f (a) and d = f (b), where
a and b are the endpoints of the interval. Now consider the sum of all the vertical
values of the triangles dz. What is their sum? It is exactly d − c!. So if we let dx
go to 0 and consider the sum of the corresponding rectangles under C, we see that
the area under C is exactly the sum of all the vertical values of the tiny triangles
b
under A, d − c. If ∫ y ⋅ dx is the area under the curve C on the interval [a, b], we
a
b
have that ∫ y ⋅ dx = d − c.
a
Next time, we will see the beauty of this result.