Section
6.1 Approximate Rate of Change in Open Form
from Exact Accumulation in Closed Form
Review of Chapter 5
Chapter 5 addressed the problem of representing
an accumulation function in open form when we have its rate of change
function in closed form. Chapter 5 addressed the first foundational
problem of calculus:
You know how fast a quantity is
changing at every moment; you want to know how much of that quantity
there is at every moment.
We will work backward from where we ended in Chapter 5 to develop techniques
that start with an exact accumulation function in closed form and derive its
exact rate of change function in closed form. We will then close the loop
regarding the relationship between exact accumulation functions in closed
form and exact rate of change functions in closed form. We do this with the
aim that we can recover one whenever we have sufficient information about
the other.
Figure 6.1.1 repeats the overview given in Chapter 5 of the journey from
rate of change functions in closed form to accumulation functions in open
form.
Figure 6.1.1. Calculate A(x) by assuming
that r is constant over intervals of size $\Delta x$.
Figure 6.1.1 is a reminder of the method we developed in Chapter 5.

Start with an exact rate of change function $r_f$ in closed form. Its
values gives the rate of change at every moment of an (unknown) exact
accumulation function f.

Use the meaning of rate of change at a moment to assume that
$r_f$ is essentially constant over small intervals of size $\Delta x$
that contained each moment.

Let dx and dy vary over each $\Delta x$sized
interval. Approximations to the exact accumulation over intervals are
then continuous. Each value of dx produces a value of
$\mathrm{d}y=m \mathrm{d}x$ as dx varies, thus defining the
approximate net accumulation function A as a piecewise linear
function that approximates the exact net accumulation function $A_f$.
The meaning of $A_f(x)$ is that any value of $A_f(x)$ is the net
accumulation (change) in the value of f from an initial
value a to any value of x.

When $\Delta x$ is so small that making it smaller has no appreciable
effect on our estimates of the net accumulation from a to x,
we said $A_f$ was the exact net accumulation function for $r_f$. We
represented $A_f$ in open form as $A_f(x) = \int_a^x r(t)dt$.

The relationship between $A_f$ and f is that for any given
value of a, $A_f(x)=f(x)f(a)$. Said another way,
every value of $A_f(x)$ gives us the net change in f from a
to x, and the net change in f from a to
x is $f(x)f(a)$. Therefore $f(x) = f(a) + A_f(x)$, which
says that f(x)—the accumulation in f
up to a value of x—is made by the accumulation in f
up to a, plus the accumulation in f from a
to x.
6.1.1 Rate of Change from Exact Accumulation
Chapter 6 will begin from where Chapter 5 ended to address the second
foundational problem of calculus:
You know how much of a quantity there
is at every moment; you want to know how fast it is changing at every
moment.
Reflection 6.1.1. Compare the first
and second foundational problems of calculus. How are they different? How
are they related?
You have dealt with exact accumulation functions in closed form since at
least 7th grade, without knowing it. Any function g that gives an amount
of quantity A in relation to an amount of quantity B is an exact
accumulation function. Any value $g(x)$ is the exact
accumulation of the quantity it represents as x varied up to its
present value.
The function f defined as $f(x)=f(a)+\int_a^x \cos(t)dt$ is an exact
accumulation function, in open form. If the function g
defined as $g(x)=x\cdot 2^{x}$ relates measures of two quantities, then g
is an exact accumulation function, in closed form.
A function’s definition is in closed form
when it is expressed explicitly as a finite number of elementary
operations on numbers, variables, or familiar functions.
A function is defined in open form when
it its definition is given in such a way that it is not apparent how
to compute a result for a specific argument. The function f
defined as $$f(x)=\sum_{k=0}^\infty \frac{x^k}{k!}$$ is in open form
because you cannot compute values of it in a finite number of steps.
The function g defined as $$g(x)=\frac{f(x+h)f(x)}{h}$$
is expressed in open form because the operations of subtraction and
division involve a function that is not specified explicitly and a
parameter that is yet to be given a value. The function F
defined as $$F(x)=F(a)+\int_a^x(t^2t)dt$$ is in open form because
you cannot compute values of it in a finite number of steps.

Example: Area of a square as an accumulation of changes
in area
$f(x) = x^2, x \ge 0$ is an amount function. Every value of f gives
an amount of area enclosed within a square of side length x. When
the square’s side length is 13.327 inches, its area is $f(13.327)=13.327^2$
, or 177.609 $\mathrm{in}^2$. The area of a square with side length x
inches is $f(x)=x^2$ square inches. .
We can also think of f as an accumulation function. For any given
side length x, we can think of the square’s side as having grown in
length from $t=0$ to $t=x$. Thinking of the side length varying means that
the square itself varies. The square’s area will grow as t varies
from 0 to x inches. Figure 6.1.2 shows f(15) in terms of
the area of the square accumulating as its side length varies from 0 to 15.
Figure 6.1.2. The square's area accumulates as its side's length grows
from 0 to 15 inches.
Do not just think of area as a function of side length. Rather, think of
area as an accumulation of changes in area that are made at some
rate with respect to changes in side length as the side length varies from
$t=0$ to $t=x$. In other words, think of $x^2$ as
$$x^2 = \int_a^x r_f(t)dt$$
where $r_f(t)$ is a function that gives the exact, momentary rate of change
of the square’s area with respect to its side length for every value of its
side length. However, we do not know the definition of $r_f$ so that $x^2 =
\int_a^x r_f(t)dt$.
The aim of this chapter is to develop methods that will allow us to start
with any accumulation function, in closed form, and derive the rate of
change function from which it came, if there is one, also in closed form.
Then you will learn that an accumulation function’s rate of change function
gives you many insights into a situation that the accumulation function
models.
You can reconceptualize any function whose values give an exact
amount of something as being an exact accumulation function. If
$g(x)$ gives an exact amount of something, then we can
reconceptualize each value of g as an amount that has
accumulated from some starting point, at some rate of change, over
some interval. In other words, if g is an amount function,
then $g(x)=g(a)+\int_a^x r_f(t)dt$ for some exact rate of change
function $r_f$. 
Not every function can be thought of as being built from a rate of change
function. The Blancmange and Weierstrass functions from Section
4.8 had no moments at which they changed at a rate that was
essentially constant. It is therefore not possible to think of any value of
either function as having accumulated. We will address this matter in more
detail later. For the meantime, we will concentrate on functions that can be
viewed as accumulation functions that come from some quantity accumulating
at some rate of change at almost every moment of their independent
variables.
Reflection 6.1.2. Remind yourself of what the word “exact” means when
we speak of an exact rate of change function and an exact
accumulation function.
Reflection 6.1.3: Remind yourself of what the terms “open form” and
“closed form” mean when we speak of a function represented in open
form or a function represented in closed form.
6.1.2 Reversing the Process of Deriving Exact
Accumulation from Exact Rate of Change
Overview of Accumulation from Rate
The focus of Chapter 5 was to develop and understand the mathematics of
determining an exact accumulation function from an exact rate of change
function given in closed form. This endeavor addressed the first
foundational problem of calculus:
You know how fast a quantity is changing
at every moment; you want to know how much of that quantity there is at
every moment
Our efforts in Chapter 5 involved this process:
This process ended with the integral, an open form representation of the
net change in the exact accumulation function f from a
to x that is built from $r_f$:
$A_f(x)=\int_a^x r_f(t)dt$
The integral function was the culmination of the 4step process. It achieved
the desired result in one step:
Reversing the Process
To get an exact rate of change function in closed form from an exact
accumulation function in closed form, we must reverse the process that we
developed in Chapter 5. In doing so, we will address the second foundational
problem of calculus:
You know how much of a quantity there is
at every moment; you want to know how fast it is changing at every
moment.
The method for reversing the process developed in Chapter 5, of going from
an exact rate of change function to an exact accumulation function, entails
these steps:
Like before, our final goal is to develop a method that accomplishes the
three steps listed above in one step. That is, we aim to derive the exact
rate function directly from a given exact accumulation function:
Here is an outline of our quest:
 We are given a function f in closed form.
 Interpret f as being an accumulation function—that each value
$f(x)$ represents an amount that has accumulated at a rate of change
$r_f(t)$ at each moment t as t varied up to the moment
x. Therefore,
$$f(x)=\int_a^x r_f(t)dt$$
 Our goal is to recover $r_f$ in closed form.
Reversing the accumulation process requires that we have the accumulation
process itself firmly in mind. You would be wise to revisit the introduction
and its description of Figure 6.1.1 before continuing.
Figure 6.1.3 gives a visual summary of the approach we will develop to
approximate an accumulation function’s rate of change function, which we
explain after the animation.
Figure 6.1.3. Moving from exact accumulation function to its approximate
rate of change function.
The highlighted points on the graph of f in Figure 6.1.3 are the
values of f at the left end of each $\Delta x$interval. Each
interval has a width of $\Delta x$. The piecewise linear function (call it g)
over each $\Delta x$interval in the left graph of Figure 6.1.3 produces the
same change over that interval as does f. Its constant rate of
change over any $\Delta x$interval containing a moment of x will be
the value of $r(x)$ once we define r appropriately.
Therefore, the conceptual definition of g is
$$g(x) = r(x)\left(x\mathrm{left}(x)) +
f(\mathrm{left}(x)\right)$$
Equation 6.1.2. The definition of the
piecewise linear function g.
The graph of g over any $\Delta x$interval passes through the
points $(\mathrm{left}(x), f(\mathrm{left}(x))$ and $(\mathrm{left}(x+\Delta
x), f(\mathrm{left}(x+\Delta x))$, and has a constant rate of change over
that interval. Therefore the rate of change of g over any $\Delta
x$interval containing a moment $x_0$ of x will be the change in f
that occurred over that $\Delta x$interval divided by $\Delta x$—the change
in x over that interval. This rate of change is the value of
$r(x_0)$, the rate of change of g (the piecewiselinear approximation of
f) over the interval containing $x_0$.
In symbols, $r(x_0)$, the constant rate of change of the function g
over any $\Delta x$interval containing the moment $x_0$, is
$$r(x_0) =
\frac{f\left(\mathrm{left}(x_0))+\Delta x\right) 
f\left(\mathrm{left}(x_0)\right)}{(\mathrm{left}(x_0)+\Delta
x)\mathrm{left}(x_0)} = \frac {f(\mathrm{left}(x_0)+\Delta x) 
f(\mathrm{left}(x_0))}{\Delta x}$$
Equation 6.1.3. The constant rate of change of g over any $\Delta
x$interval.
Therefore, the function r, the
function whose values approximate $r_f$, the (yet unknown) exact rate of
change function for f, can be defined as
$$r(x) = \frac {f(\mathrm{left}(x)+\Delta
x)  f(\mathrm{left}(x))}{\Delta x}$$
Equation 6.1.4. The approximate rate function r(x).
6.1.3 Your Turn
Part 1 — By hand
Use a printout of this
graph with the following activity. Note: An interval $a≤x≤ b$ is also
represented as $[a,b]$.
 The ordered pairs $(1, 1.7)$, $(0.1, 2.7)$ and $(0.8, 2)$ are on the
graph of $y = f(x)$, $‑1 ≤x ≤ 0.8$. $\Delta x = 0.9$.
 Highlight the two $\Delta x$intervals of length 0.9 that comprise
the interval $ [‑1, 0.8]$.
 How much does f change over each interval? Represent each
change symbolically and numerically.
 Does f vary at a constant rate of change or at a changing
rate over each interval? How can you tell?
 What is the constant rate of change over each $\Delta x$interval
that another function g would need so that it produces the same change
over a $\Delta x$interval as produced by f? Represent each
constant rate of change symbolically and numerically.
 Define the function g so that its graph coincides with the
piecewise approximation to the function f over the interval
$[0.1, 0.8]$.
 Define the function r so that its graph represents the rate
of change of g at each moment of x in $[0.1, 0.8]$.
Sketch a graph of r over the interval $[0.1, 0.8]$.
 What is the numerical value of $r(0.25)$? What does the value
$r(0.25)$ represent? Is it the same as $f(0.25)$? Explain.
Part 2 — Using GC (Windows users — type “Dx” in place of “$\Delta x$”)
Save your GC file with this work. You will use it in the next exercise
set.
The function f in Part 1 is $f(x)=e^{\cos x}$.
 In what follows, Mac users should type “\optionj x\” to get
“$\Delta x$”. Windows users should type \Dx\ whenever you see “$\Delta
x$”.
 Save your file — you will use it in Exercise Set 6.1.
 Define f as $f(x)=e^{\cos x}$.
 Enter $a = 1$, \$\Delta x$\ = 0.2 on one line (type the comma, too).
 Define the function left as left$(x) = a + \Delta x
\left\lfloor \dfrac{xa}{\Delta x} \right\rfloor$. Type “floor
(xa)/$\Delta x$” to get $\left\lfloor \dfrac{xa}{\Delta x}
\right\rfloor$. Remind yourself of what a value of left(x) means
for given values of a and $\Delta x$.
 Define g and r as in Equations 6.1.2 and 6.1.4.
 Enter $y=f(x), a ≤ x$ on one line.
 Enter $y=g(x), a ≤ x ≤ n$ on a second line.
 Enter $y'=r(x'), a ≤ x' ≤ n$ on a third line. The prime ($’$)
notation tells GC to display the graph in a second pane.
 Click “n” on the n slider. Set lower and upper values to 1
and 8, respectively.
 Click the play button on the n slider. You should see a replay
of Figure 6.1.3, except that you might see calculator drool in your
graph of $y'=r(x')$.
 Explain to yourself what is happening on your screen in terms of the
functions and graphs being displayed.
Exercise Set 6.1.3
 For
each function in ae,

Let $\Delta x = 0.25$ and a=1

Evaluate r for three values of x in
the intervals (2.25, 2.5] and (3.75, 4.0]. What do
you notice? Why?
a) $f(x) = x^2+2x5$
b)
$f(x)
= \sin(x)$
c)
$f(x)
= 3e^{5x}$
d)
$f(x)
=
\sqrt{x^25}$
e)
$f(x)
=
\dfrac{(x4)(x+2)}{(x+2)}$

What
did each value of r that you found in
Exercise 6.1.2 #2 mean? Do this for each particular
value you found until you can give a general meaning
that covers them all.
Use your file from Part 2 above to
answer exercises 36, but hide the graphs. The following exercises
assume that you defined f, g, and r
successfully.
 In GC, enter $g(0.5)$. You should get 2.3973 (to 4 decimal places).
What does this number mean?
 In GC, enter $r(0.5)$. You should get 1.14654 (to 4 decimal places).
What does this number mean?
 In GC, enter $y=g(x), a ≤ x ≤ a + \Delta x$. Interpret this graph.
 In GC, enter $y'=r(x'), a ≤ x' ≤ a + \Delta x$. Interpret this graph.
 Open a new GC file. In each of a.  d., you are given a function f.
Define g and r as in Eq 6.1.2 and Eq 6.1.4. Enter
$g(c)$ and $r(c)$ on separate lines for three different values of c.
Interpret the values that GC displays each time.
 $f(x) = \sin(x)$
 $f(x) = \sin(x) e^{\cos x}$
 $f(x) = x(x1)(x+1)$
 $f(x) = x \sqrt{\sin x}$. For what values of x are g
and r defined?
 The figure below shows graphs of f (in aqua),
g (in magenta ),
and r (in red ),
with a point on the graph of r highlighted. What does the
highlighted point represent about the graph of g?
 Give a detailed explanation of how the animation in Figure
6.1.3 works.
6.1.4 Smoothing the Rate of Change Function
The method outlined in Section 6.1.2 above, Reversing the Process,
indeed produces an approximate rate of change function r for an
exact accumulation function f. However, if f is not linear,
then r is discontinuous from one $\Delta x$interval to the next.
Consider the diagram below and think about why r is discontinuous
from one $\Delta x$interval to the next. What would the displayed graph of
r look like over the interval [1,0.8)?
The reason for the discontinuities in $r(x)$ is because when f is
not linear, values of $r(x)$ will “jump” from a constant value within one
$\Delta x$interval to a different value at the beginning of the next
$\Delta x$interval.
We can solve the problem of discontinuities in the values of r
with a simple change of technique. Instead of chopping the
domain of x into successive intervals of length $\Delta x$, we can
slide one interval of length $\Delta x$ along the axis. The value of r
for any value of x will be the constant rate of change over the
interval $[x, x+\Delta x]$ as the value of x varies. Figure
6.1.4 illustrates this change of technique. It starts with the technique of
using fixed intervals of length $\Delta x$. It then displays the “sliding
interval” technique, using intervals of length 1, 0.1, and 0.01.
Figure 6.1.4. Start with fixed intervals of length $\Delta x$; then use
a sliding interval of length $\Delta x$. The fixedinterval rate of
change function (in ) is discontinuous; the slidinginterval
rate of change function (in
Reflection 6.1.4: Why does the graph of the approximate rate of change
function defined with a sliding interval always go through the left
endpoints of the parts of the fixedinterval rate of change function’s
graph? Does this happen also in the animation when $\Delta x$ = 0.01?
Reflection 6.1.5: Consider again why in Figure 6.1.4 the graph of the
fixedinterval rate of change function (in magenta)
is discontinuous, while the graph of the sliding interval rate of change
function (in blue) is continuous.
The sliding interval technique has a huge benefit. Not only does the sliding
interval produce a continuous function, the definition of r is
simpler. Before introducing sliding intervals, the definition of r
as the approximate rate of change function for f is repeated below
as Eq. 6.1.5:
$$r(x) = \frac{f(\mathrm{left}(x))+\Delta
x)  f(\mathrm{left}(x))}{(\mathrm{left}(x)+\Delta x)\mathrm{left}(x)} =
\frac {f(\mathrm{left}(x)+\Delta x)  f(\mathrm{left}(x))}{\Delta x}$$
Equation 6.1.5. The approximate rate
function defined with static $\Delta x$ intervals.
With the sliding secant, the definition of r is of the constant rate
of change over the interval that starts at the current value of x,
not left(x), and which has a length of $\Delta x$. The
simplified definition of r is
$$r_2(x) =\frac {f(x+\Delta x)  f(x)}{(x
+ \Delta x)  \Delta x}$$, or simpified,
$$r_2(x) =\frac {f(x+\Delta x)  f(x)}{\Delta x}$$
Equation 6.1.6. The approximate rate function defined with a sliding $\Delta
x$ interval.
Reflection 6.1.6: Will the sliding
interval produce a smooth graph for $r_2(x)= \frac {f(x+\Delta x)f(x)}
{(x+\Delta x)x}$ even when
f is discontinuous? Define
f
as shown below in GC and let $\Delta x=1.0$. Graph $y = f(x)$ in GC
and graph $y'=r_2(x')$ in GC's right pane. Why does the graph of $r_2$
appear as it does? (
Hint: Imagine an interval of length 1.0 sliding
along the xaxis in the left graph. If you have trouble imagining
this, then download this file
or view this
video.)
$$f(x)= 0.5 + \left\lbrace \begin{align} &x \text{ if } x \lt 1 \\
&x^2 \text{ if } x \ge 1 \\ \end{align} \right.$$
As a look ahead, Figure 6.1.5 compares the graphs of $r_2$ and the exact
rate of change function $r_f$ as $\Delta x$ gets very small. Visually,
the graph of r appears to move atop the graph of $r_f$. This
suggests that $r_2$ produces good approximates of $r_f$ for sufficiently
small values of $\Delta x$.
Figure 6.1.5. Left: The graph of f, approximated by
$g(x)=r_2(x)(xa)+f(a)$. Right: the graph of $r_2$ (in
blue), produced by a sliding interval of length $\Delta x$,
appears to coincide with the graph of the actual exact rate of change
function $r_f$ (in magenta) when
values of $\Delta x$ become sufficiently small.
The right side of the animation in Figure 6.1.5 might suggest to you that
the graph of $r_2$ moves sideways as $\Delta x \to 0$. Examine the
definition of $r_2$ in Equation 6.6. For each value of x,
$f(x+\Delta x)$ approaches the value of $f(x)$ as $\Delta x$ gets smaller.
Therefore, the values of $r_2$ must be moving vertically (up or down) as
values of $\Delta x$ become smaller.
Figure 6.6 confirms that points on the graph of $y =r_2(x)$ move vertically
as $\Delta x \to 0$. The graph of $r_2$ does not move sideways.
Rather, it changes shape because the graph's points move up or down
as the value of $\Delta x$ changes.
Figure 6.1.6. Right: the graph of $r_2$ (in blue),
produced by a sliding interval of length $\Delta x$,
appears to coincide with the graph of the actual
exact rate of change function $r_f$ (in magenta)
when values of $\Delta x$ become sufficiently small.
However, points on the graph of $r_2$ move
vertically as the value of $\Delta x$ changes.
We
will use the definition of $r_2$ as our
standard definition of r, the approximate
rate of change function for a given
accumulation function. To repeat, from here
forward, $$r(x)=\frac{f(x+h)f(x)}{h}$$ will
be our standard definition of a function f’s
approximate rate of change function. 
Alert: From now on, we will use “h” in
place of “$\Delta x$” so that Mac and Windows users
can type the same thing.
Exercise Set 6.1.4
1. Go through the following steps for both (a) and (b).
i. Examine the displayed graph of f.
ii. Imagine what the graph of $r_f$ might look like
based on the behavior of f’s graph.
iii. Make a handdrawn sketch of the graph you
anticipate for $r_f$.
iv. Enter the definition of f and r
into GC and compare GC’s graph of r with your
handdrawn sketch.
v. Click on a point of r’s displayed graph.
Write the point’s coordinates and an explanation of
what they mean.
2. The total cost C of producing p units
of a widget depends upon p. Economists use
the term marginal cost for the rate of change of
the total cost with respect to the number of units
produced. The function C, defined as
$C(p)=200+5 \sqrt{30p}$, models the cost of producing p
items at a particular factory.
a) In GC, graph $y=C(x)$ (scale the axes
appropriately). What do you notice about the cost of
goods as p increases?
b) Define $r(p)$, the approximate rate of change
function for C (consider what would be a
logical value for h).
c) Enter $y’=r(x’)$ so that GC displays the graph of r
in its right pane. What is the approximate marginal
cost when $p=15$?
d) How does the marginal cost vary as the number of
units produced varies? Why does this make sense?
(Adapted from Kline p. 35)
3. A species of bacteria is being researched in
laboratory settings. Researchers are testing a new
disinfectant. They start a bacterial culture with a mass
of 5 grams and allow the culture to grow for several
hours. Then they release the disinfectant after 20
hours. The bacteria will begin to die off once the
disinfectants are released. The function b,
defined below, models
the number of grams of bacteria in
the petri dish t hours
after the culture is started.
$$b(t)=5(1.25^{\frac {t} {3}}d(t)) \text {
if } t \ge 0$$, where
$$d(t) = \left\lbrace \begin{align} &1.4^{\frac {t20} {2}}1 \text{
if } t \ge 20 \\ &0 \text{ if } t \lt 20 \\ \end{align} \right.$$
a) Trace the graph of $y=b(x)$ in GC to determine the
culture's maximum mass. After how many
hours does the culture reach this maximum?
b) According to this model, when will the bacteria
have vanished?
c) In GC, graph y = r(x),
the approximate rate of change function for b. What
is true about the value of r(x)
at the moment that the bacterial culture reaches its
maximum mass? Why is it reasonable that r(x)
would have that value?
4. Save your GC file for this exercise. You will use it
in the next exercise.
For
each of a) through c), use GC to graph each function’s
approximate rate of change function. Let
$h=0.01$. Each of these graphs will approximate
the graph of a function that should be familiar to
you. What is the familiar function in each case?
a)
$f(x) = \mathrm{ln}(x)$
b)
$f(x) = \cos(x)$
c)
$f(x)
= e^x$
5. Click on a point on each of the approximate rate
graphs you made in 6.a, 6.b, and 6.c. Record its
coordinates. State what the point’s ycoordinate
means in relation to the value of its xcoordinate.
6.1.5 Relating Rate of Change and Accumulation
In section 6.1.4 we determined that the function
$r(x)=\frac{f(x+h)f(x)}{h}$ approximates an exact
accumulation function f’s exact rate of change
function. Each value $r(x)$ is an approximation to f’s
rate of change at a moment x. If this is indeed
true, then the function F defined as
$F(x)=\int_a^x r(t)dt$ for some number give should
the net change in the function f from
a to x. We will check to see that this is true
and explore why it is true.
First let’s consider the meaning of a in $\int_a^x
r_f(t)dt$.
It is the initial value of x from which a change
in exact accumulation is measured. Since f is an
exact accumulation function built from its rate of
change function $r_f$, the accumulation $\int_a^x
r_f(t)dt$
is starting at $f(a)$. Likewise, we know from Chapter 5
that $\int_a^x
r_f(t)dt = f(x)f(a)$
and therefore $f(x)=f(a)+\int_a^x
r(t)dt$. These observations will be important right
away.
Let the function f be defined as $f(x)=x^2+1$.
We can think of it as an exact accumulation function,
such as the area of two squares, one of side length 1
and the other of side length x. f's
approximate rate of change function r is defined
in open form as $$r(x)=\frac{f(x+h)f(x)}{(x+h)x},$$ or
more simply as $$r(x)=\frac{f(x+h)f(x)}{h}.$$The
smaller the value of h, the better the
approximation that r is to $r_f$.
Figure 6.1.7 examines the graph of the function F
defined as $F(x)=\int_a^x
r(t)dt$
and compares it to the graph of the function f.
Figure 6.1.7. Exploring the relationship between
accumulation function f and accumulationfromrate
function F: (1) r yields
an accumulation function that is a close
approximation of the function f, so the rate of
change function r seems to be a good approximation
of f’s exact rate of change function $r_f$. (2)
Values of F and f always differ by f(a) for small
values of h regardless of the value of a.
The animation in Figure 6.1.7 suggests that r
defined as $r(x)=\frac{f(x+h)f(x)}{h}$,
for small values of h, approximate f’s
exact rate of change function $r_f$. At the beginning of
Figure 6.1.7 we graphed $y = F(x)$, where $F(x)=\int_a^x
r(t)dt$
and $a = 2$. The graph appeared to be congruent to the
graph of $y = f(x)$ but differed by a constant (it was
shifted vertically). When we graphed $y = f(a) + F(x)$,
the two graphs coincided. This says that f, the
exact accumulation function we started with, can be
recovered by integrating r. This means that r
indeed is an approximation to $r_f$, f’s exact
rate of change function.
The function r is defined in open form. At
this moment, we
do not have a closed form representation of $r_f$. In Section
6.2 we will learn to derive
closed form
representations of exact rate of change functions when
given their exact accumulation functions in closed form.
Exercise Set 6.1.5
Exercises 18 refer to the GC file displayed in
Figure 6.1.7.

In
GC, enter the statements listed in Figure 6.1.7. Now
do this:

Make
a list on
a sheet of paper
of 5 other definitions for f,
each of your choosing.

Change
the definition of f in Figure 6.1.7 to
each definition you listed, one by one.

Answer
this question: Why does the definition of f
not matter in terms of whether the graph of
$y=f(x)$ coincides with the graph of
$y=f(a)+\int_a^x r(t)dt$?

Change the second line to h =
slider(0.001,1). Change the value of h.
How does the value of h affect the graph
of F?

Enter
$y'=r(x')$. This makes GC display a graph of r
in the right graphing pane. Click on a point on
the graph of r. Write the coordinates of
that point and explain what those coordinates
mean.
Questions 7  11 are based on Figure
6.1.7.

At the beginning of the animation, the graph of
$y=F(x)$ moved up to touch the xaxis as
the value of a increased to 0, then the
graph moved down as the value of a
increased beyond 0. Why did the graph not rise
above the xaxis as the value of a
increased?

The value of a is 2 at the beginning of
the animation. When $a=2$ the graphs of $y=F(x)$
and $y=f(x)$ differ by a constant, meaning that
there is some number C so that
$f(x)F(x)=C$. What is C?

Why did the graph of $y = F(x)$ move as it did
when the value of the a slider changed.
Does the value of a effect r’s
approximation of $r_f$?

Suppose that $c < a$. Draw a diagram to show
that $f(a)+F(c)$ still represents the accumulation
of f up to the value of c.

$F(x)=\int_a^x r_f(t)dt$. The graph of
$y=f(a)+F(x)$ coincided with the graph of $y=f(x)$
regardless of the value of a. Why? What
does this imply about $\int_a^x r_f(t)dt$?