Chapter
8. More Integrals
Section 8.0 Review of Terms and Meanings
It will be useful for our future work to review terms and meanings that
we established in prior chapters.
Term

Symbol 
Term's Meaning

Moment of a variable 
Any notation that represents a value of a quantity whose value
varies 
A small interval containing a variable's value.
For example,
 Let $x$ represent the number of hours since I set up my camera
on a tripod. Let $x_0$ represent the moment that the camera's
shutter opened to take a picture.
 Let $u$ represent the number of seconds since the starter's
gun fired. Let $u_0$ represent the moment that a runner crossed
the finish line.
 Let $g$ represent the height (in cm) of water in a cylindrical
tank with radius 2 m. Let $V(g)$ represent the volume
($\text{m}^3$) of water in the tank at the moment that its
height is $g$ cm.

Differential 
$$dx, dy, dt$$ 
A small change in a variable's value. 
A variable $y$ changes at a constant rate with respect to a
variable $x$ 
$$y=mx+b$$

Changes in $y$ are proportional to changes in
$x$. Stated symbolically,$$dy=m\cdot dx$$ for some real number $m$.
If $y=f(x)$ and $f(x)=mx+b$, then
$$\begin{align}dy&=f(x+dx)f(x)\\&=\left(
m(x+dx)+b\right)(mx+b)\\&=(mx+mdx +b)(mx +b)\\&=m\cdot
dx\end{align}$$ 
Exact rate of change of a function at each moment of its
argument 
$r_f(x),
\dfrac{d}{dx}f(x),\, f'(x)$

A function f has an essentially constant rate of change
of $r_f(x)$ over a suitably small interval that contains the value
of $x$.
To represent an exact rate of change at a specific value of x,
such as $x=a$, you could write any of$$r_f(a)\qquad \qquad\left.
\dfrac{d}{dx}f(x)\right_{x=a}\quad \qquad f '(a).$$

Exact net accumulation from exact rate of change 
$$\int_a^x r_f(t)dt$$ 
Suppose that:
 Every value of a yettobeknown function $f$ is an exact
amount of a quantity in relation to an amount of another
quantity whose value is $t$.
 $f$ has a known exact rate of change of $r_f(t)$ at each value
of $t$.
Each value of $\int_a^x r_f(t)dt$ for a value of $x$ is the net
amount that has accumulated in $f$ from $t=a$ to $t=x$ as
$f$ changed at the exact rate of $r_f(t)$ over infinitesimal
intervals of length $dt$.
Determining $\int_a^x r_f(t)dt$ from a known rate of change function
$r_f$ is called integrating. The function $\int_a^x
r_f(t)dt$ is called the integral of $r_f$.
In most situations, you do not know the definition of $f$ at the
moment that you write a specific form of $\int_a^x r_f(t)dt$ in a
particular context. Instead, you will know $r_f$ and $a$.

Exact rate of change from exact accumulation 
$\dfrac{d}{dx}g(x),r_g(x),g'(x)$

Suppose that a known function $g$ gives the exact amount of a
quantity in relation to the value $x$ of another quantity. Then:
 Think of $g(x)$ as an amount that accumulated as $x$ varied
from some starting value $a$.
 Since $g(x)$ accumulated as $x$ varied, it accumulated at some
exact rate of change at each value of $x$.
 $g(x)$ can therefore be expressed as $g(x)=\int_a^x r_g(t)dt$
for some yettobeknown exact rate of change function $r_g$.
 $r_g(x)$ is the exact rate of change of $g$ with respect to
$x$ at a value of $x$. $r_g$ is called the derivative of
$g$.
Deriving $r_g$ from a known exact accumulation function $g$ is
called differentiation. The
function $r_g$ is called the derivative
of the function $g$.

Antiderivative 
${\displaystyle \int r_g(t)dt}$ 
The function $g$ has $r_g$ as its exact rate of change function. Any
function $F$ that has $r_g$ as its exact rate of change function
(derivative) is called an antiderivative
of $r_g$. The function $g$ is an antiderivative of $r_g$ by
definition. If $F$ is another antiderivative of $r_g$, then
$g(x)=F(x)+C$ for some constant $C$. 
Closed form definition of a function 

A function is defined in closed form if its definition tells us
how to compute values of the function in a finite number of
operations on values of variables or on values of familiar
functions. Notice that we say that a function's definition
is in closed form. It is an error to say that a function
is in closed form.

Open form definition of a function 

An open form definition of a function is one that gives a
conceptual outline of the function’s meaning, but it does not give
specific instructions for computing the function’s value in a finite
number of steps.
The definition of $F$ as $F(x)={\displaystyle \int_1^x t^3dt}$ is
in open form. The definition of $F$ as $F(x)=\dfrac{x^4}
{4}\dfrac{1}{4}$ is in closed form. It turns out that these open
and closed form definitions of $F$ define the same
function$F(x)$ has the same value in relation to values of $x$
regardless of which definition we use.
The open form definition of $F$ tells us explicitly that $F$ is an
accumulation function. The closed form definition does not tell us
this. But the closed form definition of $F$ does tell us how to
compute values $F(x)$ for values of $x$.

Unsimplified expression 

The expression $x(x1)(x2)$ is unsimplified. In simplified form
it is $x^33x^2+2x$.
It is often useful to leave expressions unsimplified, especially
when you construct them as models of quantitative situations. The
unsimplified expression will remind you of the quantities,
relationships, and inferences you made in modeling the situation.
A rectangular box of height $x$, width $2x$, and length $3x+5$ has a
volume of $x(2x)(3x+5)$. Leaving the expression unsimplified can
remind you of how you came up with the expression. Simplifying this
expression as $6x^3+10x^2$ loses the visual reminder that you
computed the volume by the conceptual formula
$\text{Volume}=\text{height}\cdot\text{width}\cdot\text{length}$.

The Fundamental Theorem of Calculus
The following discussion of the Fundamental Theorem of Calculus
(FTC) relies heavily on you having internalized the terms and
meanings reviewed in the table given above. 
The FTC is fundamental to calculus because it does two things:
 It relates the ideas of exact rate of change function, exact
accumulation function, and antiderivative.
 It provides a way, in principle, to directly compute values of exact
accumulations functions.
It does this by drawing on the fact that rate of change and accumulation
are two sides of a coin. When a quantity changes at some rate, the changes
accumulate. When a quantity accumulates in relation to changes in another
quantity, it changes at some rate with respect to the other quantity.
The FTC's statement is this: Given that $r_f$ is a rate of
change function:
 The accumulation function $\int_a^x r_f(t)dt$ is an antiderivative of
$r_f$.
 All antiderivatives of $r_f$ differ by at most a constant. Put another
way, if $H$ is also an antiderivative of $r_f$, then $$\int_a^x
r_f(t)dt=H(x)+C\text{, for some constant }C.$$It turns out, as we will
see below, that $C=H(a)$, so that$$\int_a^x r_f(t)dt=H(x)H(a).$$
Example
Suppose that values of $r_g(x)=3e^{2x}$ are the exact
rate of change of a function $g$ (which we do not know) in relation to all
values of $x$. We know that
$$\int_2^x 3e^{2t}dt$$
gives the exact net accumulation of g over the interval from $t=2$ to
$t=x$ for any value of $x$.
We also know that any value $g(x)$ is the accumulation of $g$ up to $x=2$
plus the accumulation of $g$ from 2 to $x$. We say this symbolically as
$$g(x)=g(2)+\int_2^x 3e^{2t}dt.$$
Although we do not know the value of $g(2)$ (because we do not know the
function $g$), we can say that
$$(*)\qquad\int_2^x 3e^{2t}dt=g(x)g(2).$$
However, without a closedform definition of $g$ we cannot compute values
$g(x)$ directly! Without a closed form definition of $g$, the best we can do
is to approximate values of $g(x)$ numerically for any value of $x$. GC does
this (creates a numerical approximation) when it evaluates $\int_2^x
3e^{2t}dt$ for any value of $x$.
However, we can make progress by making a connection between the integral
and the notion of antiderivative.
By definition of $g$, $r_g(x)=3e^{2x}$ is the exact rate of change function
for $g$, so $g$ is an antiderivative of $r_g(x)=3e^{2x}$.
But $H(x)=\frac{3}{2}e^{2x}$ is also an antiderivative of
$r_g(x)=3e^{2x}$, because $$\frac{d}{dx}\frac{3}{2}e^{2x}=3e^{2x}.$$
Since all antiderivatives of $r_g$ differ by at most a constant,
$g(x)=H(x)+C$ for some constant $C$. We therefore have $g(2)=H(2)+C$.
By (*), we now have
$$\begin{align}\int_2^x 3e^{2t}dt
&=g(x)g(2)\\[1ex]
&=\left(H(x)+C)\right)\left(H(2)+C\right)\\[1ex]
&=H(x)H(2)\\[1ex]
&=\left(\frac{3}{2}e^{2x}\right)\left(\frac{3}{2}e^{2\cdot
2}\right)\end{align},$$
We now have a way to compute values of $\int_2^x 3e^{2t}dt$ directly for
any value of $x$ because we have represented values of $\int_2^x 3e^{2t}dt$
in closed form!
This is what the FTC does for us. It allows us to represent integrals in
closed form whenever we know a closed form antiderivative of the integral's
rate of change function. Where do antiderivatives come from? They come from
us noting the accumulation functions that produce rate of change functions
of various forms. We will dwell on this more in Section 8.1.
Why integrate from a to x instead of from a
to b?
Many books introduce integrals in the context of what they call definite
integrals. A definite integral is an integral that has numbers as
lower and upper bounds of the integral. The integral $$\int_2^5 t^2dt$$is
a definite integral.
If you think of b as a variable, then integrating from a
to b is the same as integrating from a to x.
The real question then is, "Why use a variable as the upper bound of an
integral?"
The answer is that if we define the function f as $f(x)=\int_2^x
t^2dt$, then $f(5)$ is the same as $\int_2^5 t^2dt$. In other words, every
value $f(x)$ for a value of $x$ is a definite integral. Figure 8.0.1 is GC's
graph of $y=f(x)$ where $f(x)=\int_2^x t^2dt$. We can model situations with
integrals as functions instead of just answering questions about specific
states of the situation.
Figure 8.0.1. GC's graph of $y=\int_2^x t^2dt$. Each point on the
graph has coordinates $\left(x,\int_2^x t^2dt\right)$.
Here is a standard integration problem from another textbook:
A cyclist pedals along a straight road with
velocity $v(t)=2t^28t+6$ mi/hr for $0\le t \le 3$ hours. How far did the
cyclist go in the first hour? In 3 hours? From $t=1$ to $t=3$ hours?
Solution. The textbook's
solutions were $\int_0^1 v(t)dt$, $\int_0^3 v(t)dt$, and $\int_1^3
v(t)dt$.
But we could easily define the function $d$ as $d(x)=\int_0^x v(t)dt$. The
answers would then be $d(1)$, $d(3)$, and $d(3)d(1)$. Moreover, we could
graph $y=d(x)$ to get a sense of how far the cyclist was from her start at
each moment in time. See Figure 8.0.2.
Figure 8.0.2. Graph of a cyclist's distance in miles from start at
each moment in time during first 3 hours of the ride.
Exercise Set 8.0
Unsimplified Expressions
 Silver Ants can run at 50 $\mathrm{\frac{cm}{sec}}$. If a Super Silver
Ant was capable of running at a constant speed from San Diego to New
York City (2760 miles), how many days would it take? The answer is given
below as an unsimplified expression:$$\dfrac{2760}{\dfrac{50}{2.54\cdot
12 \cdot 5280}} \cdot \dfrac{1}{60\cdot 60\cdot 24}$$
Match each quantity in the first list to the expression in the second list
that calculates its measure:
 The number of minutes in a day
 The number of centimeters in a mile
 The number of hours it takes the ant to go from San Diego to New
York City
 The Super Silver Ant's speed in miles/second
 The number of seconds in a day
 The number of seconds it takes the ant to go from San Diego to New
York City
 The number of centimeters in a foot
 The number of days it takes the ant to go from San Diego to New York
City
======================================================

$\dfrac{50}{2.54⋅12⋅5280}$
 $60⋅60⋅24$
 $\dfrac{2760}{\dfrac{50}{2.54⋅12⋅5280}}⋅\dfrac{1}{60⋅60⋅24}$
 $\dfrac{2760}{\dfrac{50}{2.54⋅12⋅5280}}⋅\dfrac{1}{60⋅60}$
 $2.54⋅12⋅5280$
 $\dfrac{2760}{\dfrac{50}{2.54⋅12⋅5280}}$
 $2.54⋅12$
 $60⋅24$

The Sahara Desert has an area of approximately $9,400,000 \,
\mathrm{km^2}$. While estimates of its average depth vary, they center
around 150 m. $\mathrm{One \, cm^3}$ holds approximately 8000 grains
of sand.
 Approximately how many grains of sand are in the Sahara Desert?
Write an unsimplified expression.
 How many Sahara deserts are in the volume of 1 grain of sand? Write
an unsimplified expresson. (Hint: obviously far less than 1 Sahara
Desert)

The Grand Canyon is enormous. It is 433 km long and has an average
depth of 1.6 km. The US Forest Service estimates its volume at 4.17
trillion $(4.17⋅10^{12}) \, \mathrm{m^3}$. A small dump truck can
carry approximately $20.5 \, \mathrm{m^3}$ of sand. Suppose a long
line of dump trucks were to dump a load of sand every 30 seconds. Give
an unsimplified expression for each of the following quantities'
measure:
 The number of cubic meters of sand dumped every hour
 The number of cubic meters of sand dumped every year
 The number of dump truck loads it will take to fill the Grand Canyon
 The number of years it will take to fill the Grand Canyon
 In the figure below, what is the length of the orange segment marked
with a question mark? Write an unsimplified expression. The only numbers
you may use in your answer are "5" "9" and "2."
Function Notation
For Exercises 56: Samuel’s swimming pool has
sloped walls that are rounded at the bottom so that the area of the
water's top surface varies with the amount of water in the pool. Let p
be a function that relates the water’s height in the pool in meters with
the pool’s surface area in square meters. Let w represent the
water’s height in meters. Then p(w) represents the water’s
surface area in square meters at height w.
 Match each statement in the first list to a meaning about the swimming
pool in the second list.
 $p(1.8)$
 $p(c) = 14.3$ for some number c
 $p(0.9)  p0.7)$
 $p(1 + a)  p(a)$
 $p(b + 0.5)  p(b)$
==================================================
 the height of the water level in meters when the surface area of the
pool is 1.8 square meters
 the surface area of our pool is 14.3 square meters when the pool’s
water level is c meters
 the change in the pool’s surface area, in square meters, when the
water level increases from b meters to b+5 meters
 the surface area of our pool is c square meters when the pool’s
water level is 14.3 meters
 the surface area of our pool in square meters when the pool’s water
level is 1.8 meters
 the change in the pool’s surface area, in square meters, when the
water level increases from a meters to 1+a meters
 the surface area of the pool in square meters, when the water level
is 0.2 meters
 the change in the pool’s surface area, in square meters, when the
water level increases from .7 meters to .9 meters
 the height of the water level in meters, when the water level is 1
meter
 the height of the water level in meters, when the water level is b
meters

Essay Question: In the swimming pool context above, a student in
another class claimed that p(0.9)−p(0.7)=p(0.2)
. Do you agree?
Explain why or why not in 23 sentences.
For Exercises 78: Al Unser won the Indianapolis
500mile race in 1987. Juan Pablo Montoya won the Indianapolis 500mile
race in 2015. Unser completed his race in 3.083089 hours. Montoya
completed his race in 3.099033 hours. Imagine that they drove their exact
races against each other and that the race ended when the winner crossed
the finish line. Let t represent the number of hours since their
imaginary race started. Let U be a function that gives Unser's
distance in miles with respect to hours since the race began. Let M
be a function that gives Montoya's distance in miles with respect to hours
since the race began
 Represent the value of each of the following, either using function
notation or not as needed.

The distance that Unser traveled in the first 0.09 hours of the
race.
 The time it took for Unser to complete the race.
 The distance that Montoya travel in the first 97 minutes of the
race.
 The time it took for Montoya to get to the finish line after Unser
crossed the line.
 The distance that Unser traveled from $t=1.192$ to
$t=2.013$ hours
 The distance that Montoya traveled in every oneminute
time period.
 The distance between Unser and Montoya when Unser crossed the finish
line.

Use function notation to represent the following facts with either an
equation or an inequality.
 After 2 hours, Montoya had gone 351.47 miles.
 When the race finished, Unser was ahead of Montoya.
 From the end of the first hour to the end of the second hour, Unser
drove 176.37 miles.
 After c hours, Montoya is 0.47 miles ahead of Unser.
 When $t=1.76$ hours, Unser is at the same place as Montoya was 6
minutes earlier.

The animation below shows two baseball players, one on 1st base
(Player 1) and one on 2nd base (Player 2). Player 1 runs at $18
\mathrm{\frac{ft}{sec}}$. Player 2 runs at $26
\mathrm{\frac{ft}{sec}}$. The distance between bases is 90 ft. The
runners leave their bases simultaneously. Assume that they accelerate
immediately to their running speeds. Runner 2 stops at 3rd base the
moment he reaches it. Runner 1 stops at 2nd base the moment he reaches
it.

Define a function B that gives the distance between Player
1 and 2nd base as a function of the number of seconds since they
began running. You will need to define your function in two parts (a
piecewise function), one part for Player 1’s distance from 2nd base
while he is running and another part for the distance after he
reaches 2nd base.
 Define a function A that gives the distance between Player 2
and 2nd base as a function of the number of seconds since they began
running. You will need to define your function in two parts (a
piecewise function), one part for Player 2’s distance from 2nd base
while he is running and another part for the distance after he reaches
3rd base.
 Define a function D that gives the distance between runners
as a function of the number of seconds since they began running.
(Hint: Think carefully about how many parts this piecewise function
has.)
 Graph your function D in GC. (You will need to rescale your
axes to see GC's displayed graph.) Use GC’s tracing feature (click and
drag on the graph) to estimate the maximum distance and the minimum
distance between runners, and the number of seconds since starting at
which these distances happen. If you define your functions correctly,
the minimum distance will be 73.9973 feet and the maximum distance
will be 94.16403 feet. Print & turn in this GC file.
Moments, Differentials, and Constant Rate of Change

Let t be the time in seconds since I started measuring the volume of
my bacterial culture in mg. Write down 3 moments at t=1.476:

My investment account's balance increased from x=1,475 to x=1,477.
What is the value of dx?

Last week my investment account grew at a constant rate of 4.73
dollars/day. How long did it take my account to grow from 1,475 to
1,477?

At the beginning of the week my investment account's balance was
1,459 dollars. If it grows at a constant rate of 4.73/day, what is my
account's balance at every point in time during the week? Use $n$ to
represent the number of days since the week began.
Reviewing Open Form Definitions & Closed Form Definitions of
Functions

Identify each function definition as either being of an open form or
a closed form. Keep in mind that "$\lim_{x\to0}$" has the same meaning
as "when $x \doteq 0$", etc.
 $A(x)=7x^2+\sin(x)2$
 $g(t)= \int_2^t 4x^2 \, dx$
 $f(u)=\sum_{i=1}^u i^2$
 $t(w) = w^5 + 4w^4 + 7w^3  2.1w^2 + w  1.3$
 $h(z)= 7z^2 + \int_2^4 \sin(x) \, dx$
 $B(t) = 4t + \lim_{x\to 2} \dfrac{(x+3)(x2)}{(x2)}$ OR $B(t) = 4t
+ \dfrac{(x+3)(x2)}{(x2)}$ when $x \doteq 2$
 $n(v) = \sqrt{v+3}$
 $T(y)=\sum_{j=1}^\infty t^2+sin(y)$
 $H(q) = \lim_{x\to0} \dfrac{1}{x}$ OR $H(q) = \dfrac{1}{x}$ when $x
\doteq 0$
 $Q(x)=e^{4x^2}$
 $U(k)=\sum_{i=1}^\infty y^i$
 $h(t)=\mathrm{log}(t)$
 $m(x)=\sum_{j=1}^{25} x⋅j^2$

Essay Question: Explain the difference between an open form
definition of a function and a closed form definition of a function, in
your own words.
Reviewing Antiderivatives
 Fill in the blanks correctly with either "derivative" or
"antiderivative":

$f(x)=\mathrm{cos}(x)$ is the/an ______________ of
$g(x)=\mathrm{sin}(x)$.
 $g(t)=t^3$ is the/an ______________ of $h(t)=3t^2$.
 $g(t)=t^34$ is the/an ______________ of $h(t)=3t^2$.
 $r_f(x)$ is the/an ______________ of $f(x)$.
 $g(t)$ is the/an ______________ of $g'(t)$.
 $k(x)=e^{2x}$ is the/an ______________ of $d(x)=\int e^{2x}⋅dx$.
 $h(x)$ is the/an ______________ of $r_h(x)$.
 $\frac{d}{du}z(u)$ is the/an ______________ of $z(u)$.
 $b'(c)$ is the/an ______________ of $b(c)$.
 $m(x)$ is the/an ______________ of$\frac{d}{dx}m(x)$.
 $\int p(r)dr$ is the/an ______________ of $p(r)$.
Exact Rate of Change at a Moment
 A ball is rolling down a hill, increasing its speed as it rolls. The
distance it has rolled down the plane is modeled by the equation $d(t)$.
 Represent the ball's distance after 3 seconds:
 Represent the ball's velocity after 3 seconds, in three different
ways:

Read the following exchange. Write what you think Bill's final
response should be:
Bill: If $d'(4)=14$, that
means that at exactly 4 seconds after the ball began rolling down the
hill, its instantaneous velocity is $4\mathrm{\frac{m}{s}}$.
Bob: Okay, but what does
that velocity mean, exactly?
Bill: Just that! Its
velocity at that instant is $4\mathrm{\frac{m}{s}}$. What more do you
want?
Bob: If a ball is rolling
at a constant velocity, that tells me something. If it's
moving at a constant $7\mathrm{\frac{m}{s}}$ I know that
$dx=7⋅dt$. I can predict that it goes .7 meters in .1 seconds, and 14
meters in 2 seconds, and $7⋅.218=1.526$ meters in .218 seconds, and
generally $7⋅dt$ meters in $dt$ seconds. But your "instantaneous
velocity" doesn't help me predict anything about how far it goes.
Bill. Sure it does. It's a
velocity, so it has to tell you about how distance changes over time.
Bob: How? As soon as you
pick some change in time $dt$, the velocity has changed.
Bill: *thinks for a second,
then lightbulb goes on* Ahha! Here's how: [finish Bill's statement]
 The ball from the previous problems had its velocity measured at
several moments in time. The data is:
$r_d(0.0)=0$,
$r_d(0.2)=.75$,
$r_d(0.4)=1.05$,
$r_d(0.6)=1.35$,
$r_d(0.8)=1.65$.
Use this information to give an estimate of $d(1)$, the distance the
ball has rolled after one second.