1
Integration
An integral is an infinite sum of infinitesimal terms.
An integral is an operator.
I want to convey what I mean by the term “operator” by first
discussing an operator with which you are already familiar, namely the derivative operator.
An
operator is just a mathematical agent that acts on a function to yield another function.
The
derivative operator
dx
d
(the derivative with respect to
x
) acts on a function of
x
to yield another
function of
x
.
So, for instance, if
y
is a function of
x
then we can apply the derivative operator to
it.
We write the derivative of
y
(
x
) as
dx
d
y
or
dx
dy
.
I now write the derivative operator as
°
dx
d
to emphasize the fact that the derivative operator operates on a function and that function goes
where the box is.
The integral operator is an operator too.
Using the “box” notation the integral
operator can be written
∫
°
dx
.
(But see footnote 1.)
The integral operator operates on a
function and that function goes where the box is.
The point here, is not about the order.
(In fact,
it is perfectly okay to write the integral operator as
∫
°
dx
. )
The point is that the integration
operator includes both the
integral sign
“
∫
”
and
the
differential
“
dx
”.
Remember, an integral
is an infinite sum of terms.
An infinite sum of finite terms is infinite.
For the integral to be finite
(and thus meaningful), the terms themselves must be infinitesimal
2
.
So, given the function
f
(
x
) ,
∫
)
(
x
f
is nonsense, whereas,
dx
x
∫
)
(
f
is the integral of
f
with respect to
x
.
Let’s make that a
little more concrete.
Consider the function
2
)
(
x
x
=
f
.
The expression
∫
2
x
is nonsense,
whereas,
dx
x
∫
2
is the integral of
2
x
with respect to
x
.
The bottom line is, if you want to stick
an integral sign “
∫
” in front of something, you better make sure that something has a
differential in it, otherwise, your final expression is nonsense.
Enough about notation and jargon, let’s explore the big idea behind integration by means of an
example.
Consider an object whose velocity as a function of time is given by
2
3
5
1
)
(
t
s
m
t
.
=
v
.
Assume you need to know how far it goes during the first 4
.
0 seconds of its motion.
At first you
might be tempted to use your junior high school formula “distance is speed times time”.
The
“time” is clearly 4
.
0 seconds but what are you going to use for the speed.
If you evaluate
v
(
t
) at
t
=
4
.
0 seconds you get the speed at the 4 second mark but the object is not going that fast for the
whole foursecond time interval from 0 to 4
.
0 s.
Its speed is 0 at time zero, 1
.
5
m/s at 1
s, 6
.
0 m/s
at 2 seconds, 13
.
5 m/s at 3 seconds, and 24 m/s at 4 seconds.
The object is clearly speeding up
during the entire 4 seconds.
You might try using 12 m/s for the average speed calculated by
adding the initial velocity and the final velocity and dividing by two.
But that only works if the
1
I am using the generic variable name
x
.
The variable does not have to be
x
.
It could be
t
or
z
or anything else.
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 Winter '12
 WilliamPattersonIII
 Calculus, Derivative, 0 m/s

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