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tom.h.wilson
[email protected]
Dept. Geology and Geography
West Virginia University
Tom Wilson, Department of Geology and Geography
Discussion of problems in Chapter 8
8.13
Tom Wilson, Department of Geology and Geography
Problem 8.14
Tom Wilson, Department of Geology and Geography
cos  
3
2
x x
 x
3 2

e  5
1
e
2
x
19
x
Tom Wilson, Department of Geology and Geography
We’re really
asking “what is
the integral?”
 cos  d
 x3 x 2

  3  2  x  dx

19
e

5

d




1
e x
 x2 dx
From class notes last time:
The velocity problem
We’ve got an object traveling with velocity kt.
Since v = dx/dt, we can rearrange and say
dx  vdt And to find out how far
the object travels or where it’s going to be at
a certain time, we sum up all the little steps:


i.e. we take the integral dx  vdt.
Tom Wilson, Department of Geology and Geography
The integral is just a sum of some quantity of interest
when the steps are individually very very small.
 dx   vdt
is really no different from
x  vt
In the case of the integral we’ve
just let the steps become very very
– infinitesimally – small: i.e. dx.
We could play the mathematician and write this
n
n
n
i 1
i 1
i 1
2

k

t

x

k
(
i

t
)

t
i
 i 
Tom Wilson, Department of Geology and Geography
Going through the discrete sum - i.e. summing over steps
that are small but not infinitesimally small – we get
n(n  1)
k t  i  k t
2
i 1
n
2
2
Which for large n is
2
n

t


n
t
k t 2
k
k
2
2
2
2
Tom Wilson, Department of Geology and Geography
2
So we get the antiderivative
We’ve been asking the question: “What do you
have to differentiate to get kt ? “
kt 2
2
The guess is easy to confirm by taking its
derivative
d  kt 2 
would = kt.


dt  2 
Tom Wilson, Department of Geology and Geography
For small steps
x  vt
 dx   vdt
For very small steps
Which with substitution becomes
 dx   ktdt
What do we have to differentiate to get dx?
x
so
Tom Wilson, Department of Geology and Geography
kt 2
x
2
Is the result unique?
You ask yourself, but wait a minute –
2
kt
I can differentiate
2 and get a
result, but, I can also differentiate
kt 2
C
2
and get the same result. What does this
mean? What additional information do you
need to help you know the position x?
Tom Wilson, Department of Geology and Geography
Our solution is non-unique
This idea that we’ve just gone over
illustrates the indefinite nature of the
process. In other words, we can find
lots of things to differentiate that will
give us the same answer.

d  kt 2
 C   kt

dt  2

…. whatever C is.
Tom Wilson, Department of Geology and Geography
dx  ktdt
The little step dx has length ktdt
To get from the start to the finish …
to finish

dx 
start  x
to end time

ktdt
start time
x
t
0
0
 dx   ktdt
and this becomes

x
0
We get lazy
t
dx   ktdt
0
kt 2
C
And we know that this gives us x 
2
Tom Wilson, Department of Geology and Geography
To use this new notation, then we’d say that
kt 2
 ktdt  2  C
This is referred to as an indefinite
integral or as indefinite integration.
In the problem of finding the
moving target we have
kt 2
x   ktdt 
C
2
Tom Wilson, Department of Geology and Geography
kt 2
 ktdt  2  C
Graphically, what does this represent?
Basically it says the starting point
can vary. To know where the thing is
going to be at a certain time, you
have to know where it started.
Tom Wilson, Department of Geology and Geography
y  x2
3
x
2
x
 dx  3  C
Tom Wilson, Department of Geology and Geography

10
4
3 10
x
2
x dx 
3
4
103 43

  312
3
3
y  x2
Tom Wilson, Department of Geology and Geography
We have to know where the object started its journey. Then we
can specify the limits of our integration

T
0
2 T
kt
ktdt 
2
0
This is referred to as a definite
integral. It’s evaluated as follows

T
0
Tom Wilson, Department of Geology and Geography
2 T
kt
ktdt 
2
0
kT 2 k 02 kT 2



2
2
2
More generally 2 b
kt
a ktdt  2
b
a
kb 2 ka 2


2
2
Notice that when we introduce specific
begin and end points, for example:
y
2
1
2
x

xdx    c 
2
1
2
 22
 12

   c    c
2
 2

2  0.5  c  c
Those constants cancel out.
Tom Wilson, Department of Geology and Geography
Here are some more that you’ve already done
 cos( )d  sin( )  k
Remember this one?
 e
x
 20 x19  cos( x)  dx 
e  20x  sin( x)  k
x
When you have an indefinte integral, the
limits of integration are not specified.
Tom Wilson, Department of Geology and Geography
dy
x n 1
n
x
then
y
dx
n 1
if
Thus
n 1
x
n
x
 dx  n  1  k
When n = -1, what happens?
1
x
 dx  ln( x)  k
Tom Wilson, Department of Geology and Geography
This is just another indefinite integral.
1
x
 dx  ln( x)  c
To overcome the problem we encountered
with trying to find the moving target
kt 2
 ktdt  2  C
We need to have some additional information
when we do the integration.
We need to know the starting point to give us
a definite answer.
Tom Wilson, Department of Geology and Geography
• You should have finished reading chapter 9.
•Any questions about the problems in the text
discussed in the applications section 9.6. How
would you approach problems 9.7 and 9.8
•Also look over questions 9.9 and 9.10 & bring
questions to class this Thursday.
•Look over the in-class integrals handed out
today and be prepared to work through them
in groups this coming Thursday.
Tom Wilson, Department of Geology and Geography