Download The Evaluation Theorem

These notes closely follow the presentation of the material given in James Stewart’s
textbook Calculus, Concepts and Contexts (2nd edition). These notes are intended
primarily for in-class presentation and should not be regarded as a substitute for
thoroughly reading the textbook itself and working through the exercises therein.
The Evaluation Theorem
As we have learned by experience, it is very difficult and tedious to compute
definite integrals,
b
 a fx dx,
using the definition of the definite integral,
b
R .
 a fx dx  lim
n n
In fact, it is usually not possible to do this unless the Riemann Sum, R n , that is
involved contains “special” summation formulas such as
n
i 
i1
nn  1
2
that allow us to simplify R n in such a way that lim n R n can be computed. Some
previous examples that we studied in which we were able to compute definite integrals
using the definition were
1
 0 x 2 dx  13
and
2
 1 x 3 dx  3. 75.
Some other examples in which we were able to compute the exact value of a
definite integral by using basic geometry were
5
 0 32t  48 dt  160
and
1
 1
1  x 2 dx   .
2
The theorem that we are now about to state, which the author or our textbook,
James Stewart, calls The Evaluation Theorem, gives us a considerably easier way to
evaluate many definite integrals.
Theorem (The Evaluation Theorem) Suppose that the function f is continuous on the
interval a, b and suppose that F is an antiderivative of f on a, b. Then
1
b
 a fx dx  Fb  Fa.
Example Use the Evaluation Theorem to find the values of
1
2
5
 0 x 2 dx,  1 x 3 dx,  0 32t  48 dt,
and

 0 sinx dx.
2
Remark It is not always easy to use the Evaluation Theorem. Sometimes it can be very
difficult to find an antiderivative of the integrand, f. In fact, sometimes it is
impossible! An example of a difficult (but not impossible) problem on which the
Evaluation Theorem can be used is
1
 1
1  x 2 dx.
An example of a problem on which it is impossible to use the Evaluation Theorem
is

 0 sinx 2  dx.
3
Why Does the Evaluation Theorem Work?
Suppose that f is continuous on the interval a, b and suppose that F is an
antiderivative of f.
For any given positive integer, n, we use the usual method of subdividing the
interval a, b into n pieces:
a
x n  b 
n
x0  a
x 1  a  1  x n
x 2  a  2  x n

x n1  a  n  1  x n
x n  a  n  x n  b
and, as usual, we define
n
Rn 
 fx i x n .
i1
Now, observe that
Fb  Fa  Fx n   Fx n1 
 Fx n1   Fx n2 

 Fx 2   Fx 1 
 Fx 1   Fx 0 
n

Fx i   Fx i1 .
i1
Since the function F is differentiable on each of the subintervals x i1 , x i , then by the
Mean Value Theorem, there exists at least one point x i  x i1 , x i  such that
Fx i   Fx i1 
Fx i   Fx i1 

.
F  x i  
x i  x i1
x n
From this, we conclude that
Fx i   Fx i1   F  x i x n .
However, recall that F is an antiderivative of f (which means that F   f). Therefore,
Fx i   Fx i1   fx i x n .
We thus see that
n
Fb  Fa 
 fx i x n .
i1
Taking the limit as n   on both sides of the above equation gives us
4
Fb  Fa 
b
 a fx dx.
5
Indefinite Integrals
The Evaluation Theorem shows that there is a close connection between the
processes of “integration” and “antidifferentiation”. For this reason, we sometimes use
the integration symbol,
,
as a symbol for antidifferentiation. We write
b
 a fx dx
to denote the definite integral of f over the interval a, b, and we write
 fx dx
to denote an indefinite integral of f, meaning any antiderivative of f.
Thus, for example, we could write
 x 2 dx  13 x 3  C
and
 e x dx  e x  C.
6
Example Find the following indefinite integrals:
 x n dx
(assuming that n is a constant with n  1)
 x 1 dx
 sinx dx
 cosx dx
 sec 2 x dx
 e x dx
 1 dx
Remark The notion of an indefinite integral as defined in most textbooks (including
ours) is not technically correct. The statement that “F is an antiderivative of f”
does not make sense unless we refer to some interval. Thus, a statement that does
make sense is “F is an antiderivative of f on the interval a, b”. A more correct
notation for an indefinite integral would make reference to the interval a, b. Thus,
a more correct notation for an indefinite integral might look something like
a,b .
In most situations encountered in elementary calculus, this nuance does not really
matter (which is probably why most textbook authors ignore it). For example, the
statements
a,b x 2 dx  13 x 3  C
and
a,b cosx dx  sinx  C
are correct no matter what interval a, b we use.
However, the statement
a,b x 1 dx  lnx  C
might or might not be correct, depending on what the interval a, b is. If a, b is
an interval of positive numbers, then the statement is correct, but if a, b is an
interval of negative numbers, then the statement is not correct.
7
How To Make Up an Integration Problem That You
Can Do But Nobody Else Can (Unless They Are Very
Clever)
Want to impress you friends?
The Evaluation Theorem tells us that if F is an antiderivative of f, then
b
 a fx dx  Fb  Fa.
Therefore, the definite integral of f over the interval a, b is easy to compute if you
know what F is.
One way to make up a hard problem (that you can easily do but others will have
trouble with) is to start with some very complicated function F (that you make up), then
compute its derivative, f. Chances are that if F is very complicated–looking, then f will
also be very complicated–looking. Then ask your friends to compute
b
 a fx dx.
You will know how to compute this integral because you know what F is. However,
your friends might have a hard time.
Here is an example of this: Let’s take
Fx 
x 2  2x  3
.
x2  3
The derivative of this function is
8

F x 

x 2  3 
1
d
dx
x 2  2x  3 2

d
dx
x 2  3
1
1
x 2  2x  3  2  2x  2  x 2  2x  3 2
1
2
x 2  3
 2x
2
x 2  3
1

x 2  3 x 2  2x  3 2  x  1  x 2  2x  3 2
 2x
2
x 2  3
x1
x 2  3
 2x x 2  2x  3
x 2 2x3
2
x 2  3
x 2 3x1


x 2  3 2
1

1
 x 2  2x  3 2
x 2 2x3

2xx 2 2x3
2
x  3
x 2 2x3
2
x 2 3x12xx 2 2x3


x 2 2x3
2
x 2  3
x 2  3x  1  2xx 2  2x  3
x 2  3 2 x 2  2x  3
x 3  3x 2  9x  3 .
x 2  3 2 x 2  2x  3
What we have just discovered is that the function

Fx 
x 2  2x  3
x2  3
is an antiderivative of the function
x 3  3x 2  9x  3 .
x 2  3 2 x 2  2x  3
Now suppose you ask your friend to do the problem
3
3
 3x 2  9x  3 dx.
 1 x
x 2  3 2 x 2  2x  3
I bet they can’t do it. But you can!
fx 
9
3
1
x 3  3x 2  9x  3 dx  F3  F1
x 2  3 2 x 2  2x  3




3 2  23  3

32  3
12
 0
12
4
12
12
3
.
6
1 2  21  3
12  3
Assignment: Make up an integration problem that you can do but that you think I
can’t do. Hand in your problem (without its solution) on Monday, November 29. What
you hand in on November 29 should have this format:
Evaluate the integral:
3
3
 3x 2  9x  3 dx.
 1 x
x 2  3 2 x 2  2x  3
Then hand in the solution to your problem on December 3 (the day of the next exam).
If I can’t do the problem that you hand in (on November 29) and you show me how to
do it (in what you hand in on December 3), then I will replace your two lowest problem
scores on the next exam with 10s. If I can do the problem that you hand in on
November 29 and you show me how to do it (in what you hand in on December 3),
then I will replace your lowest problem score on the December 3 exam with a 10.
Have fun and be creative!
Note: This is a great opportunity to get a good grade on the December 3 exam, but I
expect you to put some effort into it so that you are getting credit for actually having
learned something. This assignment is not for the purpose of getting giving away
“free” credit in exchange for only a little bit of effort from you. I expect you to hand in
something that is very neatly written with correct notation, etc. If you hand in
something that looks like you did not spend much time on it, then I will probably frown
upon it.
10