Integration
Antidfferentiation
The Fundamental Theorem of Calculus (FTC)
and the Algebra of Integration
Second FTC
Theorem 4.5 The Definite Integral
as the Area of a Region
Fundamental Theorem of
Calculus
4.3
The Fundamental Theorem of
Calculus
• If a function f is continuous on the closed [a,b]
and F is an antiderivative of f on the interval
[a,b], then:

b
a
f ( x)dx  F (b)  F (a)
FTC and the Constant of
Integration
It is not necessary to include a constant of integration C in the antiderivative
because:
FTC Example

b
a
f ( x)dx F ( x)

3
1
4 3
x

4
1
b
a
 F (b)  F (a)
31 3
x
x dx 
3 1 1
3
34 14 81 1
     20
4 4 4 4
Using FTC to Find Area
Example 3 in the text: Find the area of the region bounded by the
graph of
y= 2x2-3x+2, the x-axis and the vertical lines x=0 and x=2
Note that all of the function is above the x-axis in the interval
Start by integrating the function over the closed interval [0,2]
2
Area   (2 x 2  3x  2)dx
0
Then find the antiderivative and apply the FTC
2
3
3
3
2

2
 2

  x3  x 2  2 x     23   22  2  2     03   02  2  0 
2
2
2
3
0 3
 3

Simplify.

10
3
Theorem 4.4 Continuity Implies
Integrability
Theorem 4.6 Additive Interval
Property
Absolute Value Integration
Example:
• Example 2 in the text: keep in mind the definition
of absolute value and where your function is
positive and where it would be negative:
 (2 x  1), x1/ 2
2x 1  
 2 x  1, x  1/ 2
• From this, you can rewrite the integral in two
2
1/ 2
2
parts:

0
2 x  1dx  
0
 ( x  x )
2
1/ 2
0
  x  x
2
2 x  1dx

1/ 2
(2 x  1)dx  
2
1/ 2
 1 1
1 1 5
      (0  0)  (4  2)     
 4 2
4 2 2
Theorem 4.7 Properties of
Definite Integrals
Definitions of Two Special
Definite Integrals
Average Value of a
Function
Julia S. Lucas
Definition of the Average Value of a
Function on an Interval and Figure 4.32
Let’s get started:
• You already know about functions and
how to take the average of some finite
set.
• Today we’re going to take the average
over infinitely many values (those that
the function takes on over some
interval)…which means CALCULUS!
• Before I show how to do this…let’s talk
about WHY we might want to do this?
Consider the following picture:
• How high would the water level be if the
waves all settled?
Okay! So, now that you have
seen that this an interesting
question...
Let’s forget about real life, and...
Do Some Math
Suppose we have a “nice”function and we need to
find its average value over the interval [a,b].
Let’s apply our knowledge of how to find the
average over a finite set of values to this problem:
First, we partition the interval [a,b] into n subintervals of
equal length to get back to the finite situation:
In the above graph, we have n=8
Let us set up our notation:
x  (b  a) / n
xi comes from the i-th interval
Now we can get an estimate for the
average value:
f ( x1)  f ( x 2)  ...  f ( xn)
faverage 
n
Let’s try to clean this up a little:
f ( x1)  f ( x 2)  ...  f ( xn)
faverage 
n
x

[ f ( x1)  f ( x 2)  ... f ( xn)]
ba
Since x  (b  a) / n
In a more condensed form, we now get:
n
1
faverage 
f ( xi )x

b  a i 1
But we want to get out of the finite,
and into the infinite!
How do we do this?
Take Limits!!!
In this way, we get the average value
of f(x) over the interval [a,b]:
b
1
faveragelim n 1  f ( xi ) 
f ( x ) dx

baa
b  a i 1
n
So, if f is a “nice” function (i.e. we
can compute its integral) then we have
a precise solution to our problem.
Let’s look back at our graph:
So we’ve solved our problem!
If I give you the equation f ( x)  3x  1
and ask you to find it’s average value
over the interval [0,2], you’ll all say
2
NO PROBLEM!!!!
the answer is…
1
faverage 
ba
b

f ( x)dx
a
2
1
2

3
x
 1dx

20 0
2
1 3
 [x  x ]
2
0
1
 (10  0)  5
2
Now we can answer our fish tank question!
(That is, if the waves were described by an integrable function)
Theorem 4.10 Mean Value Theorem for
Integrals and Figure 4.30
Finding the x values where we GET the
average value of the function.
First take the equation f ( x)  3x  1
and find it’s average value
over the interval [0,2],
2
NO PROBLEM!!!!
the answer is…
1
faverage 
ba
b

f ( x)dx
a
2
1
2

3
x
 1dx

20 0
2
1 3
 [x  x ]
2
0
1
 (10  0)  5
2
Finding the x values where we GET the
average value of the function.
NOW take the equation f ( x)  3x  1
and it’s average value
over the interval [0,2], average f = 5
and set that equal to the function…
And then solve for x…
2
f ( x)  3 x  1  5
2
NO PROBLEM!!!!
the answer is…
Mean Value Theorem (find the
value of x that gives you the
average value of your function)
f ( x)  3 x  1  5
2
f ( x)  3x 2  1  5
3x 2  4
4
x 
3
2
4
x
 1.547
3
Average Value of a Function
Definition:
If f is integrable on the closed interval [a,b],
then the average value of f on the interval is:
1 b
f ( x)dx

ba a
Average Value Example
• Find the average
value of
f ( x)  3x  2 x,[1, 4]
2
1 b
1 4
2
f
(
x
)
dx

3
x
 2 x dx



a
1
ba
3
4
1 3
2
 x  x 
1
3

1 3
1 3 2
2
4

4

1 1 



3
3

1
 64  16  (1  1)   16
3
Mean Value Theorem
• If f is continuous on the closed interval [a,b],
then there exists a number c in the closed
interval [a,b] such that:

b
a
f ( x)dx  f (c)(b  a)
b
1
OR :
f ( x)dx  f (c)

(b  a) a
Mean Value Theorem Example:
• Find the value(s) of c guaranteed by the Mean
Value Theorem for Integrals for the function over
the given interval: f ( x)  9 ,[1,3]
x3
1 39
1 3 3
dx   9 x dx
3

1
2 x
2 1
31 3
2 3
1 9x
9x
 

2 3  1 1
4
1
9

4 x 2
3

1
9
4  3 
2

9
4 1
2
1 9
  2
4 4
Now we need to find the x coordinate where we get this average
y value:
9
9
9
3
3
2  f (c )  2  3  x   x 
x
2
2
MVT-I
• The mean value theorem does not specify
how to find c, just that there exists at least
one number c in the interval that will give
you the average value of the function.
Theorem 4.8 Preservation of
Inequality
The Second Fundamental
Theorem of Calculus
4.4
Second Fundamental Theorem of
Calculus
• Earlier you saw that the definite integral of
f on the interval [a,b] was defined using
the constant b as the upper limit of
integration and x as the variable of
integration. However, a slightly different
situation may arise in which the variable x
is used as the upper limit of integration. To
avoid the confusion of using x in two
different ways, t is temporarily used as the
variable of integration
Theorem 4.11 The Second
Fundamental Theorem of Calculus
Definite Integral diagrams
The Definite Integral as a Function
x
  
0
6 4 3
F ( x)   cos tdt , x  0,

x
0
,
,
, and

2
cos tdt  sin t 0  sin x  sin 0  sin x
x
F (0)  sin(0)  0
    1
F   sin   
6 6 2
2
 
 
F    sin   
4
4 2
3
 
 
F    sin   
3
3 2
 
 
F    sin    1
2
2
You can think of the function F(x) as
accumulating the area under the curve
f(t)=cost from t=0 to t=x. For x=0, the area is 0
and F(x)=0. for x=pi/2, F(pi/2)=1 gives the
accumulated area under the cosine curve on
the entire interval [0,pi/2]. This interpretation
of an integral as an accumulation function is
used often in applications of integration. See
p. 288 for a graphical example of this.
Figure 4.35
The Second Fundamental Theorem
of Calculus
• If f is continuous on an open interval I containing
a, then, for every x in the interval,
d  x

f (t )dt  f ( x)


dx  a
• The proof of this is on p. 289 in the text.
Using the SFTC
2
t
 1 is continuous on the entire
• Note that
real line. So, using the SFTC you can write
d  x 2
2

t  1dt  x  1



0
dx 
• The differentiation shown in tthis example is a
straightforward application of the SFTC. The
next example shows an application of this
combined with the chain rule to find the
derivative of a function
Using the SFTC
d  x3
cos tdt 



/
2
dx 

dF du
du dx
• Chain Rule
d
du
F ( x) 
 F ( x) • Definition of dF/du
du
dx
d 
du • Substitute the integral


cos tdt 
du  
 dx
for F(x)
d 
 du • Substitute u for x3

cos
tdt
 dx
du  
 cos u  3x  • Apply the SFTC
  cos x  3x  • Rewrite as a function
of x.
F ( x ) 
x3
 /2
u
 /2
2
3
2
Find 𝐹′(𝑥) if 𝐹 𝑥 =
𝑥2
2
𝑠𝑖𝑛𝜃
𝑑𝜃
0
What is u and du?
𝑢 = 𝑥 2,
𝑑𝑢 = 2𝑥
𝑑 𝑢
𝑑𝑢
2
𝑠𝑖𝑛𝜃 𝑑𝜃
𝑑𝑢 0
𝑑𝑥
= 𝑠𝑖𝑛𝑢2 𝑑𝑢
= sin(𝑥 2 )2 (2𝑥)
= 𝑠𝑖𝑛𝑥 4 (2𝑥)
U-Substitution
Antidifferentiation of a Composite
Function
• Let g be a function whose range is an interval I
and let f be a function that is continuous on I. If g
is differentiable on its domain and F is an
antiderivative of f on I, then
 f ( g ( x)) g '( x)dx  F ( g ( x))  c
• If u = g(x), then du = g’(x)dx and
 f (u )du  F (u )  c
Pattern Recognition
In this section you will study techniques for
integrating composite functions.
This is split into two parts, pattern
recognition and change of variables.
u-substitution is similar to the techniques
used for the chain rule in differentiation.
Recognizing Patterns
 2 x( x

2
 1) dx  u  x  1  du  2 x
4
2

3x 2 x3  1 dx  u  x3  1  du  3x 2
2
2
sec
x
tan
x

3
dx

u

tan
x

3

du

sec
x




Pattern Recognition for Finding the
Antiderivative
• Find
 5cos 5xdx
• Let g(x)=5x and we have g’(x)=5
So we have f(g(x))=f(5x)=cos5x
From this, you can recognize that the integrand
follows the f(g(x))g’(x) pattern. Using the trig
integration rule, we get
  cos 5x  5dx  sin 5x  c
You can check this by differentiating the answer to
obtain the original integrand.
Multiplying and Dividing by a
Constant
• Example 3 p. 297
 xx
2
 1 dx
2
let _ u  x 2  1  du  2 xdx
but _ we _ have _ only _ x, not 2 x...so _ we _ do _ a lg ebra
1
du  xdx  now _ substitute :
2
1
 u 2 du
2
1 2
1 3
u
du

u c

2
6
3
1
now _ back _ substitute :  x 2  1  c
6

Change of Variables
Example 4 p. 298
• P. 298

2 x  1dx
let _ u  2 x  1  du  2dx
but _ we _ have _ only _ dx, not 2dx...so _ we _ do _ a lg ebra
1
du  dx  now _ substitute :
2
1
1
2
u
 2 du
1
2
1
1
2
3
2
1
1u
1u
1 32
  u du  1  c  3  c  u  c
2
2 2 1
2 2
3
3
1
now _ back _ substitute :  2 x  1 2  c
3
Change of Variables
• Example 5 p. 298
x
2 x  1dx
let _ u  2 x  1  du  2dx
solve _ for _ x : (u  1) / 2
1
 u  1  2 du
Substitute :  x 2 x  1dx   
u
 2  2
1
3
1


1
1
1  2 5/ 2 2 3/ 2 
2
2
2
   u  1u du    u  u du   u  u   c
4
4 
45
3


5
3
1
1
Back _ Substitute :  2 x  1 2   2 x  1 2  c
10
6
Change of Variables Example
• Example 6 p. 299
Guidelines for Making a Change of
Variables
1. Choose a substitution u=g(x). Usually it is best
to choose the INNER part of a composite
function, such as a quantity raised to a power.
2. Compute du=g’(x)dx
3. Rewrite the integral in terms of the variable u
4. Find the resulting integral in terms of u
5. Replace u by g(x) to obtain an antiderivative in
terms of x.
6. Check your answer by differentiating.
General Power Rule for Integration
• Theorem: The General Power Rule for
Integration
If g is a differentiable function of x, then
 g  x  
  g  x  g   x  dx  n  1
n
n 1
 c, n  
Equivalently, if u=g(x), then
u n 1
 u du  n  1  c, n  1
n
Theorem 4.14 Change of Variables
for Definite Integrals
Change of Variables for Definite
Integrals
• Theorem 4.14 p 301
Definite Integrals and Change of
Variables
• Example 8 p. 301
Definite Integrals and Change of
Variables
• Example 9 p. 302
Theorem 4.15 Integraion of Even
and Odd Functions and Figure 4.39
Integration of Even and Odd
Functions
• P. 303
Integration of an Odd Function
• Example 10 p. 303
Figure 4.41
Theorem 4.16 The Trapezoidal
Rule
Theorem 4.17 Integral of p(x)
=Ax2 + Bx + C
Theorem 4.18 Simpson's Rule
(n is even)
Theorem 4.19 Errors in the Trapezoidal
Rule and Simpson's Rule