AP CALCULUS AB
Chapter 5:
The Definite Integral
Section 5.4:
Fundamental Theorem of Calculus
What you’ll learn about





Fundamental Theorem, Part 1
x
Graphing the Function a f (t )dt
Fundamental Theorem, Part 2
Area Connection
Analyzing Antiderivatives Graphically
… and why
The Fundamental Theorem of Calculus is a
Triumph of Mathematical Discovery and
the key to solving many problems.
Section 5.4 – Fundamental Theorem of
Calculus

Fundamental Theorem of Calculus, Part 1
If f is continuous on [a, b], then the
x
function
F  x    f  t  dt
a
has a derivative at every point x in [a, b],
and
d
d x
dx
F  x 
f  t  dt  f  x 

dx
a
d x
Example:
sin t  dt  sin x


dx 0
The Fundamental Theorem of
Calculus
d
 f (t )dt  f ( x)
dx
x
a
Every continuous function f is the derivative of some other function.
Every continuous function has an antiderivative.
The processes of integration and differentiation are inverses of one another.
Example Applying the
Fundamental Theorem
d
Find
 sin tdt.
dx
x

d
 sin tdt  sin x
dx
x

Example The Fundamental
Theorem with the Chain Rule
Find dy / dx if y   sin tdt.
x2
y   sin tdt
1
x2
1
y   sin tdt and u  x .
dy dy du
Apply the chain rule:


dx du dx
 d
 du
   sin tdt  
 du
 dx
du
 sin u 
dx
 sin u  2 x
u
2
1
u
1
 2 x sin x
2
Example Variable Lower Limits of
Integration
dy
Find
if y   t sin tdt.
dx
5
x
d
d
   t sin tdt 
 t sin tdt 
dx
dx
d
    t sin tdt 
dx
  x sin x
5
x
x
5
x
5
Section 5.4 – Fundamental Theorem of
Calculus

Fundamental Theorem of Calculus, Part 2
If f is continuous at every point of [a, b],
and if F is any antiderivative of f on [a, b],
then
b

a
f  x  dx  F  b   F  a 
This is also called the Integral Evaluation
Theorem.
The Fundamental Theorem of
Calculus, Part 2
 f ( x)dx  F (b)  F (a )
b
a
Any definite integral of any continuous function f can be calculated without
taking limits, without calculating Riemann sums, and often without effort so long as an antiderivative of f can be found.
Section 5.4 – Fundamental Theorem of
Calculus

Example:
  4 x  dx  2 x
5
2
2 5

2
 2  5  2  2 
2
2
 2  25   2  4 
 50  8
 42
Example Evaluating an Integral
Evaluate   3x  1 dx using an antiderivative.
3
2
-1
  3 x  1 dx   x  x 
3
2
3
-1
3
1
  3  3    1   1 
3
 32
3
Section 5.4 – Fundamental Theorem of
Calculus

To find the Total Area Analytically
To find the area between the graph of
y=f(x) and x-axis over the interval [a, b]
analytically:
1.
Partition [a, b] with the zeros of f.
2.
Integrate f over each subdivision.
Add the absolute value of the integrals.
3.
Section 5.4 – Fundamental Theorem of
Calculus

Ex: Find the
area between
the curve
y  x  x6
2
and the x-axis
over the interval
[-4, 4].
y  x 2  x  6   x  3 x  2
Zerosof f are 3 and  2.
2
Area   f  x dx 
4

2

  x 2  x  6 dx 
4

3
2
f x dx   f x dx
 x
3
2
4
3
2

4


 x  6 dx   x 2  x  6 dx
3
  23  22
   43  42

 

 6 2  

 6 4 
2
2
 3
  3

 33 32

   23  22

 6 2 
   63  
2
 3 2
  3

 43 4 2
  33 32

   64     63
 3 2

 3 2
109
1

or 36
3
3
Section 5.4 – Fundamental Theorem of
Calculus
To find Total Area Numerically (on the
calculator)
To find the area between the graph of
y=f(x) and the x-axis over the interval
[a, b] numerically, evaluate:
On the TI-89:
nInt (|f(x)|, x, a, b, 50)
On the TI-83 or 84:
fnInt (|f(x)|, x, a, b, 50)

Section 5.4 – Fundamental Theorem of
Calculus

Application: Average Daily Inventory:
If I(x) is the number of items on hand on
day x, the average daily inventory of the
items for the period a  x  b
is
1 b
av  I  
I  x  dx

ba a
If h is the dollar cost of holding one item
per day, the average daily holding cost for
the period a  x  b
is
av  I   h