The Fundamental Theorem of Calculus
(Part I)
The overall idea behind the Fundamental Theorem of Calculus
is that differentiation and integration are operational opposites;
In the same way that multiplication and division are.
d
f (x)
dx


f ( x)dx

Graphical Proof
2 Reminders
1)
f ' ( x)  lim
h0
f ( x  h)  f ( x)
h
2) A function is like a machine that takes in an
input, manipulates it according to some rule,
and spits out an output.
Example:
s(t )  16t 2  96t  112
is a function that gives you the distance an object is from
the surface of the earth at any given time, t. Think of it as
“the distance away from earth function.” You input a time, t,
and the function spits out a distance the object is form the earth.
Start of proof…
Lets take some function, f(t), that has the following graph:
f(t)
The area under the curve
depends on where x is
(‘a’ is any constant and
x is a variable).
Therefore we can say
that the area under the
curve is a function of x.
a
Area1= F(x1) = 4.0000
Area2= F(x2) = 5.9532
Area3= F(x3) = 8.0000
x1 x2
x3
t
Let’s call it F(x) and
think of it as “the area so
far” function.
2) Now lets say we move x a small distance, h, to x+h.
f(t)
This means that the
change in our “area
so far” function,
F(x), is:
f(x)
F ( x  h)  F ( x )
h
a
x
x+h
t
which is approximately equal to: h  f (x)
(the area of the rectangle with base h and height f(x)).
So What we end up with is:
F ( x  h)  F ( x )  h  f ( x )
4) This means that if we divide both sides by h,
F ( x  h)  F ( x )  h  f ( x )
we get:
F ( x  h)  F ( x )
 f ( x)
h
When will
F ( x  h)  F ( x )
 f (x)
h
When h  0 !!!
?
Therefore we can say that
F ( x  h)  F ( x)

f
(
x
)
lim
h
h0
and we know that
is
F ' ( x)
(the definition of the derivative of F(x) )

F ( x  h)  F ( x)
f (x)

lim
h
h0
implies
F ' ( x)
 f (x)
So we know that
a)
F ' ( x)  f ( x)
Since F(x) is “the area so far” function, we can say that
x
b)
F ( x)   f (t )dt
a
If we take d/dx of both sides in part b, we get
c)
Since
d
d x
F ( x) 
f
(
t
)
dt

dx
dx a
d
F ( x)  F ' ( x) We can combine statements a and c
dx
d x
F ' ( x) 
f
(
t
)
dt

f
(
x
)

dx a
What we are saying by the statement
d x
f
(
t
)
dt

f
(
x
)

dx a
is that the derivative of the integral of a function
gives us the original function.
In other words, differentiation can undo integration, and vise versa
Examples:
d x
1)  cos(t )dt  cos( x)
dx 
d x 1
1
2)
dt

dx 0 1  t 2
1 x2
5
3) If y   3t sin( t )dt
x
dy d 5
d x
then 
3t sin( t )dt    3t sin( t )dt

x
dx dx
dx 5
 3x sin( x)
What we are also saying by the statement
d x
f
(
t
)
dt

f
(
x
)

dx a
is that the integral is the
of the original function
Example:
If f ( x)  x 2
3
x
then F ( x)   f ( x)dx   x 2 dx   the anti  derivative of x 2
3
d  x3 
because F ' ( x)     x 2  f ( x)
dx  3 
The Fundamental Theorem of Calculus
(Part II)
If f is continuous at every point on the interval [a, b],
And if F is any anti-derivative of f on [a, b] then:

b
a
f ( x)dx  F (b)  F (a)
Example 1: Evaluate

2
1
x 2 dx
Solution:
We saw earlier that a simple anti-derivative of x2 is x3/3
3 2
x
2
x dx 
1
3

2
23

3
1

(1)3 8  1
 
3
3 3
3
Example 2: Evaluate

3
1
( x3  1)dx
Solution:
4
x
A simple anti-derivative of (x3+1) is
x
4
4
3
x
3
1 ( x  1)dx  4  x 
1
3

 34
  (1) 4
  3  
 (1)  

4
  4
 81   1 
  3     1  24
4
 4 
Example 3: Find the area of the region between the curve y = 4 – x2
And the x-axis on [0, 3]
Note that the total area
3
will not simply be  (4  x 2 )dx
0
What we need to do is find
The zeros of f, separate our
interval into sub-intervals,
integrate each sub-interval
separately, and then add the absolute
value of each integral
3 2
2
x
16
2
(
4

x
)
dx

4
x


Sub-interval1 [0, 2] : 0
3
3
0
3
x3
7
2
(
4

x
)
dx

4
x


Sub-interval1 [2, 3] : 2
3 2
3
16  7 23
Area of entire shaded region = 3  3  3
3
Steps for Finding Total Area
Between Graph and x-axis on [a, b]
1) Partition [a, b] with the zeros of f.
2) Integrate f over each sub-interval
3) Add the absolute value of the integrals
Finding Area Using FNINT
Example: Find the area of the region between the curve
y  x cos( 2 x) and the x-axis on [-3, 3]
Solution: We want to find the area of the red region, but instead
of separating [-3,3] into sub-intervals, we will simply find the
integral of the absolute value of y. The absolute value of y simply
flips all the regions below the x-axis to the top.
So on your calculator type:
FNINT(abs(xcos(2x)), x, -3, 3)