Calc 1 Lecture Notes
Section 4.5
Page 1 of 5
Section 4.5: The Fundamental Theorem of Calculus
Big idea: Integration and differentiation are inverse processes. That means antiderivatives can
be used to solve integrals.
Big skill: You should be able to calculate simple definite integrals using antiderivatives of the
integrand.
Theorem 5.1: Fundamental Theorem of Calculus, Part 1
If f(x) is continuous on [a, b] and F(x) is any antiderivative of f(x), then
b
 f  x  dx  F  b   F  a  .
a
b
Alternative notation:

a
Practice:
1
1.
 x dx 
2
0
2
2.
x
3
 1 dx 
0

3.
 sin  x  dx 
0
f  x  dx  F  x  a
b
Calc 1 Lecture Notes
Section 4.5
Page 2 of 5
Proof:
b

n
f  x  dx  lim  f  ci  x
n 
a
i 1
n
 lim  F   ci  xi  xi 1 
n 
i 1
n
 lim  F  xi   F  xi 1 
n 
i 1
  F  x1   F  x0     F  x2   F  x1     F  x3   F  x2   
  F  xn   F  xn 1  
 F  xn   F  x0 
 F b   F  a 
The third limit comes from the Mean Value Theorem: If g is differentiable on [xi-1, xi], then there
g  xi   g  xi 1 
exists c  [xi-1, xi] such that g   c  
 g   c  xi  xi 1   g  xi   g  xi 1 
xi  xi 1
Notice:
b
 f  x  dx  F  b   F  a 
a
a
 f  x  dx  F  a   F  b 
b
   F b   F  a 
b


a
f  x  dx    f  x  dx
a
b
ie., REVERSING THE LIMITS OF INTEGRATION REVERSES THE SIGN OF THE
INTEGRAL
Practice:
4
1

1.   x  2  dx 
x 
1
4
2.
e
0
3 x
dx 
Calc 1 Lecture Notes
1.5
3.
Section 4.5
Page 3 of 5
2
 x dx 
0.5
x
4.
15t dt 
5
1
Theorem 5.2: Fundamental Theorem of Calculus, Part 2
x
If f(x) is continuous on [a, b] and F  x    f  t  dt , then F(x) = f(x) on [a, b].
a
Proof:
F   x   lim
h 0
 lim
F  x  h  F  x
h
F  x  h  F  a    F  x   F  a 
h 0
 lim
h
xh
x
a
a
 f  t  dt   f  t  dt
h 0
h
xh
 lim

a
f  t  dt   f  t  dt
a
x
h 0
h
a
 lim

f  t  dt 
xh
x
a
h 0
1
 lim 
h 0 h
 f  t  dt
h
xh
 f  t  dt
x
 lim f  c  , c  [ x, x  h]
h 0
 f  x
Note: the second to last step is true by the Integral Mean Value Theorem (a function takes on its
average value somewhere in an interval).
Calc 1 Lecture Notes
Section 4.5
Page 4 of 5
Notice:
d 
The chain rule dictates that

dx 
u( x)

a

f  t  dt   f  u  x    u   x 

Practice:
x
1. Compute F(x) for F  x     t 3  2t 2  3t  1 dt
1
2. Compute F(x) for F  x  
x2
 t
3
 2t 2  3t  1 dt
1
F  x 
x2
 t
3
 2t 2  3t  1 dt
1
x2
2
3
1

  t4  t3  t2  t 
3
2
4
1
4
3
2
2
3
2
3
1
 1

   x 2    x 2    x 2    x 2     14  13  12  1
3
2
3
2
4
 4

1
2
3
5
F  x   x8  x 6  x 4  x 2 
4
3
2
12
d 1
2
3
5
F   x    x8  x 6  x 4  x 2  
dx  4
3
2
12 
 2 x 7  4 x5  6 x3  2 x
F   x    x 6  2 x 4  3 x 2  1  2 x
OR
F   x   f  u  x    u  x 

 x   2  x   3 x   1  dxd  x 
2 3
2 2
2
2
F   x    x 6  2 x 4  3 x 2  1  2 x
Rate of change problems: Integrate the rate of change to find the total change in the underlying
quantity.
mph
t , where t is measured in seconds, find the change in distance after 10
Practice: If v  t   5
sec
seconds.
Calc 1 Lecture Notes
Section 4.5
Page 5 of 5