The Definite Integral
and The Fundamental
Theorem of Calculus
This method used the sum of
the area of intervals under a
curve- called Reimann Sums
The limit of the sums of intervals is the same as
a definite integral over the same interval.
• A’ (x) = f (x)
• A (a) = 0
and F (x) = A (x) + C
• A (b) = A
A (x)
b

F ( b ) - F ( a ) = é A ( b) + C ù - é A ( a ) + C ù =
ë
û ë
û

A ( b) - A ( a ) =

A-0 =

A
The Fundamental Theorem
of Calculus, Part I
b

ò
a

f ( x ) dx = F(x)] a
b
How about some
practice?
ù
3
3
3
3
2x ú
2 æ 2 2 ö 2 æ 9 2 -12 ö = 2 27 -1 = 52
x dx = ò x dx =
÷
(
)
= ç 9 -1 ÷ = ç
ú
3è
3
ø 3
3è
3
1
ø
ú1
û

9

1. ò

9

1

p

3 9
2

1
2

p

ù
cos x ù 2
sin x
1é æp ö
= - êcos ç ÷ - cos0ú = - 1 [ 0 -1] = 1
2. ò
dx = ú
5 û0
5
5ë è 2 ø
û
5
5
0
2

ù = 5 ( eln3 - e0 ) = 5 (3-1) =10
3. ò 5e dx = 5e û0
ln3

0

x

x ln3
More Examples !!!
6

Evaluate

ò f (x)dx If
0

2

ò
0

ì x 2,
ï
f (x) = í
ï3x - 2,
î

x<2
x³2

ù
x ù 3x 2
2
x dx + ò (3x - 2) dx =
- 2x ú =
ú +
3 û0 2
û2
2
6

æ8 ö
128
ç - 0 ÷ + ( 42 - 2) =
è3 ø
3

3 2

6
TOTAL AREA

A1

a

A3

A5

A2

Total area =

A4

b

ò
a

f (x) dx

b

6.3 integration by substitution

  • 1.
    The Definite Integral andThe Fundamental Theorem of Calculus
  • 2.
    This method usedthe sum of the area of intervals under a curve- called Reimann Sums
  • 3.
    The limit ofthe sums of intervals is the same as a definite integral over the same interval.
  • 6.
    • A’ (x)= f (x) • A (a) = 0 and F (x) = A (x) + C • A (b) = A A (x) b F ( b ) - F ( a ) = é A ( b) + C ù - é A ( a ) + C ù = ë û ë û A ( b) - A ( a ) = A-0 = A
  • 7.
    The Fundamental Theorem ofCalculus, Part I b ò a f ( x ) dx = F(x)] a b
  • 8.
    How about some practice? ù 3 3 3 3 2xú 2 æ 2 2 ö 2 æ 9 2 -12 ö = 2 27 -1 = 52 x dx = ò x dx = ÷ ( ) = ç 9 -1 ÷ = ç ú 3è 3 ø 3 3è 3 1 ø ú1 û 9 1. ò 9 1 p 3 9 2 1 2 p ù cos x ù 2 sin x 1é æp ö = - êcos ç ÷ - cos0ú = - 1 [ 0 -1] = 1 2. ò dx = ú 5 û0 5 5ë è 2 ø û 5 5 0 2 ù = 5 ( eln3 - e0 ) = 5 (3-1) =10 3. ò 5e dx = 5e û0 ln3 0 x x ln3
  • 9.
    More Examples !!! 6 Evaluate òf (x)dx If 0 2 ò 0 ì x 2, ï f (x) = í ï3x - 2, î x<2 x³2 ù x ù 3x 2 2 x dx + ò (3x - 2) dx = - 2x ú = ú + 3 û0 2 û2 2 6 æ8 ö 128 ç - 0 ÷ + ( 42 - 2) = è3 ø 3 3 2 6
  • 10.