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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.
b
A (x)
• A’ (x) = f (x)
• A (a) = 0 and F (x) = A (x) + C
• A (b) = A
F b
( )-F a
( )= A b
( )+C
é
ë ù
û- A a
( )+C
é
ë ù
û=
A b
( )- A a
( )= A-0 = A
The Fundamental Theorem
of Calculus, Part I
f x
( )dx
a
b
ò = F(x)]a
b
How about some
practice?
1. x dx
1
9
ò = x
1
2
dx
1
9
ò = 2x
3
2
3
ù
û
ú
ú
ú1
9
=
2
3
9
3
2
-1
3
2
æ
è
ç
ö
ø
÷ =
2
3
9
3
2
-1
3
2
æ
è
ç
ö
ø
÷ =
2
3
27-1
( ) =
52
3
2.
sin x
5
dx
0
p
2
ò =-
cosx
5
ù
û
ú
0
p
2
=-
1
5
cos
p
2
æ
è
ç
ö
ø
÷-cos0
é
ë
ê
ù
û
ú= -
1
5
0-1
[ ] =
1
5
3. 5ex
dx
0
ln3
ò =5ex ù
û0
ln3
=5 eln3
-e0
( )= 5 3-1
( )=10
More Examples !!!
Evaluate f (x)dx
0
6
ò If f (x) =
x2
, x < 2
3x -2, x ³ 2
ì
í
ï
î
ï
x2
dx
0
2
ò + 3x -2
( )dx
2
6
ò =
x3
3
ù
û
ú
0
2
+
3x2
2
-2x
ù
û
ú
2
6
=
8
3
-0
æ
è
ç
ö
ø
÷+ 42-2
( )=
128
3
TOTAL AREA
A1 A3 A5
a A2 A4 b
Total area = f (x) dx
a
b
ò
Practice Time !!!
Find the total area between the curve y = 1 – x2
and the x-axis over the interval [0, 2].
A = 1- x2
dx
0
2
ò = 1- x2
( )dx
0
1
ò + - 1- x2
( )dx
1
2
ò =
x -
x3
3
æ
è
ç
ö
ø
÷
ù
û
ú
0
1
- x -
x3
3
æ
è
ç
ö
ø
÷
ù
û
ú
1
2
= 1-
1
3
-0
æ
è
ç
ö
ø
÷- 2-
8
3
æ
è
ç
ö
ø
÷- 1-
1
3
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú=
2
3
- -
4
3
æ
è
ç
ö
ø
÷ =
6
3
= 2
The Mean Value Theorem for Integrals:
Over any interval, there exists an x value which creates a y value
that is the height of a rectangle which will equal the area under the
curve.
The Average Value:
The function value, f(c), found by the
Mean Value Theorem
Þ f c
( ) =
1
b-a
f x
( )
a
b
ò
Example
faverage =
1
2-
1
2
x2
+1
x2
æ
è
ç
ö
ø
÷
1
2
2
ò dx =
2
3
1+
1
x2
æ
è
ç
ö
ø
÷
1
2
2
ò dx =
2
3
x -
1
x
æ
è
ç
ö
ø
÷
ù
û
ú
1
2
2
=
2
3
2-
1
2
æ
è
ç
ö
ø
÷-
1
2
-2
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú=
2
3
1
1
2
æ
è
ç
ö
ø
÷- -1
1
2
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú=
2
3
×3= 2
In analyzing the graph of F(x) we would look
at the derivative:
d
dx
F x
( ) =???
F x
( )= sint
( )
a
x
ò dt =-cost]a
x
=-cosx - -cosa
( )=
d
dx
F x
( ) =
d
dx
-cosx +cosa
[ ] =
-cosx+cosa
sin x
f (x)
The Fundamental Theorem of
Calculus, Part II
d
dx
f (t)
a
x
ò dt
é
ë
ê
ù
û
ú= f (x)
1.
d
dx
tan4
t
( )dt
0
x
ò
é
ë
ê
ù
û
ú=
How about some
practice?
tan4
x
2.
d
dx
t3
dt
1
x
ò
é
ë
ê
ù
û
ú= x3
3.
d
dx
sint
t
dt
0
x
ò
é
ë
ê
ù
û
ú=
sin x
x
Integrals with Functions as
Limits of Integration
d
dx
f (t)
a
g(x)
ò dt
é
ë
ê
ù
û
ú= f g x
( )
( )×g' x
( )
Let’s Practice !!!
d
dx
1
t3
dt
2
x2
ò =
1
x2
( )
3
×2x = 1
x6
×2x =
2
x5
d
dx
1-t2
( )dt
1
sin x
ò =1-sin2
x
( )×cosx =cos2
x×cosx = cos3
x

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6.5 & 6.6 & 6.9 the definite integral and the fundemental theorem of calculus copy

  • 1. The Definite Integral and The Fundamental Theorem of Calculus
  • 2. This method used the sum of the area of intervals under a curve- called Reimann Sums
  • 3. The limit of the sums of intervals is the same as a definite integral over the same interval.
  • 4.
  • 5.
  • 6. b A (x) • A’ (x) = f (x) • A (a) = 0 and F (x) = A (x) + C • A (b) = A F b ( )-F a ( )= A b ( )+C é ë ù û- A a ( )+C é ë ù û= A b ( )- A a ( )= A-0 = A
  • 7. The Fundamental Theorem of Calculus, Part I f x ( )dx a b ò = F(x)]a b
  • 8. How about some practice? 1. x dx 1 9 ò = x 1 2 dx 1 9 ò = 2x 3 2 3 ù û ú ú ú1 9 = 2 3 9 3 2 -1 3 2 æ è ç ö ø ÷ = 2 3 9 3 2 -1 3 2 æ è ç ö ø ÷ = 2 3 27-1 ( ) = 52 3 2. sin x 5 dx 0 p 2 ò =- cosx 5 ù û ú 0 p 2 =- 1 5 cos p 2 æ è ç ö ø ÷-cos0 é ë ê ù û ú= - 1 5 0-1 [ ] = 1 5 3. 5ex dx 0 ln3 ò =5ex ù û0 ln3 =5 eln3 -e0 ( )= 5 3-1 ( )=10
  • 9. More Examples !!! Evaluate f (x)dx 0 6 ò If f (x) = x2 , x < 2 3x -2, x ³ 2 ì í ï î ï x2 dx 0 2 ò + 3x -2 ( )dx 2 6 ò = x3 3 ù û ú 0 2 + 3x2 2 -2x ù û ú 2 6 = 8 3 -0 æ è ç ö ø ÷+ 42-2 ( )= 128 3
  • 10. TOTAL AREA A1 A3 A5 a A2 A4 b Total area = f (x) dx a b ò
  • 11. Practice Time !!! Find the total area between the curve y = 1 – x2 and the x-axis over the interval [0, 2]. A = 1- x2 dx 0 2 ò = 1- x2 ( )dx 0 1 ò + - 1- x2 ( )dx 1 2 ò = x - x3 3 æ è ç ö ø ÷ ù û ú 0 1 - x - x3 3 æ è ç ö ø ÷ ù û ú 1 2 = 1- 1 3 -0 æ è ç ö ø ÷- 2- 8 3 æ è ç ö ø ÷- 1- 1 3 æ è ç ö ø ÷ é ë ê ù û ú= 2 3 - - 4 3 æ è ç ö ø ÷ = 6 3 = 2
  • 12. The Mean Value Theorem for Integrals: Over any interval, there exists an x value which creates a y value that is the height of a rectangle which will equal the area under the curve. The Average Value: The function value, f(c), found by the Mean Value Theorem Þ f c ( ) = 1 b-a f x ( ) a b ò
  • 13. Example faverage = 1 2- 1 2 x2 +1 x2 æ è ç ö ø ÷ 1 2 2 ò dx = 2 3 1+ 1 x2 æ è ç ö ø ÷ 1 2 2 ò dx = 2 3 x - 1 x æ è ç ö ø ÷ ù û ú 1 2 2 = 2 3 2- 1 2 æ è ç ö ø ÷- 1 2 -2 æ è ç ö ø ÷ é ë ê ù û ú= 2 3 1 1 2 æ è ç ö ø ÷- -1 1 2 æ è ç ö ø ÷ é ë ê ù û ú= 2 3 ×3= 2
  • 14. In analyzing the graph of F(x) we would look at the derivative: d dx F x ( ) =??? F x ( )= sint ( ) a x ò dt =-cost]a x =-cosx - -cosa ( )= d dx F x ( ) = d dx -cosx +cosa [ ] = -cosx+cosa sin x f (x)
  • 15. The Fundamental Theorem of Calculus, Part II d dx f (t) a x ò dt é ë ê ù û ú= f (x)
  • 16. 1. d dx tan4 t ( )dt 0 x ò é ë ê ù û ú= How about some practice? tan4 x 2. d dx t3 dt 1 x ò é ë ê ù û ú= x3 3. d dx sint t dt 0 x ò é ë ê ù û ú= sin x x
  • 17. Integrals with Functions as Limits of Integration d dx f (t) a g(x) ò dt é ë ê ù û ú= f g x ( ) ( )×g' x ( )
  • 18. Let’s Practice !!! d dx 1 t3 dt 2 x2 ò = 1 x2 ( ) 3 ×2x = 1 x6 ×2x = 2 x5 d dx 1-t2 ( )dt 1 sin x ò =1-sin2 x ( )×cosx =cos2 x×cosx = cos3 x