4.4
The Fundamental
Theorem of Calculus
If a function is continuous on the closed interval [a, b],
then
b
Ú f ( x)dx = F (b) - F (a)
a
where F is any function that F’(x) = f(x)
" x in [a, b].
Examples
2
2
2
˘
x
Ê8
ˆ Ê1 ˆ
- 3x˙ = Á - 6 ˜ - Á - 3 ˜ = ( x - 3)dx =
3
3
3
3
Ë
¯
Ë
¯
˚
1
1
Ú
3
2
4
2 ˘
2
!
3
x
12
˙ = 16 - 2 = 14
3 x dx = 3 x dx =
3 ˙
1
1
˚1
4
4
Ú
p
3
Ú
4
p 4
2
sec
x
dx
=
tan x ]0 =
Ú
0
1–0=1
Integration with Absolute Value
2
Ú 2 x - 1dx =
0
We need to rewrite the integral
into two parts.
12
2
0
12
1
2
Ú -(2x -1) dx + Ú (2x -1) dx
= [-x + x ]
2
12
0
+ [ x - x]
2
2
12
Ê 1 1ˆ
Ê1 1ˆ 5
= Á - + ˜ - (0 + 0) + (4 - 2) - Á - ˜ =
Ë 4 2¯
Ë4 2¯ 2
Ex. Find the area of the region bounded by y = 2x2 – 3x + 2,
the x-axis, x = 0, and x = 2.
A=
2
2
2x
Ú ( - 3x + 2) dx
0
2
˘
2 x 3x
=
+ 2 x˙
3
2
˚0
3
2
10
16
= -6+4 =
3
3
The Mean Value Theorem for Integrals
If f is continuous on [a, b], then
open interval (a, b)
'
b
Ú f ( x)dx = f (c)(b - a)
$ a number c in the
rectangle area is equal
to actual area under curve.
a
a
b
inscribed rectangle
a
b
Mean Value rect.
a
b
Circumscribed Rect
Find the value c guaranteed by the Mean Value Theorem for
Integrals for the function f(x) = x3 over [0, 2].
b
Ú f ( x)dx = f (c)(b - a)
a
2
Ú x dx = f (c)(2 - 0)
3
0
4 = 2c3
8
c3 = 2
c = 3 2 = 1.2599
2
x ˘
˙ =4
4 ˚0
4
2
Average Value
If f is continuous on [a, b], then the average value of
f on this interval is given by
b
1
f (c ) =
f ( x)dx
Ú
b-a a
Find the average value of
f(x) = 3x2 – 2x
on [1, 4].
4
1
2
(
3
x
- 2 x)dx
Ú
4 -1 1
1 3
2 4
= x -x 1
3
[
]
1
= [64 - 16 - (1 - 1)]
3
= 16
40
16
(1,1)
Ave. = 16
The Second Fundamental Theorem
of Calculus
If f is continuous on an open interval I containing a, then
for every x in the interval
x
È
˘
d
Í Ú f (t )dt ˙ = f ( x)
dx Î a
˚
Ex.
Apply the Second Fund. Thm. of Calculus
x
d
2
t + 1 dt =
Ú
dx - 2
0
x2 +1
x
d
du
d
du
1
=
2
2 = Ú
Ú
dx x 1 + u
dx 0 1 + u
1+ x2
Find the derivative of F(x) =
x3
Ú cos t dt
p 2
Apply the second Fundamental Theorem of Calculus
with the Chain Rule.
= (cos x 3 )( 3x 2 )