Transcript CHAPTER 4 SECTION 4.3 RIEMANN SUMS AND DEFINITE …

CHAPTER 4
SECTION 4.3
RIEMANN SUMS AND DEFINITE
INTEGRALS
Riemann Sum
1. Partition the interval [a,b] into n subintervals
a = x0 < x1 … < xn-1< xn = b
•
•
•
•
Call this partition P
The kth subinterval is xk = xk-1 – xk
Largest xk is called the norm, called ||  ||
If all subintervals are of equal length, the norm is
called regular.
2. Choose an arbitrary value from each
subinterval, call it ci
Riemann Sum
3. Form the sum
n
Rn  f (c1 )x1  f (c2 )x2  ...  f (cn )xn   f (ci )xi
i 1
This is the Riemann sum associated with
•
•
•
•
the function f
the given partition P
the chosen subinterval representatives
ci
We will express a variety of quantities in
terms of the Riemann sum
This illustrates that the size of ∆x is
allowed to vary
y = f (x)
x1* x2*
a
x1
x3*
x2 x3
x4*
x5*
x4
x5
Etc…
Then a < x1 < x2 < x3 < x4 ….etc. is a partition of [ a, b ]
Notice the partition ∆x does not have to be the same size
for each rectangle.
And x1* , x2* , x3* , etc… are x coordinates such that
a < x1* < x1, x1 < x2* < x2 , x2 < x3* < x3 , … and are used to
construct the height of the rectangles.
The graph of a typical continuous function y = ƒ(x) over [a, b].
Partition [a, b] into n subintervals a < x1 < x2 <…xn < b. Select any
number in each subinterval ck. Form the product f(ck)xk.
Then take the sum of these products.
n
 f (ck )xk
k 1
This is called the Riemann Sum of the
partition of x.
The width of the largest subinterval of a
partition  is the norm of the partition,
written ||x||.
As the number of partitions, n, gets larger
and larger, the norm gets smaller and
smaller.
As n, ||x|| 0 only if ||x|| are the same
width!!!!
The Riemann Sum
Calculated
• Consider the function
2x2 – 7x + 5
• Use x = 0.1
• Let the ci = left edge
of each subinterval
• Note the sum
x
2x^2-7x+5
9
9.92
10.88
11.88
12.92
14
15.12
16.28
17.48
18.72
20
21.32
22.68
24.08
25.52
27
28.52
30.08
31.68
33.32
dx * f(x)
0.9
0.992
1.088
1.188
1.292
1.4
1.512
1.628
1.748
1.872
2
2.132
2.268
2.408
2.552
2.7
2.852
3.008
3.168
3.332
Riemann sum =
40.04
4
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
5
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
The Riemann Sum
f(x) = 2x2 – 7x + 5
n
 f (c )x
i 1
i
i
 40.04
• We have summed a series of boxes
• If the x were smaller, we would have
gotten a better approximation
Finer partitions of [a, b] create more rectangles with shorter bases.
n
 f (ci )xi
i 1
n
lim  f ( ci )xi  L
 0
i 1
The Definite Integral
n
b
I   f ( x)dx 
a
lim  f  ci  xi
 0
k 1
• The definite integral is the limit of the
Riemann sum
• We say that f is integrable when
– the number I can be approximated as accurate
as needed by making ||  || sufficiently small
– f must exist on [a,b] and the Riemann sum must
exist
–   0 is the same as saying n 

Notation for the definite integral
upper limit of integration
Integration
Symbol

f  x  dx
b
a
integrand
variable of integration
lower limit of integration
Important for AP test
[ and mine too !! ]
Recognizing a Riemann Sum as a Definite integral
3i

3
lim  2(1 
)  1
n 
n
n
i 1 
3
ba
1. x 

x  dx
n
n
3i
2. 1 
 a  ix
so a  1
n
3. b  a  3 so b  4
n
4.
3i
2(1 
)  2x
n
5. Thus

4
 (2 x  1)dx
1
Recognizing a Riemann Sum as a Definite integral
5i 

lim  3 

 n
2
n
n 
i 1
5
n
5
ba
1. x 

n
n
x  dx
5i
2. a=0 b  5 since a  ix =
n
5i
3. x =
n
4. Thus


5
0
( 3 x 2 ) dx
Recognizing a Riemann Sum as a Definite integral
From our textbook
n
lim
 0
[5c 
2
i
i 1
answer:

4
1
 3 ci ] xi
over [ 1,4 ]
(5 x 2  3x )dx
Notice the text uses ∆ instead of ∆x, but it is basically
the same as our ∆x , and ci is our xi *
Try the reverse : write the integral as a
Riemann Sum … also on AP and my test

10
3
2
4
 x  x dx
10-3 7
1. a  3 b  10 so x=

n
n
7i
2. a  ix  3 
n
2


n
7i 

7i  7

3. Thus lim  4  (3 )  ( 3  )

n
n 
n
i 1
n 
 

Theorem 4.4 Continuity Implies
Integrability
Relationship between Differentiability, Continuity, and
Integrability
I
D
C
D – differentiable functions, strongest condition … all Diff ’ble
functions are continuous and integrable.
C – continuous functions , all cont functions are integrable, but not
all are diff ’ble.
I – integrable functions, weakest condition … it is possible they are
not con‘ t, and not diff ‘ble.
Evaluate the following Definite Integral

4
3xdx
2
First … remember these sums and definitions:
n
 c  cn
1
ci = a + 
i x
n
n(n  1)
k  2
1
ba
x 
n
n
3xdx  lim  f (ci )  x i
2
n

4
i 1
n
6 6
 lim  3( 2  i )
n
n n
i 1
18 n
6
 lim
( 2  i )

n  n
n
i 1
18
6 n(n  1)
 lim [ 2n  (
)]
n  n
n
2
1
 lim[ 36  54(1  )]
n 
n
 36  54  18
n
 c  cn
1
n
k 
1
n(n  1)
2
ci = a + i  x
x 
ba
n
EXAMPLE
Evaluate the definite integral by the limit
definition
n
5i
lim  f ( ci )xi  L ci  1 
n
i 1
 0
n
 5i  5
 f 1  n  n
i 1
n
5
x 
n
 5i  5
  1  
n n
i 1 
1 n
25 n
 5  2 i
n i 1
n i 1

6
1
n
x dx
 5 25i 
   2 
n 
i 1  n
Evaluate the definite integral by the
limit definition, continued

6
1
n
x dx
lim  f ( ci )xi  L
 0
i 1

6
1
L   f ( x )dx
a
1
25 n( n  1)
 5n   2
n
2
n
1 n
25 n
 5  2 i
n i 1
n i 1
25 ( n  1)
 5
n 2
b
25
 5   n  1
2n
25 25 

x dx  lim  5   
n
2 2n 

35

2
25 25
 5 
2 2n

b
a
f ( x)dx
The Definite integral above represents the Area
of the region under the curve y = f ( x) ,
bounded by the x-axis, and the vertical lines
x = a, and x = b
y = f ( x)
y
x
a
b
Theorem 4.4 Continuity Implies
Integrability
Relationship between Differentiability, Continuity, and
Integrability
I
D
C
D – differentiable functions, strongest condition … all Diff ’ble
functions are continuous and integrable.
C – continuous functions , all cont functions are integrable, but not
all are diff ’ble.
I – integrable functions, weakest condition … it is possible they are
not con‘ t, and not diff ‘ble.
Areas of common geometric shapes
3
 xdx
0
Y= x
y
0
x
A
2
3
2 3
0
9

2
Sol’n to definite integral
x
1
9
A  33 
2
2
A = ½ base * height
A Sight Integral ... An integral you should know
on sight
 a
a
2
-a
-a
x
2

1
2
dx 
a
2
a
This is the Area of a semi-circle of radius a
Special Definite Integrals
for f (x ) integrable from
1.
2.

a
a

a to b
f ( x )dx  0
b
a
a
f ( x )dx    f ( x )dx
b
EXAMPLE
2
1.
(
x

5
)
dx

0

2.
9
xdx


xdx


3
0
2
2
0
3
Additive property of integrals
If f is integrable over interval [ a , b ],
where a < c < b, then:
b
c
b
a
a
c
 f ( x)dx   f ( x)dx   f ( x)dx
y
x
a
c
b
More Properties of Integrals
For f, g integrable on [ a, b ], and k is a
constant ... , then since kf and f  g
are integrable on [ a, b ], we have :
1.
2.
b
 kf ( x)dx
a
b
 k  f ( x )dx
a
b
b
a
a
 [ f ( x)  g ( x)]dx   f ( x)dx

b
 g ( x)dx
a
EXAMPLE
Given
Solve:

2
1

2
1
7
x dx 
3
2
3
1 xdx  2
2
3( x 2  x ) dx
2
2
 3 x dx + 3 xdx 
1
1
2
7
3
9
23
 3( ) + 3( )  7  
3
2
2
2
Even – Odd Property of Integrals
For f ( x ) an even function:

a
a
a
f ( x )dx  2  f ( x )dx
0
Even function:
f ( x ) = f ( - x ) … symmetric about y - axis
For f ( x ) an odd function:

a
a
f ( x )dx  0
Finally …. Inequality Properties
If f is integrable and nonnegative on [ a, b ] :
0 
b
 f(x)dx
a
If f , g are integrable on [a, b ] , and f(x)  g(x) :
b
b
a
a
 f(x)dx   g( x)dx
END
Rules for definite integrals
Example 2:
Evaluate the using the following values:
4
3
x
   2  dx
2
4
4
4
2
2
2
3
3
x

2
dx

x
    dx    2 dx

4
x
2
4
3
4
 2  dx    x  dx  2 dx
3
2
2
= 60 + 2(2) = 64
Using the TI 83/84 to check
your answers
Find the area undery  3 x
on [1,5]
• Graph f(x)
•
•
•
•
•
Press 2nd CALC 7
Enter lower limit 1
Press ENTER
Enter upper limit 5
Press ENTER.
Set up a Definite Integral for finding the area of the shaded
region. Then use geometry to find the area.
1. f  x   4
2. f  x   x  1
6
6
4
4
2
2
5
5
Use the limit definition to find

3
1
2
3x dx
Set up a Definite Integral for finding the area of the shaded
region. Then use geometry to find the area.
1. f  x   4
2. f  x   x  1
6
6
4
4
2
2
5
5
A   4 dx
1
 4  4
 16 un2
5
6
A    x  1 dx
2
 3  4  21  4 4
rectangle
 20 un2
triangle