CHAPTER 4 SECTION 4.3 RIEMANN SUMS AND DEFINITE INTEGRALS
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CHAPTER 4SECTION 4.3
RIEMANN SUMS AND DEFINITE INTEGRALS
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Riemann Sum
1. Partition the interval [a,b] into n subintervalsa = 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 ic
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Riemann Sum3. Form the sum
This is the Riemann sum associated with• the function f• the given partition P• the chosen subinterval representatives
• We will express a variety of quantities in terms of the Riemann sum
1 1 2 21
( ) ( ) ... ( ) ( )n
n n n i ii
R f c x f c x f c x f c x
1 1 2 21
( ) ( ) ... ( ) ( )n
n n n i ii
R f c x f c x f c x f c x
ic
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This illustrates that the size of ∆x is allowed to vary
a x1 x2 x3 x4 x5
x1* x2* x3* x4* x5*
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.
y = f (x)
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.
Etc…
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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 cck.k. Form the product f(ck)xk. Then take the sum of these products.
1
( )n
k kk
f c x
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This is called the Riemann SumRiemann Sum of the partition of x.
The width of the largest subinterval of a partition is the normnorm 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!!!!
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The Riemann SumCalculated
• Consider the function2x2 – 7x + 5
• Use x = 0.1
• Let the = left edgeof each subinterval
• Note the sum
x 2x 2̂-7x+5 dx * f(x)4 9 0.9
4.1 9.92 0.9924.2 10.88 1.0884.3 11.88 1.1884.4 12.92 1.2924.5 14 1.44.6 15.12 1.5124.7 16.28 1.6284.8 17.48 1.7484.9 18.72 1.872
5 20 25.1 21.32 2.1325.2 22.68 2.2685.3 24.08 2.4085.4 25.52 2.5525.5 27 2.75.6 28.52 2.8525.7 30.08 3.0085.8 31.68 3.1685.9 33.32 3.332
Riemann sum = 40.04
x 2x 2̂-7x+5 dx * f(x)4 9 0.9
4.1 9.92 0.9924.2 10.88 1.0884.3 11.88 1.1884.4 12.92 1.2924.5 14 1.44.6 15.12 1.5124.7 16.28 1.6284.8 17.48 1.7484.9 18.72 1.872
5 20 25.1 21.32 2.1325.2 22.68 2.2685.3 24.08 2.4085.4 25.52 2.5525.5 27 2.75.6 28.52 2.8525.7 30.08 3.0085.8 31.68 3.1685.9 33.32 3.332
Riemann sum = 40.04
ic
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The Riemann Sum
• We have summed a series of boxes
• If the x were smaller, we would have gotten a better approximation
f(x) = 2x2 – 7x + 5
1
( ) 40.04n
i ii
f c x
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Finer partitions of [a, b] create more rectangles with shorter bases.
0 1
lim ( )n
i ii
f c x L
1
( )n
i ii
f c x
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The Definite Integral
• 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– is the same as saying n 0
0
1
lim( )n
k
b
a i if f c xxI dx
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b
af x dx
IntegrationSymbol
lower limit of integration
upper limit of integration
integrandvariable of integration
Notation for the definite integral
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Important for AP test [ and mine too !! ]
Recognizing a Riemann Sum as a Definite integral
lim ( )
.
.
.
. ( )
. ( )
ni
n i
n n
xn
b a
nx dx
i
na i x so a
b a so b
i
nx
Thus x dx
2 13
13
13
2 13
1
3 3 4
4 2 13
2
5 2 1
1
1
4
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Recognizing a Riemann Sum as a Definite integral
lim
.
.
.
. ( )
ni
n i
n n
xn
b a
nx dx
a= b s a i xi
n
xi
n
Thus x dx
ince =
=
35 5
15
2 0 55
35
4 3
2
1
2
0
5
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Recognizing a Riemann Sum as a Definite integral
From our textbook
lim ]
: (5 )
[
0
2
1
2
1
4
5 3
3
ii
n
i ic c x
answer x x dx
over [ 1,4 ]
Notice the text uses ∆ instead of ∆x, but it is basically the same as our ∆x , and ci is our xi *
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Try the reverse : write the integral as a Riemann Sum … also on AP and my test
4
1 3 1010 3 7
2 37
3 4 37 7
2
3
10
2
1
37
x x dx
a b x=-
n n
a i xi
n
Thusi
n nni
n i
n
.
.
. lim ( )( )
so
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Theorem 4.4 Continuity Implies Integrability
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D CI
Relationship between Differentiability, Continuity, and Integrability
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.
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Evaluate the following Definite Integral 32
4xdx
c cnn
1
k1
n
n(n 1)
2
First … remember these sums and definitions:
ci = a + i x xb a
n
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lim ( )
lim [ (( )
)]
lim[ ( )]
ni
n
n
n
ni
n
nn
n
n n
n
182
6
182
6 1
2
36 54 11
36 54 18
1
3
3 26 6
2
4
1
1
xdx f c x
in n
n ii
n
i
ni
n
lim ( )
lim ( )
c cnn
1
k1
n
n(n 1)
2
ci = a + i x
xb a
n
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EXAMPLE Evaluate the definite integral by the limit Evaluate the definite integral by the limit definitiondefinition
0 1
lim ( )n
i ii
f c x L
6
1x dx
1
5 51
n
i
if
n n
1
5 51
n
i
i
n n
21
5 25n
i
i
n n
21 1
1 255
n n
i i
in n
5x
n
51i
ic
n
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Evaluate the definite integral by the Evaluate the definite integral by the limit definition, continuedlimit definition, continued
( )b
a
L f x dx0 1
lim ( )n
ii
f c xi L
6
1x dx
2
1 25 ( 1)5
2
n nn
n n
255 1
2n
n
6
1
25 25lim 5
2 2nx dx
n
25 255
2 2n
35
2
25 ( 1)5
2
n
n
21 1
1 255
n n
i i
in n
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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)
a b
y
x
b
adxxf )(
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Theorem 4.4 Continuity Implies Integrability
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D CI
Relationship between Differentiability, Continuity, and Integrability
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.
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3
Y = x
x
y
xdx0
3
Areas of common geometric shapes
Ax
A 2
0
3
2
9
2
1
23 3
9
2
Sol’n to definite integral A = ½ base * height
0
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A Sight Integral ... An integral you should know on sight
2 2 21
2a x dx a
a
a
-
This is the Area of a semi-circle of radius a
a-a
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1 0
2
. ( )
. ( ) ( )
f x dx
f x dx f x dx
a
a
a
b
b
a
Special Definite Integrals
for f (x ) integrable from a to b
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1 5 0
29
2
2
2
3
0
0
3
. ( )
.
x dx
xdx xdx
EXAMPLE
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Additive property of integrals
If is integrable over interval [ ],
where , then:
f a , b
a < c < b
f x dx f x dx f x dxa
b
a
c
c
b( ) ( ) ( )
a bc
y
x
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More Properties of Integrals
For integrable on [ ], and is a
constant ... , then since and
are integrable on [ ], we have :
1.
f, g a, b k
kf f g
a, b
kf x dx k f x dx
f x g x dx f x dx g x dx
a
b
a
b
a
b
a
b
a
b
( ) ( )
. [ ( ) ( )] ( ) ( )2
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EXAMPLE
Given x dx xdx
Solve x x dx
x dx xdx
+ 3
) + 3(3
2
2
2
2
1 1
2
2
1
2
2
1 1
2
7
3
3
2
3
3
37
37
9
2
23
2
: ( )
( )
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Even – Odd Property of Integrals
For an even function:
For an odd function:
f ( x )
f x dx f x dx
f ( x )
f x dx
a
a
a
a
a
( ) ( )
( )
2
0
0
Even function: f ( x ) = f ( - x ) … symmetric about y - axis
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Finally …. Inequality Properties
If is integrable and nonnegative on [ ] :
0
f a, b
f(x)dxa
b
If , are integrable on [ ] , and :
f g a, b f(x) g(x)
f(x)dx g x dxa
b
a
b
( )
END
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Rules for definite integralsEvaluate the using the following values:
Example 2:
4
3
2
2x dx
4 4 4
3 3
2 2 2
2 2x dx x dx dx
4 4 4
3 3
2 2 2
2 2x dx x dx dx = 60 + 2(2) = 64
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Using the TI 83/84 to checkcheck your answers
Find the area under on [1,5]
• Graph f(x)
• Press 2nd CALC 7• Enter lower limit 1 • Press ENTER• Enter upper limit 5• Press ENTER.
3y x
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Set up a Definite Integral for finding the area of the shaded region. Then use geometry to find the area.
1. 4f x 2. 1f x x 6
4
2
5
6
4
2
5
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Use the limit definition to find
3 2
13x dx
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Set up a Definite Integral for finding the area of the shaded region. Then use geometry to find the area.
1. 4f x 2. 1f x x 6
4
2
5
6
4
2
5
5
1
4A dx 6
2
1x dA x 4 4
216 un
123 4 4 4
rectangle triangle
220 un