chapter 4 section 4.2 area. sigma (summation) notation review in this case k is the index of...
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CHAPTER 4SECTION 4.2
AREA
Sigma (summation) notation Sigma (summation) notation REVIEWREVIEW5
1
1 2 3 4 5k
k
In this case k is the index of summation
The lower and upper bounds of summation are 1 and 5
63 3 3 3 3 3 3
1
1 2 3 4 5 6
i
i
In this case i is the index of summation
The lower and upper bounds of summation are 1 and 6
63
1
1 8 27 64 125 216
i
i
Sigma notationSigma notation
4
1
1 2 3 4 163
1 2 3 4 5 60k
k
k
3
1 1 2 2 3 31
( ) ( ) ( ) ( )k kk
f x x f x x f x x f x x
Sigma Summation Sigma Summation NotationNotation
Practice with Practice with Summation NotationSummation Notation
= 3080= 3080
Practice with Summation Practice with Summation NotationNotation
Numerical Problems can be done with the Numerical Problems can be done with the TI83+/84 as was done in PreCalc AlgebraTI83+/84 as was done in PreCalc Algebra
Sum is in LIST, MATHSum is in LIST, MATH
Seq is on LIST, OPSSeq is on LIST, OPS
Area Under a Curve by Limit DefinitionArea Under a Curve by Limit Definition
The area under a curve can be approximated by the sum of rectangles. The figure on the left shows inscribed rectangles while the figure on the right shows circumscribed rectangles
This gives the upper sum.
This gives the lower sum.
211
8y x 0 4x
Left endpoint approximation:
Approximate area: 1 1 1 31 1 1 2 5 5.75
8 2 8 4
(too low)
Approximate area: 1 1 1 31 1 2 3 7 7.75
8 2 8 4
211
8y x 0 4x
Right endpoint approximation:
(too high)
Averaging the right and left endpoint approximations:
7.75 5.756.75
2
(closer to the actual value)
Approximating definite integrals:different choices for the sample points
• If xi* is chosen to be the left endpoint of the interval, then xi* = xi-1 and we have
• If xi* is chosen to be the right endpoint of the interval, then xi* = xi and we have
• Ln and Rn are called the left endpoint approximation and right endpoint approximation , respectively.
xxfRdxxfb
a
n
iin
)()(1
xxfLdxxfb
a
n
iin
)()(
11
xxfdxxfb
a
n
ii
)()(1
*
1.031251.28125
1.781252.53125
211
8y x 0 4x
Can also apply midpoint approximation:choose the midpoint of the subinterval as the sample point.
Approximate area: 6.625
The midpoint rule gives a closer approximation than the trapezoidal rule, but in the opposite direction.
Midpoint rule
],[ ofmidpoint )( and
where
)]()()([)(
112
1
21
iiiii
nn
b
a
xxxxx
n
abx
xfxfxfxMdxxf
Approximating the Area of a Plane Region
5
1
2
3
4
2/5 4/5 6/5 8/5 2
f(x) = -x2 + 5
2/5 4/5 6/5 8/5 2
f(x) = -x2 + 5
x
y
x
y
To approximate the area under each curve, you must sum the area of each rectangle.
See next slide
a. b.
5
4
3
2
1
a. The right endpoints, Mi, of the intervals are 2i/5, where i = 1, 2, 3, 4, 5. The width of each rectangle is 2/5 and the height of each rectangle can be obtained by evaluating f at the right endpoint of each interval.
[0, 2/5], [2/5, 4/5], [4/5, 6/5], [6/5, 8/5], [8/5, 10/5]
Evaluate f(x) at the right endpoints of each of these intervals.
The sum of the area of the five rectangles is
Height Width
Because each of the five rectangles lies inside the parabolic region, you can conclude that the area of the parabolic region is greater than 6.48.
5
1
5
1
2
48.625
162
5
25
5
2
5
2
5
2
i i
iif
Approximating the Area of a Plane Region for b (con’t)
b. The left endpoints of the five intervals are 2/5(i _ 1), where i = 1, 2, 3,
4, 5. The width of each rectangle is 2/5, and the height of each
rectangle can be found by evaluating f at the left endpoint of each
interval.
Height Width
Because the parabolic region lies within the union of the five rectangular
region, that the area of the parabolic region is less than 8.08.
6.48 < Area of region < 8.08
5
1
5
1
2
08.825
202
5
25
5
22
5
2
5
22
i i
iif
ON CALCULATOR
2 2( (( ((2 2) / 5) 5) , ,1,5) 8.08
5sum seq x x
2 2( (( ((2 / 5) 5) , ,1,5) 6.48
5sum seq x x
In general for the upper sum S(n) and Lower sum s(n), you use the following for curves f(x) bound between x=a and x=b.
i1
1
b-a( ) ( ) , where M = a + i( x) (the right endpoint)
n
b-a( ) ( ) , where x = m a + (i-1)( x) (the left endpoint)
n i
n
ii
n
ii
S n f M x x and
s n f m x and
Finding Upper and Lower Sums for a RegionFind the upper and lower sums for the region bounded by the graph of
f(x) = x2
and the x-axis between x = 0 and x = 2
Solution Begin by partitioning the interval [0, 2] into n subintervals, each of length
nnn
abx
202
1 2 1 2
f(x) = x2f(x) = x2
1
2
3
4 4
3
2
1
A. B.
Left endpoints Right endpoints
n
i
nimi
12210
n
i
niM i
220
2
233
3
1i1i1i
23
1i
23
2
1i1i
3
44
3
832
3
4
2
12
6
1218
128
128212
212
is sumlower theendpoints,left the UsingA.
nnnnn
n
nnnnnn
n
iin
iinnn
i
nn
ifxmfns
nnn
nn
i
nn
i
2
233
31
23
1
2
11
3
44
3
832
3
4
6
12188
2222
is sumupper theendpoints,right the UsingB.
nnnnn
n
nnn
ni
n
nn
i
nn
ifxMfnS
n
i
n
i
n
i
n
ii
2 2
Notice that for any value of the lower sum is less than
8 4 4 8 4 4(or equal to) the upper sum
3 3 3 3The difference between these two sums lessens as increases.
If you take the limit as
n
s n S nn n n n
n
8 , both the upper and lower sum approaches
3This leads us to the Limit of the Lower and Upper Sums Theorem
n
Limit of the Lower and Upper Sums
1 1
Let be continuous and nonnegative on the interval [ ]. The limits as
of both the lower and upper sums exist and are equal to each other. That is,
lim lim limn n
i in n n
i i
f a, b n
f m x f M x
where and and are the minimum and maximum
values of on the th subinterval.i i
S n
x b a n f m f M
f i
Definition of the Area of a Region in the Plane
11
Let be continuous and nonnegative on the interval [ ]. The area of
the region bounded by the graph of , the -axis, and the vertical lines
and is
Area lim , n
i in
i
f a, b
f x
x a x b
f c x x
where
ic x
x b a n
This gives the lower sum.
If the width of each of nn rectangles is x, and the height is the minimumminimum value of f in the rectangle, f(Mi), then the area is the limit of the area of the rectangles as n
Area Under a Curve by Limit DefinitionArea Under a Curve by Limit Definition
Area under a curve by limit Area under a curve by limit definitiondefinition
This gives the upper sum.
If the width of each of nn rectangles is x, and the height is the maximummaximum value of f in the rectangle, f(mi), then the area is the limit of the area of the rectangles as n
Area under a curve by limit Area under a curve by limit definitiondefinition
The limit as n The limit as n of the Upper Sum = of the Upper Sum =
The limit as n The limit as n of the Lower Sum = of the Lower Sum =
The area under the curve between x = a and x = b.The area under the curve between x = a and x = b.
Theorem 4.3 Limits of the Lower and Upper Sums
Definition of the Area of a Region in the Plane
VisualizationVisualization
iithth interval interval
f(cf(cii))
Width = Width = ΔΔxx
ccii
Example: Area under a curve by limit Example: Area under a curve by limit definitiondefinition
Find the area of the region bounded by the graph f(x) = 2x – x3 , the x-axis, and the vertical lines x = 0 and x = 1, as shown in the figure.
Area under a curve by limit definitionArea under a curve by limit definition
Why is right, endpoint Why is right, endpoint i/ni/n??
Suppose the interval from 0 Suppose the interval from 0 to 1 is divided into 10 to 1 is divided into 10 subintervals, the endpoint subintervals, the endpoint of the first one is 1/10, of the first one is 1/10, endpoint of the second one endpoint of the second one is 2/10 … so the right is 2/10 … so the right endpoint of the endpoint of the iithth is is ii/10/10..
Visualization againVisualization again
iithth interval interval
f(cf(cii))
Width = Width = ΔΔx=x=
cci i = i/n= i/n
b a
n
1
3
1
lim ( )
1lim 2
n
in
i
n
ni
Area f c x
i i
n n n
32 4
1 1
2 1lim
n n
ni i
i in n
Find the area of the region bounded by the graph Find the area of the region bounded by the graph f(x) = 2x – xf(x) = 2x – x33 on [0, 1] on [0, 1]
2 2
2 4
2 ( 1) 1 ( 1)lim
2 4n
n n n n
n n
Sum of all the rectanglesSum of all the rectangles
Right endpoint Right endpoint
Sub Sub for for x in f(x)x in f(x)
Use rules of summationUse rules of summation
ii n
c
ii n
c
…continued1
3
1
lim ( )
1lim 2
n
ini
n
ni
Area f c x
i i
n n n
32 4
1 1
2 2lim
n n
ni i
i in n
2
2
( 1) ( 1)lim
4n
n n
n n
2 2
2 4
2 ( 1) 1 ( 1)lim
2 4n
n n n n
n n
2
1 1 1 1lim 1
4 2 4n n n n
1 3
14 4
Foil & SimplifyFoil & Simplify
2
2
1 2 1lim 1
4n
n n
n n
The area of the region bounded by the graph f(x) = 2x – x3 , the x-axis, and the vertical lines x = 0 and x = 1, as shown in the figure = .75
0.750.75
Practice with LimitsPractice with Limits
2
2
1 ( )lim
2n
n n
n
1 1lim 1
2n n
2
1 ( 1)lim
2n
n n
n
1 11 0
2 2
Multiply Multiply outout
SeparateSeparate