?⃝
-
Areas and Distances
5. 1
Sigma Notation E
4
9*+10+10-1 d¥=§gai
100
a +
,
Azt
-- -
+ a
, oo
=
[ Ai
c- =\
.
this tells
①→
¥11s
'
us
to end
Ai
us
to add
this tells
n
i =\
us
to start wilt i= i
⑧iÉÉi=nk
$
=
I +2+3+4+5
=
15
wilt i=n
+
-- -
+ n
special sums
a.
i =L
s+i+i+.+ñ=☒
,
,,
,
,
, ,,
,
,
,
[= I
i= 1
proof ①
S
s
=
I
n
=
as
25
+
-
i + n
z
-
+ n
-- -
+
--
-
-
i + n
+
=l+n+Hn+1+n+---??=
n
( I + n)
s=
¥-1
Areas
we know
some
of some regions
how to find the area
regions
,
for instance
A- = tbh
b
A
w
=
W
.
L
:
Az
A
A-
=
A , + Az + A}
,
¥,
I
1
a
b
f
n
for
Question
what about the
:
area
of the regions that lies under the
graph of
Example
a
:
function f.
use
rectangles to estimate
the area under the
from 0 to 1
parabola y=
ot-E.FI =b
.
""
*
⇐
•
•
s
2
.
O
14
12
-
. .
.
.
A,
314
L
?⃝
?⃝
can
-
.
.
-1¥
%
-
UH
*k•o•0•0•
I
④#→→→
we
-
-
-
⇐ •*
,
'
-
Az
Sy
'
As
•
approximate the
area
of each strip Si
A
?⃝
by
a
rectangle Ai whose base is the
same as the
same
as
strip and whose height is the
the
is the value
right edge of the strip ( which
of the function at the right
endpoint of the sub interval that determines Si)
If we let
of these
be the sum
approximating triangles
R④= A
, +
Notice
,
we
get
Az + As + A 4
Hi
=
=
of the areas
0.46875
that
A L Ry
+
④
ai
.
instead of
the values of the
using
function fat ×
at the right endpoints
Now
,
to build
endpoints
'
heights let
,
,
we
get
us
use
the left
1""
f-( ) 02=0
o
%
.
•¥
"
④
Lg
=
=
.
A , + Az + A ] + A 4
=÷iñ+÷i÷i+÷ki+÷l÷i
=
0.21875
We have
we
can
now
↳ < A < Ry
repeat this procedure wilt larger
a
number of strips , we get the
4
0.21875
0.46875
10
0.2850
0.3850
100
0.32835
0-33835
1000
0.3328335 0.3338335
following
data
100000
From the table , it seems that
approaching f-
Rn
( and Ln)
as n→x
This inspires us to do the following
approach
divide
we
each has
Rn
=
I
§
into ① equal sub intervals
length ④
A , + Az + A , +
+
=
-
=
=
- -
+
+
An
-
-
-
÷
1¥
①
+
=É Eni
i= I
n
=÷Éc÷i=÷E÷
c- = I
i= I
n.in?.Ei=-n.E-E-ilimRn=limn--E
?⃝
n→a
new
=-3
6h3
Note ①
f is continuous
Since
.
,
we
could have used Ln and
the
Definition the
:
result
same
area
got
.
A of the regions
that lies under the graph of the continuous
function f- is the limit of the sumof#
=
of
areas
A
Iim Rn
=
n→
Iim [fcx
=
N
n→ a
N
.
=
approximating rectangles
or
-
him
n→ &
fan)D④
f-
+
-
-
f- > DID
i= ,
Hi
.
i
!
n
=
-
7
him [
n→ a
)④+
,
5
I
i
!
§€fl×iD×i=a-i☐
a
i =\
b
☐⇐
¥
Note① If f is continuous
,
the limit in
the definition always exists
.
Note ② If f is continuous , we could take
rectangle to be
the height of the i#
the value of fat
't
i
So
,
*
✗
more
a
, ,
XI
,
. ..
,
✗i
]
,
we
n
call the
In the sample points
✗
general expression for the
Lim [ f- (E) ☐✗ +
=
n→N
.
.
+
- - .
✗i i
✗i
•
•
-
-
0
,
→
If i e
of S is
area
A
any
sub interval [✗i.
numbers
number ✗¥ in the
.
i
fix:) DX]
.
.
.
1
.
exampten
:
sample points ✗¥ ✗¥
,
/ !
i
I
i
,
¥
✗
,
¥
✗
1
:
:
:
-
it
i
i
¥
✗
,
I
I
¥¥×¥×É*
1
Summary
on
:
Let
f be
the interval [ a. b)
a
S is
,
lies under the
graph of f
the
of 5
A
is
area
continuous function
.
A Lim Rn =/ im Éfcxil Dx
=
n→ N
n → do
i =\
the
region
that
and af ✗ f b
.
n
=
him Ln
n→ a
Iim Mn
I
lim[ fix :) Dx
=
n→ a
=
-
=
n ed
i=n0
,✗¥=+i
him [ f- (E) DX
n→ N
i =\
n
=
Lim E fc×¥ > DX
n→ N
i =L
Example
(a)
I
I II.
✗
.
.
i
-
"
#
sample point
midpoint
:
Estimate the area under the graph
of
_×fT-
room
to
✗ =o
✗= 4
using four approximating rectangles
and
right endpoints sketch the
.
graph and the rectangles Is your
estimate underestimate overestimate?
.
an
or
repeat (a) using left endpoints
(b)
M77
⑥ ⑤ ⑤ ④
solution
☐✗
.
9
=
I II.Eagles
,
B.
=
=
↳
=
=
I
(F) + I (F) + ikr ) # if )
6. 14626 → overestimation
1107
+ I
4.14626
(F) + I (F) +105 )
→
underestimation
the distance problem
.
Find the distance traveled by
during
object
an
certain time period if the
a
of the object is known at all times
velocity
If the
it
"
velocity
.
constant
remains
is
4414/1/11
d
=
v. t
-7
speedometer
v
:
at
constant
.
,
€ odometer
>
✗
If He
velocity
Example
we
:
not constant
is
Suppose the
odometer is broken !
want to estimate the
over
a
30
.. .
distance driven
Seconds time interval
.
We take
speedometer readings every five seconds
and record them in the
(5) time
following table
5
10
15
20
25
30
21
24
29
32
31
28
Velocity
17
In order
to have the time and the
(m
in
velocity
consistent units let's convert the velocity
,
readings to feet / second (1mi/h
(5) time
(ft / g)velocity
v
ft / s)
②
5
10
15
20
25
30
25
31
35
43
47
46
41
'
A
I.ie#':f~2o---"
we can
have
an
t
estimation of the driven
distance by adding # rectangles above
we
get
D=
25
(5) + 31157+35 (5) +43 (5) +471s )
+
=
1135
46 (s )
ft
.
t.int?--limEflx.- ⑤
④
had in
r
to
Use
find
an
expression for the area under the
graph of
as
evaluate the limit
fcx )
☐✗
✗
i =
=
¥-1
limit
a
.
1
i
a + ibx
EX C- 3
=-3
=
=
Do not
.
=/ + i
.
(E)
=i
,,b
I
a
✗,
a + DX
Xz
✗i
a -12 DX
a + c- DX
fcxi )
=
=
_É÷
I
( -45
it
+
=
(I +
A Iim
=
n..
+
I
⑤
17
.
Determine
region
a
1imÉ
☐✗
=
a- o
,
✗i
f-a)
He
-
-
=
o
tan
✗
i =L
Eb
✗i
equal to
is
hex
whose area
=
_
=
=
+ i
(E)
=
i÷
tan Cx)
calculates the
formula
given
under
the
graph
of
FCN
=
area
tank )
✗ =o →
§
another option
☐✗
4-
=
b- a-
→
+
=
i¥
1imÉ
hex
In
a
3¥
be
a
→
4-
¥
tan
c- =L
y=tan( ✗
-
,
[ 3£ _n]
,