Lesson 24: Area and Distances

Section 5.1
Areas and Distances
V63.0121.002.2010Su, Calculus I

New York University

June 16, 2010

Announcements

Quiz Thursday on 4.1–4.4

. . . . . .

Announcements

Quiz Thursday on 4.1–4.4

. . . . . .

V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 2 / 31

Objectives

Compute the area of a
region by approximating it
with rectangles and letting
the size of the rectangles
tend to zero.
Compute the total distance
traveled by a particle by
approximating it as
distance = (rate)(time) and
letting the time intervals
over which one
approximates tend to zero.

. . . . . .


Outline

Area through the Centuries
Euclid
Archimedes
Cavalieri

Generalizing Cavalieri’s method
Analogies

Distances
Other applications

. . . . . .


Easy Areas: Rectangle

Definition
The area of a rectangle with dimensions ℓ and w is the product A = ℓw.

w
.

.
.
ℓ

It may seem strange that this is a definition and not a theorem but we
have to start somewhere.

. . . . . .


Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.

.
b
.

. . . . . .



h
.

.
b
.

. . . . . .



h
.

.

. . . . . .



h
.

.
b
.

So
Fact
The area of a parallelogram of base width b and height h is

A = bh
. . . . . .


Easy Areas: Triangle
By copying and pasting, a triangle can be made into a parallelogram.

.
b
.

. . . . . .



h
.

.
b
.

. . . . . .



h
.

.
b
.

So
Fact
The area of a triangle of base width b and height h is

1
A= bh
2
. . . . . .


Easy Areas: Other Polygons

Any polygon can be triangulated, so its area can be found by summing
the areas of the triangles:

.

.

. . . . . .


Hard Areas: Curved Regions

.

???

. . . . . .


Meet the mathematician: Archimedes

Greek (Syracuse), 287 BC
– 212 BC (after Euclid)
Geometer
Weapons engineer

. . . . . .


Archimedes and the Parabola

.

Archimedes found areas of a sequence of triangles inscribed in a
parabola.

A=

. . . . . .



1
.

.

parabola.

A=1

. . . . . .



1
.
.1
8 .1
8

.

parabola.
1
A=1+2·
8

. . . . . .



1 1
.64 .64
1
.
.1
8 .1
8

1 1
.64 .64
.

parabola.
1 1
A=1+2· +4· + ···
8 64

. . . . . .



1 1
.64 .64
1
.
.1
8 .1
8

1 1
.64 .64
.

parabola.
1 1
A=1+2· +4· + ···
8 64
1 1 1
=1+ + + ··· + n + ···
4 16 4
. . . . . .


Summing the series
[label=archimedes-parabola-sum] We would then need to know the
value of the series
1 1 1
1+ + + ··· + n + ···
4 16 4

. . . . . .


Summing the series
value of the series
1 1 1
1+ + + ··· + n + ···
4 16 4
But for any number r and any positive integer n,

(1 − r)(1 + r + · · · + rn ) = 1 − rn+1

So
1 − rn+1
1 + r + · · · + rn =
1−r

. . . . . .


Summing the series
value of the series
1 1 1
1+ + + ··· + n + ···
4 16 4

(1 − r)(1 + r + · · · + rn ) = 1 − rn+1

So
1 − rn+1
1 + r + · · · + rn =
1−r
Therefore
1 1 1 1 − (1/4)n+1
1+ + + ··· + n =
4 16 4 1 − 1/4
. . . . . .


Summing the series
value of the series
1 1 1
1+ + + ··· + n + ···
4 16 4

(1 − r)(1 + r + · · · + rn ) = 1 − rn+1

So
1 − rn+1
1 + r + · · · + rn =
1−r
Therefore
1 1 1 1 − (1/4)n+1 1 4
1+ + + ··· + n = →3 =
4 16 4 1− 1/4 /4 3
as n → ∞. . . . . . .


Cavalieri

Italian,
1598–1647
Revisited the
area
problem with
a different
perspective

. . . . . .


Cavalieri's method

Divide up the interval into
2
y
. =x pieces and measure the area of
the inscribed rectangles:

. .
0
. 1
.

. . . . . .


Cavalieri's method

2
y

1
L2 =
8

. . .
0
. 1 1
.
.
2

. . . . . .


Cavalieri's method

2
y

1
L2 =
8
L3 =

. . . .
0
. 1 2 1
.
. .
3 3

. . . . . .


Cavalieri's method

2
y

1
L2 =
8
1 4 5
L3 = + =
27 27 27

. . . .
0
. 1 2 1
.
. .
3 3

. . . . . .


Cavalieri's method

2
y

1
L2 =
8
1 4 5
L3 = + =
27 27 27
L4 =
. . . . .
0
. 1 2 3 1
.
. . .
4 4 4

. . . . . .


Cavalieri's method

2
y

1
L2 =
8
1 4 5
L3 = + =
27 27 27
1 4 9 14
L4 = + + =
. . . . . 64 64 64 64
0
. 1 2 3 1
.
. . .
4 4 4

. . . . . .


Cavalieri's method

2
y

1
L2 =
8
1 4 5
L3 = + =
27 27 27
1 4 9 14
L4 = + + =
. . . . . . 64 64 64 64
0
. 1 2 3 4 1
. L5 =
. . . .
5 5 5 5

. . . . . .


Cavalieri's method

2
y

1
L2 =
8
1 4 5
L3 = + =
27 27 27
1 4 9 14
L4 = + + =
. . . . . . 64 64 64 64
1 4 9 16 30
0
. 1 2 3 4 1
. L5 = + + + =
. . . . 125 125 125 125 125
5 5 5 5

. . . . . .


Cavalieri's method

2
y

1
L2 =
8
1 4 5
L3 = + =
27 27 27
1 4 9 14
L4 = + + =
. . 64 64 64 64
1 4 9 16 30
0
. 1
. L5 = + + + =
. 125 125 125 125 125
Ln =?

. . . . . .


What is Ln ?
1
Divide the interval [0, 1] into n pieces. Then each has width .
n

. . . . . .


What is Ln ?
1
Divide the interval [0, 1] into n pieces. Then each has width . The
n
rectangle over the ith interval and under the parabola has area
( )
1 i − 1 2 (i − 1)2
· = .
n n n3

. . . . . .


What is Ln ?
1
n
( )
1 i − 1 2 (i − 1)2
· = .
n n n3
So
1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2
Ln = + 3 + ··· + =
n3 n n3 n3

. . . . . .


What is Ln ?
1
n
( )
1 i − 1 2 (i − 1)2
· = .
n n n3
So
1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2
Ln = + 3 + ··· + =
n3 n n3 n3
The Arabs knew that
n(n − 1)(2n − 1)
1 + 22 + 32 + · · · + (n − 1)2 =
6
So
n(n − 1)(2n − 1)
Ln =
6n3
. . . . . .


What is Ln ?
1
n
( )
1 i − 1 2 (i − 1)2
· = .
n n n3
So
1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2
Ln = + 3 + ··· + =
n3 n n3 n3
The Arabs knew that
n(n − 1)(2n − 1)
1 + 22 + 32 + · · · + (n − 1)2 =
6
So
n(n − 1)(2n − 1) 1
Ln = 3
→
6n 3
as n → ∞. . . . . . .


Cavalieri's method for different functions
Try the same trick with f(x) = x3 . We have
( ) ( ) ( )
1 1 1 2 1 n−1
Ln = · f + ·f + ··· + · f
n n n n n n

. . . . . .


( ) ( ) ( )
1 1 1 2 1 n−1
Ln = · f + ·f + ··· + · f
n n n n n n
1 1 1 23 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3

. . . . . .


( ) ( ) ( )
1 1 1 2 1 n−1
Ln = · f + ·f + ··· + · f
n n n n n n
1 1 1 23 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
3 3
1 + 2 + 3 + · · · + (n − 1)3
=
n4

. . . . . .


( ) ( ) ( )
1 1 1 2 1 n−1
Ln = · f + ·f + ··· + · f
n n n n n n
1 1 1 23 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
3 3
1 + 2 + 3 + · · · + (n − 1)3
=
n4
The formula out of the hat is
[ ]2
1 + 23 + 33 + · · · + (n − 1)3 = 1
2 n(n − 1)

. . . . . .


( ) ( ) ( )
1 1 1 2 1 n−1
Ln = · f + ·f + ··· + · f
n n n n n n
1 1 1 23 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
3 3
1 + 2 + 3 + · · · + (n − 1)3
=
n4
The formula out of the hat is
[ ]2
1 + 23 + 33 + · · · + (n − 1)3 = 1
2 n(n − 1)

So
n2 (n − 1)2 1
Ln = →
4n4 4
as n → ∞. . . . . . .


Cavalieri's method with different heights

1 13 1 23 1 n3
Rn = · 3 + · 3 + ··· + · 3
n n n n n n
3 3 3
1 + 2 + 3 + ··· + n 3
=
n4
1 [1 ]2
= 4 2 n(n + 1)
n
n2 (n + 1)2 1
= →
4n4 4
.
as n → ∞.

. . . . . .


Cavalieri's method with different heights

1 13 1 23 1 n3
Rn = · 3 + · 3 + ··· + · 3
n n n n n n
3 3 3
1 + 2 + 3 + ··· + n 3
=
n4
1 [1 ]2
= 4 2 n(n + 1)
n
n2 (n + 1)2 1
= →
4n4 4
.
as n → ∞.
So even though the rectangles overlap, we still get the same answer.

. . . . . .


Outline

Euclid
Archimedes
Cavalieri

Analogies

Distances
Other applications

. . . . . .


Cavalieri's method in general
.
Let f be a positive function defined on the interval [a, b]. We want to find the
area between x = a, x = b, y = 0, and y = f(x).
For each positive integer n, divide up the interval into n pieces. Then
b−a
∆x = . For each i between 1 and n, let xi be the nth step between a and
n
b. So

x0 = a
b−a
x1 = x0 + ∆x = a +
n
b−a
x2 = x1 + ∆x = a + 2 ·
n
. ······
b−a
xi = a + i ·
n
. . . . . . . . ······
a
.
. 0 . 1 . 2 . . . . i. n−1..n
xx x b b−a
x x x xn = a + n · =b
n
. . . . . .


Forming Riemann sums

We have many choices of how to approximate the area:

Ln = f(x0 )∆x + f(x1 )∆x + · · · + f(xn−1 )∆x
Rn = f(x1 )∆x + f(x2 )∆x + · · · + f(xn )∆x
( ) ( ) ( )
x0 + x1 x1 + x2 xn−1 + xn
Mn = f ∆x + f ∆x + · · · + f ∆x
2 2 2

. . . . . .


Forming Riemann sums

We have many choices of how to approximate the area:

Ln = f(x0 )∆x + f(x1 )∆x + · · · + f(xn−1 )∆x
Rn = f(x1 )∆x + f(x2 )∆x + · · · + f(xn )∆x
( ) ( ) ( )
x0 + x1 x1 + x2 xn−1 + xn
Mn = f ∆x + f ∆x + · · · + f ∆x
2 2 2

In general, choose ci to be a point in the ith interval [xi−1 , xi ]. Form the
Riemann sum
Sn = f(c1 )∆x + f(c2 )∆x + · · · + f(cn )∆x
∑
n
= f(ci )∆x
i=1

. . . . . .


Theorem of the Day

Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then

∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1

exists and is the same value no . .
a
. ..1
xb
matter what choice of ci we
made.

. . . . . .


Theorem of the Day

Theorem

∑
n
n→∞ n→∞
i=1

exists and is the same value no . . .
a
. x
.1 ..2
xb
made.

. . . . . .


Theorem of the Day

Theorem

∑
n
n→∞ n→∞
i=1

exists and is the same value no . . . .
a
. x
.1 x
.2 ..3
xb
made.

. . . . . .


Theorem of the Day

Theorem

∑
n
n→∞ n→∞
i=1

exists and is the same value no . . . . .
a
. x
.1 x
.2 x
.3 ..4
xb
made.

. . . . . .


Theorem of the Day

Theorem

∑
n
n→∞ n→∞
i=1

exists and is the same value no . . . . . .
a x x x x x
. . . . . ..
matter what choice of ci we 1 2 3 4 b 5
made.

. . . . . .


Theorem of the Day

Theorem

∑
n
n→∞ n→∞
i=1

exists and is the same value no . . . . . . .
a x x x x x x
. . . . . . ..
matter what choice of ci we 1 2 3 4 5 b 6
made.

. . . . . .


Theorem of the Day

Theorem

∑
n
n→∞ n→∞
i=1

exists and is the same value no . . . . . . . .
ax x x x x x x
. . . . . . . ..
matter what choice of ci we 1 2 3 4 5 6 b 7
made.

. . . . . .


Theorem of the Day

Theorem

∑
n
n→∞ n→∞
i=1

exists and is the same value no . . . . . . . . .
ax x x x x x x x
. . . . . . . . ..
b
matter what choice of ci we 1 2 3 4 5 6 7 8
made.

. . . . . .


Theorem of the Day

Theorem

∑
n
n→∞ n→∞
i=1

exists and is the same value no . . . . . . . . . .
ax x x x x x x x x
. . . . . . . . . ..
b
matter what choice of ci we 1 2 3 4 5 6 7 8 9
made.

. . . . . .


Theorem of the Day

Theorem

∑
n
n→∞ n→∞
i=1

exists and is the same value no . . . . . . . . . . .
ax x x x x x x x x xb
. . . . . . . . . . ..
matter what choice of ci we 1 2 3 4 5 6 7 8 9 10
made.

. . . . . .


Theorem of the Day

Theorem

∑
n
n→∞ n→∞
i=1

exists and is the same value no . . . . . . . . . . . .
ax x x x x x x x xx xb
. . . . . . . . . . . ..
made.

. . . . . .


Theorem of the Day

Theorem

∑
n
n→∞ n→∞
i=1

exists and is the same value no . ............
ax x x x x x x x xx x xb
. . . . . . . . . . . . ..
matter what choice of ci we 1 2 3 4 5 6 7 8 9 10 12
11
made.

. . . . . .


Theorem of the Day

Theorem

∑
n
n→∞ n→∞
i=1

exists and is the same value no . .............
ax x x x x x x x xx x x xb
.. . . . . . . . .. . . ..
11 13
made.

. . . . . .


Theorem of the Day

Theorem

∑
n
n→∞ n→∞
i=1

exists and is the same value no . ..............
ax x x x x x x x xx x x x xb
.. . . . . . . . .. . . . . .
matter what choice of ci we 1 2 3 4 5 6 7 8 910 12 14
11 13
made.

. . . . . .


Theorem of the Day

Theorem

∑
n
n→∞ n→∞
i=1

exists and is the same value no . ...............
a xxxxxxxxxxxxxxb
.. . . . . . . . .. . . . . . .
x1 2 3 4 5 6 7 8 910 12 14
matter what choice of ci we 11 13 15
made.

. . . . . .


Theorem of the Day

Theorem

∑
n
n→∞ n→∞
i=1

exists and is the same value no . ................
a xxxxxxxx xxxxxxb
. . . . . . . . . .. . . . . . . .
x1 2 3 4 5 6 7 8x10 12 14 16
matter what choice of ci we 9 11 13 15
made.

. . . . . .


Theorem of the Day

Theorem

∑
n
n→∞ n→∞
i=1

exists and is the same value no . .................
a xxxxxxxx xxxxxxxb
.. . . . . . . . .. . . . . . . . .
x1 2 3 4 5 6 7 8x10 12 14 16
matter what choice of ci we 9 11 13 15 17
made.

. . . . . .


Theorem of the Day

Theorem

∑
n
n→∞ n→∞
i=1

exists and is the same value no . ..................
a xxxxxxxx xxxxxxxxb
.. . . . . . . . .. . . . . . . . . .
x12345678910 12 14 16 18
x 11 13 15 17
made.

. . . . . .


Theorem of the Day

Theorem

∑
n
n→∞ n→∞
i=1

exists and is the same value no . ...................
a xxxxxxxx xxxxxxxxxb
.. . . . . . . . .. . . . . . . . . . .
x1234567891012141618
x 1113151719
made.

. . . . . .


Theorem of the Day

Theorem

∑
n
n→∞ n→∞
i=1

exists and is the same value no . ....................
axxxxxxxx xxxxxxxxxxb
.. . . . . . . . .. . . . . . . . . . . .
x123456789 1113151719
x101214161820
made.

. . . . . .


Analogies

The Tangent Problem The Area Problem (Ch. 5)
(Ch. 2–4)

. . . . . .


Analogies

(Ch. 2–4)
Want the slope of a curve

. . . . . .


Analogies

(Ch. 2–4) Want the area of a curved
Want the slope of a curve region

. . . . . .


Analogies

Only know the slope of
lines

. . . . . .


Analogies

Only know the slope of Only know the area of
lines polygons

. . . . . .


Analogies

lines polygons
Approximate curve with a
line

. . . . . .


Analogies

lines polygons
Approximate curve with a Approximate region with
line polygons

. . . . . .


Analogies

lines polygons
Approximate curve with a Approximate region with
line polygons
Take limit over better and Take limit over better and
better approximations better approximations

. . . . . .


Outline

Euclid
Archimedes
Cavalieri

Analogies

Distances
Other applications

. . . . . .


Distances

Just like area = length × width, we have

distance = rate × time.

So here is another use for Riemann sums.

. . . . . .


Application: Dead Reckoning

. . . . . .


Computing position by Dead Reckoning

Example
A sailing ship is cruising back and forth along a channel (in a straight
line). At noon the ship’s position and velocity are recorded, but shortly
thereafter a storm blows in and position is impossible to measure. The
velocity continues to be recorded at thirty-minute intervals.

Time 12:00 12:30 1:00 1:30 2:00
Speed (knots) 4 8 12 6 4
Direction E E E E W
Time 2:30 3:00 3:30 4:00
Speed 3 3 5 9
Direction W E E E

Estimate the ship’s position at 4:00pm.
. . . . . .


Solution

Solution
We estimate that the speed of 4 knots (nautical miles per hour) is
maintained from 12:00 until 12:30. So over this time interval the ship
travels ( )( )
4 nmi 1
hr = 2 nmi
hr 2
We can continue for each additional half hour and get

distance = 4 × 1/2 + 8 × 1/2 + 12 × 1/2
+ 6 × 1/2 − 4 × 1/2 − 3 × 1/2 + 3 × 1/2 + 5 × 1/2
= 15.5

So the ship is 15.5 nmi east of its original position.

. . . . . .


Analysis

This method of measuring position by recording velocity was
necessary until global-positioning satellite technology became
widespread
If we had velocity estimates at finer intervals, we’d get better
estimates.
If we had velocity at every instant, a limit would tell us our exact
position relative to the last time we measured it.

. . . . . .


Other uses of Riemann sums

Anything with a product!
Area, volume
Anything with a density: Population, mass
Anything with a “speed:” distance, throughput, power
Consumer surplus
Expected value of a random variable

. . . . . .


Surplus by picture

c
. onsumer surplus
p
. rice (p)

s
. upply

.∗ .
p . . quilibrium
e

d
. emand f(q)
. .
.∗
q q
. uantity (q)

. . . . . .


Summary

We can compute the area of a curved region with a limit of
Riemann sums
We can compute the distance traveled from the velocity with a
limit of Riemann sums
Many other important uses of this process.

. . . . . .


Lesson 24: Area and Distances

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