Lesson 22: Areas and Distances

Section 5.1
Areas and Distances
V63.0121.006/016, Calculus I

New York University

April 13, 2010

Announcements

Quiz April 16 on §§4.1–4.4
Final Exam: Monday, May 10, 12:00noon

Announcements

Quiz April 16 on §§4.1–4.4
Final Exam: Monday, May
10, 12:00noon

V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 2 / 30

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 and w is the product A = w .

w

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

A=


1


A=1


1
1 1
8 8

1
A=1+2·
8


1 1
64 64
1
1 1
8 8

1 1
64 64

1 1
A=1+2· +4· + ···
8 64


1 1
64 64
1
1 1
8 8

1 1
64 64

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


We would then need to know the value of the series
1 1 1
1+ + + ··· + n + ···
4 16 4


1 1 1
1+ + + ··· + n + ···
4 16 4
But for any number r and any positive integer n,

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

So
1 − r n+1
1 + r + · · · + rn =
1−r


1 1 1
1+ + + ··· + n + ···
4 16 4

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

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


1 1 1
1+ + + ··· + n + ···
4 16 4

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

So
1 − r n+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 pieces
y = x2 and measure the area of the
inscribed rectangles:

0 1



1
L2 =
8

0 1 1
2



1
L2 =
8
L3 =

0 1 2 1
3 3



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

0 1 2 1
3 3



1
L2 =
8
1 4 5
L3 = + =
27 27 27
L4 =

0 1 2 3 1
4 4 4



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



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



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



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
2
1 i −1 (i − 1)2
· = .
n n n3


What is Ln ?
1
n
2
1 i −1 (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
2
1 i −1 (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
2
1 i −1 (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 = →
6n3 3
as n → ∞.

Cavalieri’s method for diﬀerent functions

Try the same trick with f (x) = x 3 . 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 2 3 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 2 3 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
1+2 3 + 33 + · · · + (n − 1)3
=
n4




1 1 1 2 1 n−1
Ln = ·f + ·f + ··· + · f
n n n n n n
1 1 1 2 3 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
1+2 3 + 33 + · · · + (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 2 3 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
1+2 3 + 33 + · · · + (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
13 + 23 + 33 + · · · + n3
=
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
13 + 23 + 33 + · · · + n3
=
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).
b−a
For each positive integer n, divide up the interval into n pieces. Then ∆x = .
n
For each i between 1 and n, let xi be the nth step between a and 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 b b−a
x0 x1 x2 . . . xi xn−1xn 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 b
x1
matter what choice of ci we
made.


Theorem of the Day

Theorem
n
n→∞ n→∞
i=1

a x1 b
x2
made.


Theorem of the Day

Theorem
n
n→∞ n→∞
i=1

a x1 x2 b
x3
made.


Theorem of the Day

Theorem
n
n→∞ n→∞
i=1

a x1 x2 x3 b
x4
made.


Theorem of the Day

Theorem
n
n→∞ n→∞
i=1

a x x x x x
matter what choice of ci we 1 2 3 4 b5
made.


Theorem of the Day

Theorem
n
n→∞ n→∞
i=1

a x x x x x x
matter what choice of ci we 1 2 3 4 5 b6
made.


Theorem of the Day

Theorem
n
n→∞ n→∞
i=1

a x x x x x x x
matter what choice of ci we 1 2 3 4 5 6 b7
made.


Theorem of the Day

Theorem
n
n→∞ n→∞
i=1

ax x x x x x x x
matter what choice of ci we 1 2 3 4 5 6 7 b8
made.


Theorem of the Day

Theorem
n
n→∞ n→∞
i=1

ax x x x x x x x x
matter what choice of ci we 1 2 3 4 5 6 7 8b9
made.


Theorem of the Day

Theorem
n
n→∞ n→∞
i=1

a x 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

ax x x x x x x x x x xb
made.


Theorem of the Day

Theorem
n
n→∞ n→∞
i=1

ax x x x x x x x xx x xb
made.


Theorem of the Day

Theorem
n
n→∞ n→∞
i=1

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

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

ax x x x x x x x xx x x x x xb
matter what choice of ci we 1 2 3 4 5 6 7 8 910 12 14
11 13 15
made.


Theorem of the Day

Theorem
n
n→∞ n→∞
i=1

axxxxxxxxxxxxxxxxb
matter what choice of ci we 1 2 3 4 5 6 7 8 910 12 14 16
11 13 15
made.


Theorem of the Day

Theorem
n
n→∞ n→∞
i=1

a xxxxxxxxxxxxxxxxb
x1 2 3 4 5 6 7 8 910 12 14 16
matter what choice of ci we 11 13 15 17
made.


Theorem of the Day

Theorem
n
n→∞ n→∞
i=1

a xxxxxxxx xxxxxxxxb
x1 2 3 4 5 6 7 8x10 12 14 16 18
matter what choice of ci we 9 11 13 15 17
made.


Theorem of the Day

Theorem
n
n→∞ n→∞
i=1

a xxxxxxxx xxxxxxxxxb
x12345678910 12 14 16 18
x 11 13 15 17 19
made.


Theorem of the Day

Theorem
n
n→∞ n→∞
i=1

a xxxxxxxx xxxxxxxxxxb
x123456789101214161820
x 1113151719
made.


Analogies

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


Analogies

The Tangent Problem
(Ch. 2–4)
Want the slope of a curve


Analogies

The Tangent Problem
(Ch. 2–4) Want the area of a curved
region


Analogies

The Tangent Problem
region
Only know the slope of lines


Analogies

The Tangent Problem
region
Only know the area of
polygons


Analogies

The Tangent Problem
region
polygons
Approximate curve with a
line


Analogies

The Tangent Problem
region
polygons
Approximate region with
line
polygons


Analogies

The Tangent Problem
region
polygons
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


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
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


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 22: Areas and Distances

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