Lesson 22: Areas and Distances (handout)

V63.0121.006/016, Calculus I Section 5.1 : Areas and Distances April 13, 2010

Notes
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
Notes

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
Notes

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.


1


Outline
Notes

Area through the Centuries
Euclid
Archimedes
Cavalieri

Generalizing Cavalieri’s method
Analogies

Distances
Other applications


Easy Areas: Rectangle
Notes

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
Notes
By cutting and pasting, a parallelogram can be made into a rectangle.

h

b b

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

A = bh


2


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

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
Notes

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


Hard Areas: Curved Regions
Notes

???


3


Meet the mathematician: Archimedes
Notes

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


Notes

1 1
64 64
1
1 1
8 8

1 1
64 64

Archimedes found areas of a sequence of triangles inscribed in a parabola.
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 Notes
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
Therefore
1 1 1 1 − (1/4)n+1 1 4
1+ + + ··· + n = →3 =
4 16 4 1 − 1/4 /4 3
as n → ∞.


4


Cavalieri
Notes

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


Cavalieri’s method
Notes

Divide up the interval into pieces
y = x2 and measure the area of the
inscribed rectangles:
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 Notes
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
So
1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2
Ln = + + ··· + =
n3 n3 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 → ∞.

5


Cavalieri’s method for diﬀerent functions
Notes
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 23 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
1 + 23 + 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
Notes

1 13 1 23 1 n3
Rn = · + · + ··· + · 3
n n3 n n3 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
Notes

Euclid
Archimedes
Cavalieri

Analogies

Distances
Other applications


6


Cavalieri’s method in general
Notes
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
Notes
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
Notes

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
axxxxxxxxxxxxxxxxxxx xb
xx12x3xx25x3635x911x69158x16x19
x1x132445x6xx7x91165x101513167
matter what choice of ci we 11x22543748 58412xx1317 18
x11x315x76866710812124179195
223x4355591781111x1217
234251 7872x71314141010
134 46 x8x537x8x16129
4365 7 9 x9101311131
8 x9 13 10 153
10 276 9x 11
21362 24 3611141114 20
2 4 1091012166818
3 4 4 10131057142
6 12113 151113
8 14 1815
459 15 14
12 17
7 16
12
8
6
4
made.


7


Analogies
Notes

The Area Problem (Ch. 5)
The Tangent Problem
(Ch. 2–4) Want the area of a curved
region
Want the slope of a curve
Only know the area of
Only know the slope of 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
Notes

Euclid
Archimedes
Cavalieri

Analogies

Distances
Other applications


Distances
Notes

Just like area = length × width, we have

distance = rate × time.

So here is another use for Riemann sums.


8


Application: Dead Reckoning
Notes


Example Notes
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.


Notes
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.


9


Analysis
Notes

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
Notes

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
Notes

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.


10

Lesson 22: Areas and Distances (handout)

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