Lesson 24: Areas and Distances, The Definite Integral (handout)

. V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral
. . April 25, 2011

Notes
Sec on 5.1–5.2
Areas and Distances, The Deﬁnite
Integral
V63.0121.001: Calculus I
Professor Ma hew Leingang
New York University

April 25, 2011

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Notes
Announcements

Quiz 5 on Sec ons
4.1–4.4 April 28/29
Final Exam Thursday May
12, 2:00–3:50pm
cumula ve
loca on TBD
old exams on common
website

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Notes
Objectives from Section 5.1
Compute the area of a region by
approxima ng it with rectangles
and le ng the size of the
rectangles tend to zero.
Compute the total distance
traveled by a par cle by
approxima ng it as distance =
(rate)( me) and le ng the me
intervals over which one
approximates tend to zero.

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. 1
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. . April 25, 2011

Notes
Objectives from Section 5.2

Compute the definite integral
using a limit of Riemann sums
Es mate the definite integral
using a Riemann sum (e.g.,
Midpoint Rule)
Reason with the definite integral
using its elementary proper es.

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Notes
Outline
Area through the Centuries
Euclid
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Analogies
Distances
Other applica ons
The definite integral as a limit
Es ma ng the Definite Integral
Proper es of the integral
Comparison Proper es of the Integral
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Notes
Easy Areas: Rectangle
Defini on
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 defini on and not a theorem but
. we have to start somewhere.
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. . April 25, 2011

Notes
Easy Areas: Parallelogram
By cu ng and pas ng, a parallelogram can be made into a rectangle.

h

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

A = bh
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Notes
Easy Areas: Triangle
By copying and pas ng, a triangle can be made into a parallelogram.

h

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b
So
Fact
The area of a triangle of base width b and height h is
1
A = bh
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Notes
Easy Areas: Other Polygons
Any polygon can be triangulated, so its area can be found by
summing the areas of the triangles:

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. . April 25, 2011

Notes
Hard Areas: Curved Regions

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???

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Notes
Meet the mathematician: Archimedes

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

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Notes
Archimedes and the Parabola
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 1 1 1
A=1+2· +4· + ··· = 1 + + + ··· + n + ···
8 64 4 16 4
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. 4
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. . April 25, 2011

Notes
Summing a geometric series
Fact
For any number r and any posi ve integer n,

(1 − r)(1 + r + r2 + · · · + rn ) = 1 − rn+1 .

Proof.
(1 − r)(1 + r + r2 + · · · + rn )
= (1 + r + r2 + · · · + rn ) − r(1 + r + r2 + · · · + rn )
= (1 + r + r2 + · · · + rn ) − (r + r2 + r3 · · · + rn + rn+1 )
= 1 − rn+1
.
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Notes
Summing a geometric series
Fact
For any number r and any posi ve integer n,

(1 − r)(1 + r + r2 + · · · + rn ) = 1 − rn+1 .

Corollary

1 − rn+1
1 + r + · · · + rn =
1−r

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Notes
Summing the series
We need to know the value of the series
1 1 1
1+ + + ··· + n + ···
4 16 4
Using the corollary,

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

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. . April 25, 2011

Notes
Cavalieri

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

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Notes
Cavalieri’s method
Divide up the interval into pieces and
y = x2 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 =?
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Notes
The Square Pyramidial Numbers
Fact
Let n be a posi ve integer. Then
n(n − 1)(2n − 1)
1 + 22 + 32 + · · · + (n − 1)2 =
6

This formula was known to the Arabs and discussed by Fibonacci in
his book Liber Abaci.

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. 6
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. . April 25, 2011

Notes
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
So
1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2
Ln = 3
+ 3 + ··· + 3
=
n n n n3
So
n(n − 1)(2n − 1) 1
Ln = →
6n3 3
. as n → ∞.
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Notes
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 23 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
1 + 23 + 33 + · · · + (n − 1)3
=
n4
n2 (n − 1)2 1
= →
4n4 4
as n → ∞.
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Notes
Nicomachus’s Theorem

Fact (Nicomachus 1st c. CE, Aryabhata 5th c., Al-Karaji 11th c.)

1 + 23 + 33 + · · · + (n − 1)3 = [1 + 2 + · · · + (n − 1)]2
[1 ]2
= 2 n(n − 1)

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. 7
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. . April 25, 2011

Notes
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 s ll get the same answer.
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Notes
Outline
Euclid
Archimedes
Cavalieri
Analogies
Distances
Other applica ons
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Notes
Cavalieri’s method in general
Problem

Let f be a posi ve func on deﬁned
on the interval [a, b]. Find the
area between x = a, x = b, y = 0,
and y = f(x).
.
. x x
x0 x1. . . xi . . xn−1 n

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. 8
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. . April 25, 2011

Notes
Cavalieri’s method in general
For each posi ve integer n, divide up the interval into n pieces. Then
b−a
∆x = . For each i between 1 and n, let xi be the ith step
n
between a and b.
x0 = a
b−a
x1 = x0 + ∆x = a +
n
b−a
x2 = x1 + ∆x = a + 2 · ...
n
b−a
xi = a + i · ...
n
. b−a
. x x
x0 x1. . . xi . . xn−1 n xn = a + n · =b
n
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Notes
Forming Riemann Sums
Choose ci to be a point in the ith interval [xi−1 , xi ]. Form the
Riemann sum
∑
n
Sn = f(c1 )∆x + f(c2 )∆x + · · · + f(cn )∆x = f(ci )∆x
i=1

Thus we approximate area under a curve by a sum of areas of
rectangles.

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Notes
Forming Riemann sums
We have many choices of representa ve points to approximate the
area in each subinterval.

…even random points!

. x

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. . April 25, 2011

Notes
Theorem of the Day
Theorem
If f is a con nuous func on on
[a, b] or has ﬁnitely many jump M15 = 7.49968
discon nui es, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1

exists and is the same value no . x
ma er what choice of ci we make.

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Notes
Analogies
The Tangent Problem The Area Problem (Ch. 5)
(Ch. 2–4) Want the area of a curved
Want the slope of a curve region
Only know the slope of Only know the area of
lines polygons
Approximate curve with a Approximate region with
line polygons
Take limit over be er and Take limit over be er and
be er approxima ons be er approxima ons

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Notes
Outline
Euclid
Archimedes
Cavalieri
Analogies
Distances
Other applica ons
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. 10
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. . April 25, 2011

Notes
Distances

Just like area = length × width, we have

distance = rate × me.

So here is another use for Riemann sums.

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Notes
Application: Dead Reckoning

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Notes
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 posi on and velocity are recorded, but
shortly therea er a storm blows in and posi on is impossible to
measure. The velocity con nues to be recorded at thirty-minute
intervals.

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. 11
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. . April 25, 2011

Notes
Computing position by Dead Reckoning
Example
Time 12:00 12:30 1:00 1:30 2:00
Speed (knots) 4 8 12 6 4
Direc on E E E E W
Time 2:30 3:00 3:30 4:00
Speed 3 3 5 9
Direc on W E E E

Es mate the ship’s posi on at 4:00pm.

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Notes
Solution
Solu on
We es mate that the speed of 4 knots (nau cal miles per hour) is
maintained from 12:00 un l 12:30. So over this me interval the
ship travels ( )( )
4 nmi 1
hr = 2 nmi
hr 2
We can con nue for each addi onal 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 posi on.
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Notes
Analysis

This method of measuring posi on by recording velocity was
necessary un l global-posi oning satellite technology became
widespread
If we had velocity es mates at ﬁner intervals, we’d get be er
es mates.
If we had velocity at every instant, a limit would tell us our
exact posi on rela ve to the last me we measured it.

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. 12
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. . April 25, 2011

Notes
Other uses of Riemann sums

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

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Notes
Outline
Euclid
Archimedes
Cavalieri
Analogies
Distances
Other applica ons
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Notes

Defini on
If f is a func on defined on [a, b], the definite integral of f from a to
b is the number
∫ b ∑n
f(x) dx = lim f(ci ) ∆x
a ∆x→0
i=1

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. 13
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. . April 25, 2011

Notes
Notation/Terminology
∫ b ∑
n
f(x) dx = lim f(ci ) ∆x
a ∆x→0
i=1
∫
— integral sign (swoopy S)
f(x) — integrand
a and b — limits of integra on (a is the lower limit and b the
upper limit)
dx — ??? (a parenthesis? an infinitesimal? a variable?)
The process of compu ng an integral is called integra on or
quadrature
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Notes
The limit can be simplified

Theorem
If f is con nuous on [a, b] or if f has only finitely many jump
discon nui es, then f is integrable on [a, b]; that is, the definite
∫ b
integral f(x) dx exists.
a

So we can find the integral by compu ng the limit of any sequence
of Riemann sums that we like,

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Example
∫ 3
Notes
Find x dx
0

Solu on
3 3i
For any n we have ∆x = and for each i between 0 and n, xi = .
n n
For each i, take xi to represent the func on on the ith interval. So
∫ 3 ∑n ∑ ( 3i ) ( 3 )
n
x dx = lim Rn = lim f(xi ) ∆x = lim
0 n→∞ n→∞
i=1
n→∞
i=1
n n
9 ∑
n
9 n(n + 1) 9
= lim i = lim 2 · = ·1
n→∞ n2 n→∞ n 2 2
i=1
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. 14
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. . April 25, 2011

Example
∫ 3
Notes
Find x2 dx
0

Solu on

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Example
∫ 3
Notes
Find x3 dx
0

Solu on

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Notes
Outline
Euclid
Archimedes
Cavalieri
Analogies
Distances
Other applica ons
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. 15
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. . April 25, 2011

Notes
Estimating the Deﬁnite Integral
Example
∫ 1
4
Es mate dx using M4 .
0 1 + x2

Solu on
1 1 3
We have x0 = 0, x1 = , x2 = , x3 = , x4 = 1.
4 2 4
1 3 5 7
So c1 = , c2 = , c3 = , c4 = .
8 8 8 8

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Notes
Example
∫ 1
4
Es mate dx using M4 .
0 1 + x2

Solu on
( )
1 4 4 4 4
M4 = + + +
4 1 + (1/8)2 1 + (3/8)2 1 + (5/8)2 1 + (7/8)2
( )
1 4 4 4 4
= + + +
4 65/64 73/64 89/64 113/64
64 64 64 64
= + + + ≈ 3.1468
65 73 89 113
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Notes
Example
∫ 1
4
Es mate dx using L4 and R4
0 1 + x2

Answer

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. 16
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. . April 25, 2011

Notes
Example
∫ 1
4
Es mate dx using L4 and R4
0 1 + x2

Answer

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Notes
Outline
Euclid
Archimedes
Cavalieri
Analogies
Distances
Other applica ons
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Notes
Properties of the integral
Theorem (Addi ve Proper es of the Integral)
Let f and g be integrable func ons on [a, b] and c a constant. Then
∫ b
1. c dx = c(b − a)
∫a b ∫ b ∫ b
2. [f(x) + g(x)] dx = f(x) dx + g(x) dx.
∫a b ∫ b a a

3. cf(x) dx = c f(x) dx.
∫a b a
∫ b ∫ b
4. [f(x) − g(x)] dx = f(x) dx − g(x) dx.
a a a
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Notes
Proofs
Proofs.
When integra ng a constant func on c, each Riemann sum
equals c(b − a).
A Riemann sum for f + g equals a Riemann sum for f plus a
Riemann sum for g. Using the sum rule for limits, the integral
of a sum is the sum of the integrals.
Di o for constant mul ples
Di o for diﬀerences

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Example
∫ 3
Notes
( 3 )
Find x − 4.5x2 + 5.5x + 1 dx
0

Solu on

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Notes
More Properties of the Integral
Conven ons: ∫ ∫
a b
f(x) dx = − f(x) dx
b a
∫ a
f(x) dx = 0
a
This allows us to have
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b

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. . April 25, 2011

Notes
Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
a a b

y
∫ b ∫ c
f(x) dx f(x) dx
a b

.
a c x
b
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Notes
Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
a a b

y
∫ ∫ c
c
f(x) dx f(x) dx =
b∫
a b
− f(x) dx
. c
a c x
b
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Notes
Using the Properties
Example
Suppose f and g are func ons
with ∫
4 Find∫
f(x) dx = 4 5
(a) [2f(x) − g(x)] dx
∫0 5
∫0 5
f(x) dx = 7
(b) f(x) dx.
∫0 5 4
g(x) dx = 3.
0

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. 19
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. . April 25, 2011

Notes
Solu on

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Notes
Outline
Euclid
Archimedes
Cavalieri
Analogies
Distances
Other applica ons
.
.

Notes
Comparison Properties of the Integral
Theorem
Let f and g be integrable func ons on [a, b].
∫ b
6. If f(x) ≥ 0 for all x in [a, b], then f(x) dx ≥ 0
a
∫ b ∫ b
7. If f(x) ≥ g(x) for all x in [a, b], then f(x) dx ≥ g(x) dx
a a

8. If m ≤ f(x) ≤ M for all x in [a, b], then
∫ b
m(b − a) ≤ f(x) dx ≤ M(b − a)
a

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Notes
Integral of a nonnegative function is nonnegative
Proof.
If f(x) ≥ 0 for all x in [a, b], then for
any number of divisions n and choice
of sample points {ci }:

∑
n ∑
n
Sn = f(ci ) ∆x ≥ 0 · ∆x = 0
i=1 ≥0 i=1
. x
Since Sn ≥ 0 for all n, the limit of {Sn } is nonnega ve, too:
∫ b
f(x) dx = lim Sn ≥ 0
a n→∞
≥0
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Notes
The integral is “increasing”
Proof.
Let h(x) = f(x) − g(x). If f(x) ≥ g(x)
for all x in [a, b], then h(x) ≥ 0 for all f(x)
x in [a, b]. So by the previous h(x) g(x)
property
∫ b
h(x) dx ≥ 0 . x
a
This means that
∫ b ∫ b ∫ b ∫ b
f(x) dx − g(x) dx = (f(x) − g(x)) dx = h(x) dx ≥ 0
a a a a

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Notes
Bounding the integral
Proof.
If m ≤ f(x) ≤ M on for all x in [a, b], then by
y
the previous property
∫ b ∫ b ∫ b M
m dx ≤ f(x) dx ≤ M dx
a a a f(x)
By Property 8, the integral of a constant
func on is the product of the constant and m
the width of the interval. So:
∫ b . x
m(b − a) ≤ f(x) dx ≤ M(b − a) a b
a
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. 21
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. . April 25, 2011

Example Notes
∫ 2
1
Es mate dx using the comparison proper es.
1 x

Solu on

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Notes
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.

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Notes
Summary

The definite integral is a limit of Riemann Sums
The definite integral can be es mated with Riemann Sums
The definite integral can be distributed across sums and
constant mul ples of func ons
The definite integral can be bounded using bounds for the
func on

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. 22
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Lesson 24: Areas and Distances, The Definite Integral (handout)

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Lesson 24: Areas and Distances, The Definite Integral (handout)