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

Sec on 5.1–5.2
Areas and Distances, The Deﬁnite
Integral
V63.0121.011: Calculus I
Professor Ma hew Leingang
New York University

April 25, 2011

.

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

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.

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.

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

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.

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

.
b


h

.
b


h

.


h

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

A = bh

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

.
b


h

.
b


h

.
So b
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 (a er 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 1
8 8

.
parabola.
1
A=1+2·
8

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

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

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

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

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

Fact

(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

Fact

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

Corollary

1 − rn+1
1 + r + ··· + r =n
1−r

Summing the series
1 1 1
1+ + + ··· + n + ···
4 16 4
Using the corollary,

1 1 1 1 − (1/4)n+1
1+ + + ··· + n =
4 16 4 1 − 1/4

Summing 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

Cavalieri

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

Cavalieri’s method
Divide up the interval into pieces and
2
y=x measure the area of the inscribed
rectangles:

.
0 1

2
rectangles:
1
L2 =
8

.
0 1 1
2

2
rectangles:
1
L2 =
8
L3 =

.
0 1 2 1
3 3

2
rectangles:
1
L2 =
8
1 4 5
L3 = + =
27 27 27

.
0 1 2 1
3 3

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

2
rectangles:
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

2
rectangles:
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

2
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 2 3 4 1 L5 = + + + =
125 125 125 125 125
5 5 5 5

2
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
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 + 3 + · · · + =
n n n3 n3

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.

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 + 3 + · · · + =
n n n3 n3
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 + 3 + · · · + =
n n n3 n3
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) = 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
1 + 23 + 33 + · · · + (n − 1)3
=
n4

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)

( ) ( ) ( )
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
=
4n4

( ) ( ) ( )
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 → ∞.

Cavalieri’s method with diﬀerent heights
1 13 1 23 1 n3
Rn = · 3 + · 3 + ··· + · 3
n n n n n n
1 + 2 + 3 + ··· + n
3 3 3 3
=
n4
1 [ ]2
= 4 1 n(n + 1)
n 2
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
1 + 2 + 3 + ··· + n
3 3 3 3
=
n4
1 [ ]2
= 4 1 n(n + 1)
n 2
n2 (n + 1)2 1
. = →
4n4 4
as n → ∞.
So even though the rectangles overlap, we s ll get the same answer.

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

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

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.

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

le endpoints…
∑
n
Ln = f(xi−1 )∆x
i=1

. x


right endpoints…
∑
n
Rn = f(xi )∆x
i=1

. x


midpoints…
∑ ( xi−1 + xi )
n
Mn = f ∆x
i=1
2

. x

the maximum value on the
interval…
∑
n
Un = max {f(x)} ∆x
xi−1 ≤x≤xi
i=1

. x

the minimum value on the
interval…
∑
n
Ln = min {f(x)} ∆x
xi−1 ≤x≤xi
i=1

. x


…even random points!

. x

Theorem of the Day
Theorem
If f is a con nuous func on on
[a, b] or has ﬁnitely many jump
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.

Theorem of the Day
Theorem
[a, b] or has ﬁnitely many jump L1 = 3.0
discon nui es, then
{ n }
∑
n→∞ n→∞
i=1

ma er what choice of ci we make. le endpoints