Lesson 24: Areas, Distances, the Integral (Section 021 slides

.

Sections 5.1–5.2
Areas and Distances
The Definite Integral
V63.0121.021, Calculus I

New York University

December 2, 2010

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Final December 20, 12:00–1:50pm

. . . . . .

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V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 2 / 56

Objectives from Section 5.1

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.

. . . . . .


Objectives from Section 5.2

Compute the definite
integral using a limit of
Riemann sums
Estimate the definite
integral using a Riemann
sum (e.g., Midpoint Rule)
Reason with the definite
integral using its
elementary properties.

. . . . . .


Outline

Area through the Centuries
Euclid
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Analogies
Distances
Other applications
The definite integral as a limit
Estimating the Definite Integral
Properties of the integral
Comparison Properties of the Integral
. . . . . .


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


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

. . . . . .


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

Fact
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
1 1 1
1+ + + ··· + n + ···
4 16 4

Fact

(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
1 1 1
1+ + + ··· + n + ···
4 16 4

Fact

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


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

1
L2 =
8

.
0 1 1
2

. . . . . .


Cavalieri's method

2

1
L2 =
8
L3 =

.
0 1 2 1
3 3

. . . . . .


Cavalieri's method

2

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

.
0 1 2 1
3 3

. . . . . .


Cavalieri's method

2

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

. . . . . .


Cavalieri's method

2

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

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

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

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

Area through the Centuries
Euclid
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Analogies
Distances
Other applications
The definite integral as a limit
Estimating the Definite Integral
Properties of the integral
Comparison Properties of the Integral
. . . . . .


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 ith step between
n
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
b−a
. x xn = a + n · =b
x0 x1 . . . xi . . .xn−1xn n
. . . . . .


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

left 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
2
i=1

. x

. . . . . .



the minimum value on the
interval…

. x

. . . . . .



the maximum value on the
interval…

. x

. . . . . .



…even random points!

. x

. . . . . .



…even random points!

. x
In general, 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
. . . . . .


Theorem of the Day

Theorem

If f is a continuous function on [a, b]
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1

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

. . . . . .


Theorem of the Day

Theorem

If f is a continuous function on [a, b] L1 = 3.0
{ n }
∑
n→∞ n→∞
i=1

. x
left endpoints

. . . . . .


Theorem of the Day

Theorem

{ n }
∑
n→∞ n→∞
i=1

. x
left endpoints

. . . . . .


Theorem of the Day

Theorem

If f is a continuous function on [a, b] R1 = 12.0
{ n }
∑
n→∞ n→∞
i=1

. x
right endpoints

. . . . . .


Theorem of the Day

Theorem

{ n }
∑
n→∞ n→∞
i=1

. x
right endpoints

. . . . . .


Theorem of the Day

Theorem

If f is a continuous function on [a, b] M1 = 7.5
{ n }
∑
n→∞ n→∞
i=1

. x
midpoints

. . . . . .


Lesson 24: Areas, Distances, the Integral (Section 021 slides

Lesson 24: Areas, Distances, the Integral (Section 021 slides

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