Lesson 24: Area and Distances

Section 5.1Areas and Distances

V63.0121.002.2010Su, Calculus I

New York University

June 16, 2010

Announcements

I Quiz Thursday on 4.1–4.4

. . . . . .

. . . . . .

Announcements

I Quiz Thursday on 4.1–4.4

V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 2 / 31

. . . . . .

Objectives

I Compute the area of aregion by approximating itwith rectangles and lettingthe size of the rectanglestend to zero.

I Compute the total distancetraveled by a particle byapproximating it asdistance = (rate)(time) andletting the time intervalsover which oneapproximates tend to zero.


. . . . . .

Outline

Area through the CenturiesEuclidArchimedesCavalieri

Generalizing Cavalieri’s methodAnalogies

DistancesOther applications


. . . . . .

Easy Areas: Rectangle

DefinitionThe area of a rectangle with dimensions ℓ and w is the product A = ℓw.

..ℓ

.w

It may seem strange that this is a definition and not a theorem but wehave to start somewhere.


. . . . . .

Easy Areas: Parallelogram

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

..b

.b

.h

So

FactThe area of a parallelogram of base width b and height h is

A = bh


. . . . . .



..b

.b

.h

So


A = bh


. . . . . .



.

.b .b

.h

So


A = bh


. . . . . .



.

.b

.b

.h

So


A = bh


. . . . . .



.

.b

.b

.h

So


A = bh


. . . . . .

Easy Areas: Triangle

By copying and pasting, a triangle can be made into a parallelogram.

..b

.h

SoFactThe area of a triangle of base width b and height h is

A =12bh


. . . . . .



..b

.h


A =12bh


. . . . . .



..b

.h


A =12bh


. . . . . .

Easy Areas: Other Polygons

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

.

.


. . . . . .

Hard Areas: Curved Regions

.

.

???


. . . . . .

Meet the mathematician: Archimedes

I Greek (Syracuse), 287 BC– 212 BC (after Euclid)

I GeometerI Weapons engineer


. . . . . .





. . . . . .





. . . . . .

Archimedes and the Parabola

.

Archimedes found areas of a sequence of triangles inscribed in aparabola.

A =

1+ 2 · 18+ 4 · 1

64+ · · ·

= 1+14+

116

+ · · ·+ 14n

+ · · ·


. . . . . .


.

.1


A = 1

+ 2 · 18+ 4 · 1

64+ · · ·

= 1+14+

116

+ · · ·+ 14n

+ · · ·


. . . . . .


.

.1.18 .18


A = 1+ 2 · 18

+ 4 · 164

+ · · ·

= 1+14+

116

+ · · ·+ 14n

+ · · ·


. . . . . .


.

.1.18 .18

.164 .164

.164 .164


A = 1+ 2 · 18+ 4 · 1

64+ · · ·

= 1+14+

116

+ · · ·+ 14n

+ · · ·


. . . . . .


.

.1.18 .18

.164 .164

.164 .164


A = 1+ 2 · 18+ 4 · 1

64+ · · ·

= 1+14+

116

+ · · ·+ 14n

+ · · ·


. . . . . .

Summing the series

[label=archimedes-parabola-sum] We would then need to know thevalue of the series

1+14+

116

+ · · ·+ 14n

+ · · ·

But for any number r and any positive integer n,

(1− r)(1+ r+ · · ·+ rn) = 1− rn+1

So

1+ r+ · · ·+ rn =1− rn+1

1− rTherefore

1+14+

116

+ · · ·+ 14n

=1− (1/4)n+1

1− 1/4→ 1

3/4=43

as n → ∞.


. . . . . .

Summing the series


1+14+

116

+ · · ·+ 14n

+ · · ·


(1− r)(1+ r+ · · ·+ rn) = 1− rn+1

So

1+ r+ · · ·+ rn =1− rn+1

1− r

Therefore

1+14+

116

+ · · ·+ 14n

=1− (1/4)n+1

1− 1/4→ 1

3/4=43

as n → ∞.


. . . . . .

Summing the series


1+14+

116

+ · · ·+ 14n

+ · · ·


(1− r)(1+ r+ · · ·+ rn) = 1− rn+1

So

1+ r+ · · ·+ rn =1− rn+1

1− rTherefore

1+14+

116

+ · · ·+ 14n

=1− (1/4)n+1

1− 1/4

→ 13/4

=43

as n → ∞.


. . . . . .

Summing the series


1+14+

116

+ · · ·+ 14n

+ · · ·


(1− r)(1+ r+ · · ·+ rn) = 1− rn+1

So

1+ r+ · · ·+ rn =1− rn+1

1− rTherefore

1+14+

116

+ · · ·+ 14n

=1− (1/4)n+1

1− 1/4→ 1

3/4=43

as n → ∞.V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 12 / 31

. . . . . .

Cavalieri

I Italian,1598–1647

I Revisited theareaproblem witha differentperspective


. . . . . .

Cavalieri's method

.

.y = x2

..0

..1

.

.12

Divide up the interval intopieces and measure the area ofthe inscribed rectangles:

L2 =18

L3 =

127

+427

=527

L4 =

164

+464

+964

=1464

L5 =

1125

+4125

+9125

+16125

=30125

Ln =?


. . . . . .

Cavalieri's method

.

.y = x2

..0

..1

.

.12


L2 =18

L3 =

127

+427

=527

L4 =

164

+464

+964

=1464

L5 =

1125

+4125

+9125

+16125

=30125

Ln =?


. . . . . .

Cavalieri's method

.

.y = x2

..0

..1

.

.12

.

.13

.

.23


L2 =18

L3 =

127

+427

=527

L4 =

164

+464

+964

=1464

L5 =

1125

+4125

+9125

+16125

=30125

Ln =?


. . . . . .

Cavalieri's method

.

.y = x2

..0

..1

.

.12

.

.13

.

.23


L2 =18

L3 =127

+427

=527

L4 =

164

+464

+964

=1464

L5 =

1125

+4125

+9125

+16125

=30125

Ln =?


. . . . . .

Cavalieri's method

.

.y = x2

..0

..1

.

.12

.

.14

.

.24

.

.34


L2 =18

L3 =127

+427

=527

L4 =

164

+464

+964

=1464

L5 =

1125

+4125

+9125

+16125

=30125

Ln =?


. . . . . .

Cavalieri's method

.

.y = x2

..0

..1

.

.12

.

.14

.

.24

.

.34


L2 =18

L3 =127

+427

=527

L4 =164

+464

+964

=1464

L5 =

1125

+4125

+9125

+16125

=30125

Ln =?


. . . . . .

Cavalieri's method

.

.y = x2

..0

..1

.

.12

.

.15

.

.25

.

.35

.

.45


L2 =18

L3 =127

+427

=527

L4 =164

+464

+964

=1464

L5 =

1125

+4125

+9125

+16125

=30125

Ln =?


. . . . . .

Cavalieri's method

.

.y = x2

..0

..1

.

.12

.

.15

.

.25

.

.35

.

.45


L2 =18

L3 =127

+427

=527

L4 =164

+464

+964

=1464

L5 =1125

+4125

+9125

+16125

=30125

Ln =?


. . . . . .

Cavalieri's method

.

.y = x2

..0

..1

.

.12

.

.


L2 =18

L3 =127

+427

=527

L4 =164

+464

+964

=1464

L5 =1125

+4125

+9125

+16125

=30125

Ln =?


. . . . . .

What is Ln?

Divide the interval [0,1] into n pieces. Then each has width1n.

Therectangle over the ith interval and under the parabola has area

1n·(i− 1n

)2=

(i− 1)2

n3.

So

Ln =1n3

+22

n3+ · · ·+ (n− 1)2

n3=1+ 22 + 32 + · · ·+ (n− 1)2

n3

The Arabs knew that

1+ 22 + 32 + · · ·+ (n− 1)2 =n(n− 1)(2n− 1)

6So

Ln =n(n− 1)(2n− 1)

6n3→ 1

3as n → ∞.


. . . . . .

What is Ln?

Divide the interval [0,1] into n pieces. Then each has width1n. The

rectangle over the ith interval and under the parabola has area

1n·(i− 1n

)2=

(i− 1)2

n3.

So

Ln =1n3

+22

n3+ · · ·+ (n− 1)2

n3=1+ 22 + 32 + · · ·+ (n− 1)2

n3

The Arabs knew that

1+ 22 + 32 + · · ·+ (n− 1)2 =n(n− 1)(2n− 1)

6So

Ln =n(n− 1)(2n− 1)

6n3→ 1

3as n → ∞.


. . . . . .

What is Ln?



1n·(i− 1n

)2=

(i− 1)2

n3.

So

Ln =1n3

+22

n3+ · · ·+ (n− 1)2

n3=1+ 22 + 32 + · · ·+ (n− 1)2

n3

The Arabs knew that

1+ 22 + 32 + · · ·+ (n− 1)2 =n(n− 1)(2n− 1)

6So

Ln =n(n− 1)(2n− 1)

6n3→ 1

3as n → ∞.


. . . . . .

What is Ln?



1n·(i− 1n

)2=

(i− 1)2

n3.

So

Ln =1n3

+22

n3+ · · ·+ (n− 1)2

n3=1+ 22 + 32 + · · ·+ (n− 1)2

n3

The Arabs knew that

1+ 22 + 32 + · · ·+ (n− 1)2 =n(n− 1)(2n− 1)

6So

Ln =n(n− 1)(2n− 1)

6n3

→ 13

as n → ∞.


. . . . . .

What is Ln?



1n·(i− 1n

)2=

(i− 1)2

n3.

So

Ln =1n3

+22

n3+ · · ·+ (n− 1)2

n3=1+ 22 + 32 + · · ·+ (n− 1)2

n3

The Arabs knew that

1+ 22 + 32 + · · ·+ (n− 1)2 =n(n− 1)(2n− 1)

6So

Ln =n(n− 1)(2n− 1)

6n3→ 1

3as n → ∞.


. . . . . .

Cavalieri's method for different functions

Try the same trick with f(x) = x3. We have

Ln =1n· f(1n

)+1n· f(2n

)+ · · ·+ 1

n· f(n− 1n

)

=1n· 1n3

+1n· 2

3

n3+ · · ·+ 1

n· (n− 1)3

n3

=1+ 23 + 33 + · · ·+ (n− 1)3

n4

The formula out of the hat is

1+ 23 + 33 + · · ·+ (n− 1)3 =[12n(n− 1)

]2So

Ln =n2(n− 1)2

4n4→ 1

4as n → ∞.


. . . . . .



Ln =1n· f(1n

)+1n· f(2n

)+ · · ·+ 1

n· f(n− 1n

)=1n· 1n3

+1n· 2

3

n3+ · · ·+ 1

n· (n− 1)3

n3

=1+ 23 + 33 + · · ·+ (n− 1)3

n4


1+ 23 + 33 + · · ·+ (n− 1)3 =[12n(n− 1)

]2So

Ln =n2(n− 1)2

4n4→ 1

4as n → ∞.


. . . . . .



Ln =1n· f(1n

)+1n· f(2n

)+ · · ·+ 1

n· f(n− 1n

)=1n· 1n3

+1n· 2

3

n3+ · · ·+ 1

n· (n− 1)3

n3

=1+ 23 + 33 + · · ·+ (n− 1)3

n4


1+ 23 + 33 + · · ·+ (n− 1)3 =[12n(n− 1)

]2So

Ln =n2(n− 1)2

4n4→ 1

4as n → ∞.


. . . . . .



Ln =1n· f(1n

)+1n· f(2n

)+ · · ·+ 1

n· f(n− 1n

)=1n· 1n3

+1n· 2

3

n3+ · · ·+ 1

n· (n− 1)3

n3

=1+ 23 + 33 + · · ·+ (n− 1)3

n4


1+ 23 + 33 + · · ·+ (n− 1)3 =[12n(n− 1)

]2

So

Ln =n2(n− 1)2

4n4→ 1

4as n → ∞.


. . . . . .



Ln =1n· f(1n

)+1n· f(2n

)+ · · ·+ 1

n· f(n− 1n

)=1n· 1n3

+1n· 2

3

n3+ · · ·+ 1

n· (n− 1)3

n3

=1+ 23 + 33 + · · ·+ (n− 1)3

n4


1+ 23 + 33 + · · ·+ (n− 1)3 =[12n(n− 1)

]2So

Ln =n2(n− 1)2

4n4→ 1

4as n → ∞.


. . . . . .

Cavalieri's method with different heights

.

Rn =1n· 1

3

n3+1n· 2

3

n3+ · · ·+ 1

n· n

3

n3

=13 + 23 + 33 + · · ·+ n3

n4

=1n4

[12n(n+ 1)

]2=

n2(n+ 1)2

4n4→ 1

4as n → ∞.

So even though the rectangles overlap, we still get the same answer.


. . . . . .

Cavalieri's method with different heights

.

Rn =1n· 1

3

n3+1n· 2

3

n3+ · · ·+ 1

n· n

3

n3

=13 + 23 + 33 + · · ·+ n3

n4

=1n4

[12n(n+ 1)

]2=

n2(n+ 1)2

4n4→ 1

4as n → ∞.

So even though the rectangles overlap, we still get the same answer.


. . . . . .

Outline





. . . . . .

Cavalieri's method in general.

.

Let f be a positive function defined on the interval [a,b]. We want to find thearea between x = a, x = b, y = 0, and y = f(x).For each positive integer n, divide up the interval into n pieces. Then

∆x =b− an

. For each i between 1 and n, let xi be the nth step between a andb. So

..a .b. . . . . . ..x0 .x1 .x2 .. . . .xi.xn−1.xn

x0 = a

x1 = x0 +∆x = a+b− an

x2 = x1 +∆x = a+ 2 · b− an

· · · · · ·

xi = a+ i · b− an

· · · · · ·

xn = a+ n · b− an

= b


. . . . . .


.


∆x =b− an


..a .b. . . . . . ..x0 .x1 .x2 .. . . .xi.xn−1.xn

x0 = a

x1 = x0 +∆x = a+b− an

x2 = x1 +∆x = a+ 2 · b− an

· · · · · ·


· · · · · ·


= b


. . . . . .


.


∆x =b− an


..a .b. . . . . . ..x0 .x1 .x2 .. . . .xi.xn−1.xn

x0 = a

x1 = x0 +∆x = a+b− an

x2 = x1 +∆x = a+ 2 · b− an

· · · · · ·


· · · · · ·


= b


. . . . . .


.


∆x =b− an


..a .b. . . . . . ..x0 .x1 .x2 .. . . .xi.xn−1.xn

x0 = a

x1 = x0 +∆x = a+b− an

x2 = x1 +∆x = a+ 2 · b− an

· · · · · ·


· · · · · ·


= b


. . . . . .


.


∆x =b− an


..a .b. . . . . . ..x0 .x1 .x2 .. . . .xi.xn−1.xn

x0 = a

x1 = x0 +∆x = a+b− an

x2 = x1 +∆x = a+ 2 · b− an

· · · · · ·


· · · · · ·


= b


. . . . . .

Forming Riemann sums

We have many choices of how to approximate the area:

Ln = f(x0)∆x+ f(x1)∆x+ · · ·+ f(xn−1)∆xRn = f(x1)∆x+ f(x2)∆x+ · · ·+ f(xn)∆x

Mn = f(x0 + x1

2

)∆x+ f

(x1 + x2

2

)∆x+ · · ·+ f

(xn−1 + xn

2

)∆x

In general, choose ci to be a point in the ith interval [xi−1, xi]. Form theRiemann sum

Sn = f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x

=n∑

i=1

f(ci)∆x


. . . . . .

Forming Riemann sums

We have many choices of how to approximate the area:

Ln = f(x0)∆x+ f(x1)∆x+ · · ·+ f(xn−1)∆xRn = f(x1)∆x+ f(x2)∆x+ · · ·+ f(xn)∆x

Mn = f(x0 + x1

2

)∆x+ f

(x1 + x2

2

)∆x+ · · ·+ f

(xn−1 + xn

2

)∆x

In general, choose ci to be a point in the ith interval [xi−1, xi]. Form theRiemann sum

Sn = f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x

=n∑

i=1

f(ci)∆x


. . . . . .

Theorem of the Day

TheoremIf f is a continuous function orhas finitely many jumpdiscontinuities on [a,b], then

limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x

exists and is the same value nomatter what choice of ci wemade.

. ..x1.a .b


. . . . . .

Theorem of the Day


limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x


. ..x1

..x2.a .b


. . . . . .

Theorem of the Day


limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x


. ..x1

..x2

..x3.a .b


. . . . . .

Theorem of the Day


limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x


. ..x1

..x2

..x3

..x4.a .b


. . . . . .

Theorem of the Day


limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x


. ..x1

..x2

..x3

..x4

..x5.a .b


. . . . . .

Theorem of the Day


limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x


. ..x1

..x2

..x3

..x4

..x5

..x6.a .b


. . . . . .

Theorem of the Day


limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x


. ..x1

..x2

..x3

..x4

..x5

..x6

..x7.a .b


. . . . . .

Theorem of the Day


limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x


. ..x1

..x2

..x3

..x4

..x5

..x6

..x7

..x8.a .b


. . . . . .

Theorem of the Day


limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x


. ..x1

..x2

..x3

..x4

..x5

..x6

..x7

..x8

..x9.a .b


. . . . . .

Theorem of the Day


limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x


. ..x1

..x2

..x3

..x4

..x5

..x6

..x7

..x8

..x9

..x10.a .b


. . . . . .

Theorem of the Day


limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x


. ..x1

..x2

..x3

..x4

..x5

..x6

..x7

..x8

..x9

..x10

..x11.a .b


. . . . . .

Theorem of the Day


limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x


. ..x1..x2..x3..x4..x5..x6..x7..x8..x9.

.x10.

.x11.

.x12.a .b


. . . . . .

Theorem of the Day


limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x


. ..x1..x2..x3..x4..x5..x6..x7..x8..x9.

.x10.

.x11.

.x12.

.x13.a .b


. . . . . .

Theorem of the Day


limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x


. ..x1..x2..x3..x4..x5..x6..x7..x8..x9.

.x10.

.x11.

.x12.

.x13.

.x14.a .b


. . . . . .

Theorem of the Day


limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x


. ..x1..x2..x3..x4..x5..x6..x7..x8..x9.

.x10.

.x11.

.x12.

.x13.

.x14.

.x15.a .b


. . . . . .

Theorem of the Day


limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x


. ..x1..x2..x3..x4..x5..x6..x7..x8..x9.

.x10.

.x11.

.x12.

.x13.

.x14.

.x15.

.x16.a .b


. . . . . .

Theorem of the Day


limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x


. ..x1..x2..x3..x4..x5..x6..x7..x8..x9.

.x10.

.x11.

.x12.

.x13.

.x14.

.x15.

.x16.

.x17.a .b


. . . . . .

Theorem of the Day


limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x


. ..x1..x2..x3..x4..x5..x6..x7..x8..x9.

.x10.

.x11.

.x12.

.x13.

.x14.

.x15.

.x16.

.x17.

.x18.a .b


. . . . . .

Theorem of the Day


limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x


. ..x1..x2..x3..x4..x5..x6..x7..x8..x9.

.x10.

.x11.

.x12.

.x13.

.x14.

.x15.

.x16.

.x17.

.x18.

.x19.a .b


. . . . . .

Theorem of the Day


limn→∞

Sn = limn→∞

n∑i=1

f(ci)∆x


. ..x1..x2..x3..x4..x5..x6..x7..x8..x9.

.x10.

.x11.

.x12.

.x13.

.x14.

.x15.

.x16.

.x17.

.x18.

.x19.

.x20.a .b


. . . . . .

Analogies

The Tangent Problem(Ch. 2–4)

I Want the slope of a curveI Only know the slope oflines

I Approximate curve with aline

I Take limit over better andbetter approximations

The Area Problem (Ch. 5)

I Want the area of a curvedregion

I Only know the area ofpolygons

I Approximate region withpolygons



. . . . . .

Analogies


I Want the slope of a curve

I Only know the slope oflines



The Area Problem (Ch. 5)

I Want the area of a curvedregion





. . . . . .

Analogies


I Want the slope of a curve

I Only know the slope oflines



The Area Problem (Ch. 5)I Want the area of a curvedregion





. . . . . .

Analogies










. . . . . .

Analogies










. . . . . .

Analogies










. . . . . .

Analogies










. . . . . .

Analogies










. . . . . .

Outline





. . . . . .

Distances

Just like area = length× width, we have

distance = rate× time.

So here is another use for Riemann sums.


. . . . . .

Application: Dead Reckoning


. . . . . .

Computing position by Dead Reckoning

Example

A sailing ship is cruising back and forth along a channel (in a straightline). At noon the ship’s position and velocity are recorded, but shortlythereafter a storm blows in and position is impossible to measure. Thevelocity continues to be recorded at thirty-minute intervals.

Time 12:00 12:30 1:00 1:30 2:00Speed (knots) 4 8 12 6 4Direction E E E E WTime 2:30 3:00 3:30 4:00Speed 3 3 5 9Direction W E E E

Estimate the ship’s position at 4:00pm.


. . . . . .

Solution

SolutionWe estimate that the speed of 4 knots (nautical miles per hour) ismaintained from 12:00 until 12:30. So over this time interval the shiptravels (

4 nmihr

)(12hr)

= 2nmi

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 15.5nmi east of its original position.


. . . . . .

Analysis

I This method of measuring position by recording velocity wasnecessary until global-positioning satellite technology becamewidespread

I If we had velocity estimates at finer intervals, we’d get betterestimates.

I If we had velocity at every instant, a limit would tell us our exactposition relative to the last time we measured it.


. . . . . .

Other uses of Riemann sums

Anything with a product!I Area, volumeI Anything with a density: Population, massI Anything with a “speed:” distance, throughput, powerI Consumer surplusI Expected value of a random variable


. . . . . .

Surplus by picture

..quantity (q)

.price (p)

.

.demand f(q)

.supply

.equilibrium

..q∗

..p∗

.consumer surplus


. . . . . .

Summary

I We can compute the area of a curved region with a limit ofRiemann sums

I We can compute the distance traveled from the velocity with alimit of Riemann sums

I Many other important uses of this process.


Lesson 24: Area and Distances

Technology

easy areas

hard areas

nyu section

parabola1 archimedes

rectangle b

parallelogram of base

sequence of triangles

rectangle definition