lesson 22: areas and distances

Section 5.1Areas and Distances

V63.0121.006/016, Calculus I

New York University

April 13, 2010

Announcements

I Quiz April 16 on §§4.1–4.4

I Final Exam: Monday, May 10, 12:00noon

Announcements

I Quiz April 16 on §§4.1–4.4

I Final Exam: Monday, May10, 12:00noon

V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 2 / 30

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 as distance= (rate)(time) and lettingthe time intervals over whichone approximates tend tozero.


Outline

Area through the CenturiesEuclidArchimedesCavalieri

Generalizing Cavalieri’s methodAnalogies

DistancesOther applications


Easy Areas: Rectangle

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 wehave to start somewhere.


Easy Areas: Parallelogram

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

b

b

h

So

Fact

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

A = bh




b b

h

So

Fact


A = bh




b

b

h

So

Fact


A = bh


Easy Areas: Triangle

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

b

h

So

Fact

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

A =1

2bh


Easy Areas: Other Polygons

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


Hard Areas: Curved Regions

???


Meet the mathematician: Archimedes

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

I Geometer

I Weapons engineer


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

A =

1 + 2 · 1

8+ 4 · 1

64+ · · ·

= 1 +1

4+

1

16+ · · ·+ 1

4n+ · · ·


1


A = 1

+ 2 · 1

8+ 4 · 1

64+ · · ·

= 1 +1

4+

1

16+ · · ·+ 1

4n+ · · ·


118

18


A = 1 + 2 · 1

8

+ 4 · 1

64+ · · ·

= 1 +1

4+

1

16+ · · ·+ 1

4n+ · · ·


118

18

164

164

164

164


A = 1 + 2 · 1

8+ 4 · 1

64+ · · ·

= 1 +1

4+

1

16+ · · ·+ 1

4n+ · · ·


We would then need to know the value of the series

1 +1

4+

1

16+ · · ·+ 1

4n+ · · ·

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− r

Therefore

1 +1

4+

1

16+ · · ·+ 1

4n=

1− (1/4)n+1

1− 1/4→ 1

3/4=

4

3

as n→∞.



1 +1

4+

1

16+ · · ·+ 1

4n+ · · ·


(1− r)(1 + r + · · ·+ rn) = 1− rn+1

So

1 + r + · · ·+ rn =1− rn+1

1− r

Therefore

1 +1

4+

1

16+ · · ·+ 1

4n=

1− (1/4)n+1

1− 1/4

→ 13/4

=4

3

as n→∞.



1 +1

4+

1

16+ · · ·+ 1

4n+ · · ·


(1− r)(1 + r + · · ·+ rn) = 1− rn+1

So

1 + r + · · ·+ rn =1− rn+1

1− r

Therefore

1 +1

4+

1

16+ · · ·+ 1

4n=

1− (1/4)n+1

1− 1/4→ 1

3/4=

4

3

as n→∞.


Cavalieri

I Italian,1598–1647

I Revisited thearea problemwith adifferentperspective


Cavalieri’s method

y = x2

0 1

1

2

Divide up the interval into piecesand measure the area of theinscribed rectangles:

L2 =1

8

L3 =

1

27+

4

27=

5

27

L4 =

1

64+

4

64+

9

64=

14

64

L5 =

1

125+

4

125+

9

125+

16

125=

30

125

Ln =?



y = x2

0 11

2


L2 =1

8

L3 =

1

27+

4

27=

5

27

L4 =

1

64+

4

64+

9

64=

14

64

L5 =

1

125+

4

125+

9

125+

16

125=

30

125

Ln =?



y = x2

0 1

1

2

1

3

2

3


L2 =1

8

L3 =

1

27+

4

27=

5

27

L4 =

1

64+

4

64+

9

64=

14

64

L5 =

1

125+

4

125+

9

125+

16

125=

30

125

Ln =?



y = x2

0 1

1

2

1

3

2

3


L2 =1

8

L3 =1

27+

4

27=

5

27

L4 =

1

64+

4

64+

9

64=

14

64

L5 =

1

125+

4

125+

9

125+

16

125=

30

125

Ln =?



y = x2

0 1

1

2

1

4

2

4

3

4


L2 =1

8

L3 =1

27+

4

27=

5

27

L4 =

1

64+

4

64+

9

64=

14

64

L5 =

1

125+

4

125+

9

125+

16

125=

30

125

Ln =?



y = x2

0 1

1

2

1

4

2

4

3

4


L2 =1

8

L3 =1

27+

4

27=

5

27

L4 =1

64+

4

64+

9

64=

14

64

L5 =

1

125+

4

125+

9

125+

16

125=

30

125

Ln =?



y = x2

0 1

1

2

1

5

2

5

3

5

4

5


L2 =1

8

L3 =1

27+

4

27=

5

27

L4 =1

64+

4

64+

9

64=

14

64

L5 =

1

125+

4

125+

9

125+

16

125=

30

125Ln =?



y = x2

0 1

1

2

1

5

2

5

3

5

4

5


L2 =1

8

L3 =1

27+

4

27=

5

27

L4 =1

64+

4

64+

9

64=

14

64

L5 =1

125+

4

125+

9

125+

16

125=

30

125

Ln =?



y = x2

0 1

1

2


L2 =1

8

L3 =1

27+

4

27=

5

27

L4 =1

64+

4

64+

9

64=

14

64

L5 =1

125+

4

125+

9

125+

16

125=

30

125Ln =?


What is Ln?

Divide the interval [0, 1] into n pieces. Then each has width1

n.

The

rectangle over the ith interval and under the parabola has area

1

n·(

i − 1

n

)2

=(i − 1)2

n3.

So

Ln =1

n3+

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)

6

So

Ln =n(n − 1)(2n − 1)

6n3→ 1

3as n→∞.


What is Ln?


n. The


1

n·(

i − 1

n

)2

=(i − 1)2

n3.

So

Ln =1

n3+

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)

6

So

Ln =n(n − 1)(2n − 1)

6n3→ 1

3as n→∞.


What is Ln?


n. The


1

n·(

i − 1

n

)2

=(i − 1)2

n3.

So

Ln =1

n3+

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)

6

So

Ln =n(n − 1)(2n − 1)

6n3

→ 1

3as n→∞.


What is Ln?


n. The


1

n·(

i − 1

n

)2

=(i − 1)2

n3.

So

Ln =1

n3+

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)

6

So

Ln =n(n − 1)(2n − 1)

6n3→ 1

3as n→∞.V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 15 / 30

Cavalieri’s method for different functions

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

Ln =1

n· f(

1

n

)+

1

n· f(

2

n

)+ · · ·+ 1

n· f(

n − 1

n

)

=1

n· 1

n3+

1

n· 23

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 =1

n· f(

1

n

)+

1

n· f(

2

n

)+ · · ·+ 1

n· f(

n − 1

n

)=

1

n· 1

n3+

1

n· 23

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 =1

n· f(

1

n

)+

1

n· f(

2

n

)+ · · ·+ 1

n· f(

n − 1

n

)=

1

n· 1

n3+

1

n· 23

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 =1

n· f(

1

n

)+

1

n· f(

2

n

)+ · · ·+ 1

n· f(

n − 1

n

)=

1

n· 1

n3+

1

n· 23

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 =1

n· 13

n3+

1

n· 23

n3+ · · ·+ 1

n· n3

n3

=13 + 23 + 33 + · · ·+ n3

n4

=1

n4

[12n(n + 1)

]2=

n2(n + 1)2

4n4→ 1

4

as n→∞.

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


Cavalieri’s method with different heights

Rn =1

n· 13

n3+

1

n· 23

n3+ · · ·+ 1

n· n3

n3

=13 + 23 + 33 + · · ·+ n3

n4

=1

n4

[12n(n + 1)

]2=

n2(n + 1)2

4n4→ 1

4

as 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 the areabetween 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 − a

n.

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

a bx0 x1 x2 . . . xixn−1xn

x0 = a

x1 = x0 + ∆x = a +b − a

n

x2 = x1 + ∆x = a + 2 · b − a

n· · · · · ·

xi = a + i · b − a

n· · · · · ·

xn = a + n · b − a

n= b


Forming Riemann sums

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

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

Theorem

If 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.

x1a b


Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x


x1 x2a b


Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x


x1 x2 x3a b


Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x


x1 x2 x3 x4a b


Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x


x1 x2 x3 x4 x5a b


Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x


x1 x2 x3 x4 x5 x6a b


Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x


x1 x2 x3 x4 x5 x6 x7a b


Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x


x1x2x3x4x5x6x7x8a b


Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x


x1x2x3x4x5x6x7x8x9a b


Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x


x1x2x3x4x5x6x7x8x9x10a b


Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x


x1x2x3x4x5x6x7x8x9x10x11a b


Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x


x1x2x3x4x5x6x7x8x9x10x11x12a b


Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x


x1x2x3x4x5x6x7x8x9x10x11x12x13a b


Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x


x1x2x3x4x5x6x7x8x9x10x11x12x13x14a b


Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x


x1x2x3x4x5x6x7x8x9x10x11x12x13x14x15a b


Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x


x1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16a b


Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x


x1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16x17a b


Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x


x1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16x17x18a b


Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x


x1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16x17x18x19a b


Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x


x1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16x17x18x19x20a b


Analogies

The Tangent Problem(Ch. 2–4)

I Want the slope of a curve

I Only know the slope of lines

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

I Take limit over better andbetter approximations


Outline





Distances

Just like area = length× width, we have

distance = rate× time.

So here is another use for Riemann sums.


Application: 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 2:00

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.


Solution

We 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 nmi

hr

)(1

2hr

)= 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 15.5 nmi east of its original position.


Analysis

I This method of measuring position by recording velocity was necessaryuntil global-positioning satellite technology became widespread

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, volume

I Anything with a density: Population, mass

I Anything with a “speed:” distance, throughput, power

I Consumer surplus

I Expected value of a random variable


Summary

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

I We can compute the distance traveled from the velocity with a limitof Riemann sums

I Many other important uses of this process.


lesson 22: areas and distances

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