Lesson 29: Areas

Section 5.1Areas and Distances

Math 1a

December 5, 2007

Announcements

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I Final tentatively scheduled for January 17

Outline

Archimedes

Cavalieri

Generalizing Cavalieri’s method

Worksheet

Meet the mathematician: Archimedes

I 287 BC – 212 BC (afterEuclid)

I Geometer

I Weapons engineer

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

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− x)(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→∞.

Outline

Archimedes

Cavalieri


Worksheet

Cavalieri

I Italian,1598–1647

I Revisitedthe areaproblemwith adifferentperspective

Cavalieri’s method

Divide up the interval intopieces and measure the areaof the inscribed 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

25=

30

125

Ln =?



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

25=

30

125

Ln =?



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

25=

30

125

Ln =?



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

25=

30

125Ln =?



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

25=

30

125

Ln =?



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

25=

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.


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.


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→∞.

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

n3

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

n3


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

n3


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

n3


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 sameanswer.

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 sameanswer.

Outline

Archimedes

Cavalieri


Worksheet

Cavalieri’s method in generalLet f be a positive function defined on the interval [a, b]. We wantto 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 ∆x =b − a

n. For each i between 1 and n, let xi be the nth

step between a and b. So

a b

x0 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 the Riemann 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 on [a, b] or has finitely many jumpdiscontinuities, then

limn→∞

Sn = limn→∞

{f (c1)∆x + f (c2)∆x + · · ·+ f (cn)∆x}

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

Outline

Archimedes

Cavalieri


Worksheet

Worksheet

We will determine the area under y = ex between x = 0 and x = 1.

Lesson 29: Areas

Technology

Transcript of Lesson 29: Areas