lesson 29: areas
DESCRIPTION
We look at the area problem of finding areas of curved regions. Archimedes had a method for parabolas, Cavalieri had a method for other graphs, and Riemann generalized the whole thing. It doesn't just work for areas, any "product law" such as distance=rate x time can be generalized to a similar computationTRANSCRIPT
Section 5.1Areas and Distances
Math 1a
December 5, 2007
Announcements
I my next office hours: Today 1–3 (SC 323)
I MT II is nearly graded. You’ll get it back Friday
I Final seview sessions: Wed 1/9 and Thu 1/10 in Hall D, Sun1/13 in Hall C, all 7–8:30pm
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
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+ · · ·
118
18
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+ · · ·
118
18
164
164
164
164
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+ · · ·
118
18
164
164
164
164
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+ · · ·
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
Generalizing Cavalieri’s method
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 =?
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 =?
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 =?
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 =?
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 =?
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 =?
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
125Ln =?
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 =?
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
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?
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?
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?
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?
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→∞.
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→∞.
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→∞.
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→∞.
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)
]2
So
Ln =n2(n − 1)2
4n4→ 1
4as 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→∞.
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
Generalizing Cavalieri’s method
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
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
Generalizing Cavalieri’s method
Worksheet
Worksheet
We will determine the area under y = ex between x = 0 and x = 1.