Lesson 24: Areas, Distances, the Integral (Section 041 slides)

..
Sections 5.1–5.2
Areas and Distances
The Definite Integral
V63.0121.041, Calculus I
New York University
December 1, 2010
Announcements
Quiz 5 in recitation §§4.1–4.4
Final December 20, 12:00–1:50pm
. . . . . .

. . . . . .
Announcements
Quiz 5 in recitation
§§4.1–4.1
Final December 20,
12:00–1:50pm
cumulative
location TBD
old exams on common
website
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 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.
..
ℓ
.
w
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

. . . . . .
..
b
.
h

. . . . . .
..
h

. . . . . .
..
b
.
h
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

. . . . . .
..
b
.
h

. . . . . .
..
b
.
h
So
Fact
The area of a triangle of base width b and height h is
A =
1
2
bh

. . . . . .
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
8
.
1
8
parabola.
A = 1 + 2 ·
1
8

. . . . . .
..
1
.
1
8
.
1
8
.
1
64
.
1
64
.
1
64
.
1
64
parabola.
A = 1 + 2 ·
1
8
+ 4 ·
1
64
+ · · ·

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

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

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

. . . . . .
Summing the series
1 +
1
4
+
1
16
+ · · · +
1
4n + · · ·
Fact
(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

. . . . . .
Summing the series
1 +
1
4
+
1
16
+ · · · +
1
4n + · · ·
Fact
(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
Italian,
1598–1647
Revisited the
area
problem with
a different
perspective

. . . . . .
Cavalieri's method
..
y = x2
..
0
..
1
Divide up the interval into
pieces and measure the area of
the inscribed rectangles:

. . . . . .
Cavalieri's method
..
y = x2
..
0
..
1
..
1
2
L2 =
1
8

. . . . . .
Cavalieri's method
..
y = x2
..
0
..
1
..
1
3
..
2
3
L2 =
1
8
L3 =

. . . . . .
Cavalieri's method
..
y = x2
..
0
..
1
..
1
3
..
2
3
L2 =
1
8
L3 =
1
27
+
4
27
=
5
27

. . . . . .
Cavalieri's method
..
y = x2
..
0
..
1
..
1
4
..
2
4
..
3
4
L2 =
1
8
L3 =
1
27
+
4
27
=
5
27
L4 =

. . . . . .
Cavalieri's method
..
y = x2
..
0
..
1
..
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

. . . . . .
Cavalieri's method
..
y = x2
..
0
..
1
..
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 =

. . . . . .
Cavalieri's method
..
y = x2
..
0
..
1
..
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

. . . . . .
Cavalieri's method
..
y = x2
..
0
..
1
..
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 =?

. . . . . .
What is Ln?
Divide the interval [0, 1] into n pieces. Then each has width
1
n
.

. . . . . .
What is Ln?
1
n
. The
rectangle over the ith interval and under the parabola has area
1
n
·
(
i − 1
n
)2
=
(i − 1)2
n3
.

. . . . . .
What is Ln?
1
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

. . . . . .
What is Ln?
1
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

. . . . . .
What is Ln?
1
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
3
as 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
)

. . . . . .
. 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

. . . . . .
. 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

. . . . . .
. 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
=
[
1
2 n(n − 1)
]2

. . . . . .
. 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
=
[
1
2 n(n − 1)
]2
So
Ln =
n2(n − 1)2
4n4
→
1
4
as 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
[
1
2 n(n + 1)
]2
=
n2(n + 1)2
4n4
→
1
4
as 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
[
1
2 n(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
Euclid
Archimedes
Cavalieri
Analogies
Distances
Other applications

. . . . . .
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
∆x =
b − a
n
. For each i between 1 and n, let xi be the ith step between
a and b. So
.. x..
x0
..
x1
..
xi
..
xn−1
..
xn
.
. . .
.
. . .
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 representative points to approximate the
area in each subinterval.
left endpoints…
Ln =
n∑
i=1
f(xi−1)∆x
.. x.......

. . . . . .
right endpoints…
Rn =
n∑
i=1
f(xi)∆x
.. x.......

. . . . . .
midpoints…
Mn =
n∑
i=1
f
(
xi−1 + xi
2
)
∆x
.. 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
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 on [a, b]
or has finitely many jump
discontinuities, then
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
exists and is the same value no
matter what choice of ci we make.
.... x.

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L1 = 3.0

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L2 = 5.25

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L3 = 6.0

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L4 = 6.375

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L5 = 6.59988

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L6 = 6.75

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L7 = 6.85692

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L8 = 6.9375

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L9 = 6.99985

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L10 = 7.04958

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L11 = 7.09064

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L12 = 7.125

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L13 = 7.15332

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L14 = 7.17819

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L15 = 7.19977

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L16 = 7.21875

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L17 = 7.23508

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L18 = 7.24927

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L19 = 7.26228

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L20 = 7.27443

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L21 = 7.28532

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L22 = 7.29448

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L23 = 7.30406

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L24 = 7.3125

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L25 = 7.31944

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L26 = 7.32559

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L27 = 7.33199

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L28 = 7.33798

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L29 = 7.34372

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
left endpoints
.
L30 = 7.34882

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R1 = 12.0

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R2 = 9.75

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R3 = 9.0

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R4 = 8.625

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R5 = 8.39969

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R6 = 8.25

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R7 = 8.14236

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R8 = 8.0625

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R9 = 7.99974

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R10 = 7.94933

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R11 = 7.90868

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R12 = 7.875

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R13 = 7.84541

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R14 = 7.8209

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R15 = 7.7997

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R16 = 7.78125

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R17 = 7.76443

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R18 = 7.74907

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R19 = 7.73572

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R20 = 7.7243

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R21 = 7.7138

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R22 = 7.70335

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R23 = 7.69531

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R24 = 7.6875

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R25 = 7.67934

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R26 = 7.6715

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R27 = 7.66508

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R28 = 7.6592

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R29 = 7.65388

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
right endpoints
.
R30 = 7.64864

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M1 = 7.5

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M2 = 7.5

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M3 = 7.5

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M4 = 7.5

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M5 = 7.4998

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M6 = 7.5

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M7 = 7.4996

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M8 = 7.5

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M9 = 7.49977

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M10 = 7.49947

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M11 = 7.49966

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M12 = 7.5

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M13 = 7.49937

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M14 = 7.49954

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M15 = 7.49968

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M16 = 7.49988

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M17 = 7.49974

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M18 = 7.49916

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M19 = 7.49898

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M20 = 7.4994

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M21 = 7.49951

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M22 = 7.49889

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M23 = 7.49962

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M24 = 7.5

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M25 = 7.49939

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M26 = 7.49847

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M27 = 7.4985

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M28 = 7.4986

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M29 = 7.49878

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M30 = 7.49872

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U1 = 12.0

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U2 = 10.55685

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U3 = 10.0379

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U4 = 9.41515

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U5 = 8.96004

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U6 = 8.76895

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U7 = 8.6033

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U8 = 8.45757

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U9 = 8.34564

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U10 = 8.27084

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U11 = 8.20132

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U12 = 8.13838

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U13 = 8.0916

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U14 = 8.05139

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U15 = 8.01364

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U16 = 7.98056

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U17 = 7.9539

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U18 = 7.92815

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U19 = 7.90414

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U20 = 7.88504

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U21 = 7.86737

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U22 = 7.84958

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U23 = 7.83463

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U24 = 7.82187

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U25 = 7.80824

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U26 = 7.79504

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U27 = 7.78429

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U28 = 7.77443

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U29 = 7.76495

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
maximum points
.
U30 = 7.7558

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L1 = 3.0

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L2 = 4.44312

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L3 = 4.96208

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L4 = 5.58484

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L5 = 6.0395

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L6 = 6.23103

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L7 = 6.39577

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L8 = 6.54242

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L9 = 6.65381

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L10 = 6.72797

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L11 = 6.7979

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L12 = 6.8616

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L13 = 6.90704

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L14 = 6.94762

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L15 = 6.98575

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L16 = 7.01942

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L17 = 7.04536

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L18 = 7.07005

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L19 = 7.09364

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L20 = 7.1136

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L21 = 7.13155

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L22 = 7.14804

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L23 = 7.16441

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L24 = 7.17812

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L25 = 7.19025

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L26 = 7.2019

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L27 = 7.21265

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L28 = 7.22269

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L29 = 7.23251

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
minimum points
.
L30 = 7.24162

. . . . . .
Analogies
The Tangent Problem
(Ch. 2–4)
The Area Problem (Ch. 5)

. . . . . .
Analogies
The Tangent Problem
(Ch. 2–4)
Want the slope of a curve

. . . . . .
Analogies
The Tangent Problem
(Ch. 2–4)
Want the area of a curved
region

. . . . . .
Analogies
The Tangent Problem
(Ch. 2–4)
Only know the slope of
lines
region

. . . . . .
Analogies
The Tangent Problem
(Ch. 2–4)
lines
region
Only know the area of
polygons

. . . . . .
Analogies
The Tangent Problem
(Ch. 2–4)
lines
Approximate curve with a
line
region
polygons

. . . . . .
Analogies
The Tangent Problem
(Ch. 2–4)
lines
line
region
polygons
Approximate region with
polygons

. . . . . .
Analogies
The Tangent Problem
(Ch. 2–4)
lines
line
Take limit over better and
better approximations
region
polygons
Approximate region with
polygons
Take limit over better and
better approximations

. . . . . .
Outline
Euclid
Archimedes
Cavalieri
Analogies
Distances
Other applications

. . . . . .
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 straight
line). At noon the ship’s position and velocity are recorded, but shortly
thereafter a storm blows in and position is impossible to measure. The
velocity 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
Solution
We estimate that the speed of 4 knots (nautical miles per hour) is
maintained from 12:00 until 12:30. So over this time interval the ship
travels (
4 nmi
hr
) (
1
2
hr
)
= 2 nmi
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
This method of measuring position by recording velocity was
necessary until global-positioning satellite technology became
widespread
If we had velocity estimates at finer intervals, we’d get better
estimates.
If we had velocity at every instant, a limit would tell us our exact
position relative to the last time we measured it.

. . . . . .
Other uses of Riemann sums
Anything with a product!
Area, volume
Anything with a density: Population, mass
Anything with a “speed:” distance, throughput, power
Consumer surplus
Expected value of a random variable

. . . . . .
Outline
Euclid
Archimedes
Cavalieri
Analogies
Distances
Other applications

. . . . . .
Definition
If f is a function defined on [a, b], the definite integral of f from a to b
is the number ∫ b
a
f(x) dx = lim
∆x→0
n∑
i=1
f(ci) ∆x

. . . . . .
Notation/Terminology
∫ b
a
f(x) dx = lim
∆x→0
n∑
i=1
f(ci) ∆x

. . . . . .
∫ b
a
f(x) dx = lim
∆x→0
n∑
i=1
f(ci) ∆x
∫
— integral sign (swoopy S)

. . . . . .
∫ b
a
f(x) dx = lim
∆x→0
n∑
i=1
f(ci) ∆x
∫
f(x) — integrand

. . . . . .
∫ b
a
f(x) dx = lim
∆x→0
n∑
i=1
f(ci) ∆x
∫
f(x) — integrand
a and b — limits of integration (a is the lower limit and b the
upper limit)

. . . . . .
∫ b
a
f(x) dx = lim
∆x→0
n∑
i=1
f(ci) ∆x
∫
f(x) — integrand
upper limit)
dx — ??? (a parenthesis? an infinitesimal? a variable?)

. . . . . .
∫ b
a
f(x) dx = lim
∆x→0
n∑
i=1
f(ci) ∆x
∫
f(x) — integrand
upper limit)
dx — ??? (a parenthesis? an infinitesimal? a variable?)
The process of computing an integral is called integration or
quadrature

. . . . . .
The limit can be simplified
Theorem
If f is continuous on [a, b] or if f has only finitely many jump
discontinuities, then f is integrable on [a, b]; that is, the definite integral
∫ b
a
f(x) dx exists.

. . . . . .
The limit can be simplified
Theorem
If f is continuous on [a, b] or if f has only finitely many jump
discontinuities, then f is integrable on [a, b]; that is, the definite integral
∫ b
a
f(x) dx exists.
Theorem
If f is integrable on [a, b] then
∫ b
a
f(x) dx = lim
n→∞
n∑
i=1
f(xi)∆x,
where
∆x =
b − a
n
and xi = a + i ∆x

. . . . . .
Example: Integral of x
Example
Find
∫ 3
0
x dx

. . . . . .
Example: Integral of x
Example
Find
∫ 3
0
x dx
Solution
For any n we have ∆x =
3
n
and xi =
3i
n
. So
Rn =
n∑
i=1
f(xi) ∆x =
n∑
i=1
(
3i
n
) (
3
n
)
=
9
n2
n∑
i=1
i
=
9
n2
·
n(n + 1)
2
−→
9
2
So
∫ 3
0
x dx =
9
2
= 4.5

. . . . . .
Example: Integral of x2
Example
Find
∫ 3
0
x2
dx

. . . . . .
Example
Find
∫ 3
0
x2
dx
Solution
3
n
and xi =
3i
n
. So
Rn =
n∑
i=1
f(xi) ∆x =
n∑
i=1
(
3i
n
)2 (
3
n
)
=
27
n3
n∑
i=1
i2
=
27
n3
·
n(n + 1)(2n + 1)
6
−→
27
3
= 9
So
∫ 3
0
x2
dx = 9

. . . . . .
Example
Find
∫ 3
0
x3
dx

. . . . . .
Example
Find
∫ 3
0
x3
dx
Solution
3
n
and xi =
3i
n
. So
Rn =
n∑
i=1
f(xi) ∆x =
n∑
i=1
(
3i
n
)3 (
3
n
)
=
81
n4
n∑
i=1
i3
=
81
n4
·
n2(n + 1)2
4
−→
81
4
So
∫ 3
0
x3
dx =
81
4
= 20.25

. . . . . .
Outline
Euclid
Archimedes
Cavalieri
Analogies
Distances
Other applications

. . . . . .
Example
Estimate
∫ 1
0
4
1 + x2
dx using M4.

. . . . . .
Example
Estimate
∫ 1
0
4
1 + x2
dx using M4.
Solution
We have x0 = 0, x1 =
1
4
, x2 =
1
2
, x3 =
3
4
, x4 = 1.
So c1 =
1
8
, c2 =
3
8
, c3 =
5
8
, c4 =
7
8
.
M4 =
1
4
(
4
1 + (1/8)2
+
4
1 + (3/8)2
+
4
1 + (5/8)2
+
4
1 + (7/8)2
)

. . . . . .
Example
Estimate
∫ 1
0
4
1 + x2
dx using M4.
Solution
1
4
, x2 =
1
2
, x3 =
3
4
, x4 = 1.
So c1 =
1
8
, c2 =
3
8
, c3 =
5
8
, c4 =
7
8
.
M4 =
1
4
(
4
1 + (1/8)2
+
4
1 + (3/8)2
+
4
1 + (5/8)2
+
4
1 + (7/8)2
)
=
1
4
(
4
65/64
+
4
73/64
+
4
89/64
+
4
113/64
)

. . . . . .
Example
Estimate
∫ 1
0
4
1 + x2
dx using M4.
Solution
1
4
, x2 =
1
2
, x3 =
3
4
, x4 = 1.
So c1 =
1
8
, c2 =
3
8
, c3 =
5
8
, c4 =
7
8
.
M4 =
1
4
(
4
1 + (1/8)2
+
4
1 + (3/8)2
+
4
1 + (5/8)2
+
4
1 + (7/8)2
)
=
1
4
(
4
65/64
+
4
73/64
+
4
89/64
+
4
113/64
)
=
64
65
+
64
73
+
64
89
+
64
113
≈ 3.1468

. . . . . .
Estimating the Definite Integral (Continued)
Example
Estimate
∫ 1
0
4
1 + x2
dx using L4 and R4

. . . . . .
Estimating the Definite Integral (Continued)
Example
Estimate
∫ 1
0
4
1 + x2
dx using L4 and R4
Answer
L4 =
1
4
(
4
1 + (0)2
+
4
1 + (1/4)2
+
4
1 + (1/2)2
+
4
1 + (3/4)2
)
= 1 +
16
17
+
4
5
+
16
25
≈ 3.38118
R4 =
1
4
(
4
1 + (1/4)2
+
4
1 + (1/2)2
+
4
1 + (3/4)2
+
4
1 + (1)2
)
=
16
17
+
4
5
+
16
25
+
1
2
≈ 2.88118

. . . . . .
Outline
Euclid
Archimedes
Cavalieri
Analogies
Distances
Other applications

. . . . . .
Theorem (Additive Properties of the Integral)
Let f and g be integrable functions on [a, b] and c a constant. Then
1.
∫ b
a
c dx = c(b − a)

. . . . . .
1.
∫ b
a
c dx = c(b − a)
2.
∫ b
a
[f(x) + g(x)] dx =
∫ b
a
f(x) dx +
∫ b
a
g(x) dx.

. . . . . .
1.
∫ b
a
c dx = c(b − a)
2.
∫ b
a
[f(x) + g(x)] dx =
∫ b
a
f(x) dx +
∫ b
a
g(x) dx.
3.
∫ b
a
cf(x) dx = c
∫ b
a
f(x) dx.

. . . . . .
1.
∫ b
a
c dx = c(b − a)
2.
∫ b
a
[f(x) + g(x)] dx =
∫ b
a
f(x) dx +
∫ b
a
g(x) dx.
3.
∫ b
a
cf(x) dx = c
∫ b
a
f(x) dx.
4.
∫ b
a
[f(x) − g(x)] dx =
∫ b
a
f(x) dx −
∫ b
a
g(x) dx.

. . . . . .
Proofs
Proofs.
When integrating a constant function c, each Riemann sum
equals c(b − a).
A Riemann sum for f + g equals a Riemann sum for f plus a
Riemann sum for g. Using the sum rule for limits, the integral of a
sum is the sum of the integrals.
Ditto for constant multiples
Ditto for differences

. . . . . .
Example
Find
∫ 3
0
(
x3
− 4.5x2
+ 5.5x + 1
)
dx

. . . . . .
Example
Find
∫ 3
0
(
x3
− 4.5x2
+ 5.5x + 1
)
dx
Solution
∫ 3
0
(x3
−4.5x2
+ 5.5x + 1) dx
=
∫ 3
0
x3
dx − 4.5
∫ 3
0
x2
dx + 5.5
∫ 3
0
x dx +
∫ 3
0
1 dx
= 20.25 − 4.5 · 9 + 5.5 · 4.5 + 3 · 1 = 7.5

. . . . . .
Example
Find
∫ 3
0
(
x3
− 4.5x2
+ 5.5x + 1
)
dx
Solution
∫ 3
0
(x3
−4.5x2
+ 5.5x + 1) dx
=
∫ 3
0
x3
dx − 4.5
∫ 3
0
x2
dx + 5.5
∫ 3
0
x dx +
∫ 3
0
1 dx
= 20.25 − 4.5 · 9 + 5.5 · 4.5 + 3 · 1 = 7.5
(This is the function we were estimating the integral of before)

. . . . . .
Theorem of the Day
Theorem
lim
n→∞
Sn = lim
n→∞
{ n∑
i=1
f(ci)∆x
}
.... x.
midpoints
.
M15 = 7.49968

. . . . . .
More Properties of the Integral
Conventions: ∫ a
b
f(x) dx = −
∫ b
a
f(x) dx

. . . . . .
Conventions: ∫ a
b
f(x) dx = −
∫ b
a
f(x) dx
∫ a
a
f(x) dx = 0

. . . . . .
Conventions: ∫ a
b
f(x) dx = −
∫ b
a
f(x) dx
∫ a
a
f(x) dx = 0
This allows us to have
5.
∫ c
a
f(x) dx =
∫ b
a
f(x) dx +
∫ c
b
f(x) dx for all a, b, and c.

. . . . . .
Example
Suppose f and g are functions with
∫ 4
0
f(x) dx = 4
∫ 5
0
f(x) dx = 7
∫ 5
0
g(x) dx = 3.
Find
(a)
∫ 5
0
[2f(x) − g(x)] dx
(b)
∫ 5
4
f(x) dx.

. . . . . .
Solution
We have
(a)
∫ 5
0
[2f(x) − g(x)] dx = 2
∫ 5
0
f(x) dx −
∫ 5
0
g(x) dx
= 2 · 7 − 3 = 11

. . . . . .
Solution
We have
(a)
∫ 5
0
[2f(x) − g(x)] dx = 2
∫ 5
0
f(x) dx −
∫ 5
0
g(x) dx
= 2 · 7 − 3 = 11
(b)
∫ 5
4
f(x) dx =
∫ 5
0
f(x) dx −
∫ 4
0
f(x) dx
= 7 − 4 = 3

. . . . . .
Outline
Euclid
Archimedes
Cavalieri
Analogies
Distances
Other applications

. . . . . .
Theorem
Let f and g be integrable functions on [a, b].

. . . . . .
Theorem
6. If f(x) ≥ 0 for all x in [a, b], then
∫ b
a
f(x) dx ≥ 0

. . . . . .
Theorem
∫ b
a
f(x) dx ≥ 0
7. If f(x) ≥ g(x) for all x in [a, b], then
∫ b
a
f(x) dx ≥
∫ b
a
g(x) dx

. . . . . .
Theorem
∫ b
a
f(x) dx ≥ 0
7. If f(x) ≥ g(x) for all x in [a, b], then
∫ b
a
f(x) dx ≥
∫ b
a
g(x) dx
8. If m ≤ f(x) ≤ M for all x in [a, b], then
m(b − a) ≤
∫ b
a
f(x) dx ≤ M(b − a)

. . . . . .
The integral of a nonnegative function is nonnegative
Proof.
If f(x) ≥ 0 for all x in [a, b], then
for any number of divisions n
and choice of sample points
{ci}:
Sn =
n∑
i=1
f(ci)
≥0
∆x ≥
n∑
i=1
0·∆x = 0
Since Sn ≥ 0 for all n, the limit
of {Sn} is nonnegative, too:
∫ b
a
f(x) dx = lim
n→∞
Sn
≥0
≥ 0
.. x.......

. . . . . .
The definite integral is “increasing"
Proof.
Let h(x) = f(x) − g(x). If
f(x) ≥ g(x) for all x in [a, b],
then h(x) ≥ 0 for all x in [a, b].
So by the previous property
∫ b
a
h(x) dx ≥ 0 .. x.
f(x)
.
g(x)
.
h(x)
This means that
∫ b
a
f(x) dx −
∫ b
a
g(x) dx =
∫ b
a
(f(x) − g(x)) dx =
∫ b
a
h(x) dx ≥ 0
So
∫ b
a
f(x) dx ≥
∫ b
a
g(x) dx.

. . . . . .
Bounding the integral using bounds of the function
Proof.
If m ≤ f(x) ≤ M on for all x in
[a, b], then by the previous
property
∫ b
a
m dx ≤
∫ b
a
f(x) dx ≤
∫ b
a
M dx
By Property 8, the integral of a
constant function is the product
of the constant and the width of
the interval. So:
m(b−a) ≤
∫ b
a
f(x) dx ≤ M(b−a)
.. x.
y
.
M
.
f(x)
.
m
..
a
..
b

. . . . . .
Example
Estimate
∫ 2
1
1
x
dx using the comparison properties.

. . . . . .
Example
Estimate
∫ 2
1
1
x
dx using the comparison properties.
Solution
Since
1
2
≤ x ≤
1
1
for all x in [1, 2], we have
1
2
· 1 ≤
∫ 2
1
1
x
dx ≤ 1 · 1

. . . . . .
Summary
We can compute the area of a curved region with a limit of
Riemann sums
We can compute the distance traveled from the velocity with a
limit of Riemann sums
Many other important uses of this process.

. . . . . .
Summary
The definite integral is a limit of Riemann Sums
The definite integral can be estimated with Riemann Sums
The definite integral can be distributed across sums and constant
multiples of functions
The definite integral can be bounded using bounds for the function

Lesson 24: Areas, Distances, the Integral (Section 041 slides)

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Lesson 24: Areas, Distances, the Integral (Section 041 slides)