2. Sums and Approximations
Differentiation is the process of dividing smaller and
smaller differences of the outputs versus the
differences in the inputs to get better and better rates.
3. Sums and Approximations
Differentiation is the process of dividing smaller and
smaller differences of the outputs versus the
differences in the inputs to get better and better rates.
Integration refers to summing more and more smaller
and smaller terms to obtain better and better “totals”.
4. Sums and Approximations
The calculation for the value of π, which is the area of
the circle with radius 1, is an example of integration.
Differentiation is the process of dividing smaller and
smaller differences of the outputs versus the
differences in the inputs to get better and better rates.
Integration refers to summing more and more smaller
and smaller terms to obtain better and better “totals”.
5. Sums and Approximations
The calculation for the value of π, which is the area of
the circle with radius 1, is an example of integration.
Differentiation is the process of dividing smaller and
smaller differences of the outputs versus the
differences in the inputs to get better and better rates.
Integration refers to summing more and more smaller
and smaller terms to obtain better and better “totals”.
Inscribed Square
We find a lower bound by inscribing a square in the
circle of radius 1.
1
6. Sums and Approximations
The calculation for the value of π, which is the area of
the circle with radius 1, is an example of integration.
Differentiation is the process of dividing smaller and
smaller differences of the outputs versus the
differences in the inputs to get better and better rates.
Integration refers to summing more and more smaller
and smaller terms to obtain better and better “totals”.
Inscribed Square
We find a lower bound by inscribing a square in the
circle of radius 1. The area of the square consists of
1 hence we may conclude that 2 < π.
four right triangles each with area ½,
7. Sums and Approximations
The calculation for the value of π, which is the area of
the circle with radius 1, is an example of integration.
Differentiation is the process of dividing smaller and
smaller differences of the outputs versus the
differences in the inputs to get better and better rates.
Integration refers to summing more and more smaller
and smaller terms to obtain better and better “totals”.
Inscribed Square
We find a lower bound by inscribing a square in the
circle of radius 1. The area of the square consists of
1 hence we may conclude that 2 < π.
four right triangles each with area ½,
Next we inscribe the circle with a
hexagon for a better approximation.
10. Sums and Approximations
Inscribed Hexagon
The hexagon consists of
six equilateral triangles
with sides 1. Each triangle
has area = (√3) /4.
1
1
√3
2
11. Sums and Approximations
Inscribed Hexagon
The hexagon consists of
six equilateral triangles
with sides 1. Each triangle
has area = (√3) /4.
1
1
√3
2
So the area of the hexagon is 6(√3)/4 = 3(√3)/2
and we conclude that 3(√3)/2 ≈ 2.598.. < π.
12. Sums and Approximations
Inscribed Hexagon
The hexagon consists of
six equilateral triangles
with sides 1. Each triangle
has area = (√3) /4.
1
1
√3
2
So the area of the hexagon is 6(√3)/4 = 3(√3)/2
and we conclude that 3(√3)/2 ≈ 2.598.. < π.
Continuing in this fashion, if n = number of sides of
the regular polygons,
13. Sums and Approximations
Inscribed Hexagon
The hexagon consists of
six equilateral triangles
with sides 1. Each triangle
has area = (√3) /4.
1
1
√3
2
So the area of the hexagon is 6(√3)/4 = 3(√3)/2
and we conclude that 3(√3)/2 ≈ 2.598.. < π.
Continuing in this fashion, if n = number of sides of
the regular polygons, then
π = lim (area of the inscribed n sided reg–polygon)
http://en.wikipedia.org/wiki/Pi
n ∞
14. Sums and Approximations
Inscribed Hexagon
The hexagon consists of
six equilateral triangles
with sides 1. Each triangle
has area = (√3) /4.
1
1
√3
2
So the area of the hexagon is 6(√3)/4 = 3(√3)/2
and we conclude that 3(√3)/2 ≈ 2.598.. < π.
Continuing in this fashion, if n = number of sides of
the regular polygons, then
π = lim (area of the inscribed n sided reg–polygon)
http://en.wikipedia.org/wiki/Pi
n ∞
The Riemann sums we will define shortly are also
related to calculations of areas. However in
applications, “area” may represent concrete
measurements such as Work, Revenue etc..
15. Let I = [a, b] be the interval a < x < b.
Riemann Sums
16. Let I = [a, b] be the interval a < x < b.
A regular–partition of I is a division of I into equal
size subintervals.
Riemann Sums
17. Let I = [a, b] be the interval a < x < b.
A regular–partition of I is a division of I into equal
size subintervals. We will use n to denote the
number of the subintervals and Δx be the length
of each subinterval.
Riemann Sums
18. For example, {0, 1/3, 2/3, 1} gives a regular partition
of [0, 1] with n = 3:
Let I = [a, b] be the interval a < x < b.
0 11/3 2/3
Δx=1/3
A regular–partition of I is a division of I into equal
size subintervals. We will use n to denote the
number of the subintervals and Δx be the length
of each subinterval.
Riemann Sums
19. For example, {0, 1/3, 2/3, 1} gives a regular partition
of [0, 1] with n = 3:
Let I = [a, b] be the interval a < x < b.
0 1
{0, ¼, ½, ¾, 1} gives the regular partition of [0, 1]
with n = 4:
0 1½ ¾¼
1/3 2/3
Δx=1/3
Δx=1/4
A regular–partition of I is a division of I into equal
size subintervals. We will use n to denote the
number of the subintervals and Δx be the length
of each subinterval.
Riemann Sums
20. For example, {0, 1/3, 2/3, 1} gives a regular partition
of [0, 1] with n = 3:
Let I = [a, b] be the interval a < x < b.
0 1
{0, ¼, ½, ¾, 1} gives the regular partition of [0, 1]
with n = 4:
0 1½ ¾¼
1/3 2/3
Δx=1/3
Δx=1/4
A regular–partition of I is a division of I into equal
size subintervals. We will use n to denote the
number of the subintervals and Δx be the length
of each subinterval.
Riemann Sums
In general Δx = (b – a)/n, so if n ∞ then Δx 0.
21. Let y = f(x) be a continuous function over [a, b].
Riemann Sums
22. Let y = f(x) be a continuous function over [a, b].
a=x0 b=xn
y = f(x)
Riemann Sums
23. a=x0
Let y = f(x) be a continuous function over [a, b].
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular
partition of [a, b].
b=xn
y = f(x)
Riemann Sums
24. a=x0 x i–1 x ix1 x2
Let y = f(x) be a continuous function over [a, b].
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular
partition of [a, b].
b=xn
y = f(x)
x n–1
Riemann Sums
25. a=x0 x i–1 x ix1 x2
Let y = f(x) be a continuous function over [a, b].
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular
partition of [a, b]. By selecting an arbitrary point in
each subinterval [x i – 1, xi],
b=xn
y = f(x)
x n–1
xi
*
Riemann Sums
26. a=x0 x i–1 x ix1 x2
x1
*
Let y = f(x) be a continuous function over [a, b].
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular
partition of [a, b]. By selecting an arbitrary point in
each subinterval [x i – 1, xi],
b=xn
y = f(x)
x n–1
xi
*
Riemann Sums
27. a=x0 x i–1 x i
x2
*
x1 x2
x1
*
Let y = f(x) be a continuous function over [a, b].
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular
partition of [a, b]. By selecting an arbitrary point in
each subinterval [x i – 1, xi],
b=xn
y = f(x)
x n–1
xi
*
Riemann Sums
28. a=x0 x i–1 x i
x2
*
x1 x2
xi
*x1
*
Let y = f(x) be a continuous function over [a, b].
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular
partition of [a, b]. By selecting an arbitrary point in
each subinterval [x i – 1, xi],
b=xn
y = f(x)
x n–1
xi
*
Riemann Sums
29. a=x0 x i–1 x i
x2
*
x1 x2
xi
*
the sum f(x1)Δx+
x1
*
is called a Riemann sum.
* f(x2)Δx+ …* f(xn)Δx =*
i=1
n
∑ f(xi)Δx*
Let y = f(x) be a continuous function over [a, b].
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular
partition of [a, b]. By selecting an arbitrary point in
each subinterval [x i – 1, xi],
b=xn
y = f(x)
x n–1
xi
*
Riemann Sums
30. Riemann Sums
the sum f(x1)Δx+
is called a Riemann sum.
* f(x2)Δx+ …* f(xn)Δx =*
i=1
n
∑ f(xi)Δx
Let y = f(x) be a continuous function over [a, b].
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular
partition of [a, b]. By selecting an arbitrary point in
each subinterval [x i – 1, xi],
xi
*
If f(x) ≥ 0, this sum represents a rectangular
approximation of the area between f(x) and the
x–axis.
*
a=x0 x i–1 x i
x2
*
x1 x2
xi
*x1
*
b=xn
y = f(x)
x n–1
31. Riemann Sums
the sum f(x1)Δx+
is called a Riemann sum.
* f(x2)Δx+ …* f(xn)Δx =*
i=1
n
∑ f(xi)Δx
Let y = f(x) be a continuous function over [a, b].
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular
partition of [a, b]. By selecting an arbitrary point in
each subinterval [x i – 1, xi],
xi
*
If f(x) ≥ 0, this sum represents a rectangular
approximation of the area between f(x) and the
x–axis.
*
a=x0 b=xn
x i–1 x i
x2
*
x1 x2
xi
*
y = f(x)
x1
*
x n–1
Δx
f(x1)
32. Riemann Sums
the sum f(x1)Δx+
is called a Riemann sum.
* f(x2)Δx+ …* f(xn)Δx =*
i=1
n
∑ f(xi)Δx
Let y = f(x) be a continuous function over [a, b].
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular
partition of [a, b]. By selecting an arbitrary point in
each subinterval [x i – 1, xi],
xi
*
If f(x) ≥ 0, this sum represents a rectangular
approximation of the area between f(x) and the
x–axis.
*
a=x0 b=xn
x i–1 x i
x2
*
x1 x2
xi
*
y = f(x)
x1
*
x n–1
Δx
f(x1)
33. Riemann Sums
the sum f(x1)Δx+
is called a Riemann sum.
* f(x2)Δx+ …* f(xn)Δx =*
i=1
n
∑ f(xi)Δx
Let y = f(x) be a continuous function over [a, b].
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular
partition of [a, b]. By selecting an arbitrary point in
each subinterval [x i – 1, xi],
xi
*
If f(x) ≥ 0, this sum represents a rectangular
approximation of the area between f(x) and the
x–axis.
*
a=x0 b=xn
x i–1 x i
x2
*
x1 x2
xi
*
y = f(x)
x1
*
x n–1
Δx
f(x1)
34. Riemann Sums
the sum f(x1)Δx+
is called a Riemann sum.
* f(x2)Δx+ …* f(xn)Δx =*
i=1
n
∑ f(xi)Δx
Let y = f(x) be a continuous function over [a, b].
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular
partition of [a, b]. By selecting an arbitrary point in
each subinterval [x i – 1, xi],
xi
*
If f(x) ≥ 0, this sum represents a rectangular
approximation of the area between f(x) and the
x–axis.
*
a=x0 b=xn
x i–1 x i
x2
*
x1 x2
xi
*
y = f(x)
x1
*
x n–1
Δx
f(x1)
35.
36. Definite Integrals
Let f(x) be a continuous function over the closed
interval [a, b] and if F'(x) = f(x), we define
∫x=a
b
f(x) dx =
37. Definite Integrals
Let f(x) be a continuous function over the closed
interval [a, b] and if F'(x) = f(x), we define
∫x=a
b
f(x) dx = F(x) | = F(b) – F(a)
x=a
b
38. Definite Integrals
Let f(x) be a continuous function over the closed
interval [a, b] and if F'(x) = f(x), we define
∫x=a
b
f(x) dx = F(x) | = F(b) – F(a)
x=a
b
The upper limit
of “integration”
The lower limit
of “integration”
39. Definite Integrals
Let f(x) be a continuous function over the closed
interval [a, b] and if F'(x) = f(x), we define
∫x=a
b
f(x) dx = F(x) | = F(b) – F(a)
x=a
b
The upper limit
of “integration”
The lower limit
of “integration”
to be the “definite integral of f(x) from x = a to x = b”.
40. Definite Integrals
Let f(x) be a continuous function over the closed
interval [a, b] and if F'(x) = f(x), we define
∫x=a
b
f(x) dx = F(x) | = F(b) – F(a)
For example,(x2)' = 2x,
x=a
b
The upper limit
of “integration”
The lower limit
of “integration”
to be the “definite integral of f(x) from x = a to x = b”.
41. Definite Integrals
Let f(x) be a continuous function over the closed
interval [a, b] and if F'(x) = f(x), we define
∫x=a
b
f(x) dx = F(x) | = F(b) – F(a)
For example,(x2)' = 2x, so ∫
x=0
2x dx = x2 |
x=a
b
The upper limit
of “integration”
The lower limit
of “integration”
0
to be the “definite integral of f(x) from x = a to x = b”.
1 1
42. Definite Integrals
Let f(x) be a continuous function over the closed
interval [a, b] and if F'(x) = f(x), we define
∫x=a
b
f(x) dx = F(x) | = F(b) – F(a)
For example,(x2)' = 2x, so ∫
x=0
2x dx = x2 | = (1)2 – (0)2 = 1
x=a
b
The upper limit
of “integration”
The lower limit
of “integration”
0
to be the “definite integral of f(x) from x = a to x = b”.
1 1
43. Definite Integrals
Let f(x) be a continuous function over the closed
interval [a, b] and if F'(x) = f(x), we define
∫x=a
b
f(x) dx = F(x) | = F(b) – F(a)
For example,(x2)' = 2x, so ∫
x=0
2x dx = x2 | = (1)2 – (0)2 = 1
x=a
b
The upper limit
of “integration”
The lower limit
of “integration”
0
Note that we would obtain the same answer
regardless of what the integration constant k is, hence
we may set k = 0.
to be the “definite integral of f(x) from x = a to x = b”.
1 1
44. Definite Integrals
Let f(x) be a continuous function over the closed
interval [a, b] and if F'(x) = f(x), we define
∫x=a
b
f(x) dx = F(x) | = F(b) – F(a)
For example,(x2)' = 2x, so ∫
x=0
2x dx = x2 | = (1)2 – (0)2 = 1
x=a
b
The upper limit
of “integration”
The lower limit
of “integration”
0
Note that we would obtain the same answer
regardless of what the integration constant k is, hence
we may set k = 0. The definite integral of a function f(x)
gives the “area” bounded by f(x) and the x axis.
to be the “definite integral of f(x) from x = a to x = b”.
1 1
46. Let f(x) be a continuous function over [a, b].
Fundamental Theorem of Calculus (FTC) Part I
The Fundamental Theorem of Calculus
47. Let f(x) be a continuous function over [a, b]. The limit
of the Riemann sums of f(x) as n ∞ (Δx 0) is the
definite integral over [a, b]
Fundamental Theorem of Calculus (FTC) Part I
The Fundamental Theorem of Calculus
48. i=1
n
∑ f(xi)Δx*
Let f(x) be a continuous function over [a, b]. The limit
of the Riemann sums of f(x) as n ∞ (Δx 0) is the
definite integral over [a, b] i.e.
Fundamental Theorem of Calculus (FTC) Part I
The Fundamental Theorem of Calculus
49. lim
i=1
n
∑ f(xi)Δx
n∞
*
Let f(x) be a continuous function over [a, b]. The limit
of the Riemann sums of f(x) as n ∞ (Δx 0) is the
definite integral over [a, b] i.e.
(or Δx0)
Fundamental Theorem of Calculus (FTC) Part I
The Fundamental Theorem of Calculus
50. lim
i=1
n
∑ f(xi)Δx
n∞
*
= ∫x=a
b
Let f(x) be a continuous function over [a, b]. The limit
of the Riemann sums of f(x) as n ∞ (Δx 0) is the
definite integral over [a, b] i.e.
(or Δx0)
Fundamental Theorem of Calculus (FTC) Part I
The Fundamental Theorem of Calculus
51. lim
i=1
n
∑ f(xi)Δx
n∞
*
= ∫x=a
b
f(x)
Let f(x) be a continuous function over [a, b]. The limit
of the Riemann sums of f(x) as n ∞ (Δx 0) is the
definite integral over [a, b] i.e.
(or Δx0)
Fundamental Theorem of Calculus (FTC) Part I
The Fundamental Theorem of Calculus
52. Let f(x) be a continuous function over [a, b]. The limit
of the Riemann sums of f(x) as n ∞ (Δx 0) is the
definite integral over [a, b] i.e.
Fundamental Theorem of Calculus (FTC) Part I
lim
i=1
n
∑ f(xi)Δx
n∞
*
= ∫x=a
b
f(x) dx
(or Δx0)
The Fundamental Theorem of Calculus
53. Fundamental Theorem of Calculus (FTC) Part I
lim
i=1
n
∑ f(xi)Δx
n∞
*
= ∫x=a
b
f(x) dx
(The theorem also holds for Riemann sums of
partitions that are not regular as long as max Δx0.)
Let f(x) be a continuous function over [a, b]. The limit
of the Riemann sums of f(x) as n ∞ (Δx 0) is the
definite integral over [a, b] i.e.
(or Δx0)
The Fundamental Theorem of Calculus
54. Fundamental Theorem of Calculus (FTC) Part I
lim
i=1
n
∑ f(xi)Δx
n∞
*
= ∫x=a
b
f(x) dx
(The theorem also holds for Riemann sums of
partitions that are not regular as long as max Δx0.)
The limits of The Riemann sums have concrete
interpretations in many applied situations.
Let f(x) be a continuous function over [a, b]. The limit
of the Riemann sums of f(x) as n ∞ (Δx 0) is the
definite integral over [a, b] i.e.
(or Δx0)
The Fundamental Theorem of Calculus
55. Fundamental Theorem of Calculus (FTC) Part I
lim
i=1
n
∑ f(xi)Δx
n∞
*
= ∫x=a
b
f(x) dx
(The theorem also holds for Riemann sums of
partitions that are not regular as long as max Δx0.)
The limit of the Riemann sums has concrete
interpretations in many applied situations.
FTC passes the calculation of the limits to integrals.
Let f(x) be a continuous function over [a, b]. The limit
of the Riemann sums of f(x) as n ∞ (Δx 0) is the
definite integral over [a, b] i.e.
(or Δx0)
The Fundamental Theorem of Calculus
56. When f(x) > 0, the limit of the Riemann sums is the
area bounded between the curve y = f(x) and the
x–axis, from x = a to b.
The Fundamental Theorem of Calculus
57. When f(x) > 0, the limit of the Riemann sums is the
area bounded between the curve y = f(x) and the
x–axis, from x = a to b.
Example A. Find the area bounded by y = –x2 + 2x and
the x–axis.
The Fundamental Theorem of Calculus
58. When f(x) > 0, the limit of the Riemann sums is the
area bounded between the curve y = f(x) and the
x–axis, from x = a to b.
Example A. Find the area bounded by y = –x2 + 2x and
the x–axis.
Graph y = –x2 + 2x.
The Fundamental Theorem of Calculus
59. When f(x) > 0, the limit of the Riemann sums is the
area bounded between the curve y = f(x) and the
x–axis, from x = a to b.
Example A. Find the area bounded by y = –x2 + 2x and
the x–axis.
Graph y = –x2 + 2x.
Solve –x2 + 2x = 0, x = 0, 2
The Fundamental Theorem of Calculus
60. When f(x) > 0, the limit of the Riemann sums is the
area bounded between the curve y = f(x) and the
x–axis, from x = a to b.
Example A. Find the area bounded by y = –x2 + 2x and
the x–axis.
Graph y = –x2 + 2x.
Solve –x2 + 2x = 0, x = 0, 2
20
x
y
y = –x2 + 2x
The Fundamental Theorem of Calculus
61. When f(x) > 0, the limit of the Riemann sums is the
area bounded between the curve y = f(x) and the
x–axis, from x = a to b.
Example A. Find the area bounded by y = –x2 + 2x and
the x–axis.
Graph y = –x2 + 2x.
Solve –x2 + 2x = 0, x = 0, 2
so a = 0 and b = 2.
20
x
y
y = –x2 + 2x
The Fundamental Theorem of Calculus
62. When f(x) > 0, the limit of the Riemann sums is the
area bounded between the curve y = f(x) and the
x–axis, from x = a to b.
Example A. Find the area bounded by y = –x2 + 2x and
the x–axis.
20
Hence the bounded area is
∫
x=0
2
–x2 + 2x dx
x
y
y = –x2 + 2xGraph y = –x2 + 2x.
Solve –x2 + 2x = 0, x = 0, 2
so a = 0 and b = 2.
The Fundamental Theorem of Calculus
63. When f(x) > 0, the limit of the Riemann sums is the
area bounded between the curve y = f(x) and the
x–axis, from x = a to b.
Example A. Find the area bounded by y = –x2 + 2x and
the x–axis.
20
Hence the bounded area is
∫
x=0
2
–x2 + 2x dx
= –x3/3 + x2 |
x=0
2
x
y
y = –x2 + 2xGraph y = –x2 + 2x.
Solve –x2 + 2x = 0, x = 0, 2
so a = 0 and b = 2.
The Fundamental Theorem of Calculus
64. When f(x) > 0, the limit of the Riemann sums is the
area bounded between the curve y = f(x) and the
x–axis, from x = a to b.
Example A. Find the area bounded by y = –x2 + 2x and
the x–axis.
Graph y = –x2 + 2x.
Solve –x2 + 2x = 0, x = 0, 2
so a = 0 and b = 2.
20
Hence the bounded area is
∫
x=0
2
–x2 + 2x dx
= –x3/3 + x2 |
x=0
2
= (–8/3 + 4) – 0 =
4
3
x
y
y = –x2 + 2x
The Fundamental Theorem of Calculus
65. If f(x) < 0, the integral produces a negative answer.
The Fundamental Theorem of Calculus
66. If f(x) < 0, the integral produces a negative answer.
∫
x= –2
0
2x + x2 dx
Example B. Calculate
The Fundamental Theorem of Calculus
67. If f(x) < 0, the integral produces a negative answer.
2x + x2 dx∫
0
y
x
0–2
x= –2
Example B. Calculate y = 2x + x2
The Fundamental Theorem of Calculus
68. If f(x) < 0, the integral produces a negative answer.
y
x
0–2
2x + x2 dx
= x2 + x3/3 |
0
∫
0
x= –2
x= –2
Example B. Calculate y = 2x + x2
The Fundamental Theorem of Calculus
69. If f(x) < 0, the integral produces a negative answer.
y
x
0–2
2x + x2 dx
= x2 + x3/3 |
0
∫
0
x= –2
x= –2
= 0 – (4 – 8/3)
Example B. Calculate y = 2x + x2
The Fundamental Theorem of Calculus
70. If f(x) < 0, the integral produces a negative answer.
y
x
0–2
2x + x2 dx
= x2 + x3/3 |
0
∫
0
x= –2
x= –2
= 0 – (4 – 8/3) = –4/3
Example B. Calculate y = 2x + x2
The Fundamental Theorem of Calculus
71. If f(x) < 0, the integral produces a negative answer.
y
x
0–2
We call this the signed area of the region.
2x + x2 dx
= x2 + x3/3 |
0
∫
0
x= –2
x= –2
= 0 – (4 – 8/3) = –4/3
Example B. Calculate y = 2x + x2
The Fundamental Theorem of Calculus
72. If f(x) < 0, the integral produces a negative answer.
y
x
0–2
We call this the signed area of the region.
The negative sign indicates that part or all of the
region is below the x–axis.
2x + x2 dx
= x2 + x3/3 |
0
∫
0
x= –2
x= –2
= 0 – (4 – 8/3) = –4/3
Example B. Calculate y = 2x + x2
The Fundamental Theorem of Calculus
73. If f(x) < 0, the integral produces a negative answer.
y
x
0–2
We call this the signed area of the region.
The negative sign indicates that part or all of the
region is below the x–axis.
The definite integral gives the sum of the positive and
negative area, i.e. the signed area over the interval.
Example B. Calculate
2x + x2 dx
= x2 + x3/3 |
0
∫
0
x= –2
x= –2
= 0 – (4 – 8/3) = –4/3
y = 2x + x2
The Fundamental Theorem of Calculus
74. Example C. Calculate and interpret the answer.
∫
x=0
2
x – 1dxa.
The Fundamental Theorem of Calculus
75. ∫
x=0
2
x – 1dxa.
0 2
Example C. Calculate and interpret the answer.
The Fundamental Theorem of Calculus
76. ∫
x=0
2
x – 1dx = x2/2 – x|a.
0
2
0 2
Example C. Calculate and interpret the answer.
The Fundamental Theorem of Calculus
77. ∫
x=0
2
x – 1dx = x2/2 – x|
= 2 – 2 = 0
a.
0
2
0 2
Example C. Calculate and interpret the answer.
The Fundamental Theorem of Calculus
78. ∫
x=0
2
x – 1dx = x2/2 – x|
= 2 – 2 = 0
0 2
a.
signed area = –1/2
signed area = +1/2
0
2
Example C. Calculate and interpret the answer.
The Fundamental Theorem of Calculus
79. ∫
x=0
2
0 2
a.
signed area = –1/2
signed area = +1/2
x – 1dx = x2/2 – x|
= 2 – 2 = 0
∫
x=0
3/2
x – 1dxb.
0
2
0 3/2
Example C. Calculate and interpret the answer.
The Fundamental Theorem of Calculus
80. ∫
x=0
2
0 2
a.
signed area = –1/2
signed area = +1/2
x – 1dx = x2/2 – x|
= 2 – 2 = 0
∫
x=0
3/2
x – 1dx = x2/2 – x |
0
3/2
b.
0
2
0 3/2
Example C. Calculate and interpret the answer.
The Fundamental Theorem of Calculus
81. ∫
x=0
2
0 2
a.
0 3/2
signed area = –1/2
signed area = +1/2
x – 1dx = x2/2 – x|
= 2 – 2 = 0
∫
x=0
3/2
x – 1dx = x2/2 – x |
= 9/8 – 3/2
= –3/8
0
3/2
b.
0
2
Example C. Calculate and interpret the answer.
The Fundamental Theorem of Calculus
82. ∫
x=0
2
0 2
a.
signed area = –1/2
signed area = +1/2
x – 1dx = x2/2 – x|
= 2 – 2 = 0
∫
x=0
3/2
x – 1dx = x2/2 – x |
= 9/8 – 3/2
= –3/8
0
3/2
b.
0
2
0 3/2
signed area = –1/2
signed area = +1/8
Example C. Calculate and interpret the answer.
The Fundamental Theorem of Calculus
We use the term (absolute) area for the traditional
notion of area–measurement which is always positive.
83. The area in example a is 1, its signed area is 0.
∫
x=0
2
0 2
a.
signed area = –1/2
signed area = +1/2
x – 1dx = x2/2 – x|
= 2 – 2 = 0
∫
x=0
3/2
x – 1dx = x2/2 – x |
= 9/8 – 3/2
= –3/8
0
3/2
b.
0
2
0 3/2
signed area = –1/2
signed area = +1/8
Example C. Calculate and interpret the answer.
The Fundamental Theorem of Calculus
We use the term (absolute) area for the traditional
notion of area–measurement which is always positive.
84. We use the term (absolute) area for the traditional
notion of area–measurement which is always positive.
∫a
b
f(x) dx
The abs-area in example a is 1, its signed area is 0.
∫
x=0
2
0 2
a.
signed area = –1/2
signed area = +1/2
x – 1dx = x2/2 – x|
= 2 – 2 = 0
∫
x=0
3/2
x – 1dx = x2/2 – x |
= 9/8 – 3/2
= –3/8
0
3/2
b.
0
2
0 3/2
signed area = –1/2
signed area = +1/8
Example C. Calculate and interpret the answer.
= signed area.
The Fundamental Theorem of Calculus
∫a
b
l f(x) l dx = abs-area.
85. Example D.
a. Find the signed area bounded by y = –x2 + 2x and
the x–axis from x = 0 to x = 3.
The Fundamental Theorem of Calculus
86. Example D.
a. Find the signed area bounded by y = –x2 + 2x and
the x–axis from x = 0 to x = 3.
1
20 x
y
3
y = –x2 + 2x
The Fundamental Theorem of Calculus
87. Example D.
a. Find the signed area bounded by y = –x2 + 2x and
the x–axis from x = 0 to x = 3.
The signed area is
∫x=0
3
–x2 + 2x dx
1
20 x
y
3
y = –x2 + 2x
The Fundamental Theorem of Calculus
88. Example D.
a. Find the signed area bounded by y = –x2 + 2x and
the x–axis from x = 0 to x = 3.
The signed area is
∫x=0
3
–x2 + 2x dx
= –x3/3 + x2 |x=0
3
1
20 x
y
3
y = –x2 + 2x
The Fundamental Theorem of Calculus
89. Example D.
a. Find the signed area bounded by y = –x2 + 2x and
the x–axis from x = 0 to x = 3.
The signed area is
∫x=0
3
–x2 + 2x dx
= –x3/3 + x2 |x=0
3
= (–9 + 9) – 0 = 0
1
20 x
y
3
y = –x2 + 2x
The Fundamental Theorem of Calculus
90. Example D.
a. Find the signed area bounded by y = –x2 + 2x and
the x–axis from x = 0 to x = 3.
The signed area is
∫x=0
3
–x2 + 2x dx
= –x3/3 + x2 |x=0
3
= (–9 + 9) – 0 = 0
1
20 x
y
3
y = –x2 + 2x
b. Find the abs-area bounded by y = –x2 + 2x and the
x–axis from x = 0 to x = 3.
The Fundamental Theorem of Calculus
91. Example D.
a. Find the signed area bounded by y = –x2 + 2x and
the x–axis from x = 0 to x = 3.
The signed area is
∫x=0
3
–x2 + 2x dx
= –x3/3 + x2 |x=0
3
= (–9 + 9) – 0 = 0
1
20 x
y
3
y = –x2 + 2x
b. Find the abs-area bounded by y = –x2 + 2x and the
x–axis from x = 0 to x = 3.
To find the abs-area, we find the positive area and
the negative area separately then add their absolute
values.
The Fundamental Theorem of Calculus
92. The positive area spans from x = 0 to x = 2.
The Fundamental Theorem of Calculus
93. The positive area spans from x = 0 to x = 2. It’s
∫x=0
2
–x2 + 2x dx
The Fundamental Theorem of Calculus
94. The positive area spans from x = 0 to x = 2. It’s
∫x=0
2
–x2 + 2x dx
= –x3/3 + x2 |
x=0
2
The Fundamental Theorem of Calculus
95. The positive area spans from x = 0 to x = 2. It’s
∫x=0
2
–x2 + 2x dx
= –x3/3 + x2 |
x=0
2
= (–8/3 + 4) = 4
3
The Fundamental Theorem of Calculus
96. The positive area spans from x = 0 to x = 2. It’s
∫x=0
2
–x2 + 2x dx
= –x3/3 + x2 |
x=0
2
= (–8/3 + 4) = 4
3
3
The negative area spans from x = 2 to x = 3.
The Fundamental Theorem of Calculus
97. The positive area spans from x = 0 to x = 2. It’s
∫x=0
2
–x2 + 2x dx
= –x3/3 + x2 |
x=0
2
= (–8/3 + 4) = 4
3
∫x=2
3
–x2 + 2x dx
The negative area spans from x = 2 to x = 3. It’s
The Fundamental Theorem of Calculus
98. The positive area spans from x = 0 to x = 2. It’s
∫x=0
2
–x2 + 2x dx
= –x3/3 + x2 |
x=0
2
= (–8/3 + 4) = 4
3
∫x=2
3
–x2 + 2x dx
= –x3/3 + x2 |x=2
3
The negative area spans from x = 2 to x = 3. It’s
The Fundamental Theorem of Calculus
99. The positive area spans from x = 0 to x = 2. It’s
∫x=0
2
–x2 + 2x dx
= –x3/3 + x2 |
x=0
2
= (–8/3 + 4) = 4
3
∫x=2
3
–x2 + 2x dx
= –x3/3 + x2 |x=2
3
= (–9 + 9) – (–8/3 + 4)
The negative area spans from x = 2 to x = 3. It’s
The Fundamental Theorem of Calculus
100. The positive area spans from x = 0 to x = 2. It’s
∫x=0
2
–x2 + 2x dx
= –x3/3 + x2 |
x=0
2
= (–8/3 + 4) = 4
3
∫x=2
3
–x2 + 2x dx
= –x3/3 + x2 |x=2
3
= (–9 + 9) – (–8/3 + 4)
The negative area spans from x = 2 to x = 3. It’s
= – 4
3
The Fundamental Theorem of Calculus
101. The positive area spans from x = 0 to x = 2. It’s
∫x=0
2
–x2 + 2x dx
= –x3/3 + x2 |
x=0
2
= (–8/3 + 4) = 4
3
Hence the real area is
∫x=2
3
–x2 + 2x dx
= –x3/3 + x2 |x=2
3
= (–9 + 9) – (–8/3 + 4)
The negative area spans from x = 2 to x = 3. It’s
= – 4
3
4
3
+ | 8
3
– 4
3
| =
The Fundamental Theorem of Calculus