Areas, Riemann Sums, and the Fundamental Theorem of Calaculus

1. Sums and Approximations

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.

8. Sums and Approximations
Inscribed Hexagon
1

9. Sums and Approximations
The hexagon consists of
six equilateral triangles
with sides 1
Inscribed Hexagon
1

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)

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

45. The Fundamental Theorem of Calculus
Fundamental Theorem of Calculus (FTC) Part I

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 Δx0)
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 Δx0)
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 Δx0)
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 Δx0)
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 Δx0.)
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 Δx0)
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 Δx0.)
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 Δx0)
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 Δx0.)
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 Δx0)
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