We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.
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Lesson 24: Areas, Distances, the Integral (Section 041 slides)
1. .
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
. . . . . .
2. 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
3. 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.
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 3 / 56
4. 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.
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 4 / 56
5. 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
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 5 / 56
6. Easy Areas: Rectangle
Definition
The area of a rectangle with dimensions ℓ and w is the product A = ℓw.
w
.
ℓ
It may seem strange that this is a definition and not a theorem but we
have to start somewhere.
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 6 / 56
7. Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
.
b
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 7 / 56
8. Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
h
.
b
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 7 / 56
9. Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
h
.
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 7 / 56
10. Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
h
.
b
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 7 / 56
11. Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
h
.
b
So
Fact
The area of a parallelogram of base width b and height h is
A = bh
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 7 / 56
12. Easy Areas: Triangle
By copying and pasting, a triangle can be made into a parallelogram.
.
b
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 8 / 56
13. Easy Areas: Triangle
By copying and pasting, a triangle can be made into a parallelogram.
h
.
b
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 8 / 56
14. Easy Areas: Triangle
By copying and pasting, a triangle can be made into a parallelogram.
h
.
b
So
Fact
The area of a triangle of base width b and height h is
1
A= bh
2
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 8 / 56
15. Easy Areas: Other Polygons
Any polygon can be triangulated, so its area can be found by summing
the areas of the triangles:
.
.
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 9 / 56
16. Hard Areas: Curved Regions
.
???
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 10 / 56
17. Meet the mathematician: Archimedes
Greek (Syracuse), 287 BC
– 212 BC (after Euclid)
Geometer
Weapons engineer
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 11 / 56
18. Meet the mathematician: Archimedes
Greek (Syracuse), 287 BC
– 212 BC (after Euclid)
Geometer
Weapons engineer
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 11 / 56
19. Meet the mathematician: Archimedes
Greek (Syracuse), 287 BC
– 212 BC (after Euclid)
Geometer
Weapons engineer
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 11 / 56
20. Archimedes and the Parabola
.
Archimedes found areas of a sequence of triangles inscribed in a
parabola.
A=
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 12 / 56
21. Archimedes and the Parabola
1
.
Archimedes found areas of a sequence of triangles inscribed in a
parabola.
A=1
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 12 / 56
22. Archimedes and the Parabola
1
1 1
8 8
.
Archimedes found areas of a sequence of triangles inscribed in a
parabola.
1
A=1+2·
8
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 12 / 56
23. Archimedes and the Parabola
1 1
64 64
1
1 1
8 8
1 1
64 64
.
Archimedes found areas of a sequence of triangles inscribed in a
parabola.
1 1
A=1+2· +4· + ···
8 64
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 12 / 56
24. Archimedes and the Parabola
1 1
64 64
1
1 1
8 8
1 1
64 64
.
Archimedes found areas of a sequence of triangles inscribed in a
parabola.
1 1
A=1+2· +4· + ···
8 64
1 1 1
=1+ + + ··· + n + ···
4 16 4
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 12 / 56
25. Summing the series
We would then need to know the value of the series
1 1 1
1+ + + ··· + n + ···
4 16 4
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 13 / 56
26. Summing the series
We would then need to know the value of the series
1 1 1
1+ + + ··· + n + ···
4 16 4
Fact
For any number r and any positive integer n,
(1 − r)(1 + r + · · · + rn ) = 1 − rn+1
So
1 − rn+1
1 + r + · · · + rn =
1−r
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 13 / 56
27. Summing the series
We would then need to know the value of the series
1 1 1
1+ + + ··· + n + ···
4 16 4
Fact
For any number r and any positive integer n,
(1 − r)(1 + r + · · · + rn ) = 1 − rn+1
So
1 − rn+1
1 + r + · · · + rn =
1−r
Therefore
1 1 1 1 − (1/4)n+1
1+ + + ··· + n =
4 16 4 1 − 1/4
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 13 / 56
28. Summing the series
We would then need to know the value of the series
1 1 1
1+ + + ··· + n + ···
4 16 4
Fact
For any number r and any positive integer n,
(1 − r)(1 + r + · · · + rn ) = 1 − rn+1
So
1 − rn+1
1 + r + · · · + rn =
1−r
Therefore
1 1 1 1 − (1/4)n+1 1 4
1+ + + ··· + n = → 3 = as n → ∞.
4 16 4 1− 1/4 /4 3
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 13 / 56
29. Cavalieri
Italian,
1598–1647
Revisited the
area
problem with
a different
perspective
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 14 / 56
30. Cavalieri's method
Divide up the interval into
2
y=x pieces and measure the area of
the inscribed rectangles:
.
0 1
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 15 / 56
31. Cavalieri's method
Divide up the interval into
2
y=x pieces and measure the area of
the inscribed rectangles:
1
L2 =
8
.
0 1 1
2
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 15 / 56
32. Cavalieri's method
Divide up the interval into
2
y=x pieces and measure the area of
the inscribed rectangles:
1
L2 =
8
L3 =
.
0 1 2 1
3 3
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 15 / 56
33. Cavalieri's method
Divide up the interval into
2
y=x pieces and measure the area of
the inscribed rectangles:
1
L2 =
8
1 4 5
L3 = + =
27 27 27
.
0 1 2 1
3 3
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 15 / 56
34. Cavalieri's method
Divide up the interval into
2
y=x pieces and measure the area of
the inscribed rectangles:
1
L2 =
8
1 4 5
L3 = + =
27 27 27
L4 =
.
0 1 2 3 1
4 4 4
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 15 / 56
35. Cavalieri's method
Divide up the interval into
2
y=x pieces and measure the area of
the inscribed rectangles:
1
L2 =
8
1 4 5
L3 = + =
27 27 27
1 4 9 14
L4 = + + =
. 64 64 64 64
0 1 2 3 1
4 4 4
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 15 / 56
36. Cavalieri's method
Divide up the interval into
2
y=x pieces and measure the area of
the inscribed rectangles:
1
L2 =
8
1 4 5
L3 = + =
27 27 27
1 4 9 14
L4 = + + =
. 64 64 64 64
0 1 2 3 4 1 L5 =
5 5 5 5
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 15 / 56
37. Cavalieri's method
Divide up the interval into
2
y=x pieces and measure the area of
the inscribed rectangles:
1
L2 =
8
1 4 5
L3 = + =
27 27 27
1 4 9 14
L4 = + + =
. 64 64 64 64
1 4 9 16 30
0 1 2 3 4 1 L5 = + + + =
125 125 125 125 125
5 5 5 5
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 15 / 56
38. Cavalieri's method
Divide up the interval into
2
y=x pieces and measure the area of
the inscribed rectangles:
1
L2 =
8
1 4 5
L3 = + =
27 27 27
1 4 9 14
L4 = + + =
. 64 64 64 64
1 4 9 16 30
0 1 L5 = + + + =
125 125 125 125 125
Ln =?
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 15 / 56
39. What is Ln ?
1
Divide the interval [0, 1] into n pieces. Then each has width .
n
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 16 / 56
40. What is Ln ?
1
Divide the interval [0, 1] into n pieces. Then each has width . The
n
rectangle over the ith interval and under the parabola has area
( )
1 i − 1 2 (i − 1)2
· = .
n n n3
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 16 / 56
41. What is Ln ?
1
Divide the interval [0, 1] into n pieces. Then each has width . The
n
rectangle over the ith interval and under the parabola has area
( )
1 i − 1 2 (i − 1)2
· = .
n n n3
So
1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2
Ln = + 3 + ··· + =
n3 n n3 n3
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 16 / 56
42. What is Ln ?
1
Divide the interval [0, 1] into n pieces. Then each has width . The
n
rectangle over the ith interval and under the parabola has area
( )
1 i − 1 2 (i − 1)2
· = .
n n n3
So
1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2
Ln = + 3 + ··· + =
n3 n n3 n3
The Arabs knew that
n(n − 1)(2n − 1)
1 + 22 + 32 + · · · + (n − 1)2 =
6
So
n(n − 1)(2n − 1)
Ln =
6n3
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 16 / 56
43. What is Ln ?
1
Divide the interval [0, 1] into n pieces. Then each has width . The
n
rectangle over the ith interval and under the parabola has area
( )
1 i − 1 2 (i − 1)2
· = .
n n n3
So
1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2
Ln = + 3 + ··· + =
n3 n n3 n3
The Arabs knew that
n(n − 1)(2n − 1)
1 + 22 + 32 + · · · + (n − 1)2 =
6
So
n(n − 1)(2n − 1) 1
Ln = 3
→
6n 3
as n → ∞. . . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 16 / 56
44. Cavalieri's method for different functions
Try the same trick with f(x) = x3 . We have
( ) ( ) ( )
1 1 1 2 1 n−1
Ln = · f + ·f + ··· + · f
n n n n n n
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 17 / 56
45. Cavalieri's method for different functions
Try the same trick with f(x) = x3 . We have
( ) ( ) ( )
1 1 1 2 1 n−1
Ln = · f + ·f + ··· + · f
n n n n n n
1 1 1 23 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 17 / 56
46. Cavalieri's method for different functions
Try the same trick with f(x) = x3 . We have
( ) ( ) ( )
1 1 1 2 1 n−1
Ln = · f + ·f + ··· + · f
n n n n n n
1 1 1 23 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
3 3
1 + 2 + 3 + · · · + (n − 1)3
=
n4
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 17 / 56
47. Cavalieri's method for different functions
Try the same trick with f(x) = x3 . We have
( ) ( ) ( )
1 1 1 2 1 n−1
Ln = · f + ·f + ··· + · f
n n n n n n
1 1 1 23 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
3 3
1 + 2 + 3 + · · · + (n − 1)3
=
n4
The formula out of the hat is
[ ]2
1 + 23 + 33 + · · · + (n − 1)3 = 1
2 n(n − 1)
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 17 / 56
48. Cavalieri's method for different functions
Try the same trick with f(x) = x3 . We have
( ) ( ) ( )
1 1 1 2 1 n−1
Ln = · f + ·f + ··· + · f
n n n n n n
1 1 1 23 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
3 3
1 + 2 + 3 + · · · + (n − 1)3
=
n4
The formula out of the hat is
[ ]2
1 + 23 + 33 + · · · + (n − 1)3 = 1
2 n(n − 1)
So
n2 (n − 1)2 1
Ln = →
4n4 4
as n → ∞. . . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 17 / 56
49. Cavalieri's method with different heights
1 13 1 23 1 n3
Rn = · 3 + · 3 + ··· + · 3
n n n n n n
3 3 3
1 + 2 + 3 + ··· + n 3
=
n4
1 [1 ]2
= 4 2 n(n + 1)
n
n2 (n + 1)2 1
= →
4n4 4
.
as n → ∞.
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 18 / 56
50. Cavalieri's method with different heights
1 13 1 23 1 n3
Rn = · 3 + · 3 + ··· + · 3
n n n n n n
3 3 3
1 + 2 + 3 + ··· + n 3
=
n4
1 [1 ]2
= 4 2 n(n + 1)
n
n2 (n + 1)2 1
= →
4n4 4
.
as n → ∞.
So even though the rectangles overlap, we still get the same answer.
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 18 / 56
51. 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
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 19 / 56
52. 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
b−a
∆x = . For each i between 1 and n, let xi be the ith step between
n
a and b. So
x0 = a
b−a
x1 = x0 + ∆x = a +
n
b−a
x2 = x1 + ∆x = a + 2 · ...
n
b−a
xi = a + i · ...
n
b−a
. x xn = a + n · =b
x0 x1 . . . xi . . .xn−1xn n
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 20 / 56
53. Forming Riemann sums
We have many choices of representative points to approximate the
area in each subinterval.
left endpoints…
∑
n
Ln = f(xi−1 )∆x
i=1
. x
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 21 / 56
54. Forming Riemann sums
We have many choices of representative points to approximate the
area in each subinterval.
right endpoints…
∑
n
Rn = f(xi )∆x
i=1
. x
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 21 / 56
55. Forming Riemann sums
We have many choices of representative points to approximate the
area in each subinterval.
midpoints…
∑ ( xi−1 + xi )
n
Mn = f ∆x
2
i=1
. x
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 21 / 56
56. Forming Riemann sums
We have many choices of representative points to approximate the
area in each subinterval.
the minimum value on the
interval…
. x
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 21 / 56
57. Forming Riemann sums
We have many choices of representative points to approximate the
area in each subinterval.
the maximum value on the
interval…
. x
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 21 / 56
58. Forming Riemann sums
We have many choices of representative points to approximate the
area in each subinterval.
…even random points!
. x
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 21 / 56
59. Forming Riemann sums
We have many choices of representative points to approximate the
area in each subinterval.
…even random points!
. x
In general, choose ci to be a point in the ith interval [xi−1 , xi ]. Form the
Riemann sum
∑ n
Sn = f(c1 )∆x + f(c2 )∆x + · · · + f(cn )∆x = f(ci )∆x
i=1
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 21 / 56
60. Theorem of the Day
Theorem
If f is a continuous function on [a, b]
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
61. Theorem of the Day
Theorem
If f is a continuous function on [a, b]
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
62. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L1 = 3.0
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
63. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L2 = 5.25
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
64. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L3 = 6.0
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
65. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L4 = 6.375
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
66. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L5 = 6.59988
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
67. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L6 = 6.75
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
68. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L7 = 6.85692
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
69. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L8 = 6.9375
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
70. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L9 = 6.99985
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
71. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L10 = 7.04958
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
72. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L11 = 7.09064
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
73. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L12 = 7.125
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
74. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L13 = 7.15332
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
75. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L14 = 7.17819
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
76. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L15 = 7.19977
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
77. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L16 = 7.21875
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
78. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L17 = 7.23508
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
79. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L18 = 7.24927
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
80. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L19 = 7.26228
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
81. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L20 = 7.27443
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
82. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L21 = 7.28532
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
83. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L22 = 7.29448
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
84. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L23 = 7.30406
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
85. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L24 = 7.3125
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
86. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L25 = 7.31944
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
87. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L26 = 7.32559
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
88. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L27 = 7.33199
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
89. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L28 = 7.33798
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
90. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L29 = 7.34372
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
91. Theorem of the Day
Theorem
If f is a continuous function on [a, b] L30 = 7.34882
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
left endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
92. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R1 = 12.0
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
93. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R2 = 9.75
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
94. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R3 = 9.0
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
95. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R4 = 8.625
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
96. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R5 = 8.39969
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
97. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R6 = 8.25
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
98. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R7 = 8.14236
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
99. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R8 = 8.0625
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
100. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R9 = 7.99974
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
101. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R10 = 7.94933
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
102. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R11 = 7.90868
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
103. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R12 = 7.875
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
104. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R13 = 7.84541
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
105. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R14 = 7.8209
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
106. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R15 = 7.7997
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
107. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R16 = 7.78125
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
108. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R17 = 7.76443
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
109. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R18 = 7.74907
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
110. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R19 = 7.73572
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
111. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R20 = 7.7243
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
112. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R21 = 7.7138
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
113. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R22 = 7.70335
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
114. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R23 = 7.69531
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
115. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R24 = 7.6875
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
116. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R25 = 7.67934
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
117. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R26 = 7.6715
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
118. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R27 = 7.66508
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
119. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R28 = 7.6592
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
120. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R29 = 7.65388
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
121. Theorem of the Day
Theorem
If f is a continuous function on [a, b] R30 = 7.64864
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
right endpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56
122. Theorem of the Day
Theorem
If f is a continuous function on [a, b] M1 = 7.5
or has finitely many jump
discontinuities, then
{ n }
∑
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no
. x
matter what choice of ci we make.
midpoints
. . . . . .
V63.0121.041, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 1, 2010 22 / 56