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.
Lesson 24: Areas, Distances, the Integral (Section 021 handout)
1. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010
Sections 5.1–5.2 Notes
Areas and Distances
The Definite Integral
V63.0121.021, Calculus I
New York University
December 2, 2010
Announcements
Final December 20, 12:00–1:50pm
Announcements
Notes
Final December 20,
12:00–1:50pm
cumulative
location TBD
old exams on common
website
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 2 / 56
Objectives from Section 5.1
Notes
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.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 3 / 56
1
2. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010
Objectives from Section 5.2
Notes
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.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 4 / 56
Outline
Notes
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
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 5 / 56
Easy Areas: Rectangle
Notes
Definition
The area of a rectangle with dimensions and w is the product A = w .
w
It may seem strange that this is a definition and not a theorem but we
have to start somewhere.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 6 / 56
2
3. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010
Easy Areas: Parallelogram
Notes
By cutting and pasting, a parallelogram can be made into a rectangle.
h
b b
So
Fact
The area of a parallelogram of base width b and height h is
A = bh
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 7 / 56
Easy Areas: Triangle
Notes
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
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 8 / 56
Easy Areas: Other Polygons
Notes
Any polygon can be triangulated, so its area can be found by summing the
areas of the triangles:
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 9 / 56
3
4. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010
Hard Areas: Curved Regions
Notes
???
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 10 / 56
Meet the mathematician: Archimedes
Notes
Greek (Syracuse), 287 BC –
212 BC (after Euclid)
Geometer
Weapons engineer
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 11 / 56
Archimedes and the Parabola
Notes
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
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 12 / 56
4
5. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010
Summing the series
Notes
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 + · · · + r n ) = 1 − r n+1
So
1 − r n+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
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 13 / 56
Cavalieri
Notes
Italian,
1598–1647
Revisited the
area problem
with a
different
perspective
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 14 / 56
Cavalieri’s method
Notes
Divide up the interval into pieces
y = x2 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 =?
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56
5
6. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010
What is Ln ?
1 Notes
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
2
1 i −1 (i − 1)2
· = .
n n n3
So
1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2
Ln = + + ··· + =
n3 n3 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 = →
6n3 3
as n → ∞.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 16 / 56
Cavalieri’s method for different functions
Notes
Try the same trick with f (x) = x 3 . 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
1 + 23 + 33 + · · · + (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 → ∞.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 17 / 56
Cavalieri’s method with different heights
Notes
1 13 1 23 1 n3
Rn = · 3 + · 3 + ··· + · 3
n n n n n n
13 + 23 + 33 + · · · + n3
=
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.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 18 / 56
6
7. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010
Outline
Notes
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
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 19 / 56
Cavalieri’s method in general
Notes
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 a
n
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
xn = a + n · =b
x n
x0 x1 . . . xi . . .xn−1xn
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 20 / 56
Forming Riemann sums
Notes
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
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 21 / 56
7
8. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010
Theorem of the Day
Notes
Theorem
If f is a continuous function on [a, b]
M15 = 7.49968
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
matter what choice of ci we make. x
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
Analogies
Notes
The Area Problem (Ch. 5)
The Tangent Problem
(Ch. 2–4) Want the area of a curved
region
Want the slope of a curve
Only know the area of
Only know the slope of lines
polygons
Approximate curve with a
Approximate region with
line
polygons
Take limit over better and
Take limit over better and
better approximations
better approximations
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 23 / 56
Outline
Notes
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
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 24 / 56
8
9. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010
Distances
Notes
Just like area = length × width, we have
distance = rate × time.
So here is another use for Riemann sums.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 25 / 56
Application: Dead Reckoning
Notes
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 26 / 56
Computing position by Dead Reckoning
Notes
Example
A sailing ship is cruising back and forth along a channel (in a straight
line). At noon the ship’s position and velocity are recorded, but shortly
thereafter a storm blows in and position is impossible to measure. The
velocity continues to be recorded at thirty-minute intervals.
Time 12:00 12:30 1:00 1:30 2:00
Speed (knots) 4 8 12 6 4
Direction E E E E W
Time 2:30 3:00 3:30 4:00
Speed 3 3 5 9
Direction W E E E
Estimate the ship’s position at 4:00pm.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 27 / 56
9
10. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010
Solution
Notes
Solution
We estimate that the speed of 4 knots (nautical miles per hour) is
maintained from 12:00 until 12:30. So over this time interval the ship
travels
4 nmi 1
hr = 2 nmi
hr 2
We can continue for each additional half hour and get
distance = 4 × 1/2 + 8 × 1/2 + 12 × 1/2
+ 6 × 1/2 − 4 × 1/2 − 3 × 1/2 + 3 × 1/2 + 5 × 1/2
= 15.5
So the ship is 15.5 nmi east of its original position.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 28 / 56
Analysis
Notes
This method of measuring position by recording velocity was necessary
until global-positioning satellite technology became widespread
If we had velocity estimates at finer intervals, we’d get better
estimates.
If we had velocity at every instant, a limit would tell us our exact
position relative to the last time we measured it.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 29 / 56
Other uses of Riemann sums
Notes
Anything with a product!
Area, volume
Anything with a density: Population, mass
Anything with a “speed:” distance, throughput, power
Consumer surplus
Expected value of a random variable
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 30 / 56
10
11. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010
Outline
Notes
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
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 31 / 56
The definite integral as a limit
Notes
Definition
If f is a function defined on [a, b], the definite integral of f from a to b
is the number
b n
f (x) dx = lim f (ci ) ∆x
a ∆x→0
i=1
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 32 / 56
Notation/Terminology
Notes
b n
f (x) dx = lim f (ci ) ∆x
a ∆x→0
i=1
— integral sign (swoopy S)
f (x) — integrand
a and b — limits of integration (a is the lower limit and b the
upper limit)
dx — ??? (a parenthesis? an infinitesimal? a variable?)
The process of computing an integral is called integration or
quadrature
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 33 / 56
11
12. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010
The limit can be simplified
Notes
Theorem
If f is continuous on [a, b] or if f has only finitely many jump
discontinuities, then f is integrable on [a, b]; that is, the definite integral
b
f (x) dx exists.
a
Theorem
If f is integrable on [a, b] then
b n
f (x) dx = lim f (xi )∆x,
a n→∞
i=1
where
b−a
∆x = and xi = a + i ∆x
n
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 34 / 56
Example: Integral of x
Notes
Example
3
Find x dx
0
Solution
3 3i
For any n we have ∆x = and xi = . So
n n
n n n
3i 3 9
Rn = f (xi ) ∆x = = i
n n n2
i=1 i=1 i=1
9 n(n + 1) 9
= · −→
n2 2 2
3
9
So x dx = = 4.5
0 2
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 35 / 56
Example: Integral of x 2
Notes
Example
3
Find x 2 dx
0
Solution
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 36 / 56
12
13. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010
Example: Integral of x 3
Notes
Example
3
Find x 3 dx
0
Solution
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 37 / 56
Outline
Notes
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
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 38 / 56
Estimating the Definite Integral
Notes
Example
1
4
Estimate dx using M4 .
0 1 + x2
Solution
1 1 3
We have x0 = 0, x1 = , x2 = , x3 = , x4 = 1.
4 2 4
1 3 5 7
So c1 = , c2 = , c3 = , c4 = .
8 8 8 8
1 4 4 4 4
M4 = + + +
4 1 + (1/8)2 1 + (3/8)2 1 + (5/8)2 1 + (7/8)2
1 4 4 4 4
= + + +
4 65/64 73/64 89/64 113/64
64 64 64 64
= + + + ≈ 3.1468
65 73 89 113
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 39 / 56
13
14. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010
Estimating the Definite Integral (Continued)
Notes
Example
1
4
Estimate dx using L4 and R4
0 1 + x2
Answer
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 40 / 56
Outline
Notes
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
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 41 / 56
Properties of the integral
Notes
Theorem (Additive Properties of the Integral)
Let f and g be integrable functions on [a, b] and c a constant. Then
b
1. c dx = c(b − a)
a
b b b
2. [f (x) + g (x)] dx = f (x) dx + g (x) dx.
a a a
b b
3. cf (x) dx = c f (x) dx.
a a
b b b
4. [f (x) − g (x)] dx = f (x) dx − g (x) dx.
a a a
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 42 / 56
14
15. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010
Proofs
Notes
Proofs.
When integrating a constant function c, each Riemann sum equals
c(b − a).
A Riemann sum for f + g equals a Riemann sum for f plus a
Riemann sum for g . Using the sum rule for limits, the integral of a
sum is the sum of the integrals.
Ditto for constant multiples
Ditto for differences
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 43 / 56
Example Notes
3
Find x 3 − 4.5x 2 + 5.5x + 1 dx
0
Solution
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 44 / 56
More Properties of the Integral
Notes
Conventions:
a b
f (x) dx = − f (x) dx
b a
a
f (x) dx = 0
a
This allows us to have
c b c
5. f (x) dx = f (x) dx + f (x) dx for all a, b, and c.
a a b
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 46 / 56
15
16. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010
Example Notes
Suppose f and g are functions with
4
f (x) dx = 4
0
5
f (x) dx = 7
0
5
g (x) dx = 3.
0
Find
5
(a) [2f (x) − g (x)] dx
0
5
(b) f (x) dx.
4
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 47 / 56
Solution Notes
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 48 / 56
Outline
Notes
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
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 49 / 56
16
17. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010
Comparison Properties of the Integral
Notes
Theorem
Let f and g be integrable functions on [a, b].
6. If f (x) ≥ 0 for all x in [a, b], then
b
f (x) dx ≥ 0
a
7. If f (x) ≥ g (x) for all x in [a, b], then
b b
f (x) dx ≥ g (x) dx
a a
8. If m ≤ f (x) ≤ M for all x in [a, b], then
b
m(b − a) ≤ f (x) dx ≤ M(b − a)
a
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 50 / 56
The integral of a nonnegative function is
nonnegative Notes
Proof.
If f (x) ≥ 0 for all x in [a, b], then
for any number of divisions n and
choice of sample points {ci }:
n n
Sn = f (ci ) ∆x ≥ 0·∆x = 0
i=1 ≥0 i=1
Since Sn ≥ 0 for all n, the limit
of {Sn } is nonnegative, too:
b
x
f (x) dx = lim Sn ≥ 0
a n→∞
≥0
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 51 / 56
The definite integral is “increasing”
Notes
Proof.
Let h(x) = f (x) − g (x). If f (x)
f (x) ≥ g (x) for all x in [a, b],
then h(x) ≥ 0 for all x in [a, b]. h(x) g (x)
So by the previous property
b
h(x) dx ≥ 0
a x
This means that
b b b b
f (x) dx − g (x) dx = (f (x) − g (x)) dx = h(x) dx ≥ 0
a a a a
b b
So f (x) dx ≥ g (x) dx.
a a
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 52 / 56
17
18. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010
Bounding the integral using bounds of the function
Notes
Proof.
If m ≤ f (x) ≤ M on for all x in
[a, b], then by the previous y
property
M
b b b
m dx ≤ f (x) dx ≤ M dx
a a a
f (x)
By Property 8, the integral of a
constant function is the product
of the constant and the width of m
the interval. So:
b
x
m(b−a) ≤ f (x) dx ≤ M(b−a) a
a b
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 53 / 56
Notes
Example
2
1
Estimate dx using the comparison properties.
1 x
Solution
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 54 / 56
Summary
Notes
We can compute the area of a curved region with a limit of Riemann
sums
We can compute the distance traveled from the velocity with a limit
of Riemann sums
Many other important uses of this process.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 55 / 56
18
19. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010
Summary
Notes
The definite integral is a limit of Riemann Sums
The definite integral can be estimated with Riemann Sums
The definite integral can be distributed across sums and constant
multiples of functions
The definite integral can be bounded using bounds for the function
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 56 / 56
Notes
Notes
19