**Unformatted text preview: **© 2011–2018 by Dale Hoffman, Bellevue College.
Additional content by Jeff Eldridge, Edmonds Community College.
Author web page:
Report typographical errors to: [email protected] A free, color PDF version is available online at: This text is licensed under a Creative Commons Attribution — Share Alike 3.0
United States License. You are free:
• to Share — to copy, distribute, display and perform the work
• to Remix — to make derivative works
under the following conditions:
• Attribution: You must attribute the work in the manner specified by the author
(but not in any way that suggests that they endorse you or your work)
• Share Alike: If you alter, transform or build upon this work, you must distribute
the resulting work only under the same, similar or compatible license. Ninth preview edition of this version printed: September 10, 2018; page numbers not finalized
ISBN-13: 978-1511850452
ISBN-10: 1511850450 DALE HOFFMAN C O N T E M P O R A RY
CALCULUS Contents 4 5 6 7 The Integral
4.0 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1 Sigma Notation and Riemann Sums . . . . . . . . . . .
4.2 The Definite Integral . . . . . . . . . . . . . . . . . . . .
4.3 Properties of the Definite Integral . . . . . . . . . . . .
4.4 Areas, Integrals and Antiderivatives . . . . . . . . . . .
4.5 The Fundamental Theorem of Calculus . . . . . . . . .
4.6 Finding Antiderivatives . . . . . . . . . . . . . . . . . .
4.7 First Applications of Definite Integrals . . . . . . . . . .
4.8 Using Tables (and Technology) to Find Antiderivatives
4.9 Approximating Definite Integrals . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
. 293
293
302
314
323
333
340
350
359
369
375 .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. 389
389
400
409
420
435
443
458
468 Differential Equations
6.1 Introduction to Differential Equations
6.2 The Differential Equation y0 = f ( x ) .
6.3 Separable Equations . . . . . . . . . .
6.4 Exponential Growth and Decay . . . .
6.5 Heating, Cooling and Mixing . . . . . .
.
.
.
. .
.
.
.
. .
.
.
.
. .
.
.
.
. .
.
.
.
. .
.
.
.
. .
.
.
.
. .
.
.
.
. .
.
.
.
. .
.
.
.
. .
.
.
.
. 479
479
484
490
496
506 Transcendental Functions
7.1 One-to-One Functions . . . . . . . . . . . . . .
7.2 Inverse Functions . . . . . . . . . . . . . . . . .
7.3 Inverse Trigonometric Functions . . . . . . . .
7.4 Derivatives of Inverse Trigonometric Functions .
.
.
. .
.
.
. .
.
.
. .
.
.
. .
.
.
. .
.
.
. 513
514
519
528
539 Applications of Definite Integrals
5.1 Volumes by Slicing . . . . . . .
5.2 Volumes: Disks and Washers .
5.3 Arclength and Surface Area . .
5.4 More Work . . . . . . . . . . . .
5.5 Volumes: Tubes . . . . . . . . .
5.6 Moments and Centers of Mass
5.7 Improper Integrals . . . . . . .
5.8 Additional Applications . . . . .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. 7.5
7.6
7.7
7.8
8 Integrals Involving Inverse Trig Functions .
Calculus Done Right . . . . . . . . . . . . .
Hyperbolic Functions . . . . . . . . . . . . .
Inverse Hyperbolic Functions . . . . . . . . Integration Techniques
8.1 Finding Antiderivatives: A Review .
8.2 Integration by Parts . . . . . . . . . .
8.3 Partial Fraction Decomposition . . .
8.4 Trigonometric Substitution . . . . . .
8.5 Integrals of Trigonometric Functions
8.6 Integration Tactics . . . . . . . . . . .
8.7 MacLaurin Polynomials . . . . . . . A Answers .
.
.
.
.
.
. .
.
.
.
.
.
. .
.
.
.
.
.
. .
.
.
.
.
.
. .
.
.
. .
.
.
.
.
.
. .
.
.
. .
.
.
.
.
.
. .
.
.
. .
.
.
.
.
.
. .
.
.
. .
.
.
.
.
.
. .
.
.
. .
.
.
.
.
.
. .
.
.
. .
.
.
.
.
.
. .
.
.
. .
.
.
.
.
.
. .
.
.
. 545
550
552
554 .
.
.
.
.
.
. 557
557
564
575
587
595
603
607
A1 D Derivative Facts A75 H How to Succeed in Calculus A77 I A81 Integral Table T Trigonometry Facts A85 4
The Integral
Previous chapters dealt with differential calculus. They started with
the “simple” geometrical idea of the slope of a tangent line to a curve,
developed it into a combination of theory about derivatives and their
properties, examined techniques for calculating derivatives, and applied
these concepts and techniques to real-life situations. This chapter
begins the development of integral calculus and starts with the “simple”
geometric idea of area — an idea that will spawn its own combination
of theory, techniques and applications.
One of the most important results in mathematics, the Fundamental
Theorem of Calculus, appears in this chapter. It unifies differential and
integral calculus into a single grand structure. Historically, this unification marked the beginning of modern mathematics, and it provided
important tools for the growth and development of the sciences.
The chapter begins with a look at area, some geometric properties of
areas, and some applications. First we will examine ways of approximating the areas of regions such as tree leaves bounded by curved
edges and the areas of regions bounded by graphs of functions. Then
we will develop ways to calculate the areas of some of these regions
exactly. Finally, we will explore the rich variety of uses of “areas.” 4.0 Area The primary purpose of this introductory section is to help develop
your intuition about areas and your ability to reason using geometric
arguments about area. This type of reasoning will appear often in the
rest of this book and is very helpful for applying the ideas of calculus.
The basic shape we will use is the rectangle: the area of a rectangle is
(base) · (height). If the units for each side of the rectangle are “meters,”
then the area will have units (meters) · (meters) = “square meters”
= m2 . The only other area formulas needed for this section are for
triangles (area = 12 b · h) and for circles (area = π · r2 ). In addition, we
will use (and assume to be true) three other familiar properties of area: 294 the integral • Addition Property: The total area of a region is the sum of the areas
of the non-overlapping pieces that comprise the region: • Inclusion Property: If region B is inside region A (see margin), then
the area of region B is less than or equal to the area of region A.
• Location-Independence Property: The area of a region does not
depend on its location: Example 1. Determine the area of the region shown below left. Solution. We can easily break the region into two rectangles (shown
above right), with areas of 35 square inches and 3 square inches respectively, so the area of the original region is 38 square inches.
J
Practice 1. Determine the area of the trapezoidal region shown in the
margin by cutting it in two ways: (a) into a rectangle and triangle and
(b) into two triangles.
We can use our three area properties to deduce information about
areas that are difficult to calculate exactly. Let A be the region bounded
1
by the graph of f ( x ) = , the x-axis, and the vertical lines x = 1 and
x
x = 3. Because the two rectangles in the margin figure sit inside region
5
1 1
A and do not overlap each other, the area of the rectangles, + = ,
2 3
6
is less than the area of region A.
Practice 2. Build two rectangles, each with base 1 unit, with boundaries
that extend outside the shaded region in the margin figure and use
their areas to make another valid statement about the area of region A. 4.0 area Practice 3. What can you say about the area of region A if we use
“inside” and “outside” rectangles each with base 21 unit?
Example 2. The figure below right includes 32 dark squares, each 1
centimeter on a side, and 31 lighter squares of the same size. We can be
sure that the area of the leaf below left is smaller than what number? Solution. The area of the leaf is smaller than 32 + 31 = 63 cm2 . J Practice 4. We can be sure that the area of the leaf is at least how large?
Functions can be defined in terms of areas. For the constant function
f (t) = 2, define A( x ) to be the area of the rectangular region (top
margin figure) bounded by the graph of f , the t-axis, and the vertical
lines at t = 1 and t = x; we can easily see that A(2) = 2 (shaded region
in the second margin figure). Similarly, A(3) = 4 and A(4) = 6. In
general, A( x ) = (base)(height) = ( x − 1)(2) = 2x − 2 for any x ≥ 1.
From the graph of y = A( x ) (in the third margin figure) we can see
that A0 ( x ) = 2 for every value of x > 1.
(The fact that A0 ( x ) = f ( x ) in the preceding discussion is not a
coincidence, as we shall soon learn.)
Practice 5. For f (t) = 2, define B( x ) to be the area of the region
bounded by the graph of f , the t-axis, and vertical lines at t = 0 and
t = x (see below left). Fill in the table below with the requested values
of B. How are the graphs of y = A( x ) and y = B( x ) related?
x B( x ) 0
0.5
1
2 Sometimes it is useful to move regions around. The area of a parallelogram is obvious if we move the triangular region from one side of
the parallelogram to fill the region on the other side, resulting in with
a rectangle (see margin). 295 296 the integral At first glance, it is difficult to estimate the total area of the shaded
regions shown below left: but if we slide all of them into a single column (above right), then
becomes easy to determine that the shaded area is less than the area of
the enclosing rectangle = (base)(height) = (1)(2) = 2.
Practice 6. The total area of the shaded regions in the margin figure is
less than what number? Some Applications of “Area”
One reason “areas” are so useful is that they can represent quantities
other than sizes of simple geometric shapes. For example, if the units
of the base of a rectangle are “hours” and the units of the height are
“ miles ,” then the units of the “area” of the rectangle are:
hour
miles
(hours) ·
= miles
hour
a measure of distance. Similarly, if the base units are “pounds” and
the height units are “feet,” then the “area” units are “foot-pounds,” a
measure of work.
In the bottom margin figure, f (t) is the velocity of a car in “miles
per hour,” and t is the time in “hours.”
So
the shaded “area” will
be (base) · (height) = (3 hours) · 20 miles = 60 miles, the distance
hour
traveled by the car in the 3 hours from 1:00 p.m. until 4:00 p.m.
Distance as an “Area”
If f (t) is the (positive) forward velocity of an object at time t, then
the “area” between the graph of f and the t-axis and the vertical
lines at times t = a and t = b will equal the distance that the object
has moved forward between times a and b.
This “area as distance” concept can make some difficult distance
problems much easier. 4.0 area Example 3. A car starts from rest (velocity = 0) and steadily speeds
up so that 20 seconds later its speed is 88 feet per second (60 miles per
hour). How far did the car travel during those 20 seconds?
Solution. We could answer the question using the techniques of Chapter 3 (try this). But if “steadily” means that the velocity increases
linearly, then it is easier to use the margin figure and the concept of
“area as distance.”
The “area” of the triangular region represents the distance traveled:
1
1
ft
distance = (base)(height) = (20 sec) 88
= 880 ft
2
2
sec
The car travels a total of 880 feet during those 20 seconds. J Practice 7. A train initially traveling at 45 miles per hour (66 feet per
second) takes 60 seconds to decelerate to a complete stop. If the train
slowed down at a steady rate (the velocity decreased linearly), how
many feet did the train travel before coming to a stop?
Practice 8. You and a friend start off at noon and walk in the same
direction along the same path at the rates shown in the margin figure.
(a) Who is walking faster at 2:00 p.m.? Who is ahead at 2:00 p.m.?
(b) Who is walking faster at 3:00 p.m.? Who is ahead at 3:00 p.m.?
(c) When will you and your friend be together? (Answer in words.)
In the preceding Example and Practice problems, a function represented a rate of travel (in miles per hour, for instance) and the area
represented the total distance traveled. For functions representing other
rates, such as the production of a factory (bicycles per day) or the flow
of water in a river (gallons per minute) or traffic over a bridge (cars per
minute) or the spread of a disease (newly sick people per week), the
area will still represent the total amount of something.
“Area” as a Total Accumulation
If f (t) represents a positive rate (in units per time interval) at time
t, then the “area” between the graph of f and the t-axis and the
vertical lines at times t = a and t = b will be the total amount of
{something} that accumulates between times a and b (see margin).
For example, the figure at the top of the next page shows the flow
rate (in cubic feet per second) of water in the Skykomish River near
the town of Gold Bar, Washington. The area of the shaded region
represents the total volume (cubic feet) of water flowing past the town
during the month of October: 297 298 the integral total water = “area” = area of rectangle + area of triangle
!
!
ft3
1
ft3
≈ 2000
(30 days) +
1500
(30 days)
sec
2
sec
!
!
ft3
ft3
(30 days) = 2750
(2592000 sec)
= 2750
sec
sec ≈ 7.128 × 109 ft3
ft3 .
For comparison, the flow over Niagara Falls is about 2.12 × 105 sec 4.0 Problems
1. (a) Calculate the area of the shaded region: 2. Calculate the area of the trapezoidal region in the
figure below left by breaking it into a triangle and
a rectangle. (b) Calculate the area of the shaded region: 3. Break the region shown above right into a triangle and rectangle and
verifythat the total area of
h+H
the trapezoid is b ·
.
2 4.0 area 4. (a) Calculate the sum of the rectangular areas in
the region shown below left. 299 axis. Use two well-placed trapezoids to estimate
the area of this region.
9. Let A( x ) represent the area bounded by the graph
of the function shown below, the horizontal axis,
and vertical lines at t = 0 and t = x. Evaluate
A( x ) for x = 1, 2, 3, 4 and 5. (b) What can you say about the area of the shaded
region shown above right?
5. (a) Calculate the sum of the areas of the rectangles shown below left.
10. Let B( x ) represent the area bounded by the graph
of the function shown below, the horizontal axis,
and vertical lines at t = 0 and t = x. Evaluate
B( x ) for x = 1, 2, 3, 4 and 5. (b) What can you say about the area of the shaded
region shown above right?
6. (a) Calculate the sum of the areas of the trapezoids shown below left. 11. Let C ( x ) represent the area bounded by the graph
of the function shown below, the horizontal axis,
and vertical lines at t = 0 and t = x. Evaluate
C ( x ) for x = 1, 2 and 3, and use that information
to deduce a formula for C ( x ).
(b) What can you say about the area of the shaded
region shown above right?
7. Consider the region bounded by the graph of
y = 2 + x3 , the positive x-axis, the positive y-axis
and the line x = 2. Use two well-placed rectangles to estimate the area of this region.
8. Consider the region bounded by the graph of
y = 9 − 3x , the positive x-axis and the positive y- 300 the integral 12. Let A( x ) represent the area bounded by the graph
of the function shown below, the horizontal axis,
and vertical lines at t = 0 and t = x. Evaluate
A( x ) for x = 1, 2 and 3, and find a formula for
A ( x ). 13. The figure below shows the velocity of a car during a 30-second time frame. How far did the car
travel between t = 0 to t = 30 seconds? 15. The figure below shows the velocities of two cars.
From the time the brakes were applied:
(a) how long did it take each car to stop?
(b) which car traveled farther before stopping? 16. A speeder traveling 45 miles per hour (in a 25mph zone) passes a stopped police car, which
immediately takes off after the speeder. If the
police car speeds up steadily to 60 mph over a 10second interval and then travels at a constant 60
mph, how long — and how far — will it be before
the police car catches the speeder, who continued
traveling at 45 mph? (See figure below.) 14. The figure below shows the velocity of a car during a 30-second time frame. How far did the car
travel between t = 0 to t = 30 seconds?
17. Fill in the table with the units for “area” of a
rectangle with the given base and height units.
base height miles per second
hours
square feet
kilowatts
houses
meals seconds
dollars per hour
feet
hours
people per house
meals “area” 4.0 area 4.0 Practice Answers
1. (a) 3(6) + 12 (4)(3) = 24 (b) 12 (3)(10) + 12 (6)(3) = 24
2. outside rectangular area = (1)(1) + (1) 12 = 1.5
3. Using rectangles with base = 12 :
1 2 1 2 1
57
inside area =
=
+ + +
≈ 0.95
2 3 2 5 3
60
1
2 1 2
72
outside area =
1+ + +
=
= 1.2
2
3 2 5
60
so the area of the region is between 0.95 and 1.2.
4. The leaf’s area is larger than the area of the dark rectangles, x 32 cm2 . 5. y = B( x ) = 2x is a line with slope 2, so it is parallel to the line
y = A( x ) = 2x − 2; see margin for table.
6. Area < area of the rectangle enclosing the shifted regions = 5; see
margin figure.
7. Draw a graph of the velocity function: and then use the concept of “area as distance”:
distance = area of shaded region
1
(base)(height)
2
1
ft
= (60 sec) 66
= 1980 feet
2
sec = 8. (a) At 2:00 p.m. both are walking at the same velocity. You are ahead.
(b) At 3:00 p.m. your friend is walking faster than you, but you are
still ahead. (The “area” under your velocity curve is larger than
the “area” under your friend’s.)
(c) You and your friend will be together on the trail when the “areas”
(distances) under the two velocity graphs are equal. 0
0.5
1
2 B( x )
0
1
2
4 301 302 the integral 4.1 Sigma Notation and Riemann Sums One strategy for calculating the area of a region is to cut the region into
simple shapes, calculate the area of each simple shape, and then add
these smaller areas together to get the area of the whole region. When
you use this approach with many sub-regions, it will be useful to have
a notation for adding many values together: the sigma (Σ) notation.
summation sigma notation
5 12 + 22 + 32 + 42 + 52 ∑ k2 7 1
k how to read it
the sum of k squared,
from k equals 1 to k equals 5 k =1 1 1 1 1 1
+ + + +
3 4 5 6 7 ∑ k =3
5 ∑ 20 + 21 + 22 + 23 + 24 + 25 the sum of 1 divided by k,
from k equals 3 to k equals 7 2j the sum of 2 to the j-th power,
from j equals 0 to j equals 5 an the sum of a sub n,
from n equals 2 to n equals 7 j =0
7 ∑ a2 + a3 + a4 + a5 + a6 + a7 n =2 The variable (typically i, j, k, m or n) used in the summation is called
the counter or index variable. The function to the right of the sigma is
called the summand, while the numbers below and above the sigma
are called the lower and upper limits of the summation.
Practice 1. Write the summation denoted by each of the following:
5 ∑ (a) k3 7 (b) ∑ (−1) j j =2 k =1 1
j 4 (c) ∑ (2m + 1) m =0 In practice, we frequently use sigma notation together with the
standard function notation:
3 ∑ f ( k + 2) = f (1 + 2) + f (2 + 2) + f (3 + 2) k =1 = f (3) + f (4) + f (5)
4 ∑ f ( x j ) = f ( x1 ) + f ( x2 ) + f ( x3 ) + f ( x4 ) j =1 x f (x) g( x ) h( x ) 1
2
3
4
5 2
3
1
0
3 4
1
−2
3
5 3
3
3
3
3 5 Example 1. Use the table to evaluate ∑ k =2 5 2 · f (k ) and ∑ [5 + f ( j − 2)]. j =3 Solution. Writing out the sum and using the table values:
5 ∑ 2 · f ( k ) = 2 · f (2) + 2 · f (3) + 2 · f (4) + 2 · f (5) k =2 = 2 · 3 + 2 · 1 + 2 · 0 + 2 · 3 = 14 4.1 sigma notation and riemann sums 303 while:
5 ∑ [5 + f ( j − 2)] = [5 + f (3 − 2)] + [5 + f (4 − 2)] + [5 + f (5 − 2)] j =3 = [5 + f (1)] + [5 + f (2)] + [5 + f (3)]
= [5 + 2] + [5 + 3] + [5 + 1]
J which adds up to 21. Practice 2. Use the values in the preceding margin table to evaluate:
5 (a) ∑ 4 g(k) (b) ∑ 5 h( j) (c) j =1 k =2 ∑ [ g(k) + f (k − 1)] k =3 Example 2. For f ( x ) = x2 + 1, evaluate 3 ∑ f ( k ). k =0 Solution. Writing out the sum and using the function values:
3 ∑ f ( k ) = f (0) + f (1) + f (2) + f (3) k =0
= 02 + 1 + 12 + 1 + 22 + 1 + 32 + 1 = 1 + 2 + 5 + 10
J which adds up to 18.
Practice 3. For g( x ) = 1
, evaluate
x 4 ∑ 3 g(k ) and k =2 ∑ g ( k + 1). k =1 The summand need ...

View
Full Document