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Exploration 5-2a: Differentials, and Linearization
of a Function
Date:
Objective: Find the equation of the linear function that best fits the graph of a given
function at a given point.
The figure shows the graph of the function
f(x) H sec x
In this Exploration you will find the equation of the linear
function that best fits the graph of f when x is kept close
to 1.
f(x) or y
4. The quantity (x D 1) in Problem 1 is called the
differential of x and is abbreviated dx. The quantity
fQ(1)(x D 1) is called the differential of y, abbreviated
dy. Let dx H 0.4. Draw lines on the graph showing
what dx and dy mean.
5. What “interesting” quantity does the ratio dy ÷ dx
equal?
5
6. Tell how dx and dy are related to ∆x and ∆y, the
changes in x and f(x) as x goes from 1 to 1.4.
x
0
1
2
1. You recall the point-slope form, y D y0 H m(x D x0), of
the linear function equation. In this equation, m is
the slope and (x0, y0) is a point on the graph. If the
linear function is to fit function f when x is close
to 1, then m should equal the slope of f at x H 1,
and y0 should equal f(1). Thus the best-fitting linear
function has an equation of the form
y H f(1) C fQ(1)(x D 1)
7. The quantity ∆y D dy is the error involved in using y
in Problem 1 as an approximation for f(x). Explain
why this quantity is equal to f(x) D y. Then, on your
grapher, make a table of values of the error for
each 0.01 unit of x from x H 0.97 to x H 1.03. What
happens to the error as x gets closer to 1?
Find the particular equation of the best-fitting linear
function.
8. How close to 1 should x be kept in order for the
error in Problem 7 to be less than 0.001?
2. Plot the graph of f and the graph of the linear
function on the same screen. Sketch the linear
function on the graph above. Is the linear function
tangent to the graph of f at x H 1?
3. Zoom in on the point (1, sec 1). What words describe
the fact that the graph looks more and more like the
tangent line as you zoom?
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9. What did you learn as a result of doing this
Exploration that you did not know before?
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Exploration 5-2b: A Motion Antiderivative Problem
Date:
Objective: Given an equation for the velocity of a moving object, find
equations for the displacement and acceleration.
Tay L. Gates wants to determine the characteristics of his
new pickup truck. With special instruments he records its
velocity at 2-second intervals as he starts off from a traffic
light.
ts
Velocity, ft/s
0
0
2
4.5
4
6.9
6
8.8
8
10.4
10
11.9
4. Use the equation in Problem 3 to predict where Tay’s
truck was 10 s after he started accelerating.
5. Calculate the answer to Problem 4 directly from the
data in the table, using the trapezoidal rule.
1. Show that the power function
v(t) H 3t 0.6
fits these data closely.
6. How long does it take before the truck is 100 ft
beyond the center of the intersection?
2. Write an equation for Tay’s acceleration, a(t). Is
the acceleration getting larger or smaller as time
goes on? Tell how you figured this out.
7. What did you learn as a result of doing this
Exploration that you did not know before?
3. Velocity is the derivative of displacement. Thus,
displacement is the antiderivative of velocity. Write
an equation for x(t), the truck’s displacement from
the middle of the intersection. Use the fact that the
truck was initially at x(0) H D50 ft from the center
of the intersection at t H 0 when the light turned
green.
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Exploration 5-4a: Riemann Sums for Definite Integrals
Date:
Objective: Find an alternative to the trapezoidal rule for estimating a definite integral.
4. The sum of the areas of the rectangles mentioned
in Problem 3 is called a midpoint Riemann sum.
Evaluate this sum. How close does the answer come
to the estimate of 14
f(x) dx you got by counting
2
squares in Problem 1?
f (x )
50
40
30
20
10
x
2
0
4
6
8
10
12
14
16
1. The symbol ab f(x) dx stands for the definite integral
of f(x) from x H a to x H b.
By counting squares, find an approximation of
14
f(x) dx for the function graphed above.
2
5. Find an estimate of f(c) for these values of c.
c
f(c)
2
4
6
8
10
12
2. Find an estimate of f(c) for these values of x H c.
c
f(c)
3
6. The values of c in Problems 2 and 5 are called
sample points for a Riemann sum. Find an estimate
f(x) dx using the sample points in Problem 5.
of 14
2
5
7
9
11
13
3. On the figure above, draw rectangles between
vertical grid lines with altitudes equal to f(c), starting
at x H 2 and ending at x H 14. Tell why the sum of the
f(x) dx.
areas of these rectangles is an estimate of 14
2
7. Tell how estimating an integral using a Riemann sum
is related to doing it by trapezoidal rule.
8. What did you learn as a result of doing this
Exploration that you did not know before?
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Exploration 5-5a: The Mean Value Theorem
Date:
Objective: Without looking at the text, discover the hypotheses and conclusion
of the mean value theorem.
1. For f(x) H D0.1x 3 C 1.2x 2 D 3.6x C 5, graphed below,
there is a value of x H c between 3 and 7 at which the
tangent to the graph is parallel to the secant line
through (3, f(3)) and (7, f(7)).
3. For g(x) H 6 D 2(x D 4)2/3, graphed below,
• Draw a secant line through (1, g(1)) and (5, g(5)).
• Is g differentiable on (1, 5)?
• Is g continuous on [1, 5]?
• Draw the secant line and the tangent line.
• Tell why there is no value of x H c between x H 1
• From the graph, c M
and x H 5 at which gQ(c) equals the slope of the
secant line.
• Is f differentiable on (3, 7)?
• Is f continuous on [3, 7]?
f (x)
5
g (x )
5
x
3
7
2. Function f from Problem 1 has two values of x H c
between x H 1 and x H 7 at which fQ(c) equals the
slope of the corresponding secant line. (That is, the
tangent line parallels the secant line.)
x
1
4. Function g from Problem 3 does have a value x H c
in (1, 4) for which gQ(c) equals the slope of the
secant line through (1, g(1)) and (4, g(4)).
• Draw the secant and tangents on the graph below.
• From the graph, c M
5
and c M
• Is f differentiable on (1, 7)?
• Draw the secant line and tangent line, below.
• Is f continuous on [1, 7]?
• From the graph, c M
• Is g differentiable on (1, 4)?
f (x)
• Is g continuous on [1, 4]?
5
g(x)
5
x
1
7
x
1
4
(Over)
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Exploration 5-5a: The Mean Value Theorem
h(x) H
0.2(x D 2)
0.3(x
D 5)
2
C 2.2,
C 1,
Date:
7. The graph below is function h from Problem 5.
5. Piecewise function h is defined by
2
continued
• Draw a secant line through (1, h(1)) and (5, h(5)).
if x K 5
if x I 5
• Show that there is a point x H c in (1, 5) where hQ(c)
equals the slope of the secant line.
• Draw a secant line through (5, h(5)) and (7, h(7)).
• Is h differentiable on (1, 5)?
• Is h differentiable on (5, 7)?
• Explain why h is continuous on [1, 5], even though
• Is h continuous on [5, 7]?
there is a step discontinuity at x H 5.
• Why is there no value x H c in (5, 7) for which hQ(c)
equals the slope of the secant line?
h(x)
5
h(x)
5
x
1
5
x
5
8. The mean value theorem states:
If f is differentiable on (a, b) and
f is continuous on [a, b],
then there is a number c in (a, b) such that
7
6. The graph below is function h from Problem 5.
• Draw a secant line through (5, h(5)) and (9, h(9)).
f(b) D f (a)
fQ(c) H b Da
• Is h differentiable on (5, 9)?
• Is h continuous on [5, 9]?
i.e., the secant’s slope
For which problem(s) are
• There is a point x H c in (5, 9) where hQ(c) equals
the slope of the secant line. Draw the tangent line.
Estimate the value of c.
• the hypotheses and conclusion true?
• the hypotheses and conclusion not true?
• the conclusion true, but not the
h(x)
hypotheses?
9. The number x H c in the above problems is the
“mean” value referred to in the name “mean value
theorem.” Explain why the hypotheses are sufficient
conditions for the conclusion, but not necessary
conditions.
5
x
5
9
10. What did you learn as a result of doing this
Exploration that you did not know before?
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Exploration 5-6a: Some Very Special Riemann Sums
Date:
Objective: Calculate Riemann sums for given sets of sample points and reach a
conclusion about how the sample points were chosen.
The figure shows the graph of f(x) H x1/2. In this
Exploration you will integrate f(x) from x H 1 to 4.
f (x)
2
1
x
1
2
3
4
1. Find an estimate for the integral
IH
4
x 1/2 dx
1
4. Find a Riemann sum for I using six subintervals of
equal width, and these sample points:
k
xHc
1
1.2458051304
2
1.7470136081
3
2.2476804000
4
2.7481034438
5
3.2483958519
6
3.7486100806
How does the integral by this Riemann sum compare
with other values in this Exploration?
by trapezoidal rule with n H 3 subintervals. Write
down all the decimal places your calculator will give
you. Does this value overestimate or underestimate
the actual integral? Explain.
2. Find a midpoint Riemann sum for integral I in
Problem 1. Use n H 3 increments. Show that this sum
is not equal to the value by trapezoidal rule. Does
the midpoint sum overestimate or underestimate
the actual integral? Explain.
3. Find a Riemann sum for integral I using the
subintervals in Problem 1, but using the following
sample points. (k stands for the subinterval number.)
k
x=c
1
1.4858425557
2
2.4916102607
3
3.4940272163
5. Let g(x) H 23x3/2. Find the point in the open interval
(1, 1.5) at which the conclusion of the mean value
theorem is true for function g. Where have you seen
this number in this Exploration?
6. How is function g related to function f ?
7. Make a conjecture about the exact value of the
integral in Problem 1.
How does this sum compare with the answers to
Problem 1 and Problem 2?
8. What did you learn as a result of doing this
Exploration that you did not know before?
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Exploration 5-6b: The Fundamental Theorem
of Calculus
Date:
Objective: Prove that a definite integral can be calculated exactly, using
an indefinite integral.
1. Write the definition of
ab f(x) dx.
f (x)
2. Write the definition of g(x) H f(x) dx.
∆x
a
x1
c1
3. How can you be sure that the mean value theorem
applies to function g?
...
x2
c2
x3
x
x n–1
c3
b
cn
g(x ) D g(a)
1
, and
6. By the mean value theorem, gQ(c1) = ∆x
so on. By the definition of indefinite integral,
gQ(c1) H f(c1). By appropriate substitutions, show
that Rn from Problem 5 is equal to g(b) D g(a).
4. The figure shows function g in Problem 2. Write the
conclusion of the mean value theorem as it applies
to g on the interval from x H a to x H x1, and
illustrate the conclusion on the graph.
g(x)
∆x
x
a
x1
7. Rn from Problem 6 is independent of n, the number
of increments. Use this fact, and the fact that
Ln K Rn K Un , to prove that
b
a
f(x) dx H g(b) D g(a)
5. The figure in the next column shows the graph of
f(x) from Problem 2. Let c1, c2, c3, . . . , cn be sample
points determined by the mean value theorem as in
Problem 4. Write a Riemann sum Rn for ab f(x) dx
using these sample points and equal ∆x values.
Show the Riemann sum on the graph.
8. The conclusion in Problem 7 is called the
fundamental theorem of calculus. Show that you
understand what it says by using it to find the exact
value of 41 x 1/2 dx. Use the back of this page if
needed.
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Exploration 5-7a: Some Properties
of Definite Integrals
Date:
Objective: Illustrate by graph or by the fundamental theorem of calculus that certain
properties of definite integrals are true.
1. The graph shows f(x) H 19 D x 2. Tell why f is called
an even function.
5. The graph shows h(x) H 0.4x 3 D 3x 2 C 5x C 5.
h (x )
f (x)
19
10
x
1
x
–3
3
D3
6
Use the fundamental theorem to calculate
2. Use the fundamental theorem to calculate
3
3
(19 D x ) dx and
2
3
0
3
(19 D x ) dx
2
6
h(x) dx and
1
h(x) dx
3
Explain why the first integral is twice the second.
3. The graph shows g(x) H x 3. Explain why g is called an
odd function.
g (x)
6. Quick! Calculate
7. Given:
x
–4
61 h(x) dx.
7
4
2
u(x) dx H 29,
7
2
v(x) dx H 13,
4
2
v(x) dx H 8
a. Find 72 [u(x) C v(x)] dx.
4. Use the fundamental theorem to calculate
Explain why the answer equals zero.
4D4 x 3 dx.
b. Find 74 v(x) dx.
8. What did you learn as a result of doing this
Exploration that you did not know before?
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Exploration 5-8a: Applications of Definite Integrals
Date:
Objective: Without looking at the text, learn an interpretation of f(x) dx in a
definite integral.
Spaceship Velocity and Displacement Problem: A spaceship
is fired into orbit. As the last stage of the booster rocket
is fired, the spaceship is going 3000 ft/s (a bit under
2000 mi/h). Its velocity, v(t), at t s since the booster
was fired is given by
6. Find the distance traveled by the spaceship by
evaluating the integral in Problem 5 using the
fundamental theorem of calculus.
v(t) H 3000 C 18t 1.4
v (t)
7. In order to orbit, the spaceship must be going at
least 26,000 ft/s (about 17,500 mi/h). To the nearest
second, at what time is it going that fast?
3000
t
100
1. On the graph, pick a sample point, t, somewhere
on the t-axis between 0 and 100. Show the
corresponding point (t, v(t)) on the graph.
2. Draw a narrow vertical strip of the region in such a
way that the ordered pair (t, v(t)) is within the strip.
Label the width of the strip dt.
8. How far does the spaceship go from the time the last
stage fires till it reaches orbital velocity?
3. If the strip is narrow (i.e., dt is small), the velocity
throughout the time interval dt is not much different
from the velocity v(t) at the sample point. Explain
why the distance traveled in this time interval is
approximately equal to v(t) • dt.
4. Write a Riemann sum that represents, approximately,
the total distance the spaceship goes between t H 0
and t H 100.
9. Based on what you observed while doing this
Exploration, why do you suppose integrals are
written in differential form, f(x) dx, instead of
simply in derivative form, f(x)?
5. Explain why the Riemann sum in Problem 4 is
between the corresponding lower sum and upper
sum. Based on this fact, why can you conclude that
the limit of the Riemann sum is exactly equal to
100
v(t) dt?
0
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Exploration 5-9a: Volume by Plane Disk Slices
Date:
Objective: Calculate exactly the volume of a solid of variable cross-sectional area.
y
y
4
Disk of
volume
dV
(x, y )
(x, y)
dy
dy
x
x
2
Draw the region.
Slice a strip perpendicular to the
axis of rotation.
Rotate
Rotate the region and the strip to
form a solid. The rotating strip forms
a flat disk.
The region in Quadrant I bounded by the parabola
yH4Dx
2
4. As a rough check on your answer, compare it with
the volume of a circumscribed cylinder and with the
volume of an inscribed cone.
is rotated about the y-axis to form a solid paraboloid, as
shown above.
1. A representative slice of the region is shown in the
figure at left. As the slice turns, it generates a
plane disk, as shown in the figure at right. Let (x, y)
be a sample point on the graph within the slice.
Find the volume, dV, of the slice in terms of the
coordinates of the sample point.
2. By appropriate algebra, transform dV so that it is in
terms of y alone.
5. Based on your answer to Problem 4, make a
conjecture about the volume of a paraboloid in
relation to the volume of the circumscribed cylinder.
3. Find, exactly, the volume V of the solid by adding the
volumes of the slabs and taking the limit as the
thickness of the slabs goes to zero (i.e., integrate).
6. What did you learn as a result of doing this
Exploration that you did not know before?
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Exploration 5-9b: Volume by Plane Washer Slices
Date:
Objective: Find the volume of a solid of revolution that has a hole in it.
1. Write the area of the circular ring as a function of y1
and y2. Use the result to write the volume, dV, of the
washer.
y
dx
(x, y1)
(x , y2)
2. Transform the equation for dV so that it is in terms
of x alone.
Region R
x
1
4
3. Find V by adding up the dV values and taking the
limit (i.e., integrate). Use the fundamental theorem to
evaluate the integral exactly.
The figure above shows the region R that is bounded
above and below by the graphs of
y1 H 6eD0.2x and y2 H x
and on the left and right by the lines x H 1 and x H 4.
The figure below shows the solid generated by rotating R
about the x-axis. A strip of width dx generates a
plane washer of thickness dx whose cross section is a
circular ring of outer radius y1 and inner radius y2.
y
dx
4. Get a decimal approximation for the volume you
found in Problem 3.
(x, y )
1
(x, y )
2
5. Evaluate the integral in Problem 3 numerically.
Show that the answer is (virtually) the same as in
Problem 4.
x
1
4
W asher of
volume dV
6. Check your answer with Example 3 in Section 5-9 of
the text.
7. What did you learn as a result of doing this
Exploration that you did not know before?
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Exploration 5-10a: Introduction to Simpson’s Rule
Date:
Objective: Calculate Riemann sums for given sets of sample points and reach
a conclusion about how the sample points were chosen.
Simpson’s rule enables you to approximate the value of a
definite integral of a function by replacing the graph with
sections of parabolas. In this Exploration you will learn
how this is done, prior to proving Simpson’s rule
algebraically.
4. Draw a rectangle on the figure with the segment
from x H 3 to x H 7 as the base, and altitude equal to
the weighted average in Problem 2. How does the
rectangle confirm your response in Problem 3?
For Problems 1–4, the figure shows the parabola
y H Dx 2 C 11x D 16
1. Find the area of the region under the graph from
x = 3 to x = 7 using the fundamental theorem.
y
15
10
5
x
3
5
7
For Problems 5 and 6, the figure and table show the
velocity, v, in ft/s, of a moving object at various times t, in
seconds.
t
v
6
10
11
7
16
12
21
19
26
22
v
25
20
15
10
5
2. Find the y-values at x H 3, 5, and 7. Then find the
weighted average of these values, using the first one,
four times the middle one, and the last one.
x
y
Weighted y
3
5
t
6
11
16
21
26
5. Assume that one parabola contains the first three
points and another parabola contains the last three
points (the middle point is used twice). Apply
Simpson’s rule twice to find an estimate, S4, of
26
v dt.
6
7
3. Multiply the weighted average from Problem 2 by the
distance between x H 3 and x H 7. What do you notice
about the answer? The result is an example of
Simpson’s rule.
6. On the figure in Problem 5, sketch the two
parabolas, and the trapezoids for T4 from the
trapezoidal rule. Why would you expect Simpson’s
rule to have less error than the trapezoidal rule?
(Over)
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Exploration 5-10a: Introduction
to Simpson’s Rule continued
7. Look up Simpson’s rule in the text. Write the formula
for Simpson’s rule. Explain how the formula is
consistent with your work in Problem 5.
Date:
11. What did you learn as a result of doing this
Exploration that you did not know before?
8. Write or download two programs, one to calculate an
integral by Simpson’s rule from a set of data as in
Problem 5, and another to calculate an integral by
Simpson’s rule from an equation. Test the first
program using the data from Problem 5. Does the
program give the same answer you calculated?
9. Use the second program from Problem 8 to find a
Simpson’s rule approximation, S10 (with n H 10
increments), for
π
sin x dx
0
Use the trapezoidal rule to find another
approximation, T10, for the integral. Which
approximation is closer to the exact answer, 2?
10. Look up the proof of Simpson’s rule in your text.
Do you understand how the formula is derived?
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Exploration 5-10b: Simpson’s Rule Demonstration
Date:
Objective: Verify Simpson’s rule by actually finding the equations of the parabolas
and integrating algebraically.
The graph shows the following points for a function f.
f(x)
5
3. Plot the given data points and the two branches of h.
Use Boolean variables to restrict the domains of the
branches. Sketch the results on the given figure. Is h
continuous at x H 5? Is h differentiable at x H 5? Use
one-sided limits to justify your answers.
x
5
10
1. Find 91 f(x) dx approximately, using Simpson’s rule
with n H 4 increments. Show your work.
x
f(x)
1
2.2
3
1.2
5
1.0
7
2.6
9
5.8
Factor
f(x) • Factor
4. Find 91 h(x) dx. Does the result agree with Simpson’s
rule for 91 f(x) dx from Problem 1?
2. Let h be the piecewise function with graphs being
the two parabolas in Simpson’s rule. Find the
particular equations for the two branches of h, one
for the first three points and one for the last three
points. You may use quadratic regression.
5. Draw something on the given figure to show why you
would expect Simpson’s rule to give an answer closer
to the actual integral of f than the trapezoidal rule
would.
6. What did you learn as a result of doing this
Exploration that you did not know before?
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Chapter 5
Exploration 5-2b
1. v(t) H 3t 0.6
Exploration 5-2a
1. f(1) H sec 1 H 1.8508...
fQ(1) H sec 1 tan 1 H 2.8824...
y H 1.8508... C 2.8824... (x D 1)
2. The linear function is tangent to the curve at x H 1.
f (x) or y
t
Velocity
0
0
0.0
v(t)
2
4.5
4.5471...
4
6.9
6.8921...
6
8.8
8.7904...
8
10.4
10.4466...
10
11.9
11.9432...
Thus the values of v(t) fit the data closely.
5
2. a(t) H vQ(t) H 1.8tD0.4
a(t) is a decreasing function for all t I 0, because
aQ(t) H D7.2tD1.4 G 0 for all t > 0.
3 1.6
t C C H 5t 1.6 C C
3. x(t) H 1.6
x(0) H D50 ⇒ D50 H 0 C C ⇒ C H D50
x(t) H 5t 1.6 D 50
x
0
1
2
3. Local linearity
4. The graph shows dx H 0.4, dy being the corresponding rise to
the linear function graph, and ∆y being the corresponding
rise to the function graph.
f (x) or y
5. Trapezoidal rule:
0
11.9
x(10) H D50 C 2 • 2 C 4.5 C 6.9 C 8.8 C 10.4 C 2
H 23.1 ft
3 1.6
6. Solve 100 H 1.6 t D 50.
5
∆y
1.6
t H 1.6 3 • 150 H 15.4678... M 15.5 s
dy
7. Answers will vary.
dx
x
0
1
Exploration 5-4a
2
1. About 15.4 squares, so the area is about 308. (Equation is
f(x) H 10 C 2x C 10 cos (0.5x), so precise answer is 308.3103....)
fQ(1) • (x D 1)
5. dy O dx H H fQ(1).
(x D 1)
dy
So, dy O dx is the same as the derivative, dx .
2. c
6. dx H ∆x always. dy H fQ(x) dx
or: dy is the rise along the tangent line and ∆y is the rise
along the graph itself.
or: dy is close to ∆y if dx is small.
7.
4. x(10) H 24.6450... M 24.6 ft, or about three-fourths of the way
through the intersection.
f(c)
3
17
5
12
7
15
9
26
x
y
f(x)
Error
11
39
0.97
1.7643...
1.7689...
0.0046...
13
46
0.98
1.7931...
1.7952...
0.0020...
0.99
1.8219...
1.8225...
0.0005...
0.0
1.00
1.8508...
1.8508...
1.01
1.8796...
1.8801...
0.0005...
1.02
1.9084...
1.9107...
0.0022...
1.03
1.9372...
1.9424...
0.0051...
As x approaches 1, the error approaches zero.
8. Solving f(x) D y H 0.001 numerically, the error will be less
than 0.001 if 0.9862... G x G 1.0134..., so if x is kept within
0.0134... of 1, then the error will be less than 0.01.
9. Answers will vary.
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Solutions for the Explorations / 183
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2.
3.
f (x)
f(x)
5
50
40
30
x
20
1
7
10
c M 2.3 or 5.7 exactly 4 J 3
f is differentiable on (1, 7).
f is continuous on [1, 7].
x
0
2
4
6
8
10
12
14
16
3.
For each rectangle, the part of the graph above the rectangle
is roughly the same as the part of the rectangle above the
graph, so the area of each rectangle is about the same as
the area under the corresponding portion of the graph.
Thus the sum of the areas of the rectangles represents the
definite integral.
g(x)
5
4. Area of rectangles
H 2 • 17 C 2 • 12 C 2 • 15 C 2 • 26 C 2 • 39 C 2 • 46
H 310, which is close to the 308 in Problem 1.
5. c
f(c)
2
19
4
14
6
12
8
19
10
33
12
44
x
1
g is not differentiable on (1, 5).
g is continuous on [1, 5].
No such x H c exists. The slope of the secant line is positive.
The slope of the tangent line is greater than the slope of the
secant when x is between 1 and 4, and it is negative when x is
between 4 and 5.
6. Area of rectangles
H 2 • 19 C 2 • 14 C 2 • 12 C 2 • 19 C 2 • 33 C 2 • 46
H 282
7. Both methods approximate the height of the function within
each subinterval, then multiply that height by the width of
the subinterval.
5
4.
g(x)
5
8. Answers will vary.
Exploration 5-5a
1.
x
f (x)
1
4
5
c M 3.1
g is differentiable on (1, 4).
g is continuous on [1, 4].
x
3
7
f(3) H 2.3, f(7) H 4.3, slope of secant line H 0.5
c M 5.5 (numerically, graphically, or by solving
D0.3c 2 C 2.4c D 3.6 H 0.5)
21
c M 5.5 exactly 4 C 13
f is differentiable on (3, 7).
f is continuous on [3, 7].
184 / Solutions for the Explorations
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8. The hypotheses and conclusion are true for Problems 1, 2,
4, and 7.
The hypotheses and conclusion are not true for Problems 3
and 5.
The conclusion is true but the hypotheses are not for
Problem 6.
5.
h(x)
5
x
5
7
h is differentiable on (5, 7).
h is not continuous on [5, 7].
2.2 D 4
The slope of the secant line is 7 D 5 H D0.9. The slope of the
curve on (5, 7) is determined by 0.6(x D 5), which is positive
for all x I 5. So the slope of the curve cannot equal the slope
of the secant line.
9. If f(x) is (1) differentiable on (a, b), and (2) continuous at
x H a and x H b, then there is a number x H c between a and
b for which
f(b) D f (a)
fQ(c) H b Da
If a function satisfies the hypotheses, then that is enough
(sufficient) to make the function satisfy the conclusion.
But for a function that satisfies the conclusion, it is not
necessary for the hypotheses to be satisfied. For example, the
function in Problem 5 between x H 2 and x H 8 does not meet
the hypotheses of the mean value theorem, but these are not
necessary for c H 2.625 and c H 5.416... to satisfy the
conclusion..
10. Answers will vary.
Exploration 5-6a
6.
1. Trapezoidal rule:
I M 0.5 C 2 C 3
C 1 H 4.6462...
Underestimates I because parts of the region lie above the
trapezoids.
h (x )
5
2. Midpoint sum:
I M 1.5 C 2.5
C 3.5
H 4.6767...
Overestimates I because the part of the region above the
top of each rectangle is slightly less than the part of each
rectangle above the region.
x
5
7
3. I M 1.4858
42555
7
C 2.4916
10260
7
C 3.4940
27716
3
H 4.6666...
This estimate is between the trapezoidal and midpoint sums.
1
4. I M 2 1.2458
05130
4
C 1.7470
13608
1
C 2.2476
80400
0
C 2.7481
03443
8
C 3.2483
95851
9
C 3.7486
10080
6
H
4.6666..., same as in Problem 3.
2
5. g(1) H 3, g(1.5) H 1.5
1.5 D 23
Slope of secant line H H 6 D 4
0.5
3
h is differentiable on (5, 9).
h is not continuous on [5, 9].
c M 5.75
7.
h(x)
4
4
gQ(c) H c 1/2 H 6
D 3 ⇒ c H 6
D 3
5
2
H 1.2458051303...,
the sample point for [1, 1.5] in Problem 4!
6. gQ(x) H f(x)
7. Conjecture: I H 423
8. Answers will vary.
x
5
7
4 D 2.4
The slope of the secant line is 5 D 1 H 0.4. The slope of the
curve on (1, 4) is determined by 0.4(x D 2). So, c H 3 (by
solving 0.4 H 0.4(x D 2)).
h is differentiable on (1, 5).
h is continuous on [1, 5] even though there is a step
continuity at x H 5 because limx→5D h(x) H h(5).
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Exploration 5-6b
x
4
4.
1. See the text.
The two parts of the region are congruent and thus have the
same area. Because the left part of the region is below the
y-axis, the integral is negative and thus cancels the integral
for the right part.
2. See the text.
3. g is differentiable on (a, b) because it has a derivative f, and
g is continuous on (a, b) because differentiability implies
continuity.
1 xH4
x 4
dx H 4
H0
xHD4
3
D4
3
5.
1
4. There is a point c1 ∈ (a, x1), such that
g(x1) D g(a)
gQ(c1) H ∆x
The graph shows x H c1 with tangent line parallel to secant
line.
6
3
h(x) dx H
(0.4x 3 D 3x 2 C 5x C 5) dx
h(x) dx H
6
3
(0.4x 3 D 3x 2 C 5x C 5) dx
xH6
H (0.1x 4 D x 3 C 2.5x 2 C 5x)xH3
H 15
6
6.
1
b. 7
7. a.
2
7
∆x
3
h(x) dx =
1
2
v(x) dx H
6
h(x) dx C
[u(x) C v(x)] dx H
x
4
2
7
4
h(x) dx H 12 C 15 H 27
3
7
2
u(x) dx C
v(x) dx C
⇒ 13 H 8 C
x1
c1
3
1
xH3
H (0.1x 4 D x 3 C 2.5x 2 C 5x)xH1
H 12
g(x)
a
7
2
v(x) dx H 42
7
v(x) dx
4
7
v(x) dx
4
v(x) dx H 5
8. Answers will vary.
Exploration 5-8a
5. Rn H f(c1)(x1 D a) C f(c2)(x2 D x1) C f(c3)(x3 D x2)
C•••C f(cn)(b D xnD1)
H [f(c1) C f(c2) C f(c3) C•••C f(cn)]∆x
1.
6. Rn H [gQ(c1) C gQ(c2) C gQ(c3) C•••C gQ(cn)]∆x
g(x1) D g(a) g(x2) D g(x1)
g(b) D g(xnD1)
C C•••C ∆x
= ∆x
∆x
∆x
H g(x1) D g(a) C g(x2) D g(x1) C g(x3) D g(x2)
C•••C g(b) D g(xnD1)
H g(b) D g(a)
7. Ln K Rn K Un ⇒ Ln K g(b) D g(a) K Un
lim Ln K g(b) D g(a) K lim Un
Problem 6
Properties of limits
n→∞
n→∞
(g(b) D g(a) is a constant.)
But the two limits equal the definite integral, by definition.
N
b
a
f(x) dx H g(b) D g(a), Q.E.D.
8. g(x) H
4
1
x
1/2
x
1/2
2
2
2
dx H 3 43/2 C C D 3 13/2 D C H 43
dt
100
3. Distance H rate • time
If the rate throughout the time interval is essentially constant
at v(t), and the length of time is dt, then D = v(t) • dt.
v(ci ) dt
i=1

1
(19 D x ) dx H 19x D 3x 
2
t
t
n
1. For all x, f(Dx) H f(x).
(Even powers of x (e.g., x 2) have this property.)
3
xH3
1
2.
(19 D x 2) dx H 19x D 3x 3 xHD3 H 96
D3
0
3000
4. Rn H
Exploration 5-7a
3
(t, v (t))
2. See the graph in Problem 1.
2
dx H 3 x 3/2 C C
Note that the C ’s cancel out!
v(t)
3
xH3
xH0
= 48
The region is symmetrical with respect to the y-axis. Thus
the area to the left of the axis equals the area to the right,
meaning that the total area is twice the area to the right.
3. For all x, g(Dx) H Dg(x).
(Odd powers of x (e.g., x 1) have this property.)
186 / Solutions for the Explorations
5. Each term in the upper sum is greater than or equal to the
corresponding v(ci) dt in the Riemann sum, by definition
of an upper sum, so that Un L Rn.
Similarly, Ln K Rn. Thus, Ln K Rn K Un.
N lim Ln K lim Rn K lim Un.
n→∞
n→∞
n→∞
If the function is integrable, the limits of the upper and lower
sums equal the definite integral, by the definition of definite
integral.
By the squeeze theorem, the limit of Rn is also equal to the
definite integral. Thus,
lim Rn H
n→∞
100
v(t) dt, Q.E.D.
0
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100
6.
0

18
2.4
(3000 C 18t 1.4) dt H 3000t C 2. 4 • t
6. Same answer!
(Note that some students may already have looked in the text
and found out that this problem is Example 3. They will no
doubt think that they have “outfoxed” you by finding this
problem in the book. But really you have outfoxed them by
getting them to read the book!)
tH100
tH0
H 300,000 C 7.5 • 1002.4 H 773,218.0083...
≈ 773 thousand ft
7. Solve 3000 C 18t 1.4 H 26000
t H 165.5376... M 166 s
166
8. Distance M
0
18
2.4
(3000 C 18t 1.4) dt H 3000t C 2. 4 • t

tH166
tH0
H 498,000 C 7.5 • 1662.4 H 2,095,057.9488...
M 2.095 million ft
7. Answers will vary.
Exploration 5-10a
2. x
y
3
8
8
5
14
56
7
12
12
Exploration 5-9a
Weighted y
3. 1223 • 4 H 5023
The answer is the same as the integral from x H 3 to x H 7.
4.
y
15
π • (4 D y) dy
1
H 4πy D 2πy 
yH4
10
yH0
2
yH4
yH0
5
H 8π
4. The circumscribed cylinder has r H 2, h H 4, and
Vcyl H πr 2h H 16π.
The inscribed cone has r H 2, h H 4, and
1
Vcone H 3πr 2h H 513 • π.
The calculated volume of the paraboloid falls between these
numbers.
x
3
16
5.
6
10 C 7 • 4 C 12
• 10 = 8313
6
v dt H
12 C 19 • 4 C 22
• 10 H 18313
6
26
6. Answers will vary.
N
2. y1 H 6eD0.2x, y2 H x
dV H π(36eD0.4x D x) dx
π(36e
H D90πe
D0.4x
xH1
D0.4x
D1.6
H D90πe
26
6
v dt H 8313 C 18313 H 26623
6.
1. Awasher = Aouter circle D Ainner circle
A H πy 21 D πy 22 H π y 21 D y 22
dV H A dx H π y 21 D y 22 dx
xH4
7
v dt M
16
Exploration 5-9b
5
The area of the rectangle appears to be the same as the area
under the curve.
5. Conjecture: The volume of a paraboloid is half the volume of
the circumscribed cylinder.
3. V H

8 C 56 C 12
Weighted average H 6 H 1223
2. y H 4 D x 2 ⇒ x 2 H 4 D y, so
dV H πx 2 dy H π(4 D y) dy.
3. V H
7
1
11
(Dx 2 C 11x D 16) dx H D3x 3 C 2x 2 D 16x 3
D343 539
99
H 3 + 2 D 112 C 9 D 2 C 48 H 5023 H 50.6666...
3
9. Both the f(x) and the dx have physical meanings. The dx is
multiplied by f(x) to get the part of the value of the integral
corresponding to the vertical strip.
1. The volume of a disk is πr 2h, where r H radius and h H
thickness (think of a disk as a very short cylinder). The slice
shown is not exactly a disk, because the radius changes ever
so slightly from the top to the bottom. But the radius of this
slice is always very close to x, and the height is dx, so the
volume is approximately dV H πx 2 dy. (In the limit, the
difference between the cylinders and the actual disk volumes
approaches zero.)
7
1.
20
15
10
D x) dx

1
D 2πx 2
D0.4
C 90πe
v
25
xH4
5
t
6
11
16
21
26
xH1
D 7.5π
4. V H 108.8816...
Simpson’s rule is probably more accurate than the
trapezoidal rule because velocity is likely to be approximated
better by smooth curves than by straight line segments.
5. Numerical integration gives 108.8816..., the same answer!
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1
7. 3(∆x)(y0 C 4y1 C 2y2 C 4y3 C 2y4 C ••• C 2ynD2 C 4ynD1 C yn)
The formula is consistent because the value of y2 is used
twice and 13(∆x) is equivalent to 160.
8. See the Programs for Graphing Calculators in this Instructor’s
Resource Book. The program should give the correct answer
for Problem 5.
9. S10 H 2.000109517, T10 H 1.983523538
The Simpson’s rule approximation is closer to the exact
answer.
Chapter 6
Exploration 6-2a
tH9
1. f(9) H
tH1
tH9
2
2
52
t 1/2 dt H 3t 3/2tH1 H 3(27 D 1) H 3
2. Graph, showing f(x) H
tHx
t 1/2 dt.
tH1
y
4
10. Answers will vary.
11. Answers will vary.
f(x)
2
Exploration 5-10b
t
1. x
f(x)
Factor
f(x) • Factor
1
2.2
1
2.2
3
1.2
4
4.8
5
1.0
2
2.0
7
2.6
4
10.4
9
5.8
1
5.8
(1/3)(2)(2.2 C 4.8 C 2.0 C 10.4 C 5.8) H 16.8
0.1x D 0.9x C 3, if 1 K x K 5
2. h(x) H
0.2x 2 D 1.6x C 4, if 5 K x K 9
3.
2
2
3. f(x) H
tHx
tH1
4
6
tHx
2
2
t 1/2 dt H 3t 3/2tH1 H 3(x 3/2 D 1)
d 2 3/2
(x
4. fQ(x) H D 1) H x 1/2
dx 3
fQ(x) is the integrand of f(x) evaluated at x!
6. h(x) H 5. g(x) H
tHx
tH3
t 3 dt ⇒ gQ(x) H x 3
tHx 3
tH2
tHx 3
cos t dt H sin ttH2 = (sin x 3 H sin 2)
d
3
3
2
2
3
7. hQ(x) H dx (sin x D sin 2) H cos x • 3x H 3x sin x
f(x)
8. The pattern is: The integrand evaluated at the upper limit of
integration, times derivative of the upper limit function (the
“inside” function), which is an example of the chain rule.
d tHsin x
tan3 (t 5) dt H tan3 (sin5 x) • cos x
dx tH0
5
x
5
10
h is continuous at x H 5 because 0.1(5)2 D 0.9(5) C 3 H
0.2(5)2 D 1.6(5) C 4.
h is not differentiable because
0.2x D 0.9, if 1 K x K 5
hQ(x) H
0.4x D 1.6, if 5 K x K 9
and 0.2(5) D 0.9 ≠ 0.4(5) D 1.6.
9
4.
1
x 8
h(x) dx H
5
1
(0.1x 2 D 0.9x C 3) dx C
9
5
(0.2x 2 D 1.6x C 4) dx
H 513 C 11175 H 16.8
9. In the statement of the fundamental theorem of calculus in
the problem, g(x) H L(x), and f(x) H 1/t, so LQ(x) H 1/x.
10. The power rule for antiderivatives works only with exponents
not equal to D1.
Exploration 6-2b
1. The power rule gives 10x0 C C, which involves division by zero.
2. gQ(x) H 1/x
The definition says g(x) H f(x) dx if and only if gQ(x) H f(x)
and gQ(x) does equal f(x).
3. ln x H
This answer agrees with Problem 1.
x
1
1
dt
t
4.
y = 1/t
5.
f(x) and h(x)
4
5
2
t
–4
–2
2
4
x
5
10
Simpson’s rule connects the points more smoothly than the
trapezoidal rule.
6. Answers will vary.
188 / Solutions for the Explorations
–2
–4
t is the variable of integration. x is the value of t at the end of
the region.
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