Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 74 pps

Given an “amount” function, we can derive a “rate” function; we now shift our viewpoint and investigate the problem of how to recover an “amount” function when given a “rate” function.1I

VIII Integration: An Introduction

22

Net Change in Amount and Area: Introducing the

Definite Integral

GRAPHICAL INTERPLAY

Introduction

The derivative allows us to answer two related problems

How do we calculate the instantaneous rate of change of a quantity?

How do we calculate the slope of the line tangent to a curve at a point?

The physical and graphical questions are intertwined; the slope of the graph of an “amount” function can be interpreted as the instantaneous rate of change of the function

Given an “amount” function, we can derive a “rate” function; we now shift our viewpoint and investigate the problem of how to recover an “amount” function when given a

“rate” function.1If we know the rate of change of a quantity, how can we find the net change

in the quantity over a certain time period? Suppose, for example, that we know an object’s velocity over a specified time interval Then we ought to be able to use that information to

1 We did this when we looked at projectile motion in Section 20.7.

711

determine the object’s net change in position during that time To figure out how to do this

we begin by looking at some simple examples where the rate of change is constant and then apply what we learn to cases where the rate of change is not constant

Let’s clarify what is meant by net change If you take two steps forward and one step

back, your net change in position is one step forward If, over the course of a day, the stock market falls 100 points and then gains 40 points, the net change for the day is −60 points

We now prepare to tackle the following two questions

Given a rate function, how do we calculate the net change in amount?

How do we calculate the area under the graph of a function?

In this chapter we aim to convince you that the physical and graphical questions are closely related We’ll approach the questions using the strategy that served us so well in developing the derivative: the method of successive approximation followed by a limiting process

Calculating Net Change When Rate of Change Is Constant

EXAMPLE 22.1 Suppose that a moose is strolling along at a constant velocity The distance she travels can

be calculated by the familiar formula distance = (rate) · (time)

If the moose moves at 4 miles per hour between 1:15 p.m and 1:30 p.m., the distance she has traveled (or her net change in position) is

4 miles hour ·1

4 hours = 1 mile, where the units cancel to give us an answer in the units we expect

More generally, let s(t) be the position of an object at time t and suppose that its velocity,dsdt, is constant between times t = a and t = b.

net change in position = (rate of change of position) · (time elapsed)

tdenotes the change in time, t = (b − a), and s = s(b) − s(a)

We can represent the net change in position graphically

∆t

t

a b

ds dt

Figure 22.1

When velocity is constant, net change in position is calculated by multiplying rate of change,dsdt, by time; geometrically, this corresponds to the area of the shaded rectangle in Figure 22.1

EXAMPLE 22.2 Let W (t) be the amount of water in a swimming pool as a function of time, where time is

measured in hours Suppose the pool is being filled at a constant rate ofdW gallons per hour

between noon and 12:30 p.m In that half-hour, the net change in amount of water = rate at which water is entering · (time elapsed)

where t is one-half of an hour

Graphically, this net change can be represented as the area of the rectangle shaded in Figure 22.2

t

dW dt

noon 12:30

Figure 22.2

EXAMPLE 22.3 Let C(x) be the cost (in dollars) of producing x grams of feta cheese Suppose that for each

additional gram produced the cost increases by a constant amount; dCdx is constant.2Then,

the net change

in production cost = of cost per gramrate of change · of grams producedchange in number

Notice that the independent variable (in this case x, the number of grams of cheese produced) does not have to represent time

x

dC

dx ∆x

a b

Figure 22.3

EXAMPLE 22.4 On a certain day, the temperature in Kathmandu decreases at a constant rate of two degrees

per hour from 6:00 p.m to 10:00 p.m If we let T(t) be the temperature at time t, we can write the following

net change in temperature = rate of change of temperature · (time elapsed)

T = (−2) degreeshour · (10 − 6) hours

= −8 degrees

2

When we represent this geometrically, we have a rectangle under the t-axis We’d like

to convey the information that T is negative by attaching a sign to the area The net change

in temperature is negative because the temperature is decreasing; dTdt is negative We can say that the “height” of the rectangle is −2 (since it lies below the t-axis) and assign the rectangle a signed area of −8

t

dT dt

–2

6 10

Figure 22.4

When using areas to represent net changes, we will use the idea of “signed area”—areas are positive or negative, depending on whether the region lies above or below the horizontal axis

To illustrate further how signed area is used, suppose that we also know that the temperature in Kathmandu on this particular day increased at a rate of 1.5 degrees per hour between 3:00 p.m and 6:00 p.m

T from 3:00 p.m to 6:00 p.m = 1.5 degreeshour · (6–3) hours

= 4.5 degrees Putting the two time periods together gives a net change of 4.5◦+ (−8◦) = −3.5◦between 3:00 p.m and 10:00 p.m Graphically, this is represented by one rectangle above the t-axis (whendTdt is positive) and one rectangle below the t-axis (whendTdt is negative)

t

dT dt

6

Figure 22.5

REMARK: Although we know that the temperature dropped 3.5 degrees between 3:00 p.m and 10:00 p.m., we have no idea of what the temperature actually was at any time We need

to know one specific data point for the function T(t) before we can determine any actual values of T For example, if we knew that at 3:00 p.m the temperature was 55 degrees, then

we would know that T (10) = 55 − 3.5 = 51.5 degrees In fact, we could calculate T(t) for any time between 3:00 p.m and 10:00 p.m On the other hand, if the temperature at 3:00 p.m was 40 degrees, then T (10) = 40 − 3.5 = 36.5

As illustrated by the preceding examples, finding a quantity’s net change is straightfor-ward provided that its rate of change is constant However, the rate of change often is not constant How can we approach these situations?

Approximating Net Change When Rate of Change is Not Constant

EXAMPLE 22.5a Suppose that a cheetah’s velocity is given by v(t) = 2t + 5 meters per second on the interval

[1, 4] Approximate its net change in position over this interval by giving upper and lower bounds that differ by less than 0.1 meter

(4, 13)

(1, 7)

v (t)

t

Figure 22.6

SOLUTION The problem we face is analogous to the one confronting us when we first set out to find

the slope of a curve At that time, we knew how to find the slope of a straight line, but not the slope of a curve, so we approximated the slope of a curve at x = a by the slope of a secant line through points (a, f (a)) and (a + x, f (a + x)) To get successively better approximations, we repeatedly shortened the interval x Then, to obtain an exact value for the slope at x = a, we evaluated the limit as x approached zero

To tackle the present problem, we will follow a similar procedure, that of successive approximations If r(t) is a nonconstant rate of change on an interval [a, b], we divide the interval into smaller subintervals estimate the net change within each of those subintervals, and sum to find the accumulated change On each of these subintervals we approximate r(t)

by a constant rate of change

The Method of Successive Approximations

Take One The function v(t) is increasing throughout the interval [1, 4], so the velocity is

at least v(1) = 7 m/sec and at most v(4) = 13 m/sec

7 ≤ v(t) ≤ 13, for all t in the interval [1, 4]

If the cheetah were moving at a constant rate, then its net change in position would be given

by (rate) · (time); therefore we know the following

(7m/sec) · (3 sec) ≤ net change in position ≤ (13 m/sec)(3 sec)

21 meters ≤ net change in position ≤ 39 meters Therefore, the cheetah moved between 21 and 39 meters

v (t)

t

7 13

1 4

Figure 22.7

How can we improve on these bounds?

Take Two To get a better approximation, let’s divide the interval [1, 4] into smaller

subintervals and approximate the rate of change by a constant function on each subinterval

For the sake of convenience, let’s divide the interval into three subintervals of equal width, [1, 2], [2, 3], and [3, 4], and call the width of each subinterval t.3The table below shows the velocity of the cheetah at the endpoints of each of these subintervals

1 2 3 4

∆t ∆t ∆t

v (t)

t

7 13

1 2 3 4

Figure 22.8

Because the velocity is increasing on the interval [1, 4], the velocity is least at the beginning of each subinterval (i.e., at the left endpoint) and greatest at the end of each subinterval (the right endpoint) During the time interval [1, 2] the velocity is greater than

or equal to v(1) = 7 m/sec and less than or equal to v(2) = 9 m/sec Again using the fact that if rate is constant, (rate) · (time) = net change, we deduce the following

v(1) · (1 sec) = (7 m/sec) · (1 sec) = 7 m is a lower bound for the distance traveled in

the interval [1, 2]

v(2) · (1 sec) = (9 m/sec) · (1 sec) = 9 m is an upper bound for the distance traveled

in the interval [1, 2]

We can refer to the lower bounds and upper bounds as “underestimates,” and “overesti-mates,” respectively

We use the same method to get lower and upper bounds for the distance the object travels in each of the other subintervals Adding together the lower bounds for the three subintervals will give us a lower bound for the net change in the cheetah’s position over the entire interval [1, 4]; similarly, summing the upper bounds for the subintervals will give an upper bound for the whole interval

lower bound for the net change in position on [1, 4]

v(1) · 1 + v(2) · 1 + v(3) · 1 = 7 · 1 + 9 · 1 + 11 · 1 =

27 meters upper bound for the net

change in position on [1, 4]

v(2) · 1 + v(3) · 1 + v(4) · 1 = 9 · 1 + 11 · 1 + 13 · 1 =

33 meters The net change in the cheetah’s position on [1, 4] is more than 27 meters and less than

33 meters

3 Here t = 1, but we will soon subdivide further, and as we do so t will shrink.

v (t)

t

7

9

11

1 2 3 4 lower bound (left-hand sum)

v (t)

t

9 11 13

1 2 3 4 upper bound (right-hand sum)

Figure 22.9

When we approximate the value of a function over each subinterval by using the value

of the function at the left endpoint of each subinterval, then multiply by the length of the

subinterval and sum the results, we call the sum a left-hand sum It is denoted by Ln, where

ndenotes the number of subintervals When we use the right endpoints of each subinterval,

we call the sum a right-hand sum and denote it by Rn

In this example, because the function v(t) is increasing, the left-hand sum will be a lower bound and the right-hand sum will be an upper bound In the graphical representation

in Figure 22.9, the rectangles corresponding to the lower bound are inscribed rectangles and the rectangles corresponding to the upper bound are circumscribed rectangles

Take Three We have now improved upon our first estimates and determined that the

cheetah’s net change in position is between 27 and 33 meters To improve further on these estimates we can use even smaller subintervals This time, let’s divide the interval [1, 4] into six subintervals, each of length t =36 = 0.5 seconds Then our estimates are the following.4

lower bound = L6= v(1.0)t + v(1.5)t + v(2.0)t + v(2.5)t + v(3.0)t + v(3.5)t

= 7(0.5) + 8(0.5) + 9(0.5) + 10(0.5) + 11(0.5) + 12(0.5)

= 28.5 meters upper bound = R6= v(1.5)t + v(2.0)t + v(2.5)t + v(3.0)t + v(3.5)t + v(4.0)t

= 8(0.5) + 9(0.5) + 10(0.5) + 11(0.5) + 12(0.5) + 13(0.5)

= 31.5 meters These bounds are closer together than in our previous estimates

v (t)

t

v (t)

t

1 2 3 4 1 2 3 4 lower bound (left-hand sum) upper bound (right-hand sum)

Figure 22.10

4 These kinds of tedious sums are the sorts of tasks for which computers or programmable calculators are perfect See if you can figure out how to construct such a program Then learn how to use the technology available to you in order to compute

left-Notice that as we use more subdivisions, the graphical representation of the lower and upper bounds as the sum of areas of inscribed and circumscribed rectangles, respectively,

is getting closer to the exact area under the rate of change function; as n increases the left-hand sum approaches the area under v(t) from below and the right-left-hand sum approaches the area from above This might lead us to conjecture that the exact value of the net change

is the area under the curve

Take Four Suppose we improve upon our estimates by partioning the interval [1, 4] into

300 subintervals, each of length t =3003 =1001 seconds We need a convenient system for labeling the endpoints of these 300 subintervals We label as indicated below

1

∆t

∆t =

t0

t1t2 t101

t299

1 100

t100

t200

t300

.

The subintervals are [t0, t1], [t1, t2], , [t299, t300], where tk= 1 + k(t) = 1 + 100k

for k = 0, 1, , 300 Using this notation we have the following

lower bound = L300= v(t0)t + v(t1)t + · · · + v(t299)t

=

299



i=0

v(ti)t

= 29.97 upper bound = R300= v(ti), t + v(t2)t + · · · + v(t300)t

=

300



i=1

v(ti)t

= 30.03 These bounds differ by less than 0.1 meter

The Difference Between Left- and Right-Hand Sums

EXAMPLE 22.5b How many subdivisions were actually necessary in order to find the cheetah’s net change in

position with the specified degree of accuracy? Can we give an exact answer to the question

of the cheetah’s net change in position?

SOLUTION Let’s compare the terms of the left- and right-hand sums in Example 22.5a

3 subdivisions:

L3= v(1) · 1 + v(2) · 1 + v(3) · 1

R3= v(2) · 1 + v(3) · 1 + v(4) · 1

6 subdivisions:

L6= v(1)(.5) + v(1.5)(.5) + v(2)(.5) + v(2.5)(.5) + v(3)(.5) + v(3.5)(.5)

R6= v(1.5)(.5) + v(2)(.5) + v(2.5)(.5) + v(3)(.5) + v(3.5)(.5) + v(4)(.5)

300 subdivisions:

L300= v(t0)t +

299



i=1

v(ti)t

= v(1)t +

299



i=1

v(ti)t

R300=

299



i=1

v(ti)t + v(t300)t

=

299



i=1

v(ti)t + v(4)t

In each case, except for the first term in the left-hand sum and the last term in the right-hand sum, all the terms are shared by both Thus, the difference between the two estimates

is merely the difference between these two terms

3 subdivisions: R3− L3= v(4) · 1 − v(1) · 1 = 13 − 7 = 6

6 subdivisions: R6− L6= v(4) · (.5) − v(1) · (.5) = (.5)[v(4) − v(1)] = 5(13 − 7) = 3

300 subdivisions: R300− L300= v(4)t − v(1)t

= tv(4) − v(1) = 1

100[13 − 7] = 6

100 = 0.06

No matter how many subintervals we use, the right endpoint of the first subinterval will

be the left endpoint of the second, the right endpoint of the second interval will be the left endpoint of the third, and so on In Example 22.5, if we use n subintervals, then the width

of each subinterval is t =3n, but the values of the velocity function at the endpoints of the interval remain the same Because v(t) is increasing on [1, 4], Rnalways gives an upper bound while Lnalways gives a lower bound

The difference between right- and left-hand sums for n subdivisions is given by

Rn− Ln= v(4) · t − v(1) · t

= t[v(4) − v(1)]

=3n(13 − 7)

=18 n

For Rn − Ln to be less than 0.1 we must have18n <101, or n > 180

If we let the number of subdivisions grow without bound, then the difference between the upper and lower bounds, 18n, will approach zero

This has a nice graphical representation (See Figure 22.11.) On each subinterval, the difference between the rectangles from Rnand Lnis a small “difference” rectangle Sliding all these difference rectangles over to one side forms a total difference rectangle with height v(4) − v(1) = 13 − 7 = 6 and width t =3n As t goes to zero, the area of this rectangle representing the difference between the R and L must also go to zero

v (t)

t

v (t)

t

13

7

6

(4, 13)

(1, 6)

∆t

Difference rectangles

Figure 22.11

The rectangles for the upper bound lie above the velocity curve and the rectangles for the lower bound lie below the velocity curve If the difference between the two bounds approaches zero, the area of each set of rectangles must be approaching the area under the velocity curve

To summarize, because v(t) is increasing, Rn(the right-hand sum) will always give an upper bound for the cheetah’s net change in position and Ln(the left-hand sum) will always give a lower bound

Ln<net change in position < Rn

Similarly, because v(t) is increasing

Ln<the area under v(t) on [1, 4] < Rn But lim

n→∞(Rn− Ln) = 0, so lim

n→∞Rn= lim

n→∞Ln= the area under v(t) on [1, 4]

by the Squeeze Theorem This means that our conjecture is true To compute the net change exactly, we need to find the area under the curve v(t) between t = 1 and t = 4 In this particular example this region

is a trapezoid so we can compute its area easily

area =1

2(height1+ height2) · (base)

=1

2(7 + 13) · (3)

= 30 meters

v (t)

t

13 7

3

Figure 22.12

The cheetah’s net change in position over the time interval [1, 4] is 30 meters

REMARK: In this example the function v(t) is linear and therefore the exact area under v(t) was midway between Ln and Rn Draw a picture to convince yourself that for a nonlinear function this is generally not true