Chapter 5 Key Concept: The Definite Integral Calculus, 6th edition, Hughes-Hallett et.al., Copyright...

Chapter 5 Key Concept: The Definite Integral

Calculus, 6th edition, Hughes-Hallett et.al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Section 5.1 How Do We Measure Distance Traveled?

Section 5.2 The Definite Integral

Section 5.3 The Fundamental Theorem and Interpretations

Section 5.4 Theorems About Definite Integrals


Section 5.1

How Do We Measure Distance Traveled?


A car is moving with increasing velocity. Table 5.1 shows the velocity every two seconds:

We might estimate the distance traveled by assuming a constant velocity in each interval and use the formula

distance = velocity × timeAt least how far has the car traveled?

20 · 2 + 30 · 2 + 38 · 2 + 44 · 2 + 48 · 2 = 360 feet.At most how far has the car traveled?ining the data

30 · 2 + 38 · 2 + 44 · 2 + 48 · 2 + 50 · 2 = 420 feet.1:

Estimating the Distance a Car Travels


How Do We Improve Our Estimate?

Table 5.2 Velocity of car every second

Time (sec) 0 1 2 3 4 5 6 7 8 9 10Speed (ft/sec) 20 26 30 34 38 41 44 46 48 49 50

New lower estimate = 20 · 1 + 26 · 1 + 30 · 1 + 34 · 1 + 38 · 1 + 41 · 1 + 44 · 1 + 46 · 1 + 48 · 1 + 49 · 1= 376 feet > 360 feet

New upper estimate = 26 · 1 + 30 · 1 + 34 · 1 + 38 · 1 + 41 · 1 + 44 · 1 + 46 · 1 + 48 · 1 + 49 · 1 + 50 · 1= 406 feet < 420 feet

The difference between upper and lower estimates is now 30 feet, half of what it was before. By halving the interval of measurement, we have halved the difference between the upper and lower estimates.


Visualizing Distance on the Velocity Graph: Two-Second Data

To visualize the difference between the two estimates, look at Figure 5.1 and imagine the light rectangles all pushed to the right and stacked on top of each other, giving a difference of 30×2=60.

Figure 5.1: Velocity measured every 2 seconds


Visualizing Distance on the Velocity Graph: One-Second Data

To visualize the difference between the two estimates, look at Figure 5.2. This difference can be calculated by stacking the light rectangles vertically, giving a rectangle of the same height as before but of half the width. Its area is therefore half what it was before. Again, the height of this stack is 30, but its width is now 1, giving a difference 30.

Figure 5.2: Velocity measured every second


Figure 5.3:Velocitymeasured every½ second

If the velocity is positive, the total distance traveled is the area under the velocity curve.

Figure 5.4:Velocity measured every¼ second

Figure 5.5:Distance traveled is areaunder curve


Left- and Right-Hand Sums

Figure 5.8: Left-hand sums Figure 5.9: Right-hand sums

If f is an increasing function, as in Figures 5.8 and 5.9, the left-hand sum is an underestimate and the right-hand sum is an overestimate of the total distance traveled. If f is decreasing, as in Figure 5.10 (next slide), then the roles of the two sums are reversed.


For either increasing or decreasing velocity functions, the exact value of the distance traveled lies somewhere between the two estimates. Thus, the accuracy of our estimate depends on how close these two sums are. For a function which is increasing throughout or decreasing throughout the interval [a, b]: Difference between upper and lower estimates = |f(b)-f(a)|Δt.

Left- and Right-Hand Sums

Figure 5.10: Left and right sums if f is decreasing


Section 5.2

The Definite Integral


Suppose f(t) is a continuous function for a ≤ t ≤ b. We divide the interval from a to b into n equal subdivisions, and we call the width of an individual subdivision Δt, so

Let t0, t1, t2, . . . , tn be endpoints of the subdivisions. Both the left-hand and right-hand sums can be written more compactly using sigma, or summation, notation. The symbol Σ is a capital sigma, or Greek letter “S.” We write

The Σ tells us to add terms of the form f(ti) Δt. The “i = 1” at the base of the sigma sign tells us to start at i = 1, and the “n” at the top tells us to stop at i = n. In the left-hand sum we start at i = 0 and stop at i = n − 1, so we write

Sigma Notation

n

abt

n

iin ttfttfttfttf

121 )()()()( sum hand-Right

1

0120 )()()()( sum hand-Left

n

iin ttfttfttfttf


Taking the Limit to Obtain the Definite Integral

Suppose f is continuous for a ≤ t ≤ b. The definite integral of f from a to b, written

is the limit of the left-hand or right-hand sums with n subdivisions of a ≤ t ≤ b as n gets arbitrarily large. In other words,

and

Each of these sums is called a Riemann sum, f is called the integrand, and a and b are called the limits of integration.

1

0

)(limsum) hand-(Leftlim)(n

ii

nn

b

attfdttf

,)(b

adttf

.)(limsum) hand-(Rightlim)(1

n

ii

nn

b

attfdttf

Figure 5.21: Approximating

with n = 10


Computing a Definite Integral

Figure 5.20: Approximating

with n = 2

Figure 5.22: Shaded area is exact value of

2

1

1dtt

2

1

1dtt

2

1

1dtt

When n = 250, a calculator or computer gives 0.6921 < < 0.6941.

So, to two decimal places, we can say that

The exact value is known to be

2

1

1dtt

.69.012

1 dt

t5.22. Figure See 693147.02ln

12

1 dt

t


Figure 5.23: Area of rectanglesapproximating the area under the curve

Figure 5.24: Shaded area is the definite

integral

The Definite Integral as an Area


When f (x) Is Not Positive

When f (x) is positive for some x values and negative for others, and a < b:

is the sum of areas above the x-axis, counted positively, and areas below the x-axis, counted negatively.

.)(b

adxxf

Figure 5.26: Integral

is negative of shaded area

Figure 5.27: Integral dxx 1

1

2 1

21

2

0

2sin AAdxx


More General Riemann SumsA general Riemann sum for f on the interval [a, b] is a sum of the form

where a = t0 < t1 < · · · < tn = b, and, for i = 1, . . . , n, Δti = ti − ti−1, and ti−1 ≤ ci ≤ ti

,)(1

i

n

ii tcf

Figure 5.28: A general Riemann sum approximating b

adttf )(


Section 5.3

The Fundamental Theorem and Interpretations


The Fundamental Theorem of CalculusTheorem 5.1: The Fundamental Theorem of CalculusIf f is continuous on the interval [a, b] and f(t) = F′(t), then

b

aaFbFdttf )()()(

F(b) − F(a) = Total change in F(t) between t = a and t = b

=

In words, the definite integral of a rate of change gives the total change.

b

adttF )('

Since the terms being added up are products of the form “f(x) times a difference in x,” the unit of measurement for is the product of the units for f(x) and the units for x.

b

adttf )(


The Definite Integral of a Rate of Change: Applications of the Fundamental Theorem

Example 2 Let F(t) represent a bacteria population which is 5 million at time t = 0. After t hours, the population is growing at an instantaneous rate of 2t million bacteria per hour. Estimate the total increase in the bacteria population during the first hour, and the population at t = 1.

Solution Since the rate at which the population is growing is F (′ t) = 2t, we have

Change in population = F(1) − F(0) = Using a calculator to evaluate the integral,

Change in population =Since F(0) = 5, the population at t = 1 is given by

Population = F(1) = F(0) +

1

02 dtt

bacteriamillion 44.121

0 dtt

bacteriamillion 44.644.1521

0 dtt


Calculating Definite Integrals: Computational Use of the Fundamental Theorem

Example 5 Compute by two different methods.

Solution Using left- and right-hand sums, we can approximate this integral as accurately as we want. With n = 100, for example, the left-sum is 7.96 and the right sum is 8.04. Using n = 500 we learn

The Fundamental Theorem, on the other hand, allows us to compute the integral exactly. We take f(x) = 2x. We know that if F(x) = x2, then F (′ x) = 2x. So we use f(x) = 2x and F(x) = x2 and obtain

Notice that to use the Fundamental Theorem to calculate a definite integral, we need to know the antiderivative, F. Chapter 6 discusses how antiderivatives are computed.

3

12 dxx

008.82992.73

1 dxx

.813)1()3(2 223

1 FFdxx


Section 5.4

Theorems About Definite Integrals


Properties of the Definite Integral

Theorem 5.2: Properties of Limits of IntegrationIf a, b, and c are any numbers and f is a continuous function, then

In words:1. The integral from b to a is the negative of the integral

from a to b.2. The integral from a to c plus the integral from c to b is the

integral from a to b. (This property holds for all numbers a, b, and c, not just for those satisfying a < c < b.)

b

a

b

c

c

a

a

b

b

a

dxxfdxxfdxxf

dxxfdxxf

)()()(.2

)()(.1


Properties of the Definite Integral

Theorem 5.3: Properties of Sums and Constant Multiples of the IntegrandLet f and g be continuous functions and let c be a constant.

In words:1. The integral of the sum (or difference) of two functions is the sum (or difference) of their integrals.2. The integral of a constant times a function is that constant times the integral of the function.

b

a

b

a

b

a

b

a

b

a

dxxfcdxxfc

dxxdxxfdxxxf

)()(.2

)()()()(.1 gg


Area Between Curves

If the graph of f(x) lies above the graph of g(x) for a ≤ x ≤ b, then

Area between f and g for a ≤ x ≤ b

b

adxxxf )()( g

Example 3 Find the area of the shaded region between two parabolas in figure 5.57 to the right.

SolutionThe points of intersection must be determined first. Equating f(x) to g(x) and solving for x gives x = 1 and 3. Then, applying Theorem 5.4:

3

1

23

1

22 667.2682)54()14( dxxxdxxxxx


Using Symmetry to Evaluate Integrals

If f is even, then

If g is odd, then 0)(

)(2)(0

dxx

dxxfdxxf

a

a

aa

a

g

Figure 5.58: For an even function, Figure 5.58: For an odd function,

0)()(2)(0

dxxdxxfdxxf

a

a

aa

ag


Comparing IntegralsTheorem 5.4: Comparison of Definite IntegralsLet f and g be continuous functions.1. If m ≤ f(x) ≤ M for a ≤ x ≤ b, then

2. If f(x) ≤ g(x) for a ≤ x ≤ b, then

)()()( abMdxxfabmb

a

Figure 5.62: The area under the graph of Figure 5.63: If f(x) ≤ g(x) then f lies between the areas of the rectangles

b

a

b

adxxdxxf )()( g

b

a

b

adxxdxxf )()( g

Average value of f

from a to b


The Definite Integral as an Average

b

a)(

1dxxf

ab

How to Visualize the Average on a GraphThe definition of average value tells us that

(Average value of f) · (b − a) b

a)( dxxf

Figure 5.65: Area and average value


The Definite Integral as an Average

Example 6 Suppose that C(t) represents the daily cost of heating your house, measured in dollars per day, where t is time measured in days and t = 0 corresponds to January 1, 2008. Interpret

Solution The units for the first expression are (dollars/day) × (days) = dollars. The integral represents the total cost in dollars to heat your house for the first 90 days of 2008, namely the months of January, February, and March. The second expression is measured in (1/days)(dollars) or dollars per day, the same units as C(t). It represents the average cost per day to heat your house during the first 90 days of 2008.

90

0

90

0(

090

1 and ( t)dtCt)dtC

Chapter 5 Key Concept: The Definite Integral Calculus, 6th edition, Hughes-Hallett et.al., Copyright...

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Transcript of Chapter 5 Key Concept: The Definite Integral Calculus, 6th edition, Hughes-Hallett et.al., Copyright...