Embed Size (px)
Citation preview
An Overview of IntegrationS. F. EllermeyerJanuary 6, 2010
The Definite Integral of a Function f Over an Interval a,bSuppose that f is a continuous function defined on an interval a,b. The definiteintegral of f from a to b is denoted by
a
bfxdx.
Graphically, the definite integral of f from a to b is the “signed area” of the regionbounded by the graph of y fx, the x axis, and the vertical lines x a and x b.Parts of the region which lie above the x axis give a positive contribution to the integraland parts of the region which lie below the x axis give a negative contribution.
Figure 1
1
Figure 2In Figure 2, part of the graph of f on the interval a,b lies above the x axis and part liesbelow the x axis. If A and B are the areas of the regions pictured (meaning that A andB are both positive numbers, since areas can’t be negative), then
a
bfxdx A B.
Let us look at several specific examples of integrals that we can compute by usingelementary geometry.
Example 1Suppose that f is the constant function fx 4 and suppose that we want to computethe definite integral of f from x 1 to x 6. The graph of f is shown in Figure 3.
-1
0
1
2
3
4
5
y
1 2 3 4 5 6 7x
Figure 3The integral we want to compute is the area of the region shown shaded in Figure 4.
2
Figure 4The shaded region is a rectangle with area 20 so
1
6fxdx 20.
Another way to write this is
1
64dx 20.
Exercise 11. Let f be the constant function fx 8. By drawing a picture and using elementarygeometry, compute the definite integral
2
7fxdx.
2. Let f be the constant function fx 3. By drawing a picture and usingelementary geometry, compute the definite integral
4
8fxdx.
(Note: Remember that parts of the graph of y fx which lie below the x axis givea negative contribution to the integral.)
3. Find a general formula for computing
a
bfxdx
where f is the constant function fx K.
From Example 1 and Exercise 1, we see that it is very easy to compute the integral ofa constant function. Another kind of function for which it is easy to compute a definite
3
integral by using geometry is a linear function – that is, a function of the formfx mx b.
Example 2Suppose that we want to compute the definite integral of the function fx 4x 6from x 0 to x 4. The graph of f is shown in Figure 5.
-15
-10
-5
0
5
10
y
-1 1 2 3 4 5x
Figure 5The value of the integral we want to compute is
0
4fxdx A B
where A and B are the areas of the regions shown in Figure 6.
Figure 6Since both of these regions are triangles, we can use the formula for the area of atriangle
Area 12 base height
to find these areas. Since the graph of f intersects the x axis at x 3/2 (something thatwe discover by solving the equation 4x 6 0), we have
A 12
32 6 9
2
4
andB 1
252 10 25
2 .
Thus,
0
4fxdx 9
2 252 8.
We could also write this as
0
44x 6dx 8.
Exercise 21. Let f be the function fx 2x 5. Draw a nice picture (using graph paper) of thegraph of f over the interval 4,4. Then use your picture (and geometry) tocompute
2
3fxdx.
2. Use the picture you drew to compute
0
3fxdx.
3. Use the picture you drew to compute
4
3fxdx.
4. Suppose that m is a constant and suppose that f is the linear function fx mx.Find a general formula (involving a, b, and m) for computing
a
bfxdx.
5. Use the formula you found in question 4 above to compute
1
57xdx.
Verify that your answer is correct by drawing a picture of fx 7x and by thencomputing the above integral by using geometry.
Integrals which can be computed by using geometry (as in the previous two examples)are rare. In fact, linear functions and constant functions (which, in fact, are really linearfunctions with zero slope) are almost the only examples which we can give. InExample 3 which follows, we give one more special example of an integral which canbe computed by using geometry.
Example 3Suppose we want to compute the integral
1
11 x2 dx.
5
Since the integrand (the function whose integral we want to compute) isfx 1 x2 ,the graph of f is the curve y 1 x2 . If we square both sides of thisequation, we obtain
y2 1 x2
orx2 y2 1.
Thus the graph of f is the upper half of the unit circle (shown in Figure 7).
-1
-0.5
0
0.5
1
y
-1 -0.5 0.5 1x
Figure 7We recall (perhaps from high school geometry) that the area of a circle of radius r isr2. Thus the area of the unit circle is 12 . The value of the integral we want tocompute, since it is only half of the area of the unit circle, is thus
1
1fxdx
2 .
Of course, we can also write this as
1
11 x2 dx
2 .
Exercise 31. Use geometry to compute
0
11 x2 dx.
(Hint: In this and the rest of the exercises here, it will be helpful for you to drawpictures.)
2. Use geometry to compute
1
1 1 x2 dx.
3. Use geometry to compute
2
04 x2 dx.
4. Use geometry to compute
6
2
04 x2 6 dx.
5. Use geometry to compute
2
04 x2 8 dx.
6. Explain why it is not so easy to determine the exact value of
0
1/21 x2 dx.
Why Do We Want to Compute Integrals?Integrals are an extremely important tool in mathematics. Just to name a fewapplications of integrals (some of which will be studied in this course), they are used tocompute areas, volumes, arc lengths, work, and probabilities. The general idea is thatintegrals are the necessary tool for measuring accumulations of small continuouslychanging quantities. You are familiar with the concept of accumulation from youreveryday experience. An example which everybody is familiar with is the accumulationof rainfall. When you hear on the 11 p.m. news that the total rainfall in Atlanta over thepast three hours was 1.5 inches, you know that means that a total of 1.5 inches of rainhas fallen in Atlanta over the past three hours.So what does this have to do with integration? If you think about it, there are manydifferent ways in which 1.5 inches of rain could have fallen in Atlanta over the pastthree hours. One possibility is that there may have been a very steady rain falling atthe rate of 0.5 inches per hour. Another possibility is that it might have rained veryheavily at the rate of 1 inch per hour from 8 p.m. until 9:30 p.m. and then not rained atall from 9:30 p.m. until 11:00 p.m. Yet another possibility is that it rained somewhatsteadily during the three hour period with the rain being heavy at times, light at othertimes, and perhaps not raining at all for some periods of time. These three possibilitiesare illustrated in Figures 8, 9, and 10. In each figure, the function R is the rate ofrainfall at time t where t is measured in hours with t 0 standing for 8 p.m.
0
0.5
1
1.5
2
R(t)
-1 1 2 3 4t
Figure 8: Steady Rainfall of 0.5 in/hr.
7
0
0.5
1
1.5
2
R(t)
-1 1 2 3 4t
Figure 9: Hard steady rain for 1.5 hrs.
0
0.5
1
1.5
2
R(t)
-1 1 2 3 4t
Figure 10: Variable Rainfall RateIn each of the three different rainfall scenarios (Figures 8, 9, and 10), the total rainfallfor the three hour period from 8 to 11 p.m. is 1.5 inches. The reason is that the definiteintegrals of each of the different rainfall rate functions (R) over the time interval 0,3 isthe same. Even though the functions in Figures 8, 9, and 10 are obviously differentfunctions, each of them satisfies
0
3Rtdt 1.5.
This simple example leads the way to a basic principle of Calculus:Basic Principle: If f is a function that tells us the rate at which something ishappening, then
a
bfxdx
tells us the total accumulated effect of what happened as the independentvariable, x, varied from x a to x b.
The basic principle stated above is, admittedly, somewhat vague. To help you tounderstand it better, we provide some more examples.
Example 4
8
1. If Rt is the the rate at which rain is falling at time t where t is measured in hoursand t 0 stands for noon, then
2
8Rtdt
is the total amount of rain that fell between 2 and 8 p.m.2. If vt is the speed at which a cross-country runner is running at time t where t ismeasured in seconds, t t0 stands for the time at which the runner starts running,and t t1 stands for the time at which the runner finishes running, then
t0
t1vtdt
is the total distance the runner has run.3. If a spherical balloon is being inflated and if Vr is the volume of the balloon whenthe radius of the balloon is r, then
r1
r2 dVdr dr
is the total increase in volume of the balloon as a result of inflating the balloonfrom radius r1 to radius r2.
4. When a beaker of water is heated, the water molecules in the beaker begin tomove faster. This means that the kinetic energy of the water increases. If fT tellsus the rate at which the kinetic energy of the water is increasing when thetemperature of the water is T where T is measured in degrees Celsius, then
40
90fTdT
tells us the total increase in kinetic energy of the water as the water is heated from40C to 90C.
Exercise 41. For the rainfall rate function shown in Figure 8, explain how you can tell (bylooking at the graph) that
0
3Rtdt 1.5.
2. For the rainfall rate function shown in Figure 9, explain how you can tell (bylooking at the graph) that
0
3Rtdt 1.5.
3. For the rainfall rate function shown in Figure 10, why is it more difficult to tell (justby looking at the graph) that
0
3Rtdt 1.5?
4. Referring to the rainfall rate function shown in Figure 10,a. At what rate was the rain falling at 8 p.m.?b. At about what time was it raining the hardest?
9
c. At about what time did it momentarily quit raining?d. Did more rain fall between 8 and 9 p.m. or between 9 and 10 p.m.? How canyou tell?
5. Comparing the rainfall rate function in Figures 8 and 10:a. Which rainfall rate function indicates a greater total amount of rainfall between8 and 9 p.m.?
b. Which rainfall rate function indicates a greater amount of total rainfall between10 and 11 p.m.?
6. Suppose that ft is the rate at which the temperature in Atlanta is changing attime t where t is measured in hours and t 0 stands for 8 a.m. Suppose also thatthe temperature in Atlanta at 8 a.m. is 75F and the temperature at 10 p.m. on thesame day is 68F . Which of the following must be true?a. The temperature between 8 a.m. and 10 p.m. was continuously decreasing.b. There must have been at least some time period during which ft 0.
7. For the function f given in the problem above, what is the value of
0
14ftdt?
8. Referring to the same scenario as in problem 6 (above), suppose that Tt is thetemperature in Atlanta at time t. Complete the following:a. T0 ______b. T14 _______c. T14 T0 ________
9. Try to generalize what you learned by doing problems 6, 7, and 8 (above): If Ttis the temperature at a certain location at time t and if ft is the rate at which thetemperature at that location is changing at time t, and if t0 and t1 are two particulartimes (with t0 t1), then
t0
t1ftdt ________________.
10. Now try to generalize the rainfall situation: If At is the total amount (measuredin inches) of rainfall that has accumulated at time t and if Rt is the rate(measured in inches per hour) at which rain is falling at time t, and if t0 and t1 aretwo particular times (measured in hours, with t0 t1), then
t0
t1Rtdt ___________________.
11. Now see if you can do this one: If KT is the kinetic energy of a beaker of waterwhen the temperature of the water is T, and if fT is the rate at which the kineticenergy of the water is increasing when the temperature of the water is T, then
40
90fTdT ______________.
12. Now generalize completely: Suppose that F is a function with independentvariable x and suppose that f is a function that tells us the rate at which F ischanging with respect to the independent variable x. Suppose that a and b are twonumbers in the domain of F (with a b). Then
10
a
bfxdx ________________.
The Fundamental Theorem of CalculusThe informally stated “Basic Principle” of the previous section is actually a very centraltheme in Calculus. In fact, when stated more formally, it is called the FundamentalTheorem of Calculus. The reason for this name is that it points out the relationshipbetween the two most basic ideas of Calculus - differentiation and integration. You willrecall from your differential calculus course that the derivative of a function, F, at agiven point x, tells us the rate which the dependent variable is changing with respect tothe independent variable when the independent variable is x. In the previous section,you learned (hopefully) from doing the exercises that if f is a function that tells us therate of change of another function, F, and if a and b are two particular values of theindependent variable (with a b), then
a
bfxdx Fb Fa.
Recall that if F x fx for all x in the interval a,b, then f is called the derivative ofF and F is called an antiderivative of f. A fact that we will not prove in this course(although you might have studied a partial proof in your differential calculus course) isthat if f is any function that is continuous on the interval a,b, then f is guaranteed tohave an antiderivative. We include this fact as Part 1 of the Fundamental Theorem ofCalculus which is given here.
Theorem (The Fundamental Theorem of Calculus - Part 1): If the function f iscontinuous on the interval a,b, then f has an antiderivative on a,b. In otherwords, there exists a differentiable function F defined on a,b such that F x fxfor all x in a,b.Theorem (The Fundamental Theorem of Calculus - Part 2): If the function f iscontinuous on the interval a,b and if F is any antiderivative of f on a,b (which isguaranteed to exist by Part 1 of this theorem), then
a
bfxdx Fb Fa.
Example 5In Example 1, we saw by using simple geometry that if f is the function fx 4, then
1
6fxdx 20.
To compute this same integral using the Fundamental Theorem Calculus (FTC), weneed to find an antiderivative of f. That is, we need to find a function F such thatF x fx. Of course, from our differential calculus experience, we know that such afunction is Fx 4x (because if Fx 4x, then F x 4 fx). Using the FTC, weobtain
11
1
6fxdx F6 F1
46 41 20
Example 6In Example 2, we saw that
0
44x 6dx 8.
To do this same problem using the FTC, we first need to find an antiderivative offx 4x 6. From our knowledge of differential calculus, we know that the functionFx 2x2 6x is an antiderivative of f. Thus
0
44x 6dx F4 F0
8 0 8
Example 7What about the problem
1
11 x2 dx
which we did in Example 3 by using geometry? It is not as easy to find anantiderivative of the function fx 1 x2 as it was for the functions in the previoustwo examples! Later in this course, we will learn a procedure that will allow us to findan antiderivative the function fx 1 x2 . For the moment, let us just give theanswer: An antiderivative of f is
Fx 12 x 1 x
2 12 arcsinx.
Using the antiderivative which has been given, we compute
F1 12 1 1 12 12 arcsin1
4
and
F1 12 1 1 12 12 arcsin1
4 .
This gives us
1
11 x2 dx F1 F1
4 4
2
12
Please don’t worry if you don’t understand how the above antiderivative was obtained.You are not supposed to at this point. Indeed, one of the main topics we will study inthis course is how to find antiderivatives of certain classes of functions - a processwhich itself is also called integration.
Exercise 51. Use the Fundamental Theorem Calculus to evaluate the following definiteintegrals.a.
0
9 5dx
b. 0
93dx
c. a
b Kdx (where K is a constant)
d. 2
32x 5dx
e. 0
32x 5dx
f. 4
32x 5dx
g. a
b mxdx (where m is a constant)
h. a
bmx bdx (where m and b are constants)
2. Use the antiderivative given in Example 7 above to evaluate (approximately)
0
1/21 x2 dx.
Hint: Use the fact, which you should remember from your knowledge of the unitcircle, that arcsin1/2 /6.
3. Use the antiderivative given in Example 7 above to evaluate
1/2
11 x2 dx.
4. Add the two answers you obtained in problems 2 and 3 above. Verify that theanswer you get is the same as the answer you get when you compute
0
11 x2 dx.
Does it make sense to you that these answers should be the same? Explain.5. Suppose that we have a differentiable function y Fx and suppose that x1 andx2 are two specific points in the domain of F such that Fx1 y1 and Fx2 y2.Then
x1
x2 dydx dx _________________.
6. Graph the function y cosx on the interval 0,/2. Explain how you can tell, justby looking at this graph, that
0
/2cosxdx
2 .
13
(Hint: Compare the given integral to the area of a certain rectangle.)7. Use the Fundamental Theorem of Calculus to show that
0
/2cosxdx 1.
8. Explain how you can tell, just by looking at the graph of y cosx, that
0
cosxdx 0.
9. Verify that
0
cosxdx 0
by using the FTC.10. Based on your work in the past few problems, explain you know that there is
some number, x1, between /2 and , such that
0
x1cosxdx 0.686.
(Hint: You don’t need to do any calculations. Just reflect on your results from theprevious few problems.)
11. Are there are two different numbers, x1 and x2, each between /2 and such that
0
x1cosxdx 0.686 and
0
x2cosxdx 0.686?
Explain your answer.12. Draw the graph of a function, f, such that
1
4fxdx
1
6fxdx.
Some Important Properties of Definite IntegralsThus far we have defined and studied definite integrals,
a
bfxdx,
under the assumption that a b. It will be convenient to extend our definition of theintegral to include the possibility that a b and the possibility that a b. We do this asfollows:1. If a is any real number that is in the domain of the function f, then we define
a
afxdx 0.
(Note that this definition makes sense when we think of the interpretation of thedefinite integral as a signed area.)
2. If f is continuous on the interval b,a, then we define
a
bfxdx
b
afxdx.
Example 8The graph of a function, f, is shown in Figure 11.
14
Figure 11We can see by looking at the picture that the area between the graph of f and the xaxis on the interval 0,1 is smaller than the area between the graph of f and the x axison the interval 1,2. However, parts of the graph that lie above the x axis give positivecontributions and parts of the graph that lie below the x axis give negativecontributions when we compute an integral from left to right. Thus
0
1fxdx 0,
1
2fxdx 0, and
0
2fxdx 0.
According to the definition (2) given above, things are the opposite when we integratefrom right to left. In particular,
1
0fxdx 0,
2
1fxdx 0, and
2
0fxdx 0.
Of course, according to definition (1) given above, when we integrate any functionfrom some point to itself, the value of the integral is always 0. Thus, for example,
1.4
1.4fxdx 0.
Having extended our notion of the definite integral with definitions (1) and (2), we nowstate three important properties of integrals that hold in all cases (whether integrationsare being performed from left to right or from right to left).3. If the function f is continuous and if K is any constant, then
a
bKfxdx K
a
bfxdx.
4. If the functions f and g are continuous, then
a
bfx gxdx
a
bfxdx
a
bgxdx
5. If the function f is continuous, then
a
cfxdx
a
bfxdx
b
cfxdx.
Properties (3) and (4) are called the linearity properties of the definite integral.Property (5) is called the additivity property.
15
Exercise 6For the function f whose graph is shown in Figure 11, suppose that we know that
0
1fxdx 7
60 and 1
2fxdx 2360 .
Use this information and properties 1–5 given above to find the values of the followingintegrals.1.
0
2 fxdx _____
2. 1
0 fxdx _____
3. 2
1 fxdx _____
4. 0
1 3fxdx _____
5. 0
13fx 15dx _____
6. 0.38
0.38 fxdx _____
7. 2
0 fxdx _____
Exercise 7The graph of a function f is shown in Figure 12.
f
-1.0 -0.5 0.5 1.0 1.5 2.0
-1.5
-1.0
-0.5
0.5
Firgure 12Suppose that we know that
1
0fxdx 97105 ,
1
2fxdx 43
105 , and 1
2fxdx 2735 .
Find the values of the following integrals.1.
0
1 fxdx _____
2. 2
1 fxdx _____
3. 2
0 fxdx _____
16
4. 0
1 105fxdx _____
5. 1
212 35fxdx _____
Indefinite IntegralsThe term “indefinite integral” is often used to denote the set of all antiderivatives of
a function, f, on some interval a,b. The reason for the use of this term is thatintegration is closely related to antidifferentiation (by the FTC). We use the notation
a,b
fxdx
to refer to the indefinite integral of the function f on the interval a,b – meaning the setof all functions that are antiderivatives of f on the interval a,b. The Mean ValueTheorem can be used to prove that if a function, F, is an antiderivative of the function fon a certain interval a,b, then any antiderivative, G, of f on a,b must be of the formGx Fx C where C can be any constant. Thus, for example, since we know thatthe function Fx 1
2 x2 is an antiderivative of the function fx x on the interval
,, then we know that any antiderivative, G, of f must be of the formGx 1
2 x2 C (where C can be any constant). We describe the indefinite integral of f
as
,
xdx 12 x
2 C.
In most cases, it is not necessary to include the interval a,b in the notation for theindefinite integral because the antiderivative does not depend on the particular intervalin question. For example, it is true that
a,b
xdx 12 x
2 C
no matter what interval a,b we are considering. Thus, we can simply write
xdx 12 x
2 C.
However, there are some exceptions – one in particular that occurs frequently enoughthat we should make note of it: The function Fx lnx with domain 0, hasderivative
F x 1x
and the function Fx lnx with domain , 0 has the same derivative:F x 1
x .Hence, the statement
a,b
1x dx lnx C
is correct only if the interval a,b is a subset of the interval 0,; whereas, if theinterval a,b is a subset of the interval , 0, then the statement
a,b
1x dx lnx C
17
is correct.A commonly–used way of getting around this difficulty is to observe that the functionFx ln|x| with domain , 0 0, has derivative
F x 1x .
Thus, the statement
a,b
1x dx ln|x| C
is correct both when the interval a,b is a subset of 0, and when a,b is a subsetof , 0. Hence, we are always safe in using the formula
1x dx ln|x| C.
Example 9Suppose we wish to evaluate the integral
4
2 1x dx.
There are two ways to go about this: The first is to observe that since 4,2 is asubset of , 0, then
4,2
1x dx lnx C.
Therefore, by the FTC,
4
2 1x dx ln2 ln4 ln2 ln4.
The other way to approach this problem is to simply use the indefinite integral
1x dx ln|x| C.Using this formula, we obtain
4
2 1x dx ln|2| ln|4| ln2 ln4.
Some Basic IntegralsThe following indefinite integrals (which come from basic differentiation formulas fromdifferential calculus) should be memorized. You will need to know them as we studyintegration techniques that will allow us to compute “less basic” integrals such as 1 x2 dx.1. If K is any constant, then Kdx Kx C.2. If n is any constant except n 1, then xn dx 1
n1 xn1 C.
3. 1x dx ln|x| C.4. If a is any positive constant except a 1, then ax dx 1
lna ax C.
5. A special case of formula 4 (the nicest case) is when we use base e. In this casethe formula is
18
ex dx 1lne e
x C
or simply
ex dx ex C.6. cosxdx sinx C.7. sinxdx cosx C.8. sec2xdx tanx C.9. csc2xdx cotx C.10. secx tanxdx secx C.11. cscxcotxdx cscx C.12. 1
1x2dx arcsinx C.
13. 11x2
dx arctanx C.The last two integrals in this basic list involve inverse trigonometric functions. Sincethese two are usually the least familiar to students who have completed a differentialcalculus course, we will give a derivation of formula number 13. (The derivation offormula 12 is left as a homework exercise.)First, we recall the definition of the inverse tangent function, which we denote by“arctan”. (Another notation for the inverse tangent function is “tan1”.) The statement
y arctanxmeans that
tany xand that
2 y
2 .
(Note that for any given real number, x, there is exactly one real number y in theinterval
2 ,2 such that tany x. This is what makes the inverse tangent function
be a well–defined function.)To find the derivative with respect to x of the function
y arctanx,we use implicit differentiation. In particular, we begin with the equivalent formula
tany xand differentiate both sides of this formula with respect to x to obtain
ddx tany
ddx x.
Recalling that y is a function of x (and hence that the Chain Rule must be used indifferentiating the left hand side of the above equation), we obtain
sec2y dydx 1
or
19
dydx 1
sec2y.
By a familiar trigonometric identity, we now havedydx 1
1 tan2y,
and since tany x, we can write the above formula asdydx 1
1 x2.
This proves thatddx arctanx
11 x2
and also proves the integration formula 13 in the above list of basic integrals.
Exercise 81. Prove integration formula number 12 in the above list. To do this, you will need torecall the definition of the inverse sine function: When we say that
y arcsinx,we mean that
siny xand that
2 y
2 .
2. The inverse cosine function is defined as follows: When we say thaty arccosx,
we mean thatcosy x
and that0 y .
Find the derivative function of the inverse cosine function. Then find the indefiniteintegral that corresponds to your result. Why was it not really necessary to includethis indefinite integral in the basic list of indefinite integrals given above?
3. Evaluate the following definite integrals. To do each, you can use theFundamental Theorem of Calculus and one of the indefinite integral formulas fromthe above basic list. These problems are designed so that you should be able todo all of them without using a calculator, although you will need to use a calculatorif you want to get a numerical (decimal number) answer on some of them.
20
a) /3
17/3 cosxdx
b) 0
/4 secx tanxdx
c) 3
3 11x2
dx
d) /2
/2 sinxdx
e) 0
1/2 11x2
dx
f) /6
/3 cscxcotxdx
g) 1
e1000 1x dx
h) 5
2 1x dx
i) /3
/4 sec2xdx
j) 5
02dx
k) e
e2 4x dx
l) 0
20 ex dx
m) /4
3/4 csc2xdx
n) 3
2 x4 dx
21
