3.3 maximum and minimum values of a...

MA111 (Semester 2/2011): Prepared by Dr. Archara Pacheenburawana 33

3.3 Maximum and Minimum Values of a Function

Some of the most important applications of differential calculus areoptimization problems, inwhich we are required to find the optimal (best) way of doing something. Here are examples ofsuch problems:

• A farmer wants to choose the mix of crops that is likely to produce the largest profit.

• A doctor wishes to select the smallest dosage of a drug that will cure certain disease.

• The manufacturer would like to minimize the cost of distributing its products.

These problems can be reduced to finding the maximum or minimum values of a function. Let’sfirst explain exactly what we mean by the maximum and minimum values.

Definition 3.1 A function f has anabsolute maximum(or global maximum) at c if f(c)≥ f (x)for all x in D, where D is the domain of f . The number f(c) is called themaximum valueof f onD. Similarly, f has anabsolute minimumat c if f(c) ≤ f (x) for all x in D and the number f(c)is called theminimum valueof f on D. The maximum and minimum values of f are called theextreme valuesof f .

Figure 3.1 shows the graph of a functionf with absolute maximum atb and absolute minimumate. Note that

(b, f (b)

)is the highest point on the graph and

(e, f (e)

)is the lowest point.

x

y

a b c d e

b

b

(b, f (b)

)

(e, f (e)

)

Figure 3.1: Maximum valuef (b), minimum valuef (e)

In general, there is no guarantee that a function will actually have an absolute maximum orminimum on the given interval.

c

f (c)

y

x

b

y= f (x)

Minimum value atc, no maximum

c

f (c)

y

x

by= f (x)

Maximum value atc, no minimum


The following theorem gives conditions under which a function is guaranteed to posses ex-treme values

Theorem 3.1 (Extreme Value Theorem)

If f is continuous on a closed interval[a,b], then f attains an absolute maximum value f(c) andan minimum value f(d) at some numbers c and d in[a,b].

The Extreme Value Theorem is illustrated in the following Figure. Note that an extreme valuecan be taken on more than once.

y

x

b

b

| |

a c d b

y

x

b

b

|

a c d = b

y

x

b

b

| |

a c1 d c2 b

The following Figure show that a function need not be possessextreme values if either hy-pothesis (continuity or closed interval) is omitted from the Extreme Value Theorem.

y

x

fb

b

bc

b|

+

0 2

1

3

This function has minimum value

f (2) = 0, but no maximum value.

y

x

g

1

1 bc

0

This continuous functiong hasno maximum and minimum.

The Extreme Value Theorem says that a continuous function ona closed interval has a max-imum value and a minimum value, but it does not tell us how to find these extreme values. Westart by looking for local extreme values.

Definition 3.2 A function f has alocal maximum (or relative maximum) at c if f(c) ≥ f (x)when x is near c [This means that f(c) ≥ f (x) for all x in some open interval containing c.]Similarly, f has alocal minimum at c if f(c) ≤ f (x) when x is near c. If f has either a localmaximum or a local minimum at c, then f is said to have alocal extreme valuesat c.


y

xab

cd

b

b

b

b

Local max[ f ′(a) does not exist]

Local minf ′(b) does not exist

Local max[ f ′(c) = 0]

Local min[ f ′(d) = 0]

Figure 3.2: Local extreme values

Figure 3.2 illustrates that a local extreme value can occur at a point in the domain of function atwhich either the graph of the function has a horizontal tangent line or function is not differentiable.

Definition 3.3 A critical number of a function f is a number c in the domain of f such thatf ′(c) = 0 or f ′(c) does not exist.

Theorem 3.2 (Fermat’s Theorem) If f has a local maximum or minimum at c, then c is a criticalnumber of f .

Example 3.17 Find all critical numbers and the local extreme values of f(x) = 2x3+3x2−12x−5.

Solution . . . . . . . . .

Example 3.18 Find all critical numbers and the local extreme values of f(x) = (2x+3)2/3.

Solution . . . . . . . . .

Fermat’s Theorem does suggest that we should at least start looking for extreme values offat a critical number off .

Example 3.19 Find the critical numbers of

f (x) =x2+3x+1

.

Solution . . . . . . . . .

To find an absolute maximum or minimum of a continuous function on a closed interval, wenote that either it is local [in which case it occurs at a critical number] or it occurs at an endpointof the interval. Thus the following three-step procedure always works.


The Closed Interval Method

To find the absolute maximum and minimum values of a continuous functionf on a closed interval[a,b]:

1. Find the critical points off in (a,b)

2. Evaluatef at all the critical numbers and at the endpointsa andb.

3. The largest of the values in Step 2 is the absolute maximum value of f on [a,b] and thesmallest value is the absolute minimum.

Example 3.20 Find the absolute maximum and absolute minimum values of thefunction

f (x) = x3−3x+1

on the interval[0,3].

Solution . . . . . . . . .

Example 3.21 Find the absolute maximum and absolute minimum values of thefunction

f (x) =lnxx

1≤ x≤ 3.

Solution . . . . . . . . .

3.4 Increasing and Decreasing Functions

Definition 3.4 Let f be defined on an interval I (open, closed, or neither), and let x1 and x2denote points in I.

(a) f is increasingon I if f (x1)< f (x2) whenever x1 < x2.

(b) f is decreasingon I if f (x1)> f (x2) whenever x1 < x2.

(c) f is constanton I if f (x1) = f (x2) for all points x1 and x2.

Theorem 3.3 (Increasing/Decreasing Test)Let f be continuous on an interval I and differen-tiable at every interior point of I.

(i) If f ′(x)> 0 for all x interior to I, then f is increasing on I.

(ii) If f ′(x)< 0 for all x interior to I, then f is decreasing on I.

(iii) If f ′(x) = 0 for all x in I, then f is constant on I.

Example 3.22 Find the interval on which f(x) = 3x4 + 4x3 − 12x2 − 5 is increasing and theinterval on which it is decreasing.

Solution . . . . . . . . .


Theorem 3.4 (First Derivative Test)

Let f be continuous on an open interval(a,b) that contains a critical number c.

1. If f ′(x)> 0 for all x ∈ (a,c) and f′(x) < 0 for all x ∈ (c,b), then f(c) is a local maximumvalue of f .

2. If f ′(x) < 0 for all x ∈ (a,c) and f′(x) > 0 for all x ∈ (c,b), then f(c) is a local minimumvalue of f .

3. If f ′(x) has the same sign on both sides of c, then f(c) is not a local extreme value of f .

It is easy to remember the First Derivative Test by visualizing diagram such as those in thefollowing Figure.

y

x0 c

f ′(x)> 0 f ′(x)< 0

(a) Local maximum

y

x0 c

f ′(x)< 0 f ′(x)> 0

(b) Local minimum

y

x0 c

f ′(x)> 0

f ′(x)> 0

(c) No maximum or minimum

y

x0 c

f ′(x)< 0

f ′(x)< 0

(d) No maximum or minimum

Example 3.23 Find the local minimum and maximum values of the function f(x) = xex.

Solution . . . . . . . . .

Example 3.24 Find the local minimum and maximum values of f(x) = x(x−1)3.

Solution . . . . . . . . .

3.5 Concavity

Definition 3.5 Let f be differentiable on an open interval I. We say that f (aswell as its graph) isconcave upon I if f ′ is increasing on I and we say that f isconcave downon I if f ′ is decreasingon I.


Theorem 3.5 (Concavity Test)

Let f be twice differentiable on an open interval I.

1. If f ′′ > 0 for all x in I, then f is concave up on I.

2. If f ′′ < 0 for all x in I, then f is concave down on I.

Definition 3.6 Let f be continuous at c. We call(c, f (c)) an inflection point of the graph f if fis concave up on one side of c and concave down on the other side.

Example 3.25 Let f(x) = x4−6x2+3. Find the intervals of concavity and the inflection points.

Solution . . . . . . . . .

Theorem 3.6 (The Second Derivative Test)Let f′ and f′′ exist at every point in an open interval(a,b) containing c, and suppose f′(c) = 0.

1. If f ′′(c)< 0, then f(c) is a local maximum value of f .

2. If f ′′(c)> 0, then f(c) is a local minimum value of f .

Example 3.26 For f (x) = 2x3+6x2−18x+5, use the Second Derivative Test to identify localextrema.

Solution . . . . . . . . .

Note. The Second Derivative Test is inconclusive whenf ′′(c) = 0. In other words, at such a pointthere might be a maximum, there might be a minimum, or there beneither. This test also failswhen f ′′(c) dost not exist. In such cases the The First Derivative Test must be used. In fact, evenwhen both tests apply, the First Derivative Test is often theeasier one to use.

Example 3.27 Let f(x) = x4−4x3+10.

• Find the interval of increase and decrease.

• Find the local maximum and minimum values.

• Find the intervals of concavity and the inflection points.

• Use the above information to sketch the graph.

Solution . . . . . . . . .

Example 3.28 Let f(x) = x2/3(6−x)1/3.

• Find the interval of increase and decrease.

• Find the local maximum and minimum values.

• Find the intervals of concavity and the inflection points.

• Use the above information to sketch the graph.

Solution . . . . . . . . .


3.6 Applied Maximum and Minimum Problems

In this section we will show how the methods discussed in the preceding section can be used tosolve various applied optimization problems.

A Procedure for Solving Applied Maximum and Minimum Problem s

1. Read the problem carefully until it is clearly understood. Ask yourself: What is the un-known? What are the given quantities? What are the given conditions?

2. Draw an appropriate figure and label the quantities relevant to the problem.

3. Find a formula for the quantity to be maximized or minimized.

4. Using the conditions stated in the problem to eliminate variables, express the quantity to bemaximized or minimized as a function of one variable.

5. Find the interval of possible values for this variable from the physical restrictions in theproblem.

6. If applicable, use the techniques of the preceding section to obtain the maximum or mini-mum.

Example 3.29 A cylindrical can is to be made to hold1 L of oil. Find the dimensions that willminimize the cost of the metal to manufacture the can.

Solution . . . . . . . . .

Example 3.30 Find the radius and height of the right circular cylinder of largest volume that canbe inscribed in a right circular cone with radius6 inches and height10 inches.

Solution . . . . . . . . .

Example 3.31 A rectangular beam is to be cut from a log with circular cross section If thestrength of the beam is proportional to the product of its width and the square of its depth, findthe dimensions of the cross section that give the strongest beam.

Solution . . . . . . . . .

3.7 Rolle’s Theorem; Mean Value Theorem

In this section we will discuss a result called the Mean ValueTheorem. The theorem has so manyimportant consequences that it is regarded as one of the major principle is calculus

Theorem 3.7 (Rolle’s Theorem)Let f be a function that satisfies the following three hypotheses:

1. f is continuous on the closed interval[a,b].

2. f is differentiable on the open interval(a,b).


3. f(a) = f (b)

Then there is a number c∈ (a,b) such that f′(c) = 0.

The following Figure shows the graphs of three such functions. In each case it appears thatthere is at least one point

(c, f (c)

)on the graph where the tangent is horizontal and therefore

f ′(c) = 0.y

x

b b

a bc

y

x

b b

a bc

y

x

b b

a bc1

c2

Example 3.32 Verify that the function

f (x) = x3−3x2+2x+2, [0,1]

satisfies the three hypotheses of Rolle’s Theorem on the given interval. Then find all numbers cthat satisfy the conclusion of Rolle’s Theorem.

Solution . . . . . . . . .

Theorem 3.8 (Mean Value Theorem)Let f be a function that satisfies the following hypotheses:

1. f is continuous on the closed interval[a,b].

2. f is differentiable on the open interval(a,b).

Then there is a number c∈ (a,b) such that

f ′(c) =f (b)− f (a)

b−a

or, equivalently,f (b)− f (a) = f ′(c)(b−a)

Example 3.33 Letf (x) = x3−x2−x+1, [−1,2]

Find all numbers c that satisfy the conclusion of the Mean Value Theorem.

Solution . . . . . . . . .

Chapter 4

Integration

4.1 Antiderivatives; The Indefinite Integral

Antiderivatives

Definition 4.1 A function F is called anantiderivativeof f on an interval I if F′(x) = f (x) forall x in I.

For instance, letf (x) = x2. It isn’t difficult to discover an antiderivativeF(x) = 13 x3 because

F ′(x) = x2 = f (x). But the functionG(x) = 13 x3+100 also satisfiesG′(x) = x2. Therefore, both

F andG are antiderivatives off . Indeed, any function of the formH(x) = 13 x3+C, whereC is a

constant, is an antiderivative off .

Question! Are there any others?

Answer. No.

Thus, ifF andG are any two antiderivatives off , then

F ′(x) = f (x) = G′(x)

soG(x)−F(x) =C, whereC is a constant. We can write this asG(x) = F(x)+C, so we have thefollowing result.

Theorem 4.1 If F is an antiderivative of f on an interval I, then the most general antiderivativeof f on I is

F(x)+C

where C is an arbitrary constant.

The Indefinite Integral

The process of finding antiderivatives is calledantidifferentiation or integration. Thus, if

ddx

[F(x)] = f (x) (4.1)

41


then integrating (or antidifferentiating) the functionf (x) produces an antiderivative of the formF(x)+C. To emphasize this process, Equation (4.1) is recast usingintegral notation.

∫

f (x)dx= F(x)+C (4.2)

whereC is an arbitrary constant. For example,

∫

x2dx= 13 x3+C is equivalent to

ddx

[13 x3] = x2

Note that if we differentiate an antiderivative off (x), we obtainf (x) back again. Thus,

ddx

[∫

f (x)dx

]

= f (x) (4.3)

The expression∫

f (x)dx is called anindefinite integral. The “elongated s” that appears onthe left side of (4.2) is called anintegral sign, the functionf (x) is called theintegrand, and theconstantC is called theconstant of integration.

The differential symbol,dx, in the differentiation and antidifferentiation operations

ddx

[ ] and∫

[ ]dx

serves to identify the independent variable. If an independent variable other thanx is used, sayt,then the notation must be adjusted appropriately. Thus,

ddx

[F(t)] = f (t) and∫

[ f (t)]dx= F(t)+C

are equivalent statements. Here are some examples of derivative formulas and their equivalentintegration formulas:

Derivative EquivalentFormula Integration Formula

ddx

[x3] = 3x2∫

3x2dx= x3+C

ddx

[√

x] =1

2√

x

∫1

2√

xdx=

√x+C

ddt[tant] = sec2 t

∫

sec2 t dt = tant +C

Integration Formulas

Some of the most important integration formulas are given inthe following Table.


∫

xndx=xn+1

n+1+C

∫1x

dx= ln |x|+C

∫

exdx= ex+C∫

axdx=ax

lna+C, a> 0 áÅÐa 6= 1

∫

sinxdx=−cosx+C∫

cosxdx= sinx+C

∫

sec2xdx= tanx+C∫

csc2xdx=−cotx+C

∫

secxtanxdx= secx+C∫

cscxcotxdx=−cscx+C

∫1√

1−x2dx= sin−1x+C

∫1

1+x2 dx= tan−1x+C

∫1

|x|√

x2−1dx= sec−1x+C

Properties of the Indefinite Integral

Our first properties of antiderivatives follow directly from the simple constant factor, sum, anddifference rules for derivative.

Theorem 4.2 Let f and g have antiderivatives (indefinite integrals) and let c be a constant. Then

(i)∫

c f(x)dx= c∫

f (x)dx

(ii)∫

[f (x)+g(x)

]dx=

∫

f (x)dx+∫

g(x)dx

(iii)∫

[f (x)−g(x)

]dx=

∫

f (x)dx−∫

g(x)dx

Example 4.1 Evaluate∫

(3ex+5x2/3)dx.

Solution . . . . . . . . .


(3cosx+2sec2x)dx.

Solution . . . . . . . . .

Example 4.3 Evaluate

(a)∫

cosx

sin2xdx (b)

∫2t4− t2√t −1

t4 dt

Solution . . . . . . . . .


4.2 Integration by Substitution

In this section we shall discuss a technique, calledsubstitution, which can often be used to trans-form complicated integration problems into simpler ones.

u-substitution

The method of substitution hinges on the following formula in whichu stands for a differentiablefunction ofx.

∫ [

f (u)dudx

]

dx=∫

f (u)du (4.4)

To justify this formula, letF be an antiderivative off , so that

ddu

[F(u)] = f (u)

or, equivalently,∫

f (u)du= F(u)+C (4.5)

If u is a differentiable function ofx, the chain rule implied that

ddx

[F(u)] =ddu

[F(u)] · dudx

= f (u)dudx

or, equivalently,∫ [

f (u)dudx

]

dx= F(u)+C (4.6)

Formula (4.4) follows from (4.5) and (4.6).The following example illustrates how Formula (4.4) is used.


(x4−1)99(4x3)dx.

Solution . . . . . . . . .

In general, suppose that we are interested in evaluating∫

h(x)dx

It follows from (4.4) that if we can express this integral in the form∫

h(x)dx=∫

f (g(x))g′(x)dx

then the substitutionu= g(x) anddu/dx= g′(x) will yield∫

h(x)dx=∫ [

f (u)dudx

]

dx=∫

f (u)du

With a “good” choice ofu = g(x), the integral on the right will be easier to evaluate than theoriginal.

In practice, this substitution process is carrier out as follows:


Guideline for u-Substitution

Step 1. Make a choice foru, sayu= g(x).

Step 2. Computedu/dx

Step 3. Make the substitutionu= g(x), du= g′(x)dx

At this stage, the entire integral must be in terms ofu; nox’s should remain. If this is not the case,try a different choice ofu.

Step 4. Evaluate the resulting integral.

Step 5. Replaceu by g(x), so the final answer is in terms ofx.


x2cos(x3−2)dx.

Solution . . . . . . . . .

Example 4.6 Evaluate∫ √

2sinx+1cosxdx.

Solution . . . . . . . . .


x√1−4x2

dx.

Solution . . . . . . . . .


cos√

x√x

dx.

Solution . . . . . . . . .


1+x2x5 dx.

Solution . . . . . . . . .

4.3 The Definite Integral

Definition of Area

The first goal in this section is to give a mathematical definition of area. We begin by attemptingto find the area of the regionS that lies under the curvey= f (x) from a to b. In general, we startby subdividingS into n stripsS1,S2, . . . ,Sn of equal width as in Figure.


S1

S2

S3 Si Sn

a x1 x2 x3 . . . xi−1 xi . . . xn−1 b x

y

y= f (x)

The width of the interval[a,b] is b−a, so the width of each of then strips is

∆x=b−a

n

These strips divide the interval[a,b] into n subintervals

[x0,x1], [x1,x2], [x2,x3], . . . , [xn−1,xn]

wherex0 = a andxn = b. The right-hand endpoints of the subintervals are

x1 = a+∆x, x2 = a+2∆x, x3 = a+3∆x, . . .

Let’s approximate theith stripSi by a rectangle with width∆x and heightf (xi), which is thevalue of f at the right-hand endpoint. Then the area of theith rectangle isf (xi)∆x. What we thingof intuitively as the area ofS is approximated by the sum of the areas of these rectangles, whichis

Rn = f (x1)∆x+ f (x2)∆x+ · · ·+ f (x2)∆x

=n

∑i=1

f (xi)∆x

Notice that this approximation appears to become better andbetter as the number of strips in-creases, that is, asn→ ∞. Therefore, we define the areaA of the regionS in the following way.

Definition 4.2 TheareaA of the region S that lies under the graph of the continuous function fis the limit of the sum of the areas of approximating rectangles:

A= limn→∞

Rn = limn→∞

n

∑i=1

f (xi)∆x


We now have that a limit of the form

limn→∞

n

∑i=1

f (xi)∆x= limn→∞

[f (x1)∆x+ f (x2)∆x+ · · ·+ f (xn)∆x

]

arises when we compute an area. However this type of limit occurs in a wide variety of situationseven whenf is not necessarily a positive function. In the next Chapter we will see that limit ofthis form also arise in finding volumes of solids. We therefore give type of limit a special nameand notation.

Definition 4.3 If f is a continuous function defined for a≤ x≤ b, we divide the interval[a,b] inton subintervals of equal width∆x= (b−a)/n. We let x0(= a),x1, x2, . . ., xn(= b) be the endpointsof these subintervals and we choosesimple pointsx∗1,x

∗2, . . . ,x

∗n in these subintervals, so x∗

i lies inthe ith subinterval[xi−1,xi]. Then thedefinite integral of f from a to b is

∫ b

af (x)dx= lim

n→∞

n

∑i=1

f (x∗i )∆x (4.7)

Note:

1. The symbol∫

was introduced by Leibniz and is called anintegral sign. In the notation∫ b

a f (x)dx, f (x) is called integrand anda andb are called thelimit of integration; a isthe lower limit andb is theupper limit. The symboldx has no official meaning by itself;∫ b

a f (x)dx is all one symbol. The procedure of calculating an integral is calledintegration

2. The definite integral∫ b

a f (x)dx is a number; it does not depend onx. In fact, we could useany letter in place ofx without changing the value of the integral:

∫ b

af (x)dx=

∫ b

af (t)dt =

∫ b

af (r)dr

3. Because we have assumed thatf is continuous, it can be proved that the limit in Defini-tion 4.3 always exists and gives the same value no matter how we choose the sample pointsx∗i .

4. The sumn

∑i=1

f (x∗i )∆x

that occurs in Definition 4.3 is calledRiemann sumafter the German mathematician Bern-hard Riemann (1826 - 1866). We know that iff happens to be positive, then the Riemannsum can be interpreted as a sum of areas of approximating rectangles

(See Figure 4.1 (a)

).

By comparing Definition 4.3 with the definition of area, we seethat the definite integral∫ b

a f (x)dxcan be interpreted as the area under the curvey= f (x) from a to b(t’ÙFigure 4.1

(b)).


a b0 x

y

x∗i

∆x

(a)

a b0 x

y

y= f (x)

(b)Figure 4.1:

If f take on both positive an negative values, then the Riemann sum is the sum of areas ofthe rectangles that lie above thex-axis and the negative of the areas of the rectangles thatlie below thex-axis. When we take the limit of such Riemann sums, we get the situationillustrated in the following Figure.

a b0 x

y

y= f (x)+

−

+

A definite integral can be interpreted as anet area, that is, a difference of areas:

∫ b

af (x)dx= A1−A2

whereA1 is the area of the region above thex-axis and below the graph off andA2 is thearea of the region below thex-axis and above the graph off .

5. Although we have defined∫ b

a f (x)dx by dividing [a,b] into subintervals of equal width,there are situations in which it is advantageous to work withsubintervals of unequal width.

When we defined the definite integral, we implicitly assumed thata< b. But the definition asa limit of Riemann sums makes sense even ifa > b. Notice that if we reversea andb, than∆xchanges from(b−a)/n to (a−b)/n. Therefore

∫ a

bf (x)dx=−

∫ b

af (x)dx (4.8)

If a= b, then∆x= 0 and so∫ a

af (x)dx= 0

We now develop some basic properties of integrals. We assumethat f andg are continuouson an interval[a,b].


Theorem 4.3 (Properties of the Integral)

1.∫ b

acdx= c(b−a), where c is any constant

2.∫ b

ac f(x)dx= c

∫ b

af (x)dx, where c is any constant

3.∫ b

a

[f (x)+g(x)

]dx=

∫ b

af (x)dx+

∫ b

ag(x)dx

4.∫ b

a

[f (x)−g(x)

]dx=

∫ b

af (x)dx−

∫ b

ag(x)dx

Theorem 4.4 If f and g are continuous on an interval[a,b] and c is any constant, then

∫ b

af (x)dx=

∫ c

af (x)dx+

∫ b

cf (x)dx.

For the case wheref (x) ≥ 0 anda < c < b, this theorem can be seen from the geometricinterpretation in the following Figure.

a c b x

y

0

y= f (x)

The area undery= f (x) from a to c plus the area fromc to b is equal to the total area froma to b.

Theorem 4.5 Let f and g be continuous on an interval[a,b] and g(x) ≤ f (x) for all x in [a,b].Then

∫ b

ag(x)dx≤

∫ b

af (x)dx

Example 4.10 If it is known that∫ 8

0f (x)dx=3,

∫ 5

0f (x)dx=−4,

∫ 6

5g(x)dx=1, and

∫ 8

6g(x)dx=

−2, find∫ 8

5

(2 f (x)+g(x)

)dx.

Solution . . . . . . . . .


4.4 The Fundamental Theorem of Calculus

In the previous section we defined the concept of the definite integral but did not give any generalmethods for evaluating them. In this section we shall give a method for using antiderivatives toevaluate definite integrals.

Theorem 4.6 (The First Fundamental Theorem of Calculus).If f is continuous on[a,b], then

∫ b

af (x)dx= F(b)−F(a) = F(x)

]b

a

where F is any antiderivative of f , that is, a function such that F′ = f .

Example 4.11 Evaluate∫ 1

0

(3+x

√x)

dx.

Solution . . . . . . . . .

Example 4.12 Evaluate∫ π/4

1

1+cos2 θcos2θ

dθ .

Solution . . . . . . . . .

Theorem 4.7 (The Second Fundamental Theorem of Calculus).If f is continuous on[a,b],then the function F defined by

F(x) =∫ x

af (t)dt, a≤ x≤ b

is continuous on[a,b] and differentiable on(a,b), and F′(x) = f (x).

Using Leibniz notation for derivative, the result in Theorem 4.7 can be expressed by the for-mula

ddx

∫ x

af (t)dt = f (x)

Example 4.13 Find the derivative of the function g(x) =∫ x

0

√

1+ t2dt.

Solution . . . . . . . . .

Example 4.14 Find the derivative of the function g(x) =∫ x4

1sect dt.

Solution . . . . . . . . .

Example 4.15 Let F(x) =∫ x3

x2

(

et2+1

)

dt. Find F′(x).

Solution . . . . . . . . .


4.5 Evaluating Definite Integrals by Substitution

There is only one slight difference in using substitution for evaluating a definite integral: If youchange variables, you must also change the limits of integration to correspond to the new vari-able. That is, when you introduce the new variableu= g(x), you must also change the limits ofintegration fromx= a andx= b to the corresponding limits foru : u = g(a) andu = g(b). Wehave

∫ b

af (g(x))g′(x)dx=

∫ g(b)

g(a)f (u)du.


1x3√

x4+5dx.

Solution . . . . . . . . .

Example 4.17 Evaluate∫ e

1

lnxx

dx.

Solution . . . . . . . . .


1

cos(π/x)x2 dx.

Solution . . . . . . . . .

Chapter 5

Applications of Definite Integral

5.1 Area Between Two Curves

In this section we use integrals to find areas of regions that lie between the graphs of two functions.Consider the region that lies between two curvesy = f (x) and y = g(x) and between the

vertical linesx= a andx= b, wheref andg are continuous functions andf (x)≥ g(x) for all x in[a,b].

a b x

y

y= g(x)

y= f (x)

The areaA of this region is

A=

∫ b

a

[f (x)−g(x)

]dx (5.1)

Example 5.1 Find the area of the region enclosed by the line y= 3− x and the parabola y=x2−9.

Solution . . . . . . . . .

Example 5.2 Find the area of the region bounded by the parabolas y= x2 and y= 2x−x2.

Solution . . . . . . . . .

Example 5.3 Find the area bounded by the graphs of y= x2 and y= 2−x2 for 0≤ x≤ 2.

Solution . . . . . . . . .

52


d

cx= f (y)

x= g(y)

x

y

Some regions are best treated by regardingx as a function ofy. If a region is bounded by curveswith equationsx= f (y), x= g(y), y= c, andy= d, wheref andg are continuous andf (y)≥ g(y)for c≤ y≤ d, then its area is

A=∫ d

c

[f (y)−g(y)

]dy (5.2)

Example 5.4 Find the area of the region bounded by the graphs of y= x2, y= 2−x, and y= 0.

Solution . . . . . . . . .

Example 5.5 Find the area bounded by the graphs of x= y2 and x= 2−y2.

Solution . . . . . . . . .

Example 5.6 Find the area of the region bounded by the curves x= y2, y= x+5, y= 2, andy=−1.

Solution . . . . . . . . .

5.2 Volumes by Slicing: Disks and Washers

In this section we will use definite integral to find volumes ofsolid of revolution.

Method of Disks

Suppose thatf (x)≥ 0 andf is continuous on[a,b]. Take the region bounded by the curvey= f (x)and thex-axis, fora≤ x≤ b and revolve it about thex-axis, generating a solid.


|a

|

bx

y

y= f (x)≥ 0

|a

|

bx

y

y= f (x)

|a

|

b

We can find the volume of this solid by slicing it perpendicular to thex-axis and recognizing thateach cross section is a circular disk of radiusr = f (x). We then have that the volume of the solidis

V =∫ b

aπ [ f (x)]2︸︷︷︸

cross-sectional area =πr2

dx

Example 5.7 Find the volume of the solid obtained by rotating the region bounded by y=√

xfrom0 to 4 about the x-axis.

Solution . . . . . . . . .

In a similar way, suppose thatg(y) ≥ 0 andg is continuous on the interval[c,d]. Then,revolving the region bounded by the curvex= g(y) and they-axis, forc≤ y≤ d, about they-axisgenerates a solid.

c

b

x

y

x= g(x)

c

b

x

y

x= g(x)

Once again, notice from Figure that the cross sections of theresulting solid of revolution arecircular disks of radiusr = g(y). The volume of the solid is then given by

V =

∫ d

cπ [g(y)]2︸︷︷︸

cross-sectional area =πr2

dy

Example 5.8 Find the volume of the solid obtained by rotating the region bounded by y= 2− x2

2from x= 0 to x= 2 about the y-axis.

Solution . . . . . . . . .


Method of Washers

There are two complications that can be found in the types of volume calculations we have beenstudying. The first of these is that you may need to compute thevolume of a solid that have acavity or “hole” in it. The second of these occurs when a region is revolved about a line other thatthex-axis or they-axis.

Suppose thatf andg are nonnegative continuous function such that

g(x)≤ f (x) for a≤ x≤ b

and letR be the region enclosed between the graphs of these functionsand the linesx = a andx= b.

y= f (x)

y= g(x)

a bx

y

When this region is revolved about thex-axis, it generates a solid having annular or washer-shaped cross sections. Since the cross section atx has inner radiusg(x) and outer radiusf (x), itsvolume of the solid is

V =

∫ b

aπ{

[ f (x)]2− [g(x)]2}

dx

Suppose thatu andv are nonnegative continuous function such that

v(y)≤ u(y) for c≤ y≤ d.

If R the region enclosed between the graphs ofx= u(y) andx= v(y) and the linesy= c andy= d.

c

d

x

y

x= v(y)

x= u(y)

When this region is revolved about they-axis, it also generates a solid having annular or washer-shaped cross sections. Since the cross section aty has inner radiusv(y) and outer radiusu(y), itsvolume of the solid is

V =∫ d

cπ{

[u(y)]2− [v(y)]2}

dy


Example 5.9 The region R enclosed by the curves y= 4−x2 and y= 0. Find the volume of thesolid obtained by rotating the region R(a)about the y-axis(b) about the line y=−3 (c)about theline y= 7, and(d) about the line x= 3.

Solution . . . . . . . . .

Example 5.10 Find the volume of the solid obtained by rotating the region bounded by y= 1+x2

2and the line y= 2 about the line y=−2.

Solution . . . . . . . . .

5.3 Volumes by Cylindrical Shells

A cylindrical shellsis a solid enclosed by two concentric right-circular cylinders.

b r1

h

r2

The volumeV of a cylindrical shall having inner radiusr1, outer radiusr2, and heighth can bewritten as

V = [area of cross section] · [height]

= (πr22−πr2

1)h

= π(r2+ r1)(r2− r1)h

= 2π ·(

r2+ r1

2

)

·h · (r2− r1)

Butr2+ r1

2is the average radius of the shell andr2− r1 is its thickness, so

V = 2π · [average radius] · [height] · [thickness]

This formula can be used to find the volume of a solid of revolution.Let Rbe a plane region bounded above by a continuous curvey= f (x), bounded below by the

x-axis, and bounded on the left and right, respectively, by the linex= a andx= b.The volume of the solid generated by revolvingRabout they-axis is given by

V = 2π∫ b

ax

︸︷︷︸

radius

f (x)︸︷︷︸

height

dx︸︷︷︸

thickness

Let Rbe a plane region bounded above by a continuous curvex= g(y), they-axis, and the liney= c andy= d.


|

a|

bx

y

y= f (x)

x

y

y= f (x)

c

d

x

y

x= g(y)

The volume of the solid generated by revolvingR about thex-axis is given by

V = 2π∫ d

cy

︸︷︷︸

radius

g(y)︸︷︷︸

height

dy︸︷︷︸

thickness

Example 5.11 Find the volume of the solid obtained by rotating the region bounded by y= x andy= x2 in the first quadrant about the y-axis.

Solution . . . . . . . . .

Example 5.12 The region bounded by the line y=(

rh

)x, the x-axis, and x= h is revolved about

the x-axis, thereby generating cone (assume r> 0, h> 0). Find it volume by the shell method.

Solution . . . . . . . . .

5.4 Length of a Plane Curve

Arc Length Problem. Supposef is continuous on[a,b] and differentiable on(a,b). Find the arclengthL of the curvey= f (x) over the interval[a,b].

In order to solve this problem, we begin by partitioning the interval[a,b] into n equal pieces:

a= x0 < x1 < x2 < · · ·< xn = b,

where

xi −xi−1 = ∆x=b−a

n,


for eachi = 1,2, . . . ,n.Between each pair if adjacent points on the curve,

(xi−1, f (xi−1)) and(xi , f (xi)) we approximate the are lengthì by the straight-line distance be-tween the two points.

b

b

|xi−1

|xi

+f (xi−1)

+f (xi)

x

y

ì

From the usual distance formula, we have

ì ≈√

(xi −xi−1)2+[ f (xi)− f (xi−1)]2.

Since f is continuous on all of[a,b] and differentiable on(a,b), f is also continuous on thesubinterval[xi−1,xi ] and is differentiable on(xi−1,xi). Recall that by the Mean Value Theorem,we have

f (xi)− f (xi−1) = f ′(ci)(xi −xi−1),

for some numberci ∈ (xi−1,xi). This give us the approximation

ì ≈√

(xi −xi−1)2+[ f (xi)− f (xi−1)]2

=√

(xi −xi−1)2+[ f ′(ci)(xi −xi−1)]2

=√

1+[ f ′(ci)]2(xi −xi−1)︸︷︷︸

∆x

=√

1+[ f ′(ci)]2∆x.

Adding together the lengths of thesen line segments, we get an approximation of the total arclength,

L ≈n

∑i=1

√

1+[ f ′(ci)]2∆x.

Notice that asn gets larger, this approximation should approach the exact arc length, that is,

L = limn→∞

n

∑i=1

√

1+[ f ′(ci)]2∆x.

You should recognize this as the limit of a Riemann sum for√

1+[ f ′(x)]2, so that the arc lengthis given by the definite integral:

L =∫ b

a

√

1+[ f ′(x)]2dx,

whenever the limit exists.


Example 5.13 Find the arc length of the curve y= x3/2 from (1,1) to (4,8).

Solution . . . . . . . . .

Chapter 6

Techniques of Integration

6.1 Integration by Parts

Every differentiation rule has a corresponding integration rule. For instance, the Substitution Rulefor integration corresponds to the Chain Rule for differentiation. The rule that corresponds to theProduct Rule for differentiation is called the rule forintegration by parts.

The Product Rule state that iff andg are differentiable functions, then

ddx

[f (x)g(x)

]= f (x)g′(x)+g(x) f ′(x)

In the notation for indefinite integrals this equation becomes∫

[f (x)g′(x)+g(x) f ′(x)

]dx= f (x)g(x)

or ∫

f (x)g′(x)dx+∫

g(x) f ′(x)dx= f (x)g(x)

We can rearrange this equation as

∫

f (x)g′(x)dx= f (x)g(x)−∫

g(x) f ′(x)dx

This formula is called theformula for integration by parts. Let u = f (x) andv = g(x). Thenthe differential aredu= f ′(x)dx anddv= g′(x)dx, so, by the Substitution Rule, the formula forintegration by parts becomes

∫

udv= uv−∫

vdu

60


Example 6.1 Find∫

xcosxdx.

Solution . . . . . . . . .

Note. Our aim in using integration by parts is to obtain a simpler integral than the one we startwith. Thus, in Example 6.1 we start with

∫

xcosxdx and expressed it in terms of the simplerintegral

∫sinxdx. If we had chosenu= cosx anddv= xdx, thendu=−sinxdxandv= x2/2, so

integration by parts gives

∫

xcosxdx= (cosx)x2

2+

12

∫

x2 sinxdx

Although this is true,∫

x2sinxdx is a more difficult integral than the one we started with. Ingeneral, when deciding on a choice foru and dv, we usually try to chooseu = f (x) to be afunction that becomes simpler when differentiated (or at least not more complicated) as long asdv= g′(x)dxcan be readily integrated to givev.

Example 6.2 Find∫

xlnxdx.

Solution . . . . . . . . .

Example 6.3 Find∫

x2sin2xdx.

Solution . . . . . . . . .

Example 6.4 Find∫

e2x cosxdx.

Solution . . . . . . . . .

If we combine the formula for integration by parts with Part 2of the Fundamental Theoremof Calculus, we can evaluate definite integrals by parts. Assuming f ′ andg′ are continuous, andusing the Fundamental Theorem, we obtain

∫ b

af (x)g′(x)dx= f (x)g(x)

]b

a−

∫ b

ag(x) f ′(x)dx

That is, ifu= f (x) andv= g(x), then

∫ b

audv= uv

]b

a−

∫ b

avdu.

Example 6.5 Find∫ 1/3

0tan−13xdx.

Solution . . . . . . . . .


6.2 Trigonometric Integrals

In this section we use trigonometric identities to integrate certain combinations of trigonometricfunctions. Our first aim is to evaluate integrals of the form

∫

sinmxcosnxdx,

wherem andn are positive integers.

Case I:m or n Is an Odd Positive Integer

If m is odd, first isolate one factor of sinx. (You’ll need this fordu). Then, replace any factors ofsin2x with 1−cos2x and make the substitutionu= cosx. Likewise, if n is odd, first isolate onefactor of cosx. (You’ll need this fordu). Then, replace any factors of cos2x with 1− sin2x andmake the substitutionu= sinx.


sin3xdx.

Solution . . . . . . . . .

Example 6.7 Find∫

cos4xsin3xdx.

Solution . . . . . . . . .

Example 6.8 Find∫ √

sinxcos5xdx.

Solution . . . . . . . . .

Case II: m and n Are Both Even Positive Integers

In this case, we can use the half-angle formulas for sine and cosine to reduce the power of in theintegrand.

Half-angle formulas:

sin2 θ = 12(1−cos2θ) and cos2 θ = 1

2(1+cos2θ)

Example 6.9 Find∫

cos4xdx.

Solution . . . . . . . . .

Example 6.10 Find∫

sin2xcos4xdx.

Solution . . . . . . . . .

Our next aim is to devise a strategy for evaluating integralsof the form∫

tanmxsecnxdx,

wherem andn are integers.


Case I:m Is an Odd Positive Integer

First, isolate one factor of secxtanx. (You’ll need this fordu). Then, replace any factors of tan2xwith sec2x−1 and make the substitutionu= secx.


tan3xsec3/2 xdx.

Solution . . . . . . . . .

Case II: n Is an Even Positive Integer

First, isolate one factor of sec2x. (You’ll need this fordu). Then, replace any remaining factorsof sec2x with 1+ tan2x and make the substitutionu= tanx.


tan4xsec4xdx.

Solution . . . . . . . . .

For other cases, the guidelines are not as clear-cut. We may need to use identities, integrationby part, and occasionally a little ingenuity. We will sometimes need to be able to integrate tanxby using the formula

∫

tanxdx= ln |secx|+C

We will also need the indefinite integral of secant:

∫

secxdx= ln |secx+ tanx|+C


tan3xdx.

Solution . . . . . . . . .


sec3xdx.

Solution . . . . . . . . .

Integrals of the form∫

cotmxcscnxdxcan be found by similar methods because of the identity

1+cot2x= csc2x.

Finally, we can make use of another set of trigonometric identities: To evaluate the integrals

(a)∫

sinmxcosnxdx,

(b)∫

sinmxsinnxdx, or


(c)∫

cosmxcosnxdx,

we use the corresponding identity:

sinAcosB = 12

[sin(A−B)+sin(A+B)

]

sinAsinB = 12

[cos(A−B)−cos(A+B)

]

cosAcosB = 12

[cos(A−B)+cos(A+B)

]


sin5xcos6xdx.

Solution . . . . . . . . .


cos3xcos2xdx.

Solution . . . . . . . . .

6.3 Trigonometric Substitutions

If an integral contains a term of the form√

a2−x2,√

a2+x2 or√

x2−a2, for somea> 0, youcan often evaluate the integral by making a substitution involving a trig function.

First, suppose that an integrand contains a term of the form√

a2−x2, for somea> 0. If welet x= asinθ , where−π

2 ≤ θ ≤ π2 , then we can eliminate the square root, as follows. Notice that

we now have√

a2−x2 =√

a2− (asinθ)2 =√

a2−a2sin2θ

=

√

a2(1−sin2θ) =√

a2cos2 θ = acosθ

since for−π2 ≤ θ ≤ π

2 , cosθ > 0.


16−x2

x2 dx.

Solution . . . . . . . . .

Next, suppose that an integrand contains a term of the form√

a2+x2, for somea> 0. If welet x= atanθ , where−π

2 < θ < π2 , then we can eliminate the square root, as follows. Notice that

in this case, we have

√

a2+x2 =√

a2+(atanθ)2 =√

a2+a2 tan2 θ

= a√

1+ tan2 θ = a√

sec2 θ = asecθ

since for−π2 < θ < π

2 , secθ > 0.


Example 6.18 Evaluate the integral∫

1√4+x2

dx.

Solution . . . . . . . . .

Finally, suppose that an integrand contains a term of the form√

x2−a2, for somea > 0. Ifwe letx= asecθ , whereθ ∈

[0, π

2

)∪[π , 3π

2

), then we can eliminate the square root, as follows.

Notice that in this case, we have

√

x2−a2 =√

(asecθ)2−a2 =√

a2sec2θ −a2

= a√

sec2θ −1 = a√

tan2 θ = atanθ

since forθ ∈[0, π

2

)∪[π , 3π

2

), tanθ ≥ 0.

Example 6.19 Evaluate the integral∫ √

9x2−1x

dx.

Solution . . . . . . . . .


x√x2−6x+13

dx.

Solution . . . . . . . . .

We summarize the three trigonometric substitutions presented here in the following table.

Expression Substitution Identity√

a2−x2 x= asinθ , −π2 ≤ θ ≤ π

2 1−sin2θ = cos2 θ√

a2+x2 x= atanθ , −π2 < θ < π

2 1+ tan2 θ = sec2 θ√

x2−a2 x= asecθ , θ ∈[0, π

2

)∪[π , 3π

2

)sec2θ −1= tan2 θ

6.4 Integrating Rational Functions by Partial Fractions

In this section we show how to integrate any rational function by expressing it as a sum of simplerfractions, calledpartial fractions.

Observe that3

x+2− 2

x−5=

3(x−5)−2(x+2)(x+2)(x−5)

=x−19

x2−3x−10.

To integrate the function on the right side of this equation,we have∫

x−19x2−3x−10

dx =∫ (

3x+2

− 2x−5

)

dx

= 3ln|x+2|−2ln|x−5|+C


To see how the method of partial fractions work in general, let’s consider a rational function

f (x) =P(x)Q(x)

(6.1)

whereP andQ are polynomial. It’s possible to expressf as a sum of simpler fractions providedthe degree ofP

(deg(P)

)is less than the degree ofQ

(deg(Q)

). Such a rational function is called

proper.If f is improper, that is, deg(P)≥ deg(Q), then we must take the preliminary step of dividing

Q into P (by long division) until the remainderR(x) is obtained such that deg(R)< deg(Q). Thedivision statement is

f (x) =P(x)Q(x)

= S(x)+R(x)Q(x)

whereSandR are also polynomial.


x3+2x2−1x−2

dx

Solution . . . . . . . . .

The next step is to factor the denominatorQ(x) as far as possible. And the third step is toexpress the proper rational functionR(x)/Q(x) as a sum ofpartial factionsof the form

A(ax+b)i or

Ax+B(ax2+bx+c) j

CASE I: The denominator Q(x) is a product of distinct linear factors.

This means that we can write

Q(x) = (a1x+b1)(a2x+b2) · · ·(anx+bn)

where no factor is repeated. In this case the partial fraction theorem states that there exist constantsA1, A2, . . ., Ak such that

R(x)Q(x)

=R(x)

(a1x+b1)(a2x+b2) · · ·(akx+bk)

=A1

a1x+b1+

A2

a2x+b2+ · · ·+ Ak

akx+bk(6.2)

These constant can be determined as in the following example.


x−9x2+3x−10

dx.

Solution . . . . . . . . .


3x2−7x−2x3−x

dx.

Solution . . . . . . . . .


CASE II: Q(x) is a product of distinct linear factors, some of which are repeated.

Suppose the first linear factor(a1x+ b1) is repeatedr times; that is,(a1x+ b1)r occurs in the

factorization ofQ(x). Then instead of the single termA1/(a1x+b1) in Equation (6.2), we woulduse

A11

a1x+b1+

A12

(a1x+b1)2 + · · ·+ A1r

(a1x+b1)r . (6.3)

For example,x2−5

x2(x+1)3 =Ax+

Bx2 +

C(x+1)

+D

(x+1)2 +E

(x+1)3


5x2+20x+6x3+2x2+x

dx.

Solution . . . . . . . . .

CASE III: Q(x) contains irreducible quadratic factors, none of which is repeated.

If Q(x) has the factorax2+bx+c, whereb2−4ac< 0, then, in addition to the partial fractions inEquations (6.2) and (6.3), the expression forR(x)/Q(x) will have a term of the form

Ax+Bax2+bx+c

(6.4)

whereA andB are constants to be determined. For instance, the fraction given by

f (x) =x

(x+2)(x2+1)(x2+2)

has a partial fraction decomposition of the form

x(x+2)(x2+1)(x2+2)

=A

x+2+

Bx+Cx2+1

+Dx+Ex2+2

The term given in (6.4) can be integrate by completing the square and using the formula

∫1

x2+a2 dx=1a

tan−1(x

a

)

+C


3x2−4x+3x3+x

dx.

Solution . . . . . . . . .


6x2−3x+1(4x+1)(x2+1)

dx.

Solution . . . . . . . . .


CASE IV: Q(x) contains a repeated irreducible quadratic factors.

If Q(x) has the factor(ax2+bx+c)r , whereb2−4ac< 0, then instead of the single partial fraction(6.4), the sum

A1x+B1

ax2+bx+c+

A2x+B2

(ax2+bx+c)2 + · · ·+ Arx+Br

(ax2+bx+c)r (6.5)

occurs in the partial fraction decomposition ofR(x)/Q(x). Each of the term of (6.5) can beintegrated by first completing the square.


6x2−15x+22(x+3)(x2+2)2 dx.

Solution . . . . . . . . .

6.5 Improper Integrals

In defining a definite integral∫ b

a f (x)dx we dealt with a functionf defined on a finite interval[a,b] and we assumed thatf does not have an infinite discontinuity. In this section we extend theconcept of the definite integral to the case where the interval is infinite and also to the case wheref has an infinite discontinuity in[a,b]. In either case the integral is called animproperintegral.

Type I: Infinite Intervals

Definition 6.1

(a) If∫ t

a f (x)dx exists for every number t≥ a, then

∫ ∞

af (x)dx= lim

t→∞

∫ t

af (x)dx

provided this limit exists (as a finite number).

(b) If∫ bt f (x)dx exists for every number t≤ b, then

∫ b

−∞f (x)dx= lim

t→−∞

∫ b

tf (x)dx

provided this limit exists (as a finite number).

The improper integrals∫ t

a f (x)dx and∫ bt f (x)dx are calledconvergentif the corresponding limit

exists anddivergentif the limit does not exist.

(c) If both∫ ∞

a f (x)dx and∫ a−∞ f (x)dx are convergent, then we define

∫ ∞

−∞f (x)dx=

∫ a

−∞f (x)dx+

∫ ∞

af (x)dx

In part (c) any real numbera can be used.


Example 6.28 Determine whether the integral∫ ∞

1

1x

dx

is convergent or divergent.

Solution . . . . . . . . .


−∞xex dx.

Solution . . . . . . . . .

Example 6.30 Evaluate∫ ∞

−∞

11+x2 dx.

Solution . . . . . . . . .

Example 6.31 What for values of p is the integral∫ ∞

1

1xp dx.

convergent?

Solution . . . . . . . . .

Type II: Discontinuous Integrands

Definition 6.2

(a) If f continuous on[a,b) and is discontinuous at b, then

∫ b

af (x)dx= lim

t→b−

∫ t

af (x)dx

if this limit exists (as a finite number).

(b) If f continuous on(a,b] and is discontinuous at a, then

∫ b

af (x)dx= lim

t→a+

∫ b

tf (x)dx

if this limit exists (as a finite number).

The improper integrals∫ b

a f (x)dx is calledconvergentif the corresponding limit exists anddiver-gent if the limit does not exist.

(c) If f has a discontinuity at c, where a< c < b, and both∫ c

a f (x)dx and∫ b

c f (x)dx areconvergent, then we define

∫ b

af (x)dx=

∫ c

af (x)dx+

∫ b

cf (x)dx


Example 6.32 Find∫ 5

2

1√x−2

dx.

Solution . . . . . . . . .

Example 6.33 Determine whether the integral

∫ π/2

0secxdx

converges or diverges.

Solution . . . . . . . . .


0

1x−1

dx if possible.

Solution . . . . . . . . .


0lnxdx.

Solution . . . . . . . . .

Chapter 7

Infinite Sequence and Series

7.1 Sequences

A sequencecan be thought of as a list of numbers written in a definite order:

a1,a2,a3,a4, . . . ,an, . . .

The numbera1 is called thefirst term, a2 is thesecond term, and in generalan is thenth term.The sequence{a1,a2,a3, . . .} is also denoted by

{an} or {an}∞n=1

Example 7.1 List the first five terms of the following sequence

(a){2n}∞n=1 = {2,4,8,16,32, . . .}

(b)

{

−23,39,− 4

27,

581

,− 6243

, . . .

}

(c)

{

1,12,23,35,58, . . .

}

♣

Definition 7.1 A sequence{an} has thelimit L and we write

limn→∞

an = L or an → L as n→ ∞

if we can make the terms an as close to L as we like by taking n sufficiently large. Iflimn→∞

an exists,

we say that the sequenceconverges(or is convergent). Otherwise, we say that the sequencediverges(or is divergent).

Limit Laws for Sequences

If {an} and{bn} are convergent sequences andc is a constant, then

1. limn→∞

(an+bn) = limn→∞

an+ limn→∞

bn

71


2. limn→∞

(an−bn) = limn→∞

an− limn→∞

bn

3. limn→∞

can = c limn→∞

an

4. limn→∞

(anbn) = limn→∞

an · limn→∞

bn

5. limn→∞

an

bn=

limn→∞

an

limn→∞

bnif lim

n→∞bn 6= 0

6. limn→∞

c= c

Theorem 7.1 If limn→∞

|an|= 0, then limn→∞

an = 0.

Example 7.2 Determine whether the sequence converges of diverges. If itconverges, find thelimit.

(a) an =n

2n+1(b) an =

(−1)n

n(c) {7−5n}

Solution . . . . . . . . .

7.2 Infinite Series

If we try to add the terms of an infinite sequence{an}∞n=1 we get the expression of the form

a1+a2+a3+ · · ·+an+ · · · (7.1)

which is called aninfinite series(or just aseries) and is denoted, for short, by the symbol

∞

∑n=1

an or ∑an

To determine whether or not a general series (7.1) has the sum. We consider thepartial sum.

s1 = a1

s2 = a1+a2

s3 = a1+a2+a3

s4 = a1+a2+a3+a4...

sn = a1+a2+a3+ · · ·+an =n

∑i=1

ai

These partial sum form a new sequence{sn}, which may or may not have a limit. If limn→∞

sn = s

exist (as a finite number), then we call it the sum of the infinite series∑an.


Definition 7.2 Given a series∞

∑n=1

an = a1+a2+a3+ · · · , let sn denote it nth partial sum:

sn =n

∑i=1

ai = a1+a2+a3+ · · ·+an

If the sequence{sn} is convergent andlimn→∞

sn = s exists as a real number, then the series∑an is

calledconvergentand we write

a1+a2+a3+ · · ·+an+ · · ·= s or∞

∑n=1

an = s

The number s is called thesumof the series. Otherwise, the series is calleddivergent.

Example 7.3 Determine whether the series

12+

122 +

123 + · · ·

is convergent or divergent. If it is convergent, find its sum.

Solution Consider the partial sum:

s1 =12

s2 =12+

122 =

34

= 1− 122

s3 =12+

122 +

123 =

78

= 1− 123

s4 =12+

122 +

123 +

124 =

1516

= 1− 124

...

sn =12+

122 +

123 + · · ·+ 1

2n = 1− 12n

Therefore

limn→∞

sn = limn→∞

(

1− 12n

)

= 1,

that is, the sequence{sn} converges to 1. Hence the series12+

122 +

123 + · · · is convergent and

its sum is 1. ♣

Definition 7.3 Thegeometric seriesis the series of the form

a+ar+ar2+ar3+ · · ·+arn−1+ · · ·=∞

∑n=1

arn−1, a 6= 0

and the number r is called theratio for the series.


Here are some examples:

• 1+2+4+8+ · · ·+2n−1+ · · · ; a= 1, r = 2

• 12− 1

4+

18− 1

16+ · · ·+(−1)n+1 1

2n + · · · ; a=12, r =−1

2

• 310

+3

102 +3

103 +3

104 + · · ·+ 310n + · · · ; a=

310

, r =110

Theorem 7.2 The geometric series

∞

∑n=1

arn−1 = a+ar+ar2+ar3+ · · ·

is convergent if|r|< 1and its sum is

s=∞

∑n=1

arn−1 =a

1− r|r|< 1

If |r| ≥ 1, the geometric series is divergent.

Example 7.4 Find the sum of43+

49+

427

+481

+ · · · .

Solution . . . . . . . . .

Example 7.5 Is the series∞

∑n=1

22n31−n convergent or divergent?

Solution . . . . . . . . .

Example 7.6 Show that the series∞

∑n=1

1n(n+1)

is convergent, and find its sum.

Solution Thenth partial sum of the series is

sn =n

∑i=1

1i(i +1)

=1

1 ·2+1

2 ·3+1

3 ·4+ · · ·+ 1n(n+1)

We can simplify this expression if we use the partial fraction decomposition

1i(i +1)

=1i− 1

i +1

Thus, we have

sn =n

∑i=1

1i(i +1)

=n

∑i=1

(1i− 1

i +1

)

=

(

1− 12

)

+

(12− 1

3

)

+

(13− 1

4

)

+ · · ·+(

1n− 1

n+1

)

= 1− 1n+1


and so

limn→∞

sn = limn→∞

(

1− 1n+1

)

= 1−0= 1

Therefore, the given series is convergent and

∞

∑n=1

1n(n+1)

= 1 ♣

Definition 7.4 Theharmonic seriesis the divergent series of the form

∞

∑n=1

1n= 1+

12+

13+

14+ · · ·

Theorem 7.3 If the series∞

∑n=1

an is convergent, thenlimn→∞

an = 0.

Note. The converse of Theorem 7.3 is not true in general. If limn→∞

an = 0, we can not conclude that

∑an is convergent. Observe that for the harmonic series∑ 1n we havean = 1/n→ 0 asn→ ∞, but

the harmonic series is divergent.

7.3 Convergence Tests

In this section we will develop various tests that can be usedto determine whether a given seriesconverges or diverges.

7.3.1 The Divergence Test

Theorem 7.4 (The Divergence Test)

(a) If limn→∞

an does not exist or iflimn→∞

an 6= 0, then the series∞

∑n=1

an diverges.

(b) If limn→∞

an = 0, then the series∞

∑n=1

an may either converges or diverges.

Example 7.7 Show that the series∞

∑n=1

n2

5n2+4diverges.

Solution . . . . . . . . .

Theorem 7.5 If ∑an and ∑bn are convergent series, then so are the series∑can (where c is aconstant),∑(an+bn), and∑(an−bn), and

(i)∞

∑n=1

can = c∞

∑n=1

an


(ii)∞

∑n=1

(an+bn) =∞

∑n=1

an+∞

∑n=1

bn

(ii)∞

∑n=1

(an−bn) =∞

∑n=1

an−∞

∑n=1

bn

Example 7.8 Find the sum of the series

∞

∑n=1

(3

n(n+1)+

12n

)

.

Solution . . . . . . . . .

Theorem 7.6 The p-series∞

∑n=1

1np is convergent if p> 1 and divergent if p≤ 1.

Example 7.9

(a) The series∞

∑n=1

1n3 =

113 +

123 +

133 +

143 + · · · is convergent because it is ap-series with

p= 3> 1.

(b) The series∞

∑n=1

1

n1/3= 1+

13√

2+

13√

3+

13√

4+ · · · is divergent because it is ap-series with

p= 13 < 1. ♣

7.3.2 The Limit Comparison Test

Theorem 7.7 (The Limit Comparison Test) Suppose that∑an and∑bn are series with positiveterms. If

limn→∞

an

bn= c

where c is a finite number and c> 0, then either both series converge or both diverge.

Example 7.10 Determine whether the series converges or diverges.

(a)∞

∑n=1

2n2+3n√5+n5

(b)∞

∑n=1

12n−1

Solution . . . . . . . . .

7.3.3 The Ratio Test

Theorem 7.8 (The Ratio Test)Let ∑an be a series with positive terms and suppose that

ρ = limn→∞

an+1

an.

(i) If ρ < 1, then the series converges.


(ii) If ρ > 1 or ρ = ∞, then the series diverges.

(iii) If ρ = 1, the series may converge or diverge, so that another test must be tried.


(a)∞

∑n=1

n3

3n (b)∞

∑n=1

(2n)!4n

Solution . . . . . . . . .

7.3.4 The Root Test

Theorem 7.9 (The Root Test)Let ∑an be a series with positive terms and suppose that

ρ = limn→∞

n√

an = limn→∞

(an)1/n

(i) If ρ < 1, the series converges.

(ii) If ρ > 1 or ρ = ∞, then the series diverges.

(iii) If ρ = 1, the series may converge or diverge, so that another test must be tried.


(a)∞

∑n=1

(2n+33n+2

)n

(b)∞

∑n=2

(1

lnn

)n

Solution . . . . . . . . .

7.4 Taylor and Maclaurin Series; Power Series

Taylor and Maclaurin Series

Definition 7.5 If f has derivatives for all order at a, then we call the series

∞

∑n=0

f (n)(a)n!

(x−a)n = f (a)+ f ′(a)(x−a)+f ′′(a)

2!(x−a)2

+f ′′′(a)

3!(x−a)3+ · · ·+ f (n)(a)

n!(x−a)n+ · · ·

theTaylor series of the functionf at a (or abouta or centered ata). In the special case wherea= 0, this series becomes

∞

∑n=0

f (n)(0)n!

xn = f (0)+ f ′(0)x+f ′′(0)

2!x2+

f ′′′(0)3!

x3+ · · ·+ f (n)(0)n!

xn+ · · ·

in which case we call it theMaclaurin series.


Example 7.13 Find the Maclaurin series for

(a) f (x) = ex (b) f (x) = sinx

Solution . . . . . . . . .

Example 7.14 Find the Taylor series of f(x) =1x

centered at1.

Solution . . . . . . . . .

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