XEQ 201: Calculus II

Contents

Course description iv

References iv

Chapter 1. Applications of Differentiation 1

1.1. Mean value theorems of differential calculus 1

1.2. Using differentials and derivatives 5

1.3. Extreme Values 7

1.4. Curve Sketching 18

1.5. Linear Approximations 22

1.6. Taylor Polynomial 26

Chapter 2. Integration 31

2.1. Antiderivatives 31

2.2. The Definite Integral 33

2.3. The Fundamental Theorem of Calculus 38

iii

Course description

Application of differentiation. Taylor theorem. Mean Value theorem of

differential calculus. Methods of integration. Applications of integration.

References

1. Calculus: A complete course by Robert A. Adams and Christopher

Essex.

2. Fundamental methods of mathematical economics by Alpha C. Chi-

ang.

3. Schaums outline series: Introduction to mathematical economics

by Edward T. Dowling

iv

CHAPTER 1

Applications of Differentiation

1.1. Mean value theorems of differential calculus

Theorem 1.1.1 (Mean Value Theorem).

Suppose that the function f is continuous on the closed finite interval [a, b]

and that it is differentiable on the interval (a, b). Then a point c (a, b)such that

f (b) f (a)b a = f

(c) .

It means that the slope of the chord joining the points (a, f (a)) and (b, f (b))

is equal to the slope of the tangent line to te curve y = f (x) at the point

(c, f (c)) so that the two lines are parallel.

Fig 2.28

Example 1.1.1.

Verify the conclusion of the mean value theorem for f (x) =x on the

interval [a, b], where a x b.

Solution. We are to show that c (a, b) such that

f (b) f (a)b a = f

(c)

1

XEQ 201

so long as f is continuous on [a, b] and is differentiable on (a, b). Now

f (x) = 12x

, f (a) =a, f (b) =

b.

12c

=

bab a =

ba(

ba)(

b+a) = 1

b+a.

The above equality implies thatc =

b+a

2 so that c =(

b+a

2

)2. Since

a < b, we have

a =

(a+a

2

)2 2pi, then sinx 1 < 2pi < x. If 0 < x 2pi, then byMVT, c (0, 2pi) such that

sinx

x=

sinx sin 0x 0 = [MVT on [0, x]] =

d

dxsinx

x=c

= cos c < 1

which implies that sinx < x in this case too.

Definition 1.1.1 (Increasing and decreasing functions).increasing

decreasing

functions

Suppose that

the function f is defined on an interval I and that x1 and x2 are two points

in I.

(a) If f (x2) > f (x1) whenever x2 > x1, we say that f is increasing on

I.

(b) If f (x2) < f (x1) whenever x2 > x1, we say that f is decreasing on

I.

(c) If f (x2) f (x1) whenever x2 > x1, we say that f is non-decreasingon I.

(d) If f (x2) f (x1) whenever x2 > x1, we say that f is non-increasingon I.

Diagram Fig 2.31

2

XEQ 201

Theorem 1.1.2.

derivative of

increasing and

decreasing

functions

Let J be an open interval and let I be an interval consisting of all points

in J and possibly one or both of the end points of J . Suppose that f is

continuous on I and differentiable on J .

(a) If f (x) > 0 for all x J , then f is increasing on I.(b) If f (x) < 0 for all x J , then f is decreasing on I.(c) If f (x) 0 for all x J , then f is non-decreasing on I.(d) If f (x) 0 for all x J , then f is non-increasing on I.

Example 1.1.3.

On what intervals is the function f (x) = x3 12x+ 1 increasing? On whatintervals is it decreasing?

Solution. f (x) = 3x212 = 3 (x 2) (x+ 2). It follows that f (x) >0 when x < 2 or x > 2 and f (x) < 0 when 2 < x < 2. Therefore fis increasing on the intervals (,2) and (2,) and is decreasing on theinterval (2, 2).

Diagram Fig 2.32

Example 1.1.4.

Show that f (x) = x3 is increasing on any interval.

Solution. Let x1, x2 be any two real numbers satisfying x1 < x2. Since

f (x) = 3x2 > 0 for all x 6= 0, we have that f (x1) < f (x2) if eitherx1 < x2 0 or 0 x1 < x2. If x1 < 0 < x2, then f (x1) < 0 < f (x2). Thusf is increasing on every interval.

Theorem 1.1.3.

derivetive of

constant

function

If f is continuous on an interval I and f (x) = 0 at every interior point ofI, then f (x) = C, a constant on I.

3

XEQ 201

Theorem 1.1.4.

derivative at

interior extreme

point

If f is defined on an open interval (a, b) and achieves a maximum (or a

minimum) at the point c (a, b), and if f (c) exists, then f (c) = 0.Values of x where f (x) = 0 are called critical points of the function f .

Theorem 1.1.5 (Rolles Theorem).

Suppose that the function g is continuous on the closed finite interval [a, b]

and if it is differentiable on the open interval (a, b). If g (a) = g (b), apoint c (a, b) such that g (c) = 0.

Theorem 1.1.6 (The Generalized Mean Value Theorem).

If functions f and g are both continuous on [a, b] and differentiable on (a, b),

and if g (x) 6= 0 for every x (a, b), then a number c (a, b) such that

f (b) f (a)g (b) g (a) =

f (c)g (c)

.

Exercise 1.1.

1. Illustrate the MVT by finding any points in the open interval (a, b)

where the tangent line is parallel to the chord joining (a, f (a)) and

(b, f (b)).

(a) f (x) = x2 on [a, b]; Ans: c = b+a2 .

(b) f (x) = x3 3x+ 1 on [2, 2]; Ans: c = 23.

2. Show that tanx > x for 0 < x < pi2 .

3. Find intervals of increase and decrease of the following functions

(a) f (x) = x3 4x + 1. Ans: Increasing on(, 2

3

)and(

23,)

; decreasing on( 2

3, 2

3

).

(b) f (x) =(x2 4)2. Ans: Increasing on (2, 0) and (2,);

decreasing on (,2) and (0, 2).(c) f (x) = x3 (5 x)2. Ans: Increasing on (, 3) and (5,);

decreasing on (3, 5).

(d) f (x) = x+ sinx. Ans: Increasing on (,).4

XEQ 201

1.2. Using differentials and derivatives

Suppose dx is regarded as a new independent variable called the differ-

ential of x we can define a new dependent variable dy, called the differentail

of y as a function of x and dx by

dy =dy

dxdx = f (x) dx.

For example if y = x2, then dy = 2xdx means the same thing as dy/dx = 2x.

If f (x) = 1/x, then df (x) = (1/x2) dx.If y is a function of x, y = f (x), then denoting a small change in x by

dx instead of x, the corresponding small change in y, y is approximated

by the differential dy, i.e.

y dy = f (x) dx.Diagram Fig 2.25

Example 1.2.1.

Without using a scientific calculator, determine by a pproximately how much

the value of sinx increases as x increases from pi/3 to (pi/3) + 0.006. To 3

decimal places, what is the value of sin ((pi/3) + 0.006)?

Solution. If y = sinx, then dydx = cosx. Now x =pi3 1.0472 and

dx = 0.006. Therefore

dy =dy

dxdx = cosxdx = cos

(pi3

) 0.006 = 1

2(0.006) = 0.003.

This means that the change in the value of sinx is approximately 0.003. Now

5

XEQ 201

sin((pi

3

)+ 0.006

) sin

(pi3

)+ 0.003 = = 0.869(3 d. p.).

Suppose changes in x are measured with respect to the size of x, then

relative change in x =dx

xand percentage change in x = 100

dx

x.

Differentials and point elasticity1

For a demand function Q = f (P ), the elasticity is defined as

Q/Q

P/P=

relative change in Q

relative change in P.

point elasticity If the change in P is infinitesimal, then the expressions Q and P reduce

to the differentials dP and dQ. In that case the elasticity measure assumes

the sense of point elasticity of the demand function which is denoted by d.

Thus

d dQ/QdP/P

=dQ/dP

Q/P.

The numerator in the right hand is the derivative (or marginal2) function

of the demand function while the denominator is the average function of the

demand function. Thus the point elasticity is a ratio of the two functions.

In general, for any given total function y = f (x), the point elasticity of y

w.r.t. x is

yx =dy/dx

y/x=

marginal function

average function.

The absolute value of the point elasticity measure is used in deciding

whether the function is elastic at a particular point. In the case of a demand

function,

The demand is

elastic

of unit elasticity

inelastic

if

|d| > 1|d| = 1|d| < 1.

1Elasticity is the measure of how an economic variable responds to change in anothervariable. An elastic variable is one which responds more than proportionally to changesin other variables. An inelastic variable is one which changes less than proportionally inresponse to changes in other variables.2Marginal denotes the rate of change of a quantity with respect to a variable on which itdepends.

6

XEQ 201

at a given point.

Example 1.2.2.

Find d for the demand function Q = 100 2P . Determine the pointelasticity at P = 25.

Solution. dQdP = 2 and QP = 1002PP . Therefore

d =2

(100 P ) /P =P

P 50 .

Thus

d

P=25

= = 1.

Therefore the demand is of unit elasticity when P = 25.

Exercise 1.2.

1. Use differentials to determine approximate change in the values of

the given function as its argument changes from the given value to

the given amount. What is the approximate value of the function

after the change?

(a) y = 1/x as x increases from 2 to 2.01.

(b) h (t) = cos (pit/4) as t increases from 2 to 2 + (1/10pi).

2. Find the approximate percentage changes in the given function that

will result from an increase of 2% in the value of x.

(a) y = x2 (b) y = 1/x2

3. Given the consumption function C = a+ bY (with a > 0; 0 < b 0.

(c) Show that the consumption is inelastic at all positive income

levels.

1.3. Extreme Values

Maximum and Minimum Values

7

XEQ 201

Definition 1.3.1.absolute

extreme values

Function f has an absolute maximum value f (x0) at

x0 in its domain if f (x) f (x0) holds for every x in the domain of f .Similarly, f has an absolute minimum value f (x0) at x0 in its domain

if f (x) f (x0) holds for every x in the domain of f .

Remark 1.

1.extreme value is

unique, can

occur at several

points

A function will have only one absolute maximum (or minimum)

value if it exists. However the value can occur at many points. For

example, f (x) = sinx has absolute maximum of 1 but it occurs at

every point pi2 + 2npi, n Z.2.existence of

extreme value

not guaranteed

A function need not have any extreme value. The function f (x) =1x becomes arbitrarily large as x approaches 0 from the right, and

so has no finite absolute maximum value.

Theorem 1.3.1.

existence of

extreme values

for closed finite

intervals

If the domain of the function f is a closed, finite interval or a union of

finitely many such intervals, and if f is continuous on that domain, then f

must have an absolute maximum value and absolute minimum value.

Definition 1.3.2.local extreme

values

Function f has a local maximum value (loc. max.)

f (x0) at the point x0 in its domain provided a number h > 0 such thatf (x) f (x0) whenever x is in the domain of f and |x x0| < h.

Similarly, f has a local minimum value (loc. min.) f (x1) at the point x1

in its domain provided a number h > 0 such that f (x) f (x1) wheneverx is in the domain of f and |x x1| < h.

Diagram Fig 4.17

8

XEQ 201

From the above figure we see that local extreme values can occur at any

of the following points.

critical, singular

or end points

(i) critical points of f ; points x D (f) where f (x) = 0.(ii) singular points of f ; points x D (f) where f (x) is not defined.

(iii) endpoints of the domain of f ; points that do belong to D (f) but

are not interior points of D (f).

In the figure above, x1, x3, x4 are critical points, x2 and x5 are singular

points and a and b are endpoints.

Theorem 1.3.2.

exteme values

occur at

critical, singular

or end points

If the function f is defined on an interval I and has a local maximum (or

local minimum) value at the point x = x0 in I, then x0 must be either a

critical point of f , a singular point of f , or and endpoint of I.

Example 1.3.1.

Find the maximum and minimum values of the function g (x) = x3 3x2 9x+ 2 on the interval 2 x 2.

Solution. Since g is a polynomial it can never have a singular point.

For critical points we calculate

g (x) = 3x2 6x 9 = 3 (x+ 1) (x 3) = 0.Thus x = 1 or x = 3. But x = 3 is not in the domain of g and so weignore it. We then investigate the endpoints x = 2 and x = 2 and criticalpoint x = 1.

g (2) = 0, g (1) = 7, g (2) = 20The maximum value of g on 2 x 2 is at the critical point x = 1,

and the minimum value is at the endpoint x = 2.

Diagram Fig 4.19

9

XEQ 201

Example 1.3.2.

Find the maximum and minimum values of h (x) = 3x2/32x on the interval[1, 1].

Solution. The derivative of h is

h (x) = 2x1/3 2.Note that x1/3 is not defined at x = 0 in D (h), so x = 0 is a singular pointof h. Also h (x) = 0 at x1/3 = 1, that is at x = 1, which also happens tobe an endpoint of the domain of h. We therefore examine the values of h at

endpoints x = 1 and x = 1 and at the singular point x = 0.h (1) = 5, h (0) = 0, h (1) = 1The function h has a maximum value 5 at the endpoint x = 1 and a

minimum value 0 at the singular point x = 0.

Diagram Fig 4.20

Theorem 1.3.3 (The first derivative test).

first derivative

test Part I: Testing interior critical points and singular points.

Suppose f is continuous at x0, and x0 is not an endpoint of the

domain of f .

(a) If an open interval (a, b) containing x0 such that f (x) > 0 on(a, x0) and f

(x) < 0 on (x0, b), then f has a local maximumvalue at x0.

(b) If an open interval (a, b) containing x0 such that f (x) < 0on (a, x0) and f

(x) > 0 on (x0, b), then f has a local minimumvalue at x0.

10

XEQ 201

Part II: Testing endpoints of the domain.

Suppose a is a left endpoint of the domain of f and f is right

continuous at a.

(c) If f (x) > 0 on some interval (a, b) , then f has a local mini-mum value at a.

(d) If f (x) < 0 on some interval (a, b) , then f has a local maxi-mum value at a.

Suppose b is a right endpoint of the domain of f and f is left

continuous at b.

(e) If f (x) > 0 on some interval (a, b) , then f has a local maxi-mum value at b.

(f) If f (x) < 0 on some interval (a, b) , then f has a local mini-mum value at b.

Remark 2.

If f is positive (or negative) on both sides of a critical or singular point,then f has neither a maximum nor a minimum value at that point.

Example 1.3.3.

Find the local and absolute extreme values of f (x) = x4 2x2 3 on theinterval [2, 2]. Sketch the graph of f .

Solution. f (x) = 4x3 4x = 4x (x2 1) = 4x (x+ 1) (x 1).The critical points are x = 1, x = 0, x = 1. The corresponding

values are f (1) = 4, f (0) = 3, f (1) = 4. There are no singularpoints. The values of f at the endpoints 2 and 2 are f (2) = 5 andf (2) = 5.We summarize the positive/negative properties of f (x) and the

implied increasing/decreasing behaviour of f (x) in a chart form below;

EP CP CP CP EP

x 2 1 0 1 2f 0 + 0 0 +f max min max min max

Diagram Fig 4.21

11

XEQ 201

Theorem 1.3.4 (Existence of extreme values on open intervals).

extreme values

in open

intervals

If f is continuous on an open interval (a, b), and if

limxa+

f (x) = L and limxb

f (x) = M

then the following conclusions hold

(i) If f (u) > L and f (u) > M for some u (a, b), then f has anabsolute maximum value on (a, b).

(ii) If f (u) < L and f (u) < M for some u (a, b), then f has anabsolute minimum value on (a, b).

In this theorem a may be and b may. Also, either or both L andM may be or .

Example 1.3.4.

Show that f (x) = x + 4x has an absolute minimum value on the interval

(0,), and find that minimum value.

Solution. We have

limx0+

f (x) = and limx f (x) =.

Since f (1) = 5

XEQ 201

Concavity and Inflections

Definition 1.3.3. We say that a function f is concave up on an open

interval I if it is differentiable there and and if f is an increasing functionon I. Similarly, f is concave down on I if f exists and is a decreasingfunction on I.

Diagram Fig 4.26

Definition 1.3.4. inflection pointWe say that the point (x0, f (x0)) is an inflection

point of the curve y = f (x) if the following two conditions hold

(a) the graph of y = f (x) has a tangent line at x = x0, and

(b) the concavity of f is opposite on opposite sides of x0.

Diagram Fig 4.27, 4.28, 4.29

Theorem 1.3.5 (Concavity and the second derivative).

concavity and

second

derivative test(a) If f (x) > 0 on interval I, then f is concave up on I.(b) If f (x) < 0 on interval I, then f is concave down on I.

13

XEQ 201

(c) If f has an inflection point at x0 and f (x0) exists, then f (x0) = 0

Proof.

(a) and (b) follow from applying Theorem 1.1.2 to the derivative

f of f .(c) If f has an inflection point at x0 and if f

(x0) exists, then f mustbe differentiable in an open interval containing x0. Since f

isincreasing on one side of x0 and decreasing on the other side, it

must have a local maximum or minimum value at x0. Thus f (x0)

must be zero.

Example 1.3.5.

Determine the intervals of concavity of f (x) = x6 10x4 and the inflectionpoints of its graph.

Solution. We have

f (x) = 6x540x3, f (x) = 30x4120x2 = 30x2 (x 2) (x+ 2)Thus f (x) = 0 at x = 2, 0. These three values of x induce/generate thefollowing four intervals on which we test concavity.

Consider the interval (,2). Here for x = 3,f (3) = 30 (3)2 (5) (1) > 0

and so f is concave up.

For x = 1 (2, 0), we havef (1) = 30 (1)2 (3) (1) < 0

and so f is concave down.

For x = 1 (0, 2), we havef (1) = 30 (1)2 (1) (3) < 0

and so f is concave down.

For x = 3 (2,), we havef (3) = 30 (3)2 (1) (5) > 0

and so f is concave up. We summarize this information in the following

chart.

14

XEQ 201

x 2 0 2f + 0 0 0 +f ^ infl _ _ infl ^

Diagram Fig 4.31

Example 1.3.6.

Determine the intervals of increase and decrease, the local extreme values

and the concavity of f (x) = x42x3 + 1. Use the information to sketch thegraph of f .

Solution.

f (x) = 4x36x2 = 2x2 (2x 3) , f (x) = 12x212x = 12x (x 1)Thus f (x) = 0 when x = 0, 1.5 and f (x) = 0 when x = 0, 1. The

critical points generates the intervals (, 0), (0, 1.5) and (1.5,) on whichwe test whether the function is increasing or decreasing.

For the interval (, 0) take x = 1 and observe that f (1) < 0implying that the function is decreasing there. For the interval (0, 1.5) take

x = 1 and observe that f (1) < 0 implying that the function is decreasingthere. For the interval (1.5,) take x = 2 and observe that f (2) > 0implying that the function is increasing there.

On the other hand the values for which f = 0 generate the followingthe intervals (, 0), (0, 1) and (1,) on which we test concavity. For theinterval (, 0), take x = 1 and observe that

f (1) = 12 (1) (2) > 0so that f is concave up. For the interval (0, 1), take x = 0.5 and obtain

f (0.5) = 12 (0.5) (0.5) < 015

XEQ 201

so that f is concave up there. For the interval (1,), take x = 2 and obtainf (0.5) = 12 (2) (1) > 0

so that f is concave up there.

CP CP

x 0 1 1.5

f - 0 0 +f + 0 0 + +f infl infl min

^ _ ^ ^

Diagram Fig 4.32

Theorem 1.3.6 (The second derivative test).

second

derivative test

of extreme

values

(a) If f (x0) = 0 and f (x0) < 0 then f has a local maximum valueat x0.

(b) If f (x0) = 0 and f (x0) > 0 then f has a local minimum value atx0.

(c) If f (x0) = 0 and f (x0) = 0 no conclusion can be drawn; fmay have a local maximum or a local minimum or it may have an

inflection point instead.

Proof.

(a) Suppose that f (x0) = 0 and f (x0) < 0. Since

limh0

f (x0 + h)h

= limh0

f (x0 + h) f (x0)h

= f (x0) < 0

16

XEQ 201

it follows that f (x0 + h) < 0 for sufficiently small positive h andf (x0 + h) > 0 for sufficiently small negative h. By the first deriv-ative test, f must have a local maximum value at x0.

(b) Exercise

(c) No conclusion can be made.

Example 1.3.7.

Find and classify the critical points of f (x) = x2ex.

Solution.

f (x) = 2xex x2ex = xex (2 x) = 0 at x = 0 or 2.

f (x) = = (2 4x+ x2) ex.Therefore f (0) = 2 > 0 and f (2) = 2e2 < 0. It follows that f has alocal minimum value at x = 0 and a local maximum value at x = 2.

Diagram Fig 4.33

Exercise 1.3.

1. Determine whether the given function has any local or absolute

extreme values, and find those values if possible

(a) f (x) = x+ 2 on [1, 1].(b) f (x) = x+ 2 on [1, 1).

(c) f (x) = x2 1 on [2, 3].(d) f (x) = x3 +x42 on [a, b].

17

XEQ 201

2. Determine the intervals of concavity of the given function, and

locate any inflection points.

(i) f (x) =x.

(ii) f (x) = x2 + 2x+ 3.

(iii) f (x) = 10x3 3x5.

(iv) f (x) =(3 x2)2.

(v) f (x) =(x2 4)3.

3. Classify the critical points of the following functions using the sec-

ond derivative test whenever possible

(i) f (x) = x (x 2)2 + 1.(ii) f (x) = x3 + 1x .

(iii) f (x) = x1+x2

.

(iv) f (x) = x lnx.

(v) f (x) =(x2 4)3.

1.4. Curve Sketching

When sketching the graph y = f (x) of a function f , we have the follow-

ing three sources of useful information;

1. the function f itself:- from this we determine the coordinates of

some points on the graph, the symmetry of the graph and any

asymptotes.

2. the first derivative, f :- from this we determine the intervals ofincrease and decrease and the location of any local extreme values.

3. the second derivative, f :- from this we determine the the concavityand inflection points, and sometimes the extreme values.

Other important points besides critical points, singular points and points of

inflection are the intercepts.

Asymptotes

Asymptotes are straight lines to which a curve draws arbitrarily close as

it recedes to infinite distance from the origin.

Definition 1.4.1. The graph of y = f (x) has a vertical asymptote at

x = a if either

limxa

f (x) = or limxa+

f (x) = or both.

18

XEQ 201

Definition 1.4.2. The graph of y = f (x) has a horizontal asymptote

at y = L if either

limx f (x) = L or limx f (x) = L or both.

Definition 1.4.3. The straight line y = ax + b (a 6= 0) is an obliqueasymptote of the graph of y = f (x) if either

limx (f (x) (ax+ b)) = 0 or limx (f (x) (ax+ b)) = 0 or both.

Suppose that f (x) = Pm (x) /Qn (x), where Pm and Qn are polynomials

of degree m and n respectively. Suppose also that Pm and Qn have no

common linear factor. Then

(a) The graph of f has a vertical asymptote at every position x such

that Qn (x) = 0.

(b) The graph of f has a two-sided horizontal asymptote y = 0 if

m < n.

(c) The graph of f has two-sided horizontal asymptote y = L, (L 6= 0)if m = n where L is the quotient of the coefficients of the highest

terms in Pm and Qn.

(d) The graph of f has two-sided oblique asymptotes if m = n + 1.

This asymptote can be found by dividing Qn by Pm to obtain a

linear quotient, ax+ b, and a remainder R, a polynomial of degree

at most n 1, i.e.

f (x) = ax+ b+R (x)

Qn (x).

The oblique asymptote is y = ax+ b.

(e) The graph f has no horizontal or oblique asymptote if m > n+ 1.

Example 1.4.1.

Find the oblique asymptote of y = x3

x2+x+1.

Solution. In this case m = 3 = 2 + 1 = n+ 1 and so by long division

we obtain

x3

x2 + x+ 1= x 1 + 1

x2 + x1.

Therefore the oblique asymptote is y = x 1.19

XEQ 201

Example 1.4.2.

Sketch the graph of y = x2+2x+42x .

Solution.

y =x2 + 2x+ 4

2x=x

2+ 1 +

2

x.

y (x) = 12 2x2

and y (x) =4

x3.

From y

Domain: all x 6= 0.Vertical asymptote: x = 0

Oblique asymptote: y = x2 + 1 since y (x2 + 1

)= 2x 0 as x

x2 + 2x+ 4 = (x+ 1)2 + 3 3 for all x. y is not defined at x = 0.From y

Critical points: x242x2

= 0 x = 2 and y is not defined at x = 0.From y

y is not defined at x = 0 and y = 0 nowhere.Facts about the graph are summarized in the chart below

CP ASY CP

x 2 0 2y + 0 undef 0 +y 0 undef + +y max undef min

_ _ ^ ^Diagram Fig 4.39

20

XEQ 201

Example 1.4.3.

Sketch the graph of f (x) = x21x24 .

Solution.

f (x) = = 6x(x2 4)2 and f

(x) = = 6(3x2 + 4

)(x2 4)3 .

From f

Domain: all x 6= 2.Vertical asymptotes: x = 2 where the denominator x2 4 = 0Horizontal asymptotes: lim

x f (x) =1 01 0 = 1

horizontal asymptote is y = 1Symmetry: is about the yaxis since the function is even.Intercepts: When y = 0, x = 1, and when x = 0, y = 14

Hence the intercepts are (0, 1/4), (1, 0) and (1, 0).The two vertical asymptotes divide the graph into three components. The

outer components require points with |x| > 2. WOLG we consider the points(3, 8/5) and (3, 8/5).From f

Critical point is at x = 0, and f is not defined at x = 2.From f

f (x) = 0 nowhere and f not defined at x = 2.ASY CP ASY

x 2 0 2f undef + 0 undef f + undef undef +f undef max undef

^ _ _ ^Diagram Fig 4.40

21

XEQ 201

Exercise 1.4.

Sketch the graphs of the given functions

1. f (x) =(x2 1)3.

2. f (x) = 2xx .3. f (x) = x

3

1+x .

4. f (x) = 12x2 .

5. f (x) = x3

x2+1.

1.5. Linear Approximations

same direction

as the curve

The tangent line to the graph y = f (x) at P (a, f (a)) is the best linear

approximation of the graph near the point since it goes through P in the

same direction as the curve y = f (x).

Diagram Fig 4.63

The height L to the tangent line gives approximate values of f (x) for values

of x near a. The tangent line has equation y = f (a) + f (a) (x a).

Definition 1.5.1. The linearization of the function f about a is the

function L defined by

L (x) = f (a) + f (a) (x a) .We say that L (x) provides a linear approximation for value of f near a.

Example 1.5.1.

Find the linearization of

(a) f (x) =

1 + x about x = 0 and

(b) g (t) = 1/t about t = 1/2.

Solution.

22

XEQ 201

(a) f (0) = 1. Also

f (x) =1

2

1 + xso that f (0) =

1

2.

Therefore L (x) = f (0) + f (0) (x 0) = 1 + 12x.(b) g (1/2) = 2, g (t) = 1

t2, g (1/2) = 4. Therefore

L (t) = g

(1

2

)+ g

(1

2

)(t 1/2) = = 4 4t.

The linearization can be given in terms of x as follows; recall

differentialsf (a+ x) L (a+ x) = f (a) + f (a) x.

Example 1.5.2.

A ball of ice melts so that its radius decreases from 5 cm to 4.92 cm. By

approximately how much does the volume of the ball decreases?

Solution. The volume of the ball is V = 4pir3

3 , so that

dV

dr= 4pir2 and L (r + r) = V (r) + 4pir2r.

Thus

V 4pir2r.For r = 5 and r = 0.08, we have

V 4pi (52) (0.08) = 8pi 25.13.The volume of the ball decreases by about 25 cm3.

Example 1.5.3.

Use the linearization forx about x = 25 to find an approximate value for

26.

Solution. Let f (x) =x. Then the linearization of f (x) is L (x) =

f (25)+f (25) (x 25). Now f (25) = 5, f (x) = 1/ (2x) so that f (25) =1/10.

26 = f (26) L (26) = 5 + 110

(26 25) = 5.1.Error Analysis

The error emanating from an approximation of f (x) using L (x) is

E (x) = f (x) L (x) = f (x) f (a) f (a) (x a) .23

XEQ 201

It is the vertical distance between the graph of f and the tangent line to

that graph at x = a as shown below.

Diagram Fig 4.64

Observe that if x is near a, then E (x) is small compared to the horizontal

distance between x and a.

Theorem 1.5.1 (An error formula for linearization).

If f (t) exists for all t in an interval containing a and x, then some points between a and x such that the error E (x) = f (x) L (x) in the linearapproximation f (x) L (x) = f (a) + f (a) (x a) satisfies

E (x) =f (s)

2(x a)2 .

Proof. Assume that x > a. Then

E (t) = f (t) f (a) f (a) (t a) , E (t) = f (t) f (a) .Applying the Generalized MVT to the two functions E (t) and (t a)2 on[a, x] bearing in mind that E (a) = 0 one obtains u (a, x) such that

E (x)

(x a)2 =E (x) E (a)

(x a)2 (a a)2

=

[MV T

on [a, x]

]=

E (u)2 (u a) =

f (u) f (a)2 (u a) =

[MV T

on [a, u]

]=

1

2f (s)

for some s (a, u); by second application Generalized MVT, now to f on[a, u]. Thus

E (x) =f (s)

2(x a)2 .

The following three corollaries are consequences of the above theorem

24

XEQ 201

Corollary 1. If f (t) has constant sign between a and x, then theerror E (x) in the approximation f (x) L (x) in Theorem 1.5.1 has thesame sign; i.e. if f (t) > 0 between a and x, then f (x) > L (x) and iff (t) < 0 between a and x, then f (x) < L (x).

Corollary 2. If |f (t)| < K for all t between a and x (where K is aconstant), then |E (x)| < (K/2) (x a)2.

Corollary 3. If f (t) satisfies M < f (t) < N for all t between aand x where M and N are constants, then

L (x) +M

2(x a)2 < f (x) < L (x) + N

2(x a)2 .

If M and N have the same sign, a better approximation to f (x) is given by

the midpoint of this interval containing f (x);

f (x) L (x) + M +N4

(x a)2 .For this approximation the error is less than half the length of the interval:

|E (x)| < N M4

(x a)2 .Example 1.5.4.

Determine the sign and estimate the size of the error in the approximation26 5.1. Use these to give a small interval that you can be sure contains26.

Solution. For f (t) = t1/2, we have

f (t) =1

2t1/2 and f (t) = 1

4t3/2.

Thus for 25 < t < 26, we havef (t) < 0, so

26 = f (26) < L (26) = 5.1

implying that E (26) < 0. Since t > 25, we have t3/2 > 253/2 = 125, sof (t) = 14 1t3/2 < 14 1125 = 1500 ,

and

|E (26)|Col 3 with N= 1

500 L (26)0.001 = 5.099 and 26 lies in the interval (5.099, 5.1).

25

XEQ 201

Exercise 1.5.

Find the linearization of the given function about the given point.

1. x2 about x = 3

2.

4 x about x = 03. 1/ (1 + x)2 about x = 2

4. sinx about x = pi

5. sin2 x about x = pi/6

1.6. Taylor Polynomial

If f (n) (x) exists in an open interval containing x = a, then the polyno-

mial

Pn (x) = f (a)+f (a)

1!(x a)+f

(a)2!

(x a)2+ +f(n) (a)

n!(x a)n

matches f and its first n derivatives at x = a;

Pn (a) = f (a) , Pn (a) = f

(a) , P n (a) = f (a) , , P (n) (a) = f (n) (a) ,

and so describes f (x) near a better than any other polynomial of degree at

most n. Pn is called the nthorder Taylor polynomial for f about a.

When a = 0, the corresponding polynomial is called Maclaurin polyno-

mial.

Example 1.6.1.

Find the following Taylor polynomials

(a) P2 (x) for f (x) =x about x = 25.

(b) P3 (x) for g (x) = lnx about x = e.

Solution.

(a) f (x) = 12x12 , f (x) = 14x

32 . Thus

P2 (x) = f (25) + f (25) (x 25) + f

(25)2!

(x 25)2

= 5 +1

10(x 25) 1

1000(x 25)2 .

26

XEQ 201

(b) g (x) = 1x , g (x) = 1

x2, g (x) = 3

x3. Thus

P3 (x) = g (e) + g (e) (x e) + g

(e)2!

(x e)2 + g (e)3!

(x e)3

= 1 +1

e(x e) 1

2e2(x e)2 + 1

3e3(x e)3 .

Example 1.6.2.

Find the nthorder Maclaurin polynomial Pn (x) for ex. Use P0 (1), P1 (1),P2 (1), to calculate approximate values for e = e1. Stop when the valueis correct to 3 decimal places.

Solution. Since every derivative of ex is ex and so is 1 at x = 0, the

nthorder Maclaurin polynomial for ex is

Pn (x) = 1 +x

1!+x2

2!+x3

3!+ + x

n

n!.

Thus

P0 (1) = 1

P1 (1) = P0 (1) +1

1!= 2

P2 (1) = P1 (1) +1

2!= = 2.5

P3 (1) = P2 (1) +1

3!= = 2.6667

P4 (1) = P3 (1) +1

4!= = 2.7083

P5 (1) = P4 (1) +1

5!= = 2.7166

P6 (1) = P5 (1) +1

6!= = 2.7180

P7 (1) = P6 (1) +1

7!= = 2.7182

Thus e 2.718 to 3 decimal places.Diagram Fig 4.65

27

XEQ 201

Theorem 1.6.1.

If the (n+ 1)st-order derivative, f (n+1) (t), exists for all t in an interval

containing a and x, and if Pn (x) is the nth-order Taylor polynomial for

f about a, then the error En (x) = f (x) Pn (x) in the approximationf (x) Pn (x) is given by

En (x) =f (n+1) (s)

(n+ 1)!(x a)n+1 ,

where s is some number between a and x. The resulting formula

f (x) = f (a)+f (a)

1!(x a)+f

(a)2!

(x a)2+ +f(n) (a)

n!(x a)n+f

(n+1) (a)

(n+ 1)!(x a)n+1 ,

for some s between a and x, is called Taylors formula with Lagrange re-

minder; the Lagrange reminder term is the explicit formula given above for

En (x).

Proof. When n = 0 we have

f (x) = P0 (x) + E0 (x) = f (a) +f (s)

1!(x a)

which is just the MVT

f (x) f (a)x a = f

(s)

for some s between a and x.

When n = 1 we obtain

f (x) = f (a) +f (a)

1!(x a) + f

(s)2!

(x a)2

so that

E (x) = f (x) (f (a) + f (a)) = f (s)2!

(x a)2

which is the error formula for linearization given in Theorem 1.5.1.

We prove the rest of the theorem by mathematical induction. Thus

suppose that the theorem holds for n = k 1, where k 1 is an integer.Then we are assuming that if f is any function whose derivative of order k

exists on an interval containing a and x, then

Ek1 (x) =f (k) (s)

k!(x a)k ,

where s is any number between a and x. To prove that the error formula is

true for n = k, assume WLOG that x > a and apply the Generalized MVT

to the functions Ek (t) and (t a)k+1 on [a.x]. Since Ek (a) = 0, we obtain28

XEQ 201

u (a, x) such thatEk (x)

(x a)k+1 =Ek (x) Ek (a)

(x a)k+1 (a a)k+1 = [GMV T ] =Ek (u)

(k + 1) (u a)k .

Now

Ek (u) =d

dt

(f (t) f (a) f (a) (t a) f

(a)2!

(t a)2

f(k) (a)

k!(t a)k

)t=u

= f (u) f (a) f (a) (u a) f(k) (a)

(k 1)! (u a)k1 .

The last expression can be written as

kf (k) (a)

k (k 1)! (u a) (u a)k1

(u a) =k

(u a)f (k) (a)

k!(u a)k .

This is just Ek1 (u) for the function f instead of f . By induction assump-tion it is equal to

(f )(k) (s)k!

(u a)k = f(k+1) (s)

k!(u a)k ,

for some s between a and u. Therefore

Ek (x) =f (k+1) (s)

(k + 1)!(x a)k+1 .

Example 1.6.3.

Use the second order Taylor polynomial forx about x = 25 found in

Example 1.6.1(a) to approximate

26. Estimate the size of the error, and

specify the interval that you can be sure contains

26.

Solution. In Example 1.6.1(a) we calculated f (x) = (1/4)x3/2and obtained the Taylor polynomial

P2 (x) = 5 +1

10(x 25) 1

1000(x 25)2 .

The required approximation is

26 = f (26) P2 (26) = 5+ 1

10(26 25) 1

1000(26 25)2 = 5.099.

29

XEQ 201

Now f (x) = (3/8)x5/2. For 25 < s < 26, we havef (s) 325, 000 6 (26 25)

3 =1

50, 000= 0.00002.

Therefore,

26 lies in the interval (5.09898, 5.09902).

Exercise 1.6.

Find the indicated Taylor polynomials for the given functions using the

definition of Taylor polynomial.

1. ex about x = 0, order 4.2. lnx about x = 2, order 4.

3.x about x = 4, order 3.

4. 1/ (2 + x) about x = 1, order n.

Use second order Taylor polynomials P2 (x) for the given function about the

point specified to approximate the indicated value. Estimate the error, and

write the smallest interval you can be sure contains the value.

5. f (x) = x1/3 about 8; approximate 91/3.

6. f (x) = 1x about 1; approximate1

1.02 .

7. f (x) = ex about 0; approximate e0.5.

30

CHAPTER 2

Integration

2.1. Antiderivatives

Definition 2.1.1. An antiderivative of a function f on an interval I is

another function F satisfying

F (x) = f (x) for x in I.

Example 2.1.1.

(a) F (x) = x is an antiderivative of the function f (x) = 1 on any

interval since F (x) = 1 = f (x) everywhere.(b) G (x) = 12x

2 is an antiderivative of g (x) = x on any interval since

G (x) = x = g (x) everywhere.(c) R (x) = 13 cos (3x) is an antiderivative of r (x) = sin (3x) on any

interval since R (x) = sin (3x) = r (x) everywhere.(d) F (x) = 1x is an antiderivative of f (x) = 1x2 on any interval not

containing x = 0 since F (x) = 1x2

= f (x) everywhere x 6= 0.We note that the antiderivative of a function is not unique. For suppose

F (x) and G (x) are antiderivatives of the function f (x), on an interval I,

then

ddx (G (x) F (x)) = f (x) f (x) = 0

G (x) F (x) = C G (x) = F (x) + C on I.Thus for a given function f (x) on an interval I, different antiderivatives can

be obtained by adding constants to a given antiderivative.

Definition 2.1.2. The indefinite integral of a function f (x) on an

interval I isf (x)dx = F (x) + C on I,

31

XEQ 201

provided F (x) = f (x) for all x in I. Here C is called the constant ofintegration.

Example 2.1.2.

(a) Since ddx(12x

2)

= x, thenxdx =

1

2x2 + C.

(b) Since ddx(14x

4 53x3 + 7x)

= x3 5x2 + 7, then (x3 5x2 + 7)dx = 1

4x4 5

3x3 + 7x+ C.

(c) Since ddx( 1x + 4x) = 1x2 + 2x for all x > 0, we have (

1

x2+

2x

)dx = 1

x+ 4x+ C.

(d) Since ddx sinx = cosx it follows that(cosx)dx = sinx+ C.

Example 2.1.3.

Find the function f (x) whose derivative is f (x) = 6x21 for all real x andfor which f (2) = 10.

Solution. Since f (x) = 6x2 1, then

f (x) =

(6x2 1)dx = 2x3 x+ C.

Now f (2) = 2 23 + C = 10 C = 4 so that f (x) = 2x3 x 4.

Example 2.1.4.

Find the function g (t) whose derivative is (t+ 5) /t3/2 and whose graph

passes through the point (4, 1).

Solution. We have

g (t) =

t+ 5

t32

dt =

t

12 + 5t

32dt = 2t

12 10t 12 + C.

Now g (4) = 2 4 12 10 4 12 + C = 1 implies that C = 2 so thatg (t) = 2t

12 10t 12 + 2, t > 0.

32

XEQ 201

Exercise 2.1.

Find the given indefinite integrals

1.

5dx.

2.

(2x1/2 + 3x1/3

)dx.

3.

x3dx.

4.

tanx cosxdx.

5.

(a2 x2)dx.

6.

xdx.

2.2. The Definite Integral

Let P be the set of points arranged in order between a and b on the real

line;

P = {x0, x1, . . . , xn}where a = x0 < x1 < < xn = b. Such a set P is called a partition of[a, b] and it divides the interval into n subintervals [xi1, xi]. Note that ndepends on P i.e. n = n (P ). The length of the subinterval is

xi = xi xi1, i = 1, 2, . . . , n.The greatest of these lengths is called the norm of P and is denoted by P,i.e.

P = max1in

xi.

Since f is continuous on the closed interval [xi1, xi], it attains a maxi-mum and a minimum on each of the subintervals. Let f (ui) and f (li) denote

the respective maximum and minimum values of f on each subinterval.

Fig 5.10

33

XEQ 201

By considering the signed value of f (x) should f (x) < 0, we note that

f (ui) xi and f (li) xi represent areas of rectangles whose base is xi and

whose heights are f (ui) and f (li) respectively. If Ai represent the area

under the curve y = f (x) between xi1 and xi, then

f (ui) xi Ai f (li) xi.

Definition 2.2.1. The lower Riemann sum L (f, P ) and the upper

Riemann sum U (f, P ) for the function f and the partition P are defined

by

L (f, P ) =ni=1

f (li) xi and U (f, P ) =ni=1

f (ui) xi.

Example 2.2.1.

Calculate the upper and lower Riemann sums for the functions f (x) = 1/x

on the interval [1, 2] corresponding to the partition P of [1, 2] into four

subintervals of equal length.

Solution. The partition P consists of the points

xi = x0 + ih, i = 0, 1, 2, 3, 4, where x0 = 1, h =2 1

4=

1

4.

x0 = 1, x1 =5

4, x2 =

6

4, x3 =

7

4, x4 = 2.

The function 1/x is decreasing on [1, 2] and so the maximum and minimum

values of f on [xi1, xi] are 1/xi1 and 1/xi respectively. Thus the upperand lower Riemann sums are

U (f, P ) =4i=1

f (ui) xi =1

4=

{1 +

4

5+

4

6+

4

7

}=?.

L (f, P ) =4i=1

f (li) xi =1

4=

{4

5+

4

6+

4

7+ 2

}=?.

Definition 2.2.2. Suppose there is exactly one number I such that for

every partition P of [a, b], we have

L (f, P ) I U (f, P ) .34

XEQ 201

Then we say that f is integrable on [a, b] and we call I the definite inte-

gral of f on [a, b], denoted as

I =

baf (x)dx.

In this notation

(i)

is the integral sign, it resembles the letter S since it represent

the limit of a sum.

(ii) a and b are the lower and upper limits of integration respec-

tively.

(iii) f is called the integrand and x the variable of integration.

(iv) dx is the differential of x. It replaces x in the Riemann sums

when the limit is determined.

Note that the definite integral is a number and not a function. x is a

dummy variable and can be replaced with another variable without affecting

the value of the definite integral.

If in the definition of the upper and lower Riemann sums we use a point

ci [xi1, xi], then the sum

R (f, P, c) =ni=1

f (ci) xi

is called Riemann sum of f on [a, b] corresponding to the partition P and

tags C.

Remark 3.

1. R (f, P, c) is a sum of signed areas of rectangles between the xaxisand the curve y = f (x).

2. R (f, P, c) satisfies the inequality

L (f, P ) R (f, P, c) U (f, P ) .3 If f is integrable on [a, b], then its integral is the limit of such

Riemann sums when n so that P 0, i.e.

limnR (f, P, c) =

baf (x)dx.

35

XEQ 201

Example 2.2.2.

Express the following limits as definite integrals

(a) limn

ni=1

2

n

(1 +

2i 1n

)1/3(b) lim

n

ni=1

1

n

i

n.

Solution.

(a) We are to interpret the sum as a Riemann sum of f (x) = (1 + x)1/3.

From the factor 2/n we deduce that the length of the interval is 2

and it has been divided into n subintervals of equal width xi =

2/n. Let ci = (2i 1) /n, i = 1, 2, . . . , n. Then

c1 =2 1n

=1

n 0 as n and cn = 2n 1

n=

2 1/n1

2 as n.

Hence the interval is [0, 2]. The partition is given by xi =2in ,

i = 0, 1, . . . , n. Thus

xi1 = xi xi = 2in 2n

=2i 2n