math104 calculus 2 ready

THE KENYA METHODIST UNIVERSITY

DEPARTMENT OF PURE AND

APPLIED SCIENCES

MATHS 104: CALCULUS II

COURSE INSTRUCTORALICE LUNANI MURWAYI (MSc – Mathematics)

DEPARTMENT OF PURE AND APPLIED SCIENCESP.O. Box 267

MERU

MATHS 104: CALCULUS II

INTRODUCTION

This course will introduce the students to the concepts and skills of Integral calculus essential for

application finding areas, length of arc, Volume and Surface area of Solids of revolution.

COURSE OBJECTIVES

During the course, the undergraduate student will be able to:.

BROAD AREAS OF COVERAGE

1. The antiderivatives

2. Use basic formula to integration of algebraic functions

Power

Sum and difference

Constant time a function

3. Integration of exponential functions and trigonometric functions using basic formulae

4. Integration using substitutions

5. Integration by parts

6. Integration using Trigonometric Identities

7. Integration using Partial Fractions

8. Definite integrals and the fundamental theorem of Calculus

9. Area under a curve

10. Improper Integrals

11. Length of an arc of a curve

12. Volume and Surface Area of Solid of Revolution

13. Trapezoidal, Mid ordinate and Simpson’s rule

TEACHING AND LEARNING METHOD’S

Lecture method in which definitions are stated and illustrate, theorems are stated, proved

and problems solved as examples.

Questions and answer method.

Tutorials in which problems are solved test the understanding.

Take away assignments.

STUDENTS ASSESSMENT

Assignment

Course Examination

Individual student’s marks of these assignments will contribute to the final examination mark

and grade for the course.

COURSE OUTLINE FOR THE TRIMESTER

WEEK 1

The antiderivatives

Use basic formula to integration of algebraic functions

Power

Sum and difference

Constant time a function

WEEK 2

Integration of exponential functions and trigonometric functions using basic formulae

WEEK 3

Integration using substitutions

WEEK 4

Integration by parts

WEEK 5

Integration using Trigonometric Identities

WEEK 6

Integration using Partial Fractions

WEEK 7

Definite integrals and the fundamental theorem of Calculus

WEEK 8

Area under a curve

WEEK 9

Improper Integrals

WEEK 10

Length of an arc of a curve

WEEK 11

Volume and Surface Area of Solid of Revolution

WEEK 12

Trapezoidal, Mid ordinate and Simpson’s rule

WEEK 14 AND 15

Assignments

Cat 1-------------------------------15%

Cat 2-------------------------------15%

Total Cats-------------------------30%

Course Examinations---------------------70%

Grant Total---------------------------------100%

LEARNING RESOURCES

Larry, J. G, et al: Calculus and it Applications, Prentice – Hall International Press, London, 1993

Swokowski, E. W.: Calculus with Analytic Geometry, Alternate, Edition PWS Publishers, 1983

Thomas, G. G and Finney, R. L.: Calculus and Analytic Geometry, Narosa Publishing House, 6th Edition, 1998

TABLE OF CONTENTS

CHAPTER ONE: INTEGRAL CALCULUS..................................................................................7

CHAPTER TWO: EXTENSION FORMULAE...........................................................................15

CHAPTER THREE: INTEGRATION USING SUBSTITUTIONS.............................................18

CHAPTER FOUR: INTEGRATION OF TRIGONOMETRIC FUNCTIONS..........................23

CHAPTER FIVE: PARTIAL FRACTIONS.................................................................................36

CHAPTER SIX : INTEGRATION BY PARTS...........................................................................44

CHAPTER SEVEN:FACTORIZATION BY COMPLETING THE SQUARE...........................48

CHAPTER EIGHT : INTEGRATION THE INVERSE TRIGONOMETRIC FUNCTIONS......50

CHAPTER NINE : CHANGE OF VARIABLE...........................................................................61

CHAPTER TEN: DEFINITE INTEGRALS.................................................................................65

CHAPTER ELEVEN: : AREA UNDER A CURVE....................................................................71

CHAPTER TWELVE: NUMERICAL INTEGRATION..............................................................85

CHAPTER THIRTEEN :VOLUME AND SURFACE AREA OF SOLIDS OF REVOLUTION.......................................................................................................................................................97

CHAPTER FOURTEEN : LENGTH OF ARC OF A CURVE..................................................112

CHAPTER ONE: INTEGRAL CALCULUS In Differential calculus, you studied the different methods of differentiation and its

applications. Now in integral calculus we shall learn the different methods of integration and its

applications

1.1: CHAPTER OBJECTIVES

By the end of the chapter, the student should be able to:

1. Explain the meaning of integration

2. Explain the meaning of the constant of integration

3. State and use the rule for Integrating

a. A constant function

b. An algebraic function

c. A constant times a function

d. Sum of functions

e. Functions of the form

f. Exponential functions

1.2: MEANING OF INTEGRATION

The process of integration is defined as the reverse or converse process of differentiation

For example

, we say that the integration of . This is written in symbols as

, we say the integral of

The sign ’’means the integral of ’’ and is called the integral sign. Means

integration of the function with respect to . Indicates the variable of

integration. C is the constant of integration.

1.3: THE CONSTANT OF INTEGRATION

The differentiation of a constant is always equal to zero; therefore several functions have the

same derivative because any constant will disappear in the process of differentiation. After

differentiation we always add a constant c or k as we can see in the two examples. Generally,

when gives all the anti-derivatives of , Here c is the constant

of integration. An integral like in which the constant c is not known/ cannot be

determined is called an indefinite integral

1.3: THE INTEGRAND

In the integral sign cannot be divorced from if the integral is with respect to . The

function in is called the integrand of the integral. For example in , is

called the integrand of the integral.

1.4: THE GENERAL RULE FOR INTEGRATING ALGEBRAIC FUNCTIONS.

Examples

Find the integrals of the following functions

1.5: INTEGRATION OF A CONSTANT TIMES A FUNCTION

EXAMPLE

Integrate 2x5

Solution

1.6: INTEGRATION OF A CONSTANT

Example

Evaluate the integral of 4 with respect to x

Solution

1.7: INTEGRATION OF THE SUM OF FUNCTIONS

The integral of sum of two functions is the sum of their integrals. This extends to the

sum of any finite number of functions.

Example 2

Find

Solution

We expand and then integrate

Example 3

Find

Solution

First we separate the terms by dividing each term by x and then integrate

Example 4

Find

Solution

We simplify the integrant and the integrate

1.8: INTEGRATION OF

Recall

For all values when using definite integrals

1.9: INTEGRATION OF EXPONENTIAL FUNCTIONS

Exponential functions are functions in which the variable is the index or the power .These

are functions of the form ex and ax

1)

Generally,

Example

2. Find

Solution

3. Integrate

Solution

EXERCISE

Integrate the following functions with respect to

1)

2)

3)

4)

5)

6)

7)

8)

9)

10)

11)

12)

13)

14) If find x.

15) If , find the

values of x

CHAPTER TWO: EXTENSION FORMULAE



1) Write the extension formulae corresponding to the basic formulae for integrating

algebraic and exponential functions.

2) Use the extension formulae to integrate expressions /functions of first degree in x.

2.2: EXTENSION FORMULAE FOR INTEGRATING ALGEBRAIC

FUNCTIONS CONTAINING FIRST DEGREE EXPRESSIONS

This is the extension of the integration rules seen in chapter one. It is one of

the techniques of integration in which the variable x in the basic formulae is

replaced by a first degree expression of the form (ax +b). It entails finding

the integrals like:

, where real numbers

And

is a linear function of .

Generally

EXAMPLES

2.3: INTEGRATING FUNCTIONS OF THE FORM

Generally,

Examples

2.4: INTEGRATING EXPONENTIAL FUNCTIONS USING EXTENDED

FORMULAE

These are functions of the form and .

The extended formulae for the functions of the form and are:

1)

2)

Example

EXERCISE

Integrate the following functions with respect to

1)

2)

3)

4)

5)

6)

7) ) 8).

CHAPTER THREE: INTEGRATION USING SUBSTITUTIONSWhen the integrand consists of two functions one of which is the derivative of the other function,

we use the substitution technique of integration. The main function is u and the derivative u’ or

so that u’ dx becomes du. The substitution transforms the integral into one of the basic

formula of integration in the new variable u only. We then integrate using the basic formula.



1) Recognize a function u and its derivative u’ or in the integrand

2) Adjust the derivative so that the integrand consists of one function u and its derivative

3) Substitute u and u’ or in the integrand and integrate

3.2: THE SUBSTITUTION METHOD OF INTEGRATION

This involves the changing of the variable so as to simplify integration. There are different types

of integrands to consider like:

Type 1

Examples

1)

Solution

Do this problem by making the substitution

2)

Solution

3)

Solution

4)

Solution

Type 3

EXERCISE

Evaluate the following

1)

2)

3)

CHAPTER FOUR: INTEGRATION OF TRIGONOMETRIC FUNCTIONS



1) Integrate trigonometric functions using the six basic formulae

2) Integrate trigonometric functions with first degree expressions using the Extension

functions

3) Use trigonometric Identities to integrate trigonometric functions

4) Use Substitution to integrate trigonometric functions

4.2: INTEGRATION OF TRIGONOMETRIC FUNCTIONS USING THE

BASIC FOMULAE

The following is a list of the basic trigonometric formula. They are also in the Advanced

Mathematical tables.

1)

2)

3)

4)

5)

6)

Examples

1) Evaluate

Solution

2) Find

Solution

3) Find

Solution

EXERCISE

Evaluate

1)

2)

4.3:INTEGRATION OF TRIGONOMETRIC FUNCTIONS USING

SUBSTITUTIONS

Example

Find

Solution

Example

Find

Solution

NB.

Example

Exercise

1) Show that

2) Use a suitable substitution to show that:

3)

4)

5) Use a suitable substitution to show that

a)

b)

c)

4.4: INTEGRATION OF TRIGONOMETRIC FUNCTIONS USING

TRIGONOMETRIC IDENTITIES

The Basic Trigonometric Identities

Examples

4.5: INTEGRATION OF TRIGONOMETRIC FUNCTIONS USING

MULTIPLE ANGLES

Multiple angle Identities

EXERCISE

1)

2)

3)

4)

5)

6)

7)

8)

9)

10)

11)

SOLUTIONS

CHAPTER FIVE: PARTIAL FRACTIONS5.1: CHAPTER OBJECTIVES


1. Decompose compound fractions into partial fractions

2. Use appropriate method to integrate the partial fractions

5.2: INTRODUCTION TO PARTIAL FRACTIONS

In this section we show how to disintegrate compound fractions byexpressing them as a sum of simpler fractions, called partial fractions. To

illustrate the method, observe that by taking the fractions and

to a common denominator we obtain

If we now reverse the procedure, we see how to integrate the function on the right side ofthis equation:

To see how the method of partial fractions works in general, let’s consider a rationalFunction

Where and are polynomials. It’s possible to express as a sum of simpler fractionsprovided that the degree of is less than the degree of . Such a rational function is calledproper. Recall that if

Where , then the degree of is and we write deg = .

If is improper, that is, , then we must take the preliminary stepof dividing into (by long division) until a remainder is obtained such that

. The division statement is

Where and are also polynomials.As the following example illustrates, sometimes this preliminary step is all that is required.

EXAMPLE 1

Find

SOLUTION Since the degree of the numerator is greater than the degree of the denominator,we first perform the long division.

This enables us to write

5.3: Denominator with linear factors examples.

(1).Evaluate

Examples

5.4: DENOMINATOR WITH A QUADRATIC FACTOR

But

Each numerator of a partial fraction must be assumed to be a polynomial of degree 1 less than

that of the corresponding denominator.

Relating the numerators

5.5: DENOMINATOR WITH REPEATED LINEAR FACTORS

EXAMPLES

Exercise

Show that

1)

2)

3)

4)

CHAPTER SIX : INTEGRATION BY PARTS6.1: CHAPTER OBJECTIVES


1. Derive the formula for integration by parts

2. Use the formula to integrate the product of two functions

6.2: INTEGRATION BY PARTS

This method applies to integrals of the form in which can be differentiated

repeatedly to become zero, while is integrated repeatedly without much difficulty.

Integration by parts is equivalent to the product rule and it states that:

Integrate with respect to x

EXAMPLES

Exercise

Use integration by parts.

1) 2).

3). 4).

5). 6).

7). xdx 1sin 8). xdxex sin

9). 10).

CHAPTER SEVEN:FACTORIZATION BY COMPLETING THE SQUARE7.1: CHAPTER OBJECTIVES


1) Factorize Quadratic expressions by completing the square

2) Use substitution to integrate the factorized function

7.2:INTEGRALS INVOLVING

By completing the square of where a

where u is a function of x evaluate by integration

Examples

1.

Completing square

lets u= differentiating both side dx = cos ydy

2.Evaluate

CHAPTER EIGHT : INTEGRATION THE INVERSE TRIGONOMETRIC FUNCTIONS8.1: CHAPTER OBJECTIVES

By the end of the chapter ,the student should be able to:

1) Define and find the inverse of a given function

2) Draw graphs of inverse trigonometric functions

3) Combine Differentiation and integration to prove identities involving

Inverse trigonometric functions.

8.2: THE INVERSE TRIGONOMETRIC FUNCTIONS

If is a function whose inverse if then

Definition:

The inverse sure function is denoted by is defined by

if and only if where .

The domain and range are very important

One to one function – inverse can be easily obtained

The Trigonometric functions are not one to one and therefore have no inverse. However, if there

domains are restricted then there is an inverse.

The inverse of the sine function

The graph of can be draw from the equation

2

1

0

-1

-2

2

1

0

-1

-2

f x

2

1

0

-1

-2

4

3

2

1

0

1

-1-1

-2

f x

21

32

342

262

10062

142

232

32

1

)(sin 1

xx

2

2

x

y

1-1

The inverse of the Cosine function

Definition:

The inverse of cosine function denoted by is defined by

x

yy = cos(x)

x

yy = arccos(x)

The inverse of the Tangent function

Definition:

The inverse of tangent function is defined by . If where

8.3. DIFFERENTIATION AND INTEGRATION

Exercise

Show that

a)

b)

c)

CHAPTER NINE : CHANGE OF VARIABLE9.1: CHAPTER OBJECTIVES

By the end of the lesson, the student should be able to:

1) Evaluate integrals by changing the variable to

2) Evaluate integrals by changing the variable to

9.2: THE CHANGE OF VARIABLE It’s used in functions with even powers of sin x and cosxExample

Find

Solution

The change of variable makes the integration easier or possible. It is important to simplify the expression as much as possible using the trigonometric identities.

Exercise

1) 2).

3). 4).

9.3: THE CHANGE OF VARIABLE

Recall trigonometric identities and change of variables

Examples

1. Find

Solution

Exercise 1)

2)

3)

CHAPTER TEN: DEFINITE INTEGRALS



1. State the fundamental theories of Calculus

2. Use the theories in Solving problems involving Calculus

3. Calculate definite integrals.

10.2:FUNDAMENTAL THEORIES OF CALCULUS1. let be continuous on closed interval

Theorem If a function F is defined by For all in then F is an antiderivative on 2. If s continuous on and is an antiderivative of on then

F F =

x

y

Substitution in definite integrals Example

1)

Examples Evaluate

Exercise

1)

2)

3)

4)

5) Find the area between and the

Solutions

Exercise on definite integralsEvaluate the following integrals.

1)

2)

3)

4)

5)

6)

CHAPTER ELEVEN: AREA UNDER A CURVE


By the end of the chapter the student should be able to:

1. Find the area under a curve

2. Calculate the area between two curves

12.2 :AREA UNDER THE CURVE

Suppose we wish to find the area under the curve from to .1. Divide the area into thin vertical strips; find their area and them up to find the

approximate area under the curve.2. Consider one strip PP’Q, Q where P(x, y).The length of the strip = y and the width can be

considered as a small increase in x ( ). The area of the strip = y ( )3. If A is the area of the region up to PP’ then the area of PPQ”Q can be considered as a

small increase in

Area of strip =

x=a x=b

y f x

P’ Q’

y

x

pQ

The accuracy of the area increases as decreases.

Total area=

from (1)

From (2) and(3)

The area of the curve is the integration of the function. Since the boundary values of defining the total area are from Then the total area is given by

is called the definite integral of over the integral

PROPERTIES

1)

2) direction interchanges the

intervals and hence the sign.

3)

11.3: FUNDAMENTAL THEORIES OF CULCULUS1. Let be continuous on a closed interval [a,b]

If a function F is defined by for all x in [a, b], then F is an antiderivative

on [a ,b].2. If is continuous on [a,b ] and F is an antiderivative of on [a,b] then,

Example1) Find the area under the curve enclosed between .line and where

are positive.

Solution

Sketch the graph

y

x

Area of shaded region

=

= = =

y

x

y mx

2 (a) Find the area between the curve and the -axis from to

SolutionSketch the graph

b) Find the area of the region bound by and the from to x=2

SolutionSketch the graph to identify the region

2 4y x

y

x

3.Find the area of the region enclose by the curve

SolutionFactorize the function

Recall curve sketch in Calculus 1

Curve cuts the x-axis when y = 0 ;

Curve cuts the x-axis at three points Find the turning points by differentiating and determine their nature by differentiating the 2nd

time.

24y x

y

x-2

2

When min. point at

When max. at

y

x-2 2

3 4y x x

Interpretation: The area under the x-axis is equal to that above the x-axis.

4. Find the area bonded by the curve and the

SolutionSketch the graphFind the y-intercepts and the x-intercepts

y

x

2

-2

4

24x y

When the function is in terms of y, integrate with respect to y and the intervals are in the form y= a and y = b.

EXERCISEFind the area bounded by given curve and given vertical line

1)2)3)

4)

5)

6)

7)

8)9) Find the area bounded by the co-ordinate axis and the line .10) Find the area bounded by and co-ordinate axis11) Find the area between the curve and the

12) Find the area between and the x-axis

11.4: AREA BETWEEN TWO CURVESRecall area under a curve as below.

y f x

a b

c

d

x g y

Suppose are continuous in and that

then the graph of lies above on the interval .

Divide the shaded area into vertical strips each of width

Consider one inch strip as shown Area of strip =

If sum of vertical strips

Defining the area between two curves If through out interval

1y f x

2y f x

x= a x= b

Area between and from to is

Examples: 1.Find the area of the region bounded by the parabola and the line Make a sketch

SolutionSketch the graphs of the functions

.These are the x-co-ordinates of the points of intersectionOf the two functions. They are the intervals.

2.Find the area of the region bound by the graphs

Solution Sketch the graphs to identify the region

2y x

21 2y x

If area is bounded by and the curves for all in cd

Then

2

24

21x y

2

22

2 4

2 4

y x

or x y

ExerciseFind the area of the regions bounded by the following curves

Exercise on definite integralsEvaluate the following integrals.

7)

8)

9)

10)

11)

12)

13)

CHAPTER TWELVE: NUMERICAL INTEGRATION 12.1: CHAPTER OBJECTIVES By the end of the chapter, the student should be able to:

1. Derive the Trapezoidal rule2. Use the Trapezoidal rule to find approximate area under a curve3. Derive the Simpson’s rule4. Find the approximate area using the Simpson’s rule

12.2: TRAPEZOIDAL RULE.

Used to approx. areas under the curves whose integrals can’t be found easily.It estimates the integral by dividing the area into trapeziums instead of rectangles consider the area under the curve.

from to

Divide the area into is trapeziums of equal width h such that

y f x

xn= bx0= a x1 x2

y0 y1 y3

Consider the shaded trapezium

Total area (TA) = sum of area of all trapeziums

Example

Using trapezoidal rule with

Solution

The more the number of trapezia the region is divided into, and approximate area exact value

This means as increase decrease and the Total area approaches exact value. Taking large enough we can make the difference between the exact values of the area (the error ) as small as possible.However, in practice, its not possible as we can’t tell how large n should be:-If is the continuous on where is 2nd derivative of it can be shown that.

Where c is a value between Thus as the error

Use the inequality

Maximum absolute value 0f evaluated at a and b .It gives us the upper

bound of the magnitude of error error.max .In practice, the exact value of can’t be obtained. We therefore estimate an upper bound value for it instead.

If M is the upper bound for the value then the inequality becomes

Definition: If is continuous and m is any upper bound for the values of on the interval

the error in the trapezoidal approx of the integral of from to statistics the inequality

That is if M is a real number such that for all x in then the error in using

the trapezoidal rule is not greater than

Examples

1. Find the upper bound on the error (max error) in the approximation for

SolutionFrom previous example

The max value of on occurs at :

.

2. Approximate by using Trapeziodal rule with n=10. Estimate the max error

in the approximation.

Solution

Max value of on occurs at 1x

12.3: SIMPSON’S RULE-This involves Approximating area under curves using parabolas.Consider the area under curve from to

Any 3 non-collinear points can’t be joined/ fitted into a parabola Divide the area into n parabolas

Consider the following co-ordinates axis

x

y f x

y

x=a x=b

Let the equation of parabola be of the form

Since the points satisfy the equation

xx=-h x=h

0y 1y 2y

y

Solve for A,B and CAdd (i) and (iii)

Simpson’s rule results from applying the formula for AP to successive pieces of the curve between and

To approx. we sum all the areas

Where n is even

x

y f x

y

x=a x=b

0y ny1ny

1y

Error Estimation for Simpson’s rule.If is continuous and is any upper bound for the values of that is,

for all in the error in the Simpson’s rule approximation of the

satisfies the inequality

Examples

1. Approximate Using Simpson’s rule with .What is the max error in

the estimate?

Solution

Using Simpson’s

Estimating the error.

Max. Occurs at any point

Both Simpson’s and Trapezoidal rule can be used to estimate

,Where is not known but the data is given .Suppose its found experimentally

that 2 physical variables are related as shown.

Assuming that is continuous estimate using (a) Trapezoidal rule.

(b) Simpson’s rule.

Solutionb). Simpson’s rule.

ExerciseUse Trapezoidal & Simpson’s rules together with max-errors of Estate to approximate the following definite integrals for the stated value of n.

a) c).

b) d).

c) e).

CHAPTER THIRTEEN :VOLUME AND SURFACE AREA OF SOLIDS OF REVOLUTION

13.1: CHAPTER OBJECTIVES By the end of the topic the student should be able to:

1.Use the formula for the volume of the solid of revolution to the volume of common solids of revolution

2.Find the surface area of common solids of revolution using the formula for the surface area of solid of revolution.

13.2: VOLUMES OF SOLID OF REVOLUTION

Consider the region bounded by ,lines and

Revolve this region about the .If a region in a plane is rotated about a line in the plane, the resulting solid is called a solid of revolution and solid is said to be generated by the region. The line about which the rotation occurs is called the axis of revolution.Consider the region bounded by

Definition:

x=a x=b

y f x

2

-2

55

The volume obtained by rotating the curve about the between the

ordinates is given by

Means rotation about or any line parallel to . But

ExampleIf ,find the volume of the solid of revolution under this graph from to

about the .

Solution Sketch the graph showing he region and the solid after rotation

Definition Volume obtained by rotating the curve about the between the ordinates and and the .

2 1y x

-11

x

y

Example. The region bound by the between and is evolved about the

.find the volume of the solid of revolution. SolutionSketch the graph

c

d

x g y

y = 1

y= 8

3y x

x

y

Definition.If the region bounded by the two curves from and is revolved about the

Its volume is given by.

Example1. If the region bound by is revolved about . Find the volume of the solid of revolution.

x = a x = b

1f x

Solution

Rewrite the equations in terms of y,that is

NOTEIf the region is rotated about:

1. Volume=

2. : Volume =

1 12y x

2 2y x

x = 1

Examples 2. Find the volume obtained by revolving the region bounded by and lines and about the line

Solution

3. The region in the first quadrant bounded by is revolved about .Find the

volume of the solid of revolution.

Solution

The equations are rewritten in terms of x

4. The region bounded by the graphs is revolved About the line y = 3. Find the volume of the solid of revolution.

Solution

1

1

2x y

1 31 2x y

EXERCISE Sketch the regions bounded by the following curves. Find the volumes generated about the given lines.Exercise

1). 2).

3). 4).

5). 6).

7). 8).

9). 10). 11). 12).

13). 14).

15).

1 12y x

2 2y x

x = 1

13.4: THE SURFACE OF SOLID OF REVOLUTION Definition: surface area.

1) The area of the surface swept by revolving the curve from to about the

2) If the axis of revolution is y-axis from y = c and y = d

3) Using parametric equation.If the curve that sweeps out the surface is given in parametric form with and as functions of a variable t that varies from to then

Examples 1)Find the area of the surface obtained by revolving the curve

Solution

2) The line segment from to is revolved about the y axis generated

a cone. Find its surface area.

2

y x

x

y

Solution

3)Find the area of the sphere generated by revolving the circle about.

The top semi circle is enough to generate the required area.Parametrically the equation of the circle is given by:

REMARKS

1) If revolved about

y

x

y

x

a

a

a

2) If revolved about

EXERCISEFind the area of the surface generated by revolving the given curve about the given lines.

1)2)

3)

4)

5)

6)

7)8)

CHAPTER FOURTEEN : LENGTH OF ARC OF A CURVE


By the end of the chapter the student should be able to:1. Derive the formula for finding the length of an arc /curve if the the equation of the

curve is known2. Accurately calculate the length of a curve using the formula.

14.2: THE LENGTH OF A CURVE IN CARTESIAN EQUATION

The length can be approximated by short line segments. The more the line segments the more accurate the approximation. Suppose that the curve whose length we want to find is

between and .The arc the polygonal path APQB.The length of the arc is defined to be the limit of the length of successively finer polygonal approximations. Divide the curve into n parts and connect the successive end points with line segments. Consider a typical segment . The length of

by Pythagoras for n segments the length of the curve from and

is approx. by the sum

The approximation improves as n increases and length of segment approaches the exact value.

x = a x = b

AB y f x

P

Q

14.3: MEAN VALUE THEOREM.

If a function is continuous on the closed interval and is differentiable on the

open interval , then there exists a number C in such that

the derivative of at C)

Suppose that has a derivative that is continuous at every point then by the mean value theorem their exists a point on the curve between such that

From (1)

Examples.

Find the length of the curve

14.4: THE LENGTH OF A CURVE IN PARAMETRIC EQUATIONSuppose

Examples

1. Find the distance traveled between and by a particle whose position at time

is given by and

Solution

At a point on the curve where fails to exist may exist. Then the arc length of

is;

2.

Solution

Sketch the curve

EXERCISEFind the length of the curves.

a). e).

2 3y x

8-1 x

y

b). f).

c). g).

d). h).

math104 calculus 2 ready

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