calculus - core course for ii semester bsc mathematics

C A L C U L U SII SEMESTER

CORE COURSEFor

B.Sc.MATHEMATICS(2014 Admission onwards)

(CUCBCSS)

UNIVERSITY OF CALICUTSCHOOL OF DISTANCE EDUCATION

Calicut university P.O, Malappuram Kerala, India 673 635.

351-A

School of Distance Education

Calculus

UNIVERSITY OF CALICUTSCHOOL OF DISTANCE EDUCATION

STUDY MATERIAL

C A L C U L US

Core COURSE

For

B.Sc.Mathematics

II Semester

Prepared by:

Module I:Gafoor. I,Asst. Professor,Department of Mathematics,NAMCollege, Kallikkandy

Module : IIAboobacker PAsst. ProfessorDepartment of MathematicsW M O College, Muttil

Module IIIShajeeb P.U.Assistant ProfessorDepartment of MathematicsGovt.Victoria College, Palakkad

Module : IVBushra Beevi.KK,Asst. Professor,Dept. of Mathematics,PSMO College,Tirurangadi

Scrutinized by : Dr. D Jayaprasad,Principal,Sree Krishna College, GurvayoorChairman, Board of Studies in MathematicsMathematics ( UG)

Layout:Computer Section, SDE

©Reserved


Calculus

CONTENTS

MODULE I - 05-09

MODULE II - 10-28

MODULE III - 29-37

MODULE –IV - 38-45


Calculus


Calculus

Module1

Chapter1Extreme value of FunctionsDefinition:-Let be a function with domain , then has an absolute maximumvalue on at point if ( ) ≤ ( ) for all ∈and an absolute minimum value on at point if( ) ≥ ( ) for all ∈Remark:- Absolute maximum and minimum values are called absolute extremaor Global extrema

Example:- If = [0,2] then y attain the absolute maximum value 4 at = 2and absolute minimum 0 at = 0 .

A function which is continuous at every point of a closedinterval has absolute maximum and absolute minimum value on theinterval.when we try to graph the function, we can locate these points clearly

Theorem IIf is continuous at every point of a closed interval , then assumes both an

absolute maximum value M and absolute minimum value somewhere in . Thatis, there are numbers & in with ( ) = and ( ) = and ≤ ( ) ≤ forevery other in .

Example :- let = [− , ] and ( ) = then ( ) takes maximum value ‘1’ at= 0 and minimum value ‘0’ at two points = ∓Definition:- A function has a local maximum value at an interior point of itsdomain if ( ) ≤ ( ) for all in some open interval containing .

A function has a local minimum value at an interior point of its domain if( ) ≥ ( ) for all in some open interval containing .

This local maximum and minimum values are called local extrema

Remark: If we consider the set of all local local maximum values , then the globalmaximum (if there is )will be one of the element of this set. Similarly If weconsider the set of all local local minimum values , then the global minimum (ifthere is )will be one of the element of this set

Theorem II

If has a local maximum or minimum value at an interior point of its domain ,and is defined at then ( ) = 0


Calculus

Proof: to prove ( ) = 0 at the local extremum , it is enough to prove ( ) isniether negative nor positive.

So suppose has a local maximum at = , then ( ) − ( ) ≤ 0 for all values ofnear enough to .since of ’s domain , ( ) is defined by the

two sided limit lim → ( ) ( )This means that the right hand side limit and left hand side limit are exist at= and equal to ( )When we examine these limits seperatly we find that( ) = lim → ( )− ( )− ≤ 0 (because ( − ) > 0 & ( ) ≤ ( )) ….(1)

Similarly, ( ) = lim → ( )− ( )− ≥ 0 (because ( − ) < 0 & ( ) ≤ ( )) ….(2)

From (1) & (2) we have ( ) = 0Similarly we can prove the case of local minimum value.

Definition:- an interior point of the domain of a function where orundefined is known as critical point of

Remark: - the only domain points where a function can assume extreme valuesare critical points and end points.

Example:-Find absolute maximum and minimum values of ( ) = on [-2, 1]

Solution:- the function is differentiable over its entire domain, so the critical pointis where ( ) = 2 = 0

Which gives = 0At = 0 , ( ) = 0

At the endpoints = −2 ( ) = 4= 1 ( ) = 1The function has an absolute maximum value 4 at = −2 and an absoluteminimum value 0 at = 0

Chapter 2

The mean value TheoremsTheorem III: Rolle’s theorem:-

Suppose that = ( ) is continuous at every point of the close interval [a, b]and differentiable at every point of its interior (a, b). If ( ) = ( ) = 0, then there isatleast one number c in (a, b) at which ( ) = 0Proof:- Given is continuous , so assume absoute maximum and minimumvalues on [a, b]


Calculus

Since exist at every interior points , the absolute extrema will attain in criticalpoint or end points.

If either the maximum or minimum occurs at a point c inside the interval then( ) = 0 and hence the proof.

If both maximum and minimum are at then is constant ( since ( ) =( )) which imply that = 0 anywhere in the interval.

Which completes the proof.

Theorem IV : Mean Value TheoremSuppose ( )is continuous in a closed inteval [a, b] and differentiable on

the interval’s interior (a,b). Then there is at least one point c in (a, b) at which( ) ( ) = ( )Proof:- consider two points ( , ( ) & ( , ( ) on the graph of ( ).Now draw astright line joining these two points.its equation is( )( ) ( ) == ( ) + ( ) ( ) ( − )

If we set = ( )We have ( ) = ( ) + ( ) ( ) ( − )

Now the vertical distance between the grapphs of & is given byℎ( ) = ( ) − ( )= ( ) − ( ) − ( ) ( ) ( − )Clearly ℎ( ) = ℎ( ) = 0Note that ℎ( ) satisfy the hypotheses of Roll’es theorem in [a, b] .There fore thereexist a point ∈ ( , ) such that ℎ ( ) = 0Which implies ( ) − ( ) − ( ) ( ) ( − ) = 0implies ( ) − ( ) ( ) = 0ie, ( ) = ( ) ( )

hence the proof.

Corollory 1If ( ) = 0 at each of an interval I then ( ) = for all in I , where c is constant

Proof: - to prove is constant , we show that if & are any two points of I then( ) = ( )


Calculus

Suppose & are any two points in I, such that < . Then satisfy the thehypothesis of the Mean Value Theorem in [x, y] , hence the exist a value cbetween & such that ( ) = ( ) ( )Since = 0 for all , we have ( ) = 0 = ( ) ( )

Implies ( ) − ( ) = 0ie, ( ) = ( )

Hence the proof

Corollary 2If ( ) = ( ) at each point of an interval I then there exists a constant C

such that ( ) = ( ) + for all ∈Proof:-at each point in I, define ℎ( ) = ( ) − ( )

Then ℎ ( ) = ( ) − ( ) = 0Then by corollary 1 ℎ( ) = , implies ( ) = ( ) + .

Example:- find the function ( ) whose derivative is and whose graph passesthrough (0, 2)

Solution:- since ( ) has same derivative as ( ) = − , we have ( ) = − +for some constant C, where C is determind by the condition (0) = 2

ie, (0) = − cos 0 + = 2gives , = 3

so ( ) = − +3 is the desired function

Defition:-Let be a function defined on an interval I and let & are any twopoints in .

increases on I if < ⟹ ( ) < ( ) decreases on I if < ⟹ ( ) < ( )

Corollary 3:-The first derivative Test for increasing and decreasing

Suppose is continuous on [a,b] and differentiable on (a, b)

If > 0 at each point of (a,b) then [ , ] If < 0 at each point of (a,b) then [ , ]

Proof:- Let & are any two points in [a, b] with <Then by mean value theorem , ( ) − ( ) = ( )( − ) for some c between &

Now suppos ( ) > 0 implies ( ) > ( ), is increasing

Similarly if ( ) < 0 implies ( ) < ( ), is decreasing


Calculus

The First derivative Test for Local Extremum ValuesTheorem 5: (First Derivative Test)

The following test applies to a continuous funcion ( )At critical point c,

1. If changes from positive to negative at ( > 0 < & < 0 > 0), then has a local maximum value at c.

2. If changes from negative to positive at c( < 0 < & > 0 > 0) then has a local minimum value at

3. If doesnot change sign at c ( has same sign on both sides of c )thenhas no extremum value at

At left end point a1. If < 0 > then ℎ local maximum value at2. If > 0 > then ℎ local minimum value at

At right end point1. If < 0 < then ℎ local minimum value at2. If < 0 < then ℎ local maximum value at

Example : Find the interval on which ( ) = − + 12 + 5, − 3 ≤ ≤ 3 isincreasing and decreasing.where does the function assume the extreme valuesand what are these values

Solution: The function in its domain. The first derivative is( ) = −3 + 12= −3( − 4)= −3( + 2)( − 2)Which is defined for all points of [-3,3], and is zero at = −2 & = 2This two ponts are the critical points

These critical points devide the domain of in to intervals on which is eitherpositive or negative

By substitutions, we can observe that has a local maxima at = −3 & = 2.Similarly local minima at = −2 & = 3The corresponding value of ( )

Local maxima : (−3) = −4 & (2) = 21Local minima : (−2) = −11 & (3) = 14

Since is defined in a closed interval , we have (−2) = −11 is the absoluteminimum and (2) = 21 is the absolute maximum.


Calculus

Module IICHAPTER 1

OPTIMIZATIONSolving Optimization Problem

Introduce variables: List every relation in the problem as an equation or algebraicexpression.

Identify the unknown:Write an equation for it. If you can, express the unknown as afunction of a single variable or in two equations in two unknowns.

Test the critical points and endpoints. The points where either first derivative is zeroor not defined

Example1 :Find two positive numbers whose sum is 20 and whose product is as large aspossible.

Solution:If one number is x, then the other is (20 - x). Their product is f (x) = x(20 - x) =20x - x2.

We want the value or values of x that make f (x) as large as possible. The domain of f is theclosed interval 0 ≤x ≤20.

We evaluate f at the critical points and endpoints. The first derivative,

fI(x) =20 - 2x,is defined at every point of the interval 0 ≤ x ≤ 20 and is f(x) = 0 only at x = 10.Listing the values of f at this one critical point and the endpoints gives

Critical-point value : f (10) = 20(10) - (10)2 = 100 Endpoint values : f (0) = 0, f (20) = 0.

We conclude that the maximum value is f (10) =100. The corresponding numbers are x = 10

And (20 -10) = 10

Example 2 : A rectangle is to be inscribed in a semicircle of radius 2. What is the largestarea the rectangle can have, and what are its dimensions?

Solution:To describe the dimensions of the rectangle, we place the semicircle andrectangle in the coordinate plane ( The length, height, and area of the rectangle can thenbe expressed in terms of the position x of the lower right hand corner:

Length: 2x , Height:√4-x2 Area: 2x √4-x2

Notice that the values of x are to be found in the interval 0 ≤ x ≤ 2.

Now, our practical problem reduces to the mathematical problem of finding absolutemaximum value of the continuous function

A(x) = 2x√4 - x2


Calculus

on the domain [0, 2]. We do this by examining the values of A at the critical points andendpoints. The derivative

=√ +2√4 − x2 ……………….(1)

is not defined when x = 2 . The derivative (1) is equal to zero when√ +2√4 − x2 =0

Multiplying both sides by √4 – x2. ,we get -2x2 + 2(4 – x2) = 0 8- 4x2 = 0 X2= 2

Ie x = -√2 or x = √2

Of the two zeros, x = √2 and x = - √2, only x =√2 lies in the interior of A's domain andmakes the critical-point list. The values of A at the endpoints and at this one critical pointare:

Critical-point value: A (√2) = 2√2√4 - 2 = 4

Endpoint values : A(0) = 0, A(2) = 0.

The area has a maximum value of 4 when the rectangle is √4 – x2 =√2 units high and

x = 2√2 units long.

Example: An open-top box is to be made by cutting small congruent squares from thecorners of a 12-by-12-in. sheet of tin and bending up the sides. How large should thesquares cut from the corners be to make the box hold as much as possible?

Solution

We take the corner squares are x inches on a side and hence the height of the open topbox is x. Note that the volume of the box is V = hlw,

where h is the height, l is the length and w is the width. Here h = x;

l= 12 - x - x = 12 - 2x;

w = 12 - x - x = 12 - 2 x.

Hence the volume of the box is a function of the variable x:

V(x) = x(12 - 2x)(12-2x) = 144x - 48x2 + 4x3.

Since the sides of the sheet of tin are only 12 in. long, 2x cannot exceed 12 and hence x £ 6and the domain of V(x) is the interval 0 < x < 6.

A minimum value of 0 at x = 0 and x = 6 and a maximum near x = 2. To locate the exactpoint, we examine the first derivative of with respect to x:

= 144 - 96x + 12x2 = 12(12 - 8x + x2) = 12(2 - x)(6 - x).

At the critical point dV = 0 , which gives x = 2 and x = 6 . Only x = 2 lies in the interior of thefunction domain and is the required criticalpoint.


Calculus

Critical-point value: V (2) = 128.

Endpoint values: V(0) = 0, V(6) = 0

The maximum volume is 128 in3. The cut-out squares should be 2 in on a side.

COST AND REVENUE IN ECONOMICS

Here we want to point out two of the many places where calculus makes a contribution toeconomic theory. The first has to do with the relationship between point, revenue (moneyreceived), and cost.

Suppose that

r (x) = the revenue from selling x items

c( x)=the cost of producing the x items

p(x) = r(x) - c(x) = the profit from selling x items.

The marginal revenue and marginal cost at this production level (items) are

= marginal revenue

= marginal cost.

Next theorem is about the relationship of the profit to these derivatives.

Theorem: Maximum profit (if any) occurs at a production level at which marginal revenueequals marginal cost.

Example: The cost and revenue functions at a soft drink company are c(x) = x3 - 6x2+15x,

and r(x) = 9x , where x represents thousands of soft drink bottles. Is there a productionlevel that will maximize company's profit? If so, what is it?

Solution

r(x) = 9x, c(x) = x3 - 6x2+15x

Differentiating with respect to x, we obtain r'(x) = 9, C'(x) = 3x2-12x+15

By the Theorem, maximum profit occurs when

r'( x) = c'( x), which implies 3x2-12x +15 = 9 3x2 -12x + 6 = 0 3x2-12x + 6 = 0 x2 -4x + 2 = 0

Solving the above quadratic equation, we obtain x =( √ ) = 2-√2

The possible production levels for maximum profit are x = 2 + √2 thousand units and x = 2 -√2 thousand units.

x = 2 + √2 to be a point of maximum profit and x = 2 - √2 to be a local maximum for loss.


Calculus

Theorem

The production level (if any) at which average cost is smallest is a level at whichthe average cost equals the marginal cost.Example:The cost function at a soft drink company is c(x) = x3 - 6x2 +15x (x in thousandsof units). Is there a production level that minimizes average cost? If so, what is it?Solution

Cost : c(x) = x3 - 6x2 +15x

Marginal cost : c'(x) = 3x2 -12x + 15

Average cost : ( ) = x2 - 6x +15

By the Theorem, the production level is minimum at which average cost equals marginalcost. Hence

X2 - 6x +15 = 3x2 - 12x +15 => 2x2 - 6x = 0 =>2x(x - 3) = 0 => x = 0 or x = 3

Since x > 0, the only production level that might minimize average cost is three thousandunits.

We check the derivatives:

Average cost ( ) = x2-6x+15

( ( ) ) = 2x-6

( ( ) ) =2 >0

The second derivative is positive, so x = 3000 units gives an absolute minimum.

LINEARIZATION AND APPROXIMATIONS

Definitions

If f is differentiable at x = a, then the approximating function

L(x)= f(a)+ f'(a) (x-a)……………….. (1) is the linearization of f at a . The approximation

f (x)≅L(x) of f by L is the standard linear approximation of f at a. The point x = a is thecenter of the approximation

Example: Find the linearization of f (x) =√1 + x at x = 3.

Solution

f (x) =√1 + x

f'(x) = √Then ,f (3) = 2, f '(3) = 1/4, and the linearization is


Calculus

L(x) = 2 +1/4 (x - 3) = 5/4 + x/4

At x = 3.2 , the linearization in Example gives

√1+x = √1 + 3.2 ≅ 5/4 + 3.2/4 = 2.050

Example 2:Find the linearization of f(x)=x4 at x =1

L(x)= f(1)+(x-1) f' (1)

f(x)=x4 , f' (x) = 4 x3

f(1)= 1 , f' (1)= 4

L(x)=1+4(x-1)= 4x-3

Differentials

Definitions Let y = f (x) be a differentiable function. The differential dx is an independent

variable. The differential dy is

dy = f'(x) dx.

Example: Find dy if

a) y = x5 + 37 x b) y = sin3x

Solution

a) dy = (5x4 + 37)dx b) dy = (3cos3x)dx

We sometimes write

df = f'(x) dx in place of dy = f'(x) dx, and call df the differential of f . For instance, if f (x) =3x2 - 6, then

df = d(3x2 - 6) = 6xdx.

Every differentiation formula like ( ) has corresponding differential form like d (u + v)= du + dv,

obtained by multiplying both sides by dx.

Formulas for differentials

dc = 0 , d (cu) = cdu

d (u + v) = du + dv d (uv) = udv + vdu

d( )=( )d (sin u) = cosudu d(cosu) = - sinudu

d (tan u) = sec2udu d (cot u) = -csc2udu

d(secu) = secu tan udu d(cscu) = - cscu cot udu


Calculus

Example

d (tan 2 x) = sec2 (2x)d(2x) = 2sec22xdx

d( ) = ( ) ( )( ) ] = dx/(x+1)

The Differential Estimate of Change

Let f (x) be differentiable at x = x0. The approximate change in the value of f when xchanges from x0 to x0 + dx is

df = f ’(x0) dx.

Example: The radius r of a circle increases from r0 =10 m to 10.1m. Estimate the increasein the circle's area A by calculating dA . Compare this with the true change D A

Solution

Since A = πr2, the estimated increase is

dA = A'(r0)dr = 2πr0dr = 2π (10)(0.1) = 2π m2.

The true change is

DA =π (10.1)2 -π102 = (102.01 - 100) π = 2.01π

Absolute, Relative, and Percentage Change

As we move from x0 to a nearby point x0 + dx, we can describe the change is f in threeways:

True Estimate

Absolute change Df = f (x0 + dx) - f (x0) df = f'(x0)dx

Relative Change Df /f(x0) df/ f(x0)

Percentage Change Df /f(x0) .100 df/ f(x0).100

Example :Show that the relative error in computing the volume of a sphere due to anerror in measuring the radius is approximately equal to three times the relative error in theradius.

Solution

Let V be the volume and r be the radius of the sphere. Then they are related by theequation

V = πr3. ………... . . (1)

Taking logarithms on both sides of (1), we obtain

logV = log ( )π + 3 log r


Calculus

Now taking differentials, we get

= 3

since the differential of the constant log (3/4)π is 0, the error relation is

=3 . . ………….. . (2)

where is the relative error in computing the volume and is the relative error incomputing the radius.

The error relation (2) simply says that the relative error in computing the volume of asphere due to an error in measuring the radius is approximately equal to three times the

relative error in the radius.

Example: About how accurately should we measure the radius r of a sphere to calculatethe surface area S = 4πr2 within 1% of its true value?

Solution

We want any inaccuracy in our measurement to be small enough to make thecorresponding increment DS in the surface area satisfy the inequality

| DS | = =

We replace DS in this inequality with

dS = = 8πrdr.

dS = dr= 8πrdr

This gives |8πrdr| = 4πr2/100 dr=r/200

We should measure r with an error dr that is no more than 0.5% of the true value.

Example :The volume V of fluid flowing through a small pipe or tube in a unit of time at afixed pressure is a constant time the fourth power of the tube's radius r. How will a 10%increase in r affect V?

Solution

Here we can take V = kr4.

dV = dr = 4kr3dr = = 4

The relative change in V is 4 times relative change in r, so a 10% increase in r will produce a40% increase in the flow.

Example :The radius of a sphere is found to be 10 cms, with a possible error of 0.02 cms.What is the relative error in the computed volume?


Calculus

Solution

If V is the volume and r is the radius of the sphere,

V =(4/3)πr3

Taking logarithms, we get,

log V = log(4/3)π + 3 log r.

Taking differentials on both sides,

= 3

The error relation is, = 3 approximately.

Given Dr = 0.02, when r = 10. , = 3. (002/10)= .006

Hence, relative error in the volume = 0.006.

Exercises

In exercises 1-3, find the linearization L(x) of f (x) at x = a

1.f (x) = sinx at x=π

2.f (x) = x3 - x at x = 1

3.f (x) =√x at x = 4

We want linearizations that will replace the functions in Exercise 4-5, over intervals thatinclude the given points x0. To make our subsequent work as simple as possible, we wantto center each linearization not at x0 but at a nearby integer x = a at which the givenfunction and its derivative are easy to evaluate. What linearization do we use in each case?

4.f (x) = x2 + 2x, x0 = 0.1

5.f (x) = 2x2 + 4x - 3, x0 =1

6. The diameter of a tree was 10 inch. During the following year the circumferenceincreases 2 inch. About how much did the tree diameter grow? About how much the tree’scross section area grow?

7.Estimate allowable percentage error in measuring the Diameter D of a sphere if thevolume is calculated correctly to within 3% ?

Answers1.π-x 2 .2x-2 3 . x +1 4. 2x 5. -56. 2/ π inch ; 10 sq inch 7. 1%


Calculus

CHAPTER 2

INTEGRATION - I

Sigma Notation for Finite sums

Definitions (Sigma Notation for Finite Sums)

The symbol ∑ denotes the sum a1+ a2 + ……… + an. The ai ’s are the terms of the sum:a1 is the first term, a2 is the second term, ak is the k th term, and an is the n th and lastterm. The variable k is the index of summation. The values of k run through the integersfrom 1 to n. The number 1 is the lower limit of summation; the number n is the upper limitof summation.

Illustrative Example∑ 2 = 12 + 22 + 32 +42 = 1 + 4 + 9 + 16 =30∑ = + + =Algebraic Rules for Finite Sums∑ (ak + bk) = ∑ ak + ∑ bk∑ (ak − bk) = ∑ ak - ∑ bk∑ (C ak) = C∑ akRiemann Sums

Given an arbitrary continuous function y = f (x) on an interval [a, b] , we partition theinterval into n subintervals by choosing n -1 points, say x1 , x2 , ……….xn-1; between a and bsubject only to the condition that

a< x1 < x2< ••• < xn-1<b .

To make the notation consistent, we usually denote a by x0 and b by xn . The set

P = { x1 , x2 , ……….xn}

is called a partition of [a, b].

The partition P defines n closed subintervals

[ x0 x1], [ x1, x2], ……… …..[ xn-1 xn].

The typical closed subinterval [xk-1 xk] is called the k th subinterval of P.

The length of the k th subinterval is Dxk = xk – xk-1.

In each subinterval [xk-1 xk], we select a point ck and construct a vertical rectangle from thesubinterval to the point (ck, f (ck)) on the curve y = f (x). The choice of ck does not matter aslong as it lies in [xk-1 xk].


Calculus

If f (ck) is positive, the number f (ck)Dxk = height • base is the area of the rectangle. If f (ck)is negative, then f (ck)Dxk is the negative of the area. In any case, we add the n products f(ck)Dxk to form the sum

Sp = ∑ f (ck)DxkThis sum, which depends on P and the choice of the numbers ck , is called a Riemann sumfor f on the interval [a, b].

As the partitions of [a,b] become finer, the rectangles defined by the partitionapproximate the region between the x -axis and the graph of f with increasing accuracy. Sowe expect the associated Riemann sums to have a limiting value. To test this expectation,we need to develop a numerical way to say that partitions become finer and to determinewhether the corresponding sums have a limit. We accomplish this with the followingdefinitions.

The norm of a partition P is the partition's longest subinterval length. It is denoted by

||P|| (read "the norm of P").

The way to say that successive partitions of an interval become finer is to say that thenorms of these partitions approach zero. As the norms go to zero, the subintervals becomeshorter and their number approaches infinity.

Example : The set P = {0, 0.2, 0.6,1,1.5, 2} is a partition of [0, 2]. There are five subintervalsof P: [0, 0.2], [0.2,0.6], [0.6,1], [1,1.5], and [1.5, 2].

The lengths of the subintervals are Dx1 = 0.2, Dx2 = 0.4, Dx3 = 0.4, Dx4 = 0.5, and Dx5 = 0.5.The longest subinterval length is 0.5, so the norm of the partition is ||P|| = 0.5. In thisexample, there are two subintervals of this length.

Definition (The Definite integral as a Limit of Riemann Sums)

Let f (x) be a function defined on a closed interval [a, b]. We say that the limit of theRiemann sums Sp = ∑ f (ck)Dxk on [a, b] as ||P|| => 0 is the number I if thefollowing condition is satisfied:

Given any number € > 0, there exists a corresponding number d> 0 such that for everypartition P of [a, b]with ||P||<d => | ∑ f (ck)Dxk - L|< €

for any choice of the numbers ck in the subintervals [xk-1;xk ].

If the limit exists, we writeLim|| || ∑ f (ck)Dxk = L

We call L the definite integral of f over [a, b], we say that f is enterable over [a, b], andwe say that the Riemann sums of f on [a, b] converge to the number L.

We usually write L as ∫ ( ) which is read "integral of f from a to b ." Thus, if thelimit exists,


Calculus

Lim → ∑ f (ck)Dxk = ∫ ( )Upper and Lowe Riemann Sum

Riemann upper sum : Up = ∑ Maxf(x) Dxk

Riemann lower sum: Lp= ∑ Minf(x) Dxk

If f(x) is Riemann integrable then Up =Lp as ||P||0

Theorem (The Existence of Definite Integrals)

All continuous functions are integrable. That is, if a function f is continuous on aninterval [a, b], then its definite integral over [a, b] exists.

Functions with No Riemann Integral

The function

f(x)= 1, when x is rational,

0, when x s irrational

has no Riemann integral over [0,1]. For any partition P of [0,1], the upper and lower sumsare

Up = ∑Maxf(x)Dxk= ∑1 Dxk= ∑Dxk= 1,

(sum of length of all sub intervals) as every subinterval contains a rational number andhence maxf(x) = 1

Lp = ∑_Minf(x)Dxk= ∑0 Dxk = 0,

as every subinterval contains an irrational number and hence Minf(x) = 0 .

For the integral of f to exist over [0,1], Up and Lp would have to have the same limit as||P|| 0. But they do not:

limLp = 0 while lim Up = 1.

Therefore, f has no integral on [0,1].

Example :Express the limit of Riemann sumslim → (3ck2 − 2ck + 5)as an integral if P denotes a partition of the interval [ -1, 3].

Solution: The function being evaluated at ck in each term of the sum is f (x) = 3x2 - 2x + 5 .The interval being partitioned is [ -1, 3]. The limit is therefore the integral of f from -1 to 3:lim → (3ck2 − 2ck + 5) = ∫ 3x2 − 2x + 5Constant Functions

Result : If f (x) has the constant value c on [a, b], then ∫ ( ) =c(b-a)


Calculus

Example∫ 3dx = 3(4 - (-1)) = (3)(5) = 15∫ (−3) = -3(4 - (-1)) = (-3)(5) = -15

The Area Under the Graph of a Nonnegative Function

Definition Let f (x) t 0 be continuous on [a, b]. The area of the region between the graph off and the x -axis is

A = ∫ ( )Whenever we make a new definition, as we have here, consistency becomes an issue.Does the definition that we have just developed for nonstandard shapes give correctresults for standard shapes? The answer is yes, but the proof is complicated and we willnot go into it.

Example :Using an area, evaluate the definite integral ∫ 0<a<b

Solution :We sketch the region under the curve y = x, a <x < b, and see that it is atrapezoid with height (b - a) and bases a and b . The value of the integral is the area of thistrapezoid:∫ = (b - a) = −In particular,∫ = = ( ) − ( ) = = − = 12

Notice that = is an antiderivative of x, further evidence of a connection betweenantiderivatives and summation.Example: Use a definite integral to find the area of the region between the parabola y = x2

and x -axis on the interval [0, b].Solution: We evaluate the integral for the area as a limit of Riemann sums.We sketch the region (a nonstandard shape) and partition [0, b] into n subintervals oflength

Dx =( ) = . The points of the partition are

X0 = 0, x1 = Dx, x2 = 2Dx, ..........., xn-1 = (n -l)Dx, xn = nDx = b.

We are free to choose the ck 's any way we please. We choose each ck to be the right-handendpoint of its subinterval, a choice that leads to manageable arithmetic. Thus, c1 = x1, c2 =x2, and so on. The rectangles defined have areas f (c1)Dx = f (Dx)Dx = (Dx)2 Dx = (12)(Dx)3

f (c2)Dx = f (2Dx)Dx = (2Dx)2Dx = (22)(Dx)3

………………………………………………………………………………………………………

f (cn )Dx = f (nDx)Dx = (nDx)2 Dx = (n2)(Dx)3


Calculus

The sum of these areas is

Sn = ∑ f (ck)Dxk= ∑ k2 (Dxk )3= ( k)3∑ k2= ( )( ) , as Dxk= .

=

= (2+ + ) ……………………….. (6)

We can now use the definition of definite integralLim → ∑ f (ck)Dxk = ∫ ( )to find the area under the parabola from x = 0 to x = b as∫ = Lim → Sn ,In this example, → 0 is equivalent to n0.

= Lim→ (2+ + )

= (2+ 0+0) =

With different values of b we get∫ = ( ) =Exercises

Write the sums in Exercises 1-3 without sigma notation. Then evaluate them.

1.∑ 2. ∑Express the sums in Exercises 3-4 in sigma notation. The form of your answer will dependon your choice of the lower limit of summation.

3.1 + 2 + 3 + 4 + 5 + 6 4. ½ + ¼ + 1/8+ 1/16

5.Suppose that .∑ = -5 and .∑ = 6. Find the values of

a) ∑ 3 a) ∑ c) .∑ ( − )6.Evaluate the sums .

a) .∑ b) .∑ 2 c) .∑ (3 − 2)In Exercises 7-8, graph each function f (x) over the given interval. Partition the interval intofour subintervals of equal length. Then add to your sketch the rectangles associated withthe Riemann sum∑ f (ck)Dxk . given that ck is the (a) left-hand endpoint, (b) right-hand


Calculus

endpoint, (c) midpoint of the k th subinterval. (Make a separate sketch for each set ofrectangles.)

7. f (x) = x2 -1, [0,2] 8. f (x) = sin x, [-p, p]

9.Find the norm of the partition P = [0,1.2,1.5, 2.3, 2.6, 3].

Express the limits in Exercises 18-21 as definite integrals.10. Lim → ∑ ck2Dxk ,where P is the partition of [0 2]11. Lim → ∑ (ck2 − 3ck)Dxk , where P is the partition of [-7 5]

12.Use areas to evaluate the integrals

a) ∫ , b>0 b) ∫ 2RULES FOR INTEGRATION

Rules for definite integrals

1.Zero: ∫ ( ) = 02.Order of Integration ∫ ( ) = −∫ ( )3.Constant Multiples ∫ ( ) = ∫ ( )4.Sum and Differences ∫ ( ( ) − ( )) = ∫ ( ) −∫ ( )5.Additivity: ∫ ( ) +∫ ( ) = ∫ ( )

6.Max-Min Inequality: If max f and min f are the maximum and minimum values of fon

[a, b], then

min f (b - a) ≤ ∫ ( ) ≤ max f (b - a)

Example: Suppose that∫ ( ) = 5, ∫ ( ) = -2, ∫ ( ) = 7.

Then∫ ( ) = -∫ ( ) = -(-2) = 2 Using Rule 2∫ [2 ( ) + 3ℎ( )] =2∫ ( ) + 3∫ ℎ( ) =2(5) + 3(7) =31 , Using Rules 3 and4∫ ( ) =∫ ( ) + ∫ ( ) = 5 + (-2) = 3 Using Rule 5

Example: ∫ ( − 7 + 5)


Calculus

Solution∫ ( − 7 + 5) = ∫ 2 + 7∫ + ∫ (5)== ( ( ) - ( ) ) -7( ( ) − ( ) ) + 5(2-0)

Example: ∫ 2Solution :We cannot apply equation (3) directly because the lower limit of integration isdifferent from 0. We can, however, use the Additivity Rule to express ∫ as adifference

of two integrals that can be evaluated with equation (3):∫ 2 + ∫ 2 =∫ 2 Rule 5∫ 2 =∫ 2 − ∫ 2=( ) − ( )

= − =

The Max-Min Inequality for definite integrals (Rule 6) says that min f (b - a) is a lowerbound for the value of ∫ ( ) and that max f (b - a) is an upper bound.

Example:Show that the value of √1 + cos xdx cannot possibly be 2.

Solution:The maximum value of √1 + cos x on [0,1] is √1+I = √2, so√1 + cos xdx ≤ ma (√1 + cos x) (1 - 0), using max-min inequality

Ie √2.1=√2The integral cannot exceed 46, so it cannot possibly equal 2.Example: Use the inequality cos x ≠ (1- ) which holds for all x, to find a lower bound

for the value of ∫Solution∫ ≠ ∫ (1 − )dx

≠ ∫ 1dx − ∫ ) dx Rules 3 and 4

20 ≠1 (1-0) - ( ) =

(1 _ 0) _ 2 ^3- = 5 » 0.83 . The value of the integral is at least 5.

The Average Value of an Arbitrary continuous FunctionDefinition: If f is integrable on [a, b], its average (mean) value on [a, b] is


Calculus

av(f) = ( )∫ ( )Example : Find the average value of f (x) = 4 - x2 on [0,3]. Does f actually take on this valueat some point in the given domain?Solution

av(f) = av(f) = ( )∫ ( ) = ( )∫ (4 − 2) = (4 (3-0) –( ) ) = (12 − 9) = 1The average value of f (x) = 4 – x2 over the interval [0, 3] is 1. The function assumes thisvalue when

4 – x2 = 1 or x = −√3 . Since one of these points, x =√3 lies in [0, 3], the function doesassume its average value in the given domain .The Mean Value Theorem for Definite Integrals Theorem (The Mean value theorem for Definite Integrals)

If f is continuous on [a, b], then at some point c in [a, b],( ) = ( )∫ ( )The continuity of f is important here. A discontinuous function can step over its averagevalue.

Example: Show that if f is continuous on [a, b], and if ∫ ( ) = 0, then f (x) = 0 at leastonce in [a, b].

Solution

The average value of f on [a, b] is

av(f) = ( )∫ ( ) = = ( ) . 0 = 0By Theorem, f assumes this value at some point c in [a, b].

Exercises

1.Suppose that f and g are continuous and that∫ ( ) = -4, ∫ ( ) = 6, ∫ ( ) = 8 ,

Use the rules at the beginning of this Chapter to find

a) ∫ ( ) b) ∫ ( )c) ∫ 3 ( ) d) ∫ ( )e) ∫ ( ( ) − ( )) f) ∫ [4 ( ) − ( )]

2. Suppose that f is continuous and that ∫ ( ) = 3 and ∫ ( ) = 7a) ∫ ( ) b) ∫ ( )


Calculus

3. Evaluate the integrals

a) ∫ 7 b) ∫ (2 − 3) c) ∫ |1 + |d) ∫ 3 2Answers

1. a) 0 b) -8 c) -12 d) 10 e) -2 f) 16

2. a) 4 b) -4

3. a) -14 b) -2 c) -7/4 d)7

CHAPTER 6

INTEGRATION IIFUNDAMENTAL THEOREMS OF CALCULUS

In this chapter we discuss two parts of fundamental theorem of calculus.Theorem (The Fundamental Theorem of Calculus, Part 1)

If f is continuous on [a, b], then F(x) = ∫ ( ) has a derivative at every point of [a, b]

and = ∫ ( ) = f (x), a ≤ x ≤ b………….. (1)

Example : if y = ∫ costdtSolution: Notice that the upper limit of integration is not x but x: To find we musttherefore treat y as the composite of

y = ∫ costdt and u = x2

and apply the Chain Rule:== ∫ costdt d Substitute the formula for y.

=cosu Eq. (1) with f (t) = cos t

=cosx2 2x

Example :Express the solution of the following initial value problem as an integral.

Differential equation: = tan x

Initial condition : y(1) = 5


Calculus

Solution

The function F(x) = ∫ tantdtis an antiderivative of tan x. Hence the general solution of the equation is

y =∫ tantdt + C

As always, the initial condition determines the value of C :

5 = ∫ tantdt + C y(1) = 5

5 = 0 + C ……………………….. (2)

C = 5.

The solution of the initial value problem is y = ∫ tantdt + 5 .

How did we know where to start integrating when we constructed F (x) ? We could havestarted anywhere, but the best value to start with is the initial value of x (in this case x =1).Then the integral will be zero when we apply the initial condition (as it was in Eq. 2) and Cwill automatically be the initial value of y .

The Evaluation of Definite Integrals using FTC2

Theorem (The Fundamental Theorem of Calculus, Part 2)

If f is continuous at every point of [a, b] and F is any antiderivative of f on [a, b], then∫ f(x)dx = F(b) - F(a) ……………………… (3)

Example∫ cosxdxП = [sinx] = sin П - sin0 = 0 -0 = 0∫ secxtanxdxП = [sec x] = sec0 -sec (− П) = 1 -√2

We can now see that without any restriction on the signs of a and b ,∫ xdx = = − becuase is an antiderivative of x

Example:Find the area of the region between the x -axis and the graph of

f (x) = x3 - x2 - 2 x, -1 ≤ x ≤ 2.

Solution:First find the zeros of f . Since

f (x) = x2 - x2 - 2x = x(x2 - x - 2) = x(x +1)(x - 2), the zeros are x = 0, -1, and 2 Thezeros partition [ -1, 2] into two subintervals: [ -1, 0], on which f < 0 and [0, 2], on which f >0. We integrate f over each subinterval and add the absolute values of the calculatedvalues.


Calculus

Integral over [ -1, 0]: ∫ ( 3 − 2 − 2 ) =[ + − ] = 0 – [ + − ] =

Integral over [0, 2]: ∫ ( 3 − 2 − 2 ) =[ + − ] = [ 4- -4] -0 =

Enclosed area: Total enclosed area = + | | =

Exercises

Evaluate the integrals in Exercises 1-13.

1. ∫ (2x + 5)dx 2. ∫ (x2 + √x)dx 3. ∫ ( − )du√ 4.. ∫ |x|dxIn Exercises 5- 8, use a substitution to find an antiderivative and then apply theFundamental Theorem to evaluate the integral.

5. . ∫ (1 − 2x3)dx 6. . ∫ t√1 + t2 dt 7. . ∫ (sin2 cos )dxIn Exercises 8-10, find the total area between the region and the x -axis.

8. y = -x2 - 2x, -3 ≤ x ≤ 2

9.y = x3 - 3x2+ 2x, 0 ≤ x ≤ 2

10.y = x3, -1 ≤ x ≤8

Answers

1. 6 2. 8 3. 1 4. 16 5. 0 6. (2√2 − 1) 7. √2/3


Calculus

MODULE III

Chapter 1

APPLICATIONS OF INTEGRALS

Things like areas between curves, volumes and surface areas of solids, lengths of curves etc can becalculated with integrals. We define all of these as limits of Riemann sums of continuous functionson closed intervals, that is, as integrals, and evaluate these limits using calculus. In this module wediscuss some of the above mentioned applications.

Areas Between CurvesHere we discuss how to find the areas of regions in the coordinate plane by integrating thefunctions that define the regions boundaries.

Definition: If and are continuous functions with ( ) ≥ ( ) ; ∀ ∈ [ , ], then the area ofthe region between the curves = ( ) = ( ) is the integral of − from

a to b : = ∫ [ ( ) − ( )]Example1: Find the area between = = from 0 4.Solution:

Step1: Identifying upper curve and lower curve

We know that ≥ ∈ [0, ]. Therefore take ( ) = ( ) =Step2 : Finding the limits of integration

The limits of integration are already given: = 0 , =Step3: Write a formula for ( ) − ( )Here ( ) − ( ) = −Step4: Applying formula

The required area is= ∫ [ ( ) − ( )] = ∫ ( − ) = [ + ] = √Example2: Find the area of the region enclosed by the parabola= 2 − ℎ = −Solution:


Since the curve = 2 − ℎ = − ,the upper curve is ( ) = 2 − andthe lower curve is ( ) = −Step2 : Finding the limits of integration


Calculus

We find the limits of integration by equating the two curves:2 − = −− + 2 = 0 ( ℎ )( + 1)( − 2) = 0 ( ℎ )= −1, = 2 ( ℎ )The limits of integration are = −1 , = 2Step3: Write a formula for ( ) − ( )( ) − ( ) = (2 − )— = 2 − +Step4: Applying formula

= [ ( ) − ( )] = (2 − + ) =Example 3:

Find the area of the region in the first quadrant that is bounded above by = √ and below bythe -axis and the line = − 2Solution :


The sketch shows that the region’s upper boundary is the curve ( ) = √ and the lowerboundary changes from ( ) = 0 0 ≤ ≤ 2 ( ) = − 2 2 ≤ ≤ 4. We divide theregion in to two subregions A and B.

Step2 : Finding the limits of integration

The limits of integration for region A are = 0 = 2.The left hand limit for region B is = 2.To find the right-hand limit, we solve the equations = √ = − 2 simaltaneously for x:√ = − 2= ( − 2)= − 4 + 4− 5 + 4 = 0( − 1)( − 4) = 0= 1, = 4


Calculus

Only the value = 4 satisfies the equation √ = − 2 . So the right hand limit is = 4Step3: Write a formula for ( ) − ( )For 0 ≤ ≤ 2 ∶ ( ) − ( ) = √ − 0 = √For 2 ≤ ≤ 4 ∶ ( ) − ( ) = √ − ( − 2) = √ − + 2Step4: Applying formula

Total Area = ∫ √ + ∫ (√ − + 2) =Remark: The above problem can be easily solved with the help of geometry as follows

The required area in the problem (example3) is the area between the curve = √ , 0 ≤ ≤ 4,and the -axis, minus the area of a triangle with base 2 and height 2

= √ − 12 (2)(2) = 103Example4:

Find the area of the region in example 3 by integrating with respect to y

Solution :

Step1: We sketch the region and a typical horizontal rectangle based on a partition of an intervalof y-values . The regions right hand boundary is the line = + 2, so ( ) = + 2.The left handboundary is the curve = ( ) =Step2: The lower limit of integration is = 0.The upper limit is obtained by solving:+ 2 =− − 2 = 0( + 1)( − 2) = 0= −1, = 2The upper limit of integration is = 2Step3: ( ) − ( ) = + 2 −


Calculus

Step4:= ∫ [ ( ) − ( )] = ∫ [ + 2 − ] = .

Exercises:

Find the areas of the regions enclosed by the following curves:

1. = − 2 = 22. = = 83. = = − + 44. = − 4 + 4 =5. = | | 5 = + 66. = 2 , = 0 = 37. − 4 = 4 4 − = 168. = 2 = 2 , 0 ≤ ≤9. = = , = − , =10. = 3 = 0 , 0 ≤ ≤

Answers

1. 2. 3. 4. 8 5. 6. 18

7. 8. 4 9. 10. 2

Chapter 2

Volumes of Solids of RevolutionSolids of revolution are solids whose shapes can be generated by revolving plane regions

about axes

Volume of a solid of Revolution (Rotation about the x-axis)

The volume of the solid generated by revolving about the x-axis the region between the x-axis andthe graph of the continuous function = ( ), ≤ ≤ is= ∫ ( ) = ∫ ( ( ))Example1: The region between the curve = √ , 0 ≤ ≤ 4, and the X-axis is revolved aboutthe X-axis to generate a solid. Find its volume.

Solution : The Volume is= ∫ ( ( )) = ∫ [ ] = ∫ = 8


Calculus

Example2: Find the volume of the solid generated by revolving the region bounded by = √and the lines = 1, = 4 about the line = 1Solution: Here the radius function is ( ) = √ − 1

The volume is

= [ ( )] = [√ − 1] = − 2√ + 1 = 76Volume of a solid of Revolution (Rotation about the Y-axis)

The volume of a solid of generated by revolving about the Y-axis the region between the Y-axis and the graph of the continuous function = ( ), ≤ ≤ is = ∫ ( ) =∫ [ ( )]Example3: Find the volume of the solid generated by revolving the region between the Y-axisand the curve = ,1 ≤ ≤ 4 about the Y-axis.

Solution: Here the rotation is about the Y-axis and the radius function is ( ) =The volume is = ∫ [ ( )] = ∫ = 3

Example4: Find the volume of the solid generated by revolving the region between theparabola = + 1 and the line = 3 about the line = 3.

Solution: Here the rotation is about the line = 3, which is parallel to the Y-axis

The radius function is ( ) = 3 − ( + 1) = 2 −The volume is = ∫ [ ( )] = ∫ [2 − ] = ∫ [4 − 4 + ] = √√√√√Exercises

Find the volumes of the solids generated by revolving the regions bounded by thefollowing curves about the X-axis

1. = , = 0, = 22. = √9 − , = 03. = √ , 0 ≤ ≤ , = 0, = 0

Find the volumes of the solids generated by revolving the regions bounded by the following curvesabout the X-axis

4. = √5 , = 0, = −1, = 15. = 2 2 , 0 ≤ ≤ , = 06. Find the volume of the solid generated by revolving the region bounded by = √ , and

the lines = 2 = 0 about:a) The X-axisb) The Y-axis


Calculus

c) The line = 2d) The line = 4

Answers:

1. 2. 36 3. 4.2 5.2 6.(a):8 (b) : (c) : (d)

Lengths of Plane CurvesDefinition:

A function with a continuous first derivative is said to be smooth and its graph is called asmooth curve

Definition:

If is smooth on [ , ] then the length of the curve = ( ) from is defined by the

integral = ∫ 1 + ( ) = ∫ 1 + ( ( ))Definition:

The length of the smooth curve = ( ) from is defined by the formula

= 1 + ( ) = 1 + ( ( ))Example1: Find the length of the curve = √ – 1 , 0 ≤ ≤ 1Solution: = 4√2 3 23 – 1⟹ = 2√2 ⟹ ( )2 = 8

The length of the curve from = 0 = 1 is= ∫ 1 + ( ) = ∫ √1 + 8 = (The last integral can be evaluated using the

substitution = 1 + 8 )

Example2:

Find the length of the curve = ( 2) from = 0 = 2.

Solution : The derivative = is not defined at = 0 , so we cannot use the firstformula. We therefore rewrite the equation to express in terms of := ( 2) ⟹ = 2 ⟹ = 2Thus the required length is the length of the curve = 2 = 0 = 1


Calculus

The derivative = 3 is continuous on [0,1] . Therefore the required length is given by the

formula = ∫ 1 + ( ) = ∫ 1 + 9 = 10√10 − 1 ≈ 2.27Exercises :Find the l1engths of the following curves:

1. = ( + 2) = 0 = 32. = 3 + 1 (4 ) from = 1 = 3 (Hint : 1 + is a perfect square)

3. = 4 + 1 (8 ) = 1 = 2 (Hint : 1 + is a perfect square)

4. = 3 4 − 3 8 + 5 ,1 ≤ ≤ 85. = ∫ √ − 1 ,− ≤ ≤

Answers:

1. 12 2. 3. 123/32 4. 99/8 5.2

Chapter 3

Areas of surfaces of revolution:Surface Area Formula for the revolution about the X-axis:

If the function ( ) ≥ 0 is smooth on [ , ] then the area of the surface generated byrevolving the curve = ( ) about the X-axis is= ∫ 2 1 + = ∫ 2 ( ) 1 + ( ( ))Surface Area Formula for the revolution about the Y-axis:

If the function ( ) ≥ 0 is smooth on [ , ] then the area of the surface generated byrevolving the curve = ( ) about the Y-axis is= ∫ 2 1 + = ∫ 2 ( ) 1 + ( ( ))Example1: Find the area of the surface generated by revolving the curve= 2√ , 1 ≤ ≤ 2 , about the X-axis

Solution : Here = 2√ , = 1, = 2= 1√ ⟹ = 1


Calculus

= 2 1 + = 2 . 2√ 1 + 1 = 4 √1 += 83 3√3 − 2√2

Example2: The line segment = 1 − , 0 ≤ ≤ 1, is revolved about the Y-axis to generate acone. Find its lateral surface area

Solution: Here = 0, = 1, = 1 − ⟹ = −1 ⟹ 1+ = √2= 2 1 + = 2 (1 − )√2 = 2√2 (1 − ) = √2

Remark: The above calculation agrees with the formula from geometry;= 2 × ℎ ℎ = √2The Short Differential Form

The equations = ∫ 2 1 + , = ∫ 2 1 + are often written in

terms of the arc length differential = ( ) + ( ) as:= ∫ 2 = ∫ 2 .

In the first of these, is the distance from the X-axis to an element of the arc length .In thesecond , is the distance from the Y-axis to an element of the arc length .Both integrals havethe form: = ∫2 ( )( ℎ) = ∫2 , ℎ is the radius from the axis ofrevolution to an element of arc length

Example: Find the area of the surface generated by revolving the curve = , 0 ≤ ≤ ,about the X-axis

Solution : We start with the short differential form:= ∫2 = ∫2 = ∫2 ( ) + ( ) (For revolution about the X-axis the radius function is )

We then decide whether to express in terms of or in terms of .The original form ofthe equation, = , makes it easier to express in terms of , so we continue thecalculation with = ⟹ = 3 , ( ) + ( ) = √1 + 9With these substitutions, becomes the variable of integration and= ∫ 2 ( ) + ( ) = ∫ 2 √1 + 9 = (the last integral is evaluatedusing the substitution = 1 + 9 )


Calculus

Exercises:

Find the areas of the surfaces generated by revolving the following curves about the indicatedaxes:

1. = , 0 ≤ ≤ 2 ; −2. = √2 − , 0.5 ≤ ≤ 1.5 ; −3. = ,0 ≤ ≤ 1 ; −4. = 2 4 − ; 0 ≤ ≤ ; −5. = + , 1 ≤ ≤ 2 ; − (Hint: Express = ( ) + ( ) in terms of

, and evaluate the integral = ∫2 with appropriate limits)

6. Find the lateral surface area of the cone generated by revolving the line segment= , 0 ≤ ≤ 4, about the X-axis. Also check your answer with the geometry formula

Answers:

1. 98 81 2. 2 3. (√8 − 1) 9 4. 35 √5 3 5. 253 20 6. 4 √5


Calculus

Module IV

Chapter I

Moments and Center of Mass

Introduction

Many structures and mechanical systems behave as if their masses were concentrated at asingle point called the center of mass.

Masses along a Line

Let , are masses on a rigid -axis supported by a fulcrum at the origin .Each massexerts a downward force equal to the magnitude of the mass times the acceleration of

gravity. Each of these forces has a tendency to turn the -axis about the origin. This turning effectis called a torque. It is measured by multiplying the force by the signed distance from thepoint of application to the origin. Masses to the left of the origin exert negative (counterclockwise) torque. Masses to the right of the origin exert positive (clockwise) torque.

The sum of the torques measures the tendency of the system to rotate about the origin.This sum is called system toque.system toque = m gx + m gx + m gxRemark

The system will balance if and only if torque is zero. System torque= ( + + )

,the torque is the product of the gravitational acceleration ,which is a feature ofenvironment.

Definition

The number + + is called the moment of the system about the origin.=Moment of the system about the origin=∑RemarkThe torque of each mass about the fulcrum in special location isTorque of about = ( )

=( − )An equation we can solve for ∑( − ) =0 (sum of the torques equals zero)∑( − ) =0 (by constant multiple rule)∑( − ) =0 as g≠0


Calculus

∑( − )=0∑ − ∑ = 0 ⇒ ∑ = ∑ ⇒ = ∑∑ =

The point is called the system’s center of mass.

Wires and thin Rods

Imagine along strip lying along x-axis from x=a to x=b and cut into small pieces of mass∆ by a partition of the interval [ , ].The piece is ∆ units long and lies approximatelyunits from the origin. Now observe the three things .First, the strips center of mass is nearly thesame as that of the system of point masses we would by attaching each masses ∆ to thepoint : ≈ …………………….(1)

Second, the moment of each piece of the strip about the origin is approximately ∆ ,so thesystem moment is approximately the sum of the ∆ .

i.e. System moment ≈∑ ∆ ………………………….(2)

Third, if the density of the strip at is ( ) expressed in terms of mass per unit length and iscontinuous, then ∆ is approximately equal to ( )∆ (mass per unit length times length):∆ ≈ ( )∆ ……………………………………….(3)

Combining these three observations gives ≈ ≈ ∑ ∆∆ ≈ ∑ ( )∆∑ ( )∆ ……………………….(4)

Note that ∑ ( )∆ is Riemann sum for the continuous function ( ) over the closedinterval [ , ] and ∑ ( )∆ is a Riemann sum for the function ( ) over this interval. Weexpect the approximations in (4) to improve as the strip partitioned more finely and we led to theto the equation

= ∫ ( )∫ ( )Moment, Mass and Center of mass of a thin rod or strip along the x-axis with density function ( )

Moment about the origin = = ∫ ( )Mass=M=∫ ( )

Center of mass = =Example:1

Show that the center of mass of a straight, thin, strip or rod of constant density lies halfway between two ends.


Calculus

Solution:We model the strip a portion of the x-axis from x=a to x=b.We want to show that =

.Given that density having constant value. So we take the function ( ) as a constant call it δ.

= ( ) = = = = = 2 = 2 ( − )= ∫ ( ) = ∫ = ∫ = [ ] = ( − ) = =

Example:2

The 10m-long rod, the thickness from left to right so that its density is ( ) = 1 + .Findthe rod’s center of mass.

Solution:

The rod’s moment about origin= ∫ ( ) = ∫ 1 + = ∫ + = + =50 + = 250 .

The rod’s mass

, = ∫ ( ) = ∫ ( ) = ∫ 1 + = + = 10 + = 15The center of mass = = = ∗ = ≈ 5.56Masses Distributed over a plane Region

Suppose we have a finite collection of masses located in the plane, with mass at thepoint ( , ).The mass of the system is

System mass = M= ∑ .

Each mass has a moment about each axis is and it’s moment about y-axis isis .The moments of the entire system about the two axes are

Moment about x-axis: = ∑Moment about y-axis: = ∑ .

The x-coordinate of system’s center of mass is defined to be = = ∑∑ ………………………..(5)

The y-coordinate of the system’s center of mass is defined to be

ӯ= = ∑∑ ………………………..(6)


Calculus

The point ( , ӯ) is called the system’s center of mass.

Thin, Flat plates

Imagine the plate occupying a region in the xy-plane, cut into thin strips parallelone of the axes. The center of mass of a typical strip is (x,y) .Which treat the strips mass ∆m as if itwas concentrated at (x,y).The moment of the strip about the y-axis is x∆m.The moment of thestrip about the x-axis is y∆m.

Now equation (5) and (6)implies = = ∑ ∆∑∆ӯ= = ∑ ∆∑∆

As in the one dimensional case, the sums are Riemann sums for integrals and approach theseintegrals as limiting values of the strips in to which the plate is cut into narrower and narrower.Then we get = ∫ ∫ And ӯ= ∫ ∫Moments, Mass and center of mass of a thin plate covering a region in the xy-plane.

Moment about x-axis: = ∫ Moment about y-axis: ∫

Mass=M=∫Center of mass= = and ӯ=

Remark

To evaluate these integrals, we picture the plate in the coordinate plane and sketch a stripof mass parallel to one of the coordinate axes. Then we express the strip’s mass dm and thecoordinates (x,y) of the strip’s center of mass in terms of x or y.Finally we integrate and and dm between limits of the integration determined by the plate’s location in the plane.

Example:3

The triangular plate is obtained by joining (0,0),(1,0) and (1,2) has a constant density ofδ=3 / .Find

a) the plate’s moment about the y-axis)

b) the plate’s mass M

c) the x-coordinate of the plate’s center of mass.

Solution:

Consider a vertical strip as shown in figure

a)The moment :The typical vertical strip has center of mass(c.m)= (x,y)=(x,x).


Calculus

length=2x

Area=dA=2xdx

Width=dx

Mass = dm=δdA=3.2Xdx=6xdx

Distance of c.m from y-axis =x=x

The moment of the strip about y-axis is, xdm=x.6x=6 dx

The moment of the plate about y-axis is,= = 6 = [2 ] = 2 .b) the plate’s mass M =∫ = ∫ 6 = [3 ] = 3

c)the x-coordinate of the plate’s center of mass = = = . . =

Example:4

Find the center of mass of a wire of constant density δ shaped like a semi circle of radius a

Solution:

We model the wire with the semi circle = √ − .The distribution of mass issymmetric about y-axis ,so = 0.To find ӯ imagine the wire divided into short segments. Thetypical segment has

Length=ds=adθ

Mass=dm=δds=δadθ

Distance of c.m to x-axis=y=asinθ

Hence ,ӯ= ∫ ∫ = ∫ ∫ = [ ] = .

Centroids:

When the density function is constant, it cancels out of the numerator and denominator ofthe formulas and ӯ.Thus when the density is constant the location of the center of masses afeature of the geometry the object and not of the material from which it is made. In such casesthe center of mass is called the centroid of the shape

Exercise

1. An 80-lb and a100-lb child are balancing on a seasaw.The 80-lb child is 5ft from thefulcrum. How far from the fulcrum is the 100-lb child?

2. The ends of a log are placed on two scales. One scale reads 100 kg and the other 200kg.Where is the log’s center of mass?

3. The ends of two thin steel rods of equal length are welded together to make a rightangled frame. Locate the frame’s center of mass.


Calculus

Exercise 4-6 give density functions of thin rods lying along various intervals of the x-axis. Find each rod’s moment about the origin, mass, and center of mass.

4. ( ) = 4 , 0 ≤ ≤ 25. ( ) = 1 + , 0 ≤ ≤ 36. ( ) = 1 + √ , 1 ≤ ≤ 4

Chapter 2

WORKIn everyday life, work means an activity that requires muscular or mental effort.Work Done by a constant ForceWhen a body moves distance d along a straight line as a result of being acted on by aforce of constant magnitude F in the direction of motion. We calculate the work Wdone by the force on the body with the formula

W=Fd (constant force formula for work)…………………………….(1)From this equation we see that the unit of work is in any system is the unit of forcemultiplied by the unit of distance. So the unit of work is nN.m.This combination iscalled Joule.Work done by a variable force

DefinitionThe work done by a variable force F(x) directed along the x axis from x=a to x=b isW=∫ ( )Example:1

The work done by a force of F(X)= along the x-axis from x=1m to x=10m isW=∫ ( )=∫ = = + 1 = 0.9Example:2A leaky 5lb bucket is lifted from the ground into the air by pulling in 20ft of rope at a constantspeed. The rope weighs 0.08lb/ft.The bucket starts with 2gal of water (16lb) and leaks at aconstant rate. It finishes draining just as reaches the top. How much work has spent?

a) Lifting the water alone.b) Lifting the water and bucket together.c) Lifting the water, bucket and rope.


Calculus

Solution:a) The water alone: The force required to lift the water is equal to the water’s weight whichvaries steadily from 16 to 0 lb over the 20ft lift. When the bucket is x-ft of the ground thewater weighs ( ) = 16 20 −20 = 16 1 − 20 = 16 − 45The work done is W=∫ ( ) =∫ 16 − = 16 − 2= 320 − 160 = 160b) The water and bucket together. According to equation (1),it takes 5*20=100 ft lb to a 5lb weight20ft.

So,160+100=260ft.lbof work were spent lifting the water and bucket together

d) The water, bucket and rope. :Now the total weight at level x is( ) = 16 − 45 + 5 + (0.08)(20 − )

The work lifting the top is

Work on rope =∫ (0.08)(20 − ) = ∫ (1.6 − 0.08 ) = 16 −The total work for the water bucket and rope is 160+100+16=276ft-lb.

Hooke’s Law for springs: =Hooke’s law says that the force it takes to stretch or compress a spring x length units from

its natural (unstressed) length is proportional to x. In symbols F=kx .The constant k is measuredin force units per length, it is called force constant spring constant of the spring

Example: 3

Find the work required to compress a spring from its natural length of 1ft to a length of0.75 ft .If the force constant is k=16lb/ft.

Solution:

We assume the uncompressed spring laid along the x-axis with its movable end at theorigin and its fixed end at x=1.This enables us to describe the force required to compress thespring from 0 to x with the formula F=16x.To compress the spring from 0 to 0.25ft the force mustincreases from F (0) =16*0=0lb to F(0.25)=16(0.25)=4lb.

The work done by F over this interval is

W=∫ 16 = 0.5 −.Example: 5

A spring has a natural length of 1m.A force of 24N stretches the spring to a length of 1.8m.


Calculus

a) Find the force constant k

b) How much work will take to stretch the spring 2m beyond it natural length.

c) How far will a 45-N force stretch the spring?

Solution:

a) The force constant: We have, F=kx

F=24N, x=1.8-1=0.08m, so k=24/0.8=30N/m

b) The work to stretch the spring 2m: We imagine the unstressed spring hanging along thex-axis with its free end at x=0.The force required to stretch the spring x m beyond itsnatural length is the force required to pull the free end of the spring x units from theorigin. Hooke’s law with k=30 says that this force is

F(X)=30x.

The work done by F on the spring from x=0 to x=2 m is= 30 = [5 ] = 60c) How far will a 45-N force stretch the spring:

We substitute F=45m equation F=30x to find

45=30x or x=1.5m

Exercises

1. A mountain climber is about to haul up a 50-m length of hanging rope. How much work will ittake if the rope weights 0.624N/m?

2. An electric elevator with a motor at the top has a multistoried cable weighing 4.5 lb/ft.

a) When the car is at the first floor, 180 ft of cable are paid out, and effectively 0 ft is out

b) When the car is at the top floor. How much work does the motor do just lifting the cable

c) When it takes the car from the first floor to the top?

calculus - core course for ii semester bsc mathematics

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