mathematics_notes.pdf

7/27/2019 mathematics_notes.pdf

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Chapter 1

1.1 INTRODUCTION

The notion of rate or rate of change was invented almost simultaneously by

Leibnitz and Newton as a tool for solving problems in geometry and astronomy. It

soon became an invaluable aid in many other branches of mathematics and

physics. Many familiar experiences in daily life have at their basis this crucial

concept - the derivative or rate of change.

An important question in daily life and applications is HOW FAST? How fast is

a population growing? How fast is money flowing in an economy? How fast is a

car moving? How fast is the heart beating?

All of these questions are part of the topicRATE OF CHANGE or DERIVATIVE(as it is called in Calculus an important branch of Mathematics). The study of this

topic is the major aim of this chapter.

1.2 CHANGE

As a prelude to studying the concept rate of change, we start with the

component notion CHANGE.

We are all familiar with what is meant by change.

The temperature in the evening is different form that in the morning, which

means there is a change in the temperature.

Marks scored by you in this week is different form your marks for last week. Thus

there is a change in your marks or performance.

More examples are given below. Study them carefully.

1 2.1 Example

The direction of the front wheels of a car is determined by the steering wheel. If

the steering wheel is turned through an angle x degrees from the neutral

position, the front wheels turn through an angle y degree. Thus the angle y

through which the front wheels turn depends on the angle x through which the

steering wheel turns. That is, y is a function of x or y = f(x).


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Now, for a small change in x, there will be a small change in y; to be precise,

consider two positions of the steering wheel given by x and x + x.

NOTE : x denotes a small change in the value of x.

Corresponding to these two positions, the front wheel will have two positions

given by y and y + y, where

y = f(y) and y + y = f(x + x.)

Thus y = y + y - y

= f(x x f x

measures the change in the position of the front wheels.

1.2.2. Example (Economics)

If x is the rate of production of an object manufactured for sale (for example, x

number of objects produced per day), and y is the production cost (total cost per

day), then y will change as x changes. Thus, if x changes f then y may increase

by Rs. 2000. (in economics this increase in y is called the marginal cost). Thus, y

is a function of x, y= f(x).

Then, when the rate of production for two days are x and x x the corresponding

costs will be y and y y, where, as above,

y= f x x f x

measures the change in the cost or production.

1.2.3 Physics

Suppose a car is driven on a straight road, then the position (denoted by y) of the

car will change with time (denoted by x). Thus y = f(x). For two instants of time x

and x x, the corresponding positions will be y and y y; where, as in 1.2.1.

y = f x x f x


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measures the change in the position.

1.2.4. Biology

Suppose a population (of organizm or human beings) is observed for a length of

time. Then the number of its members (denoted by y) will change with time

(denoted by x). Thus y = f(x). The number of members at two instants of time x

and x x will be y and y y,

So that

y = f x x f x

measures the change in the population

1.2.5 Geometry

Consider a function y = f(x). Then for two values of the variable x, namely x and

x x we have two values of y, namely y and y y where

y f x x f x

measures the change in the value of the function.

1.2.6

In all the above examples the underlying theme was same CHANGE. We note

two important points:

POINT 1: The change x may be positive or negative (but is never zero).

POINT 2: The change y may be positive or negative. (or zero sometimes)

NOTE CAREFULLY: when x is positive, y may be positive or negative.

Similarly, when x is negative, y may be positive or negative.

For example, go back to Example 1.2.4. Suppose our population consists of

healthy members. Then as time passes, the population grows: i.e., as time

changes from x to x x (where x 0) the change in the population y is

positive, because f x x is greater than f(x). (recall that f x represents the

number of members of our population).


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Average Rate of Change =x

y

in xChange

yinChange

= Change in y corresponding to a unit change in x.

We now illustrate this concept with some examples.

1.3.1: Example

Let us go back to Example 1.2.3. Suppose it is known that f (x) = 16x2. Then x

0 because x represents time (and time cannot be negative). Now, the average

rate of change (or average velocity as it is called in this case) of position of the

car over the first 10 seconds of motion in given by

ft/sec.16010

16x01016x

?x

f?xf

?x

?y 22xx

because we are observing the motion from x = 0 sec. to 10xx sec s.

Similarly, the average velocity of the car from x = 2 sec s to 4xx sec s. is

given by

ft/sec.962

16x216x422

xy

1.3.2: Example

Go back to Example 1.2.5. Consider .106)( 2 xxxf Now x can assume any

valuepositive, negative or zero. So, taking x = 2 and 3xx , we get.

.1

1

21

1

23 ff

x

y

However, If we take x = 4 and 1.4xx , we get

1.21.0

221.2

1.0

41.4 ff

x

y


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Thus the average rate of change can be either positive or negative, depending

upon the function and the values of x.

Sometimes xy can be zero also (even though x is not zero). For example,

consider x = 2 and .4xx

Than

022)2()4( ffy

but 2x

Thus .0x

y

When we deal with functions and their graphs,x

yis called the slope of

(appropriate) secant. For example, 1x

yis the slope of the secant joining the

points ))2(,2( fA and ).)3(,3( fB see the figure given below:


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EXERCISE 1.3

1. A function is given by the following table:

x 2.00 2.02 2.04 2.06 2.08

f (x) 1.000 1.020 1.042 1.064 1.086

a) For x = 2.02 and ,04.2xx evaluate xy

b) For x = 2.02 and ,06.2xx evaluate xy

c) For x = 2.04 and ,04.2xx evaluate xy

d) Verify that the value of xy in (b) is the same. As the values of

xy in (a) and (c).

e) For x = 2.04 and 00.2xx , evaluate xy .

2. If the production cost of an item is given by

232)( 2 xxxfy

find xy when x changes from 100 units to 102 units.

3. Let 106)( 2 xxxf and x vary from 0 to 4 with increment 5.0x (that

is, when x = 0, 5.0xx and when ,5.0x 0.1xx etc). Look up a

computer program which will evaluate the values of xy .

1.4 INSTANTANEOUS RATE OF CHANGE

In this section we develop the main idea RATE OF CHANGE. The passage

from the concept of average rate of change to the instantaneous rate of

change is a subtle one involving the concept of limit.

We explain this first by means of the following example:

1.4.1: Example

Go back to the Example 1.3.1. The distance traveled by a car, in x seconds, is

given by .16)( 2xxf


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We are interested in finding the instantaneous velocity of the car when x = 2

secs. To do this, we consider small changes in x namely 1.0x sec or

01.0x sec. or 001.0x sec or 1.0x sec or 01.0x sec. or 001.0x

sec etc.

NOTE : Even though x cannot take negative values here (WHY?) x can

very well take negative values.

Go back to POINT: 1 of 1.2).

The following table gives the values of xy for the various values of x

x 2 2 2 2 2 2

xx 2.1 2.01 2.001 1.9 1.99 1.999

xy 65.6 64.16 64.016 62.4 63.84 63.984

From the above table, it is clear that as xx take values close to

xyx ,2 becomes very close to the value 64.

Also, when xx takes values close enough to x = 2, x takes values close

enough to zero.

Thus, when x = 2 we see that as x assumes values close enough to zero,

xy assumes values close enough to 64, we express this fact by saying

that as x tends to zero, xy tends to 64 or we write

0

.64lim

x

x

y

we have a short hand notation for the left hand side. This isdx

dyThus

x

y

xdx

dy

0

lim

Hence we can write .64dx

dy


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Remember that all this happens when x = 2. We express this fact by

.64)( 2xdx

dy

The value 2)( xdx

dy

is called the (instantaneous) rate of change at x = 2 or

the (instantaneous) velocity at x = 2 or more generally as the rate of

change at x = 2 or the derivative of f (x) at x = 2.

The general setup is as follows: given a function )(xfx and a value x = a.

The rate of change of f(x) at x = aor the derivation of f(x) at x = a is

defined as

x

afxaf

xdx

dyax

)()(

0

lim)(

Remember the following:

1. It is customary to say rate of-change rather than instantaneous rat-of-

change even though the second name is the most appropriate.

2. The words rate-of-change and derivative stand for the same idea. They

will be used interchangeably.

3. The derivative of f (x) at x = a is given by adx

dy

)(

1.4.2: For the above example, we can now find the derivative of f(x) at, x = a,

where a can be any value (of course non-negative). The reasoning is

simple.

x

afxaf

xdx

dyax

)()(

0

lim)(

xaxa

x

22

16)(160

lim

x

xax

x

2)(16

0

lim 2


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)2(160

limax

x

a32

Thus we get .32)( 2 adx

dy

Since the above equation is true for all (non-negative) values of a, it is customary

to write it in the following form:

.32)( xdx

dy

Thus, in particular, when x = 2 we get

.64232)( xdx

dya

agreeing with what we got in 1.4.1.

Remember : if ,216)( xxf then

.32xdx

dy

This is also written as .32)16( 2 xxdx

d

1.4.3. FORMULAS

(a) Let y )(xf where .)( xxf Thus we have

y = x

hence we get

x

xxx

xfxxfy ).()(


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Therefore 1x

yand hence

.10

lim

x

y

xdx

dy

1FORMULAxyif1

dx

dy

(b) More generally, suppose )(xfy where .)( nxxf where n can take

any one of the values 2,3,4, Then it can be shown that

1

.n

xndx

dy

Thus we have

1)( nn nxxdx

d

FORMULA - 21)( nn nxx

dx

dfor 2,3,4,.

Memorize the above formulas.

1.4.4: EXAMPLE

Go back to Example 1.3.2. Here f(x) =x2 - 6x + 10. We wish to finddx

dyfor x = 2,

x = 3 and x = 4. One way of doing this is to proceed as in Example 1.4.1. but

then we must do the same kind of computations three times once for x = 2,

once for x = 3, and once for x = 4. Instead of this, a better way is to proceed as in

1.4.2 and get the value of axdx

dy)( for any value of a. This can be done, but is not

a wise way because we may have to repeat the same process all over again

once the above function is changed to f(x) = x3 + 9x2 2x +71 All these troubles

can be avoided if the above two formulas are committed to memory. These


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formulas are like computer programmes. You plug in the value and get the

required result without any extra effort.

1.4.5:

Even though the two formulas are general, still they are not sufficient for us to

handle our function f(x) = x2 6x + 10. So we need some more formulas which

we list below:

FORMULAS3 ,0)(kdx

dwhere k is a constant

For example, 0)10(,0)1(

dx

d

dx

dand

.0)3(dx

d

FORMULA4 constantanyiskwhere(x)),(fdx

dk(kf(x))

dx

d

For example, ),(2)2( 22 xdx

dx

dx

d

)(9)9( 55 xdx

dx

dx

detc.

FORMULA - 5))(())(())()(( xg

dx

dxf

dx

dxgxf

dx

d

For example, )()()( 3232 xdx

dx

dx

dxx

dx

d


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COMMIT THE ABOVE FORMULAS TO MEMORY

1.4.6: (Example 1.4.4 continued) consider our function ,106)( 2 xxxf we

have

01.)6(2

5)10()6()(

)106())((

2

2

x

formulabydx

dx

dx

dx

dx

d

xxdx

dxf

dx

d

FOR-2 FOR-4 FOR-1 FOR-3

62x

Thus we get

2622)( 2xdx

dy

0632)( 3xdx

dy

2642)( 4xdx

dy

Now can you perceive the POWER of these formulas?

1.4.7. So far as a function is concerned,

axdx

dy)(

is called the SLOPE of the tangent to the curvey = f (x) at x = a. see the figure :


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Tangent at x = a is a straight line touching the curve y = f(x) at the point

A(a, f(a)).

axdx

dy)( is the slop of this line.

SUMMARY

1. Suppose y = f(x). Then the rate-of-change (of y with respect to x) at x =a is

given by

axdx

dy)(

This is also called DERIVATIVE of f (x) at x = a or SLOPE of the tangent

line to f(x) at x = a or VELOCITY of the motion y = f (x), when x = a

2. Rate-of-change =

.dx

dy

3. Note that rate-of-change at x = a is a number, whereas rate-of-change is

a function.

4. To compute.dx

dyfollow the steps given below :


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(a)Compute f(x + x) for x 0

(b)Compute f(x) and f(x + x)f(x)

(c)Compute the ratio x

xfxxf )()(

(d) Compute the limit of the above ratio as x tends to 0: i.e. take

smaller and smaller values of x and find the corresponding values

of the above ratio. These values will stabilize at a number and this

number is.dx

dy.

EXERCISE 1. 4

Find.dx

dy of y where

1.5

242 23 xxxy

2. xxxy9

8

18

7

3

4 38

3.

35

35 xxy

4. )1()5( xxy (Hint: Multiply the righthandside)

5. 126 46 xxy

6. Find the slop of the tangent to the curve

1

532 245

xat

xxxxy

7. The position of an object at time x (in seconds) is given by y (in meters)

where

216128 xxy

(a)Find the velocity of the object.


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(b)Find the velocity of the object at time 4x secs.

(c) Verify that the velocities of the object at 2x secs and 6x secs

are equal in magnitude and opposite in sign.

8. Fill up the blanks:

xxxxxdx

d 223 )1005

13(

1.5 MORE EXAMPLE OF RATE-OF-CHANGE

In this section, we show how to make use of the rate-ofchange idea in solving

real-life problems.

1.5.1 Example

A tank of water is filled in such a way that in x hours there are xx

22

2

litres of

water in the tank. The person filling the tank is instructed to turn off the water

when the water is entering the tank at the rate of 15 liters per hour. When should

he turn off the water?

Here our function is y xx

22

2

(lit). The rate of-change is .2222

1xxxdx

dy

Give that the water is to be turned off when the rate is 15 liters per hour.

ratedx

dy15 lit per hour

i.e., x+2 = 15

or x =13 hours.

So the water should be turned off after 13 hours.

Before we give the next example, we take another, look at the derivative (or rate-

of-change).


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Sincedx

dyis the rate-of-change of y with respect to x, we have that y changes

dx

dytimes as fast as x.

Familiarize yourself with this idea.

1.5.2 Example

The profit of a retail store is Rs. y (in hundred) if when Rs. x are spent daily on

advertising where 2

5

1362500 xxy . Use the derivative to determine if it would

be profitable for the daily advertising budget to be increased if the current daily

budget is (i) Rs. 60 and (ii) Rs. 100.

From 2

5

1362500 xxy we get

xdx

dy

5

236

Thus, when x = Rs. 60, ( 12) 60xdx

dy

And when x = Rs. 100, 4)( 100xdxdy

So, when x = 60, the profit y is changing 12 times as fast as the expenses x;

whereas, when x = 100, the profit y changes4 times as fast as the expenses x.

Thus, it would be profitable for the daily advertising budget to be increased when

the current budget is Rs. 60. (Note that the budget should never be Rs. 100).

EXERCISE 1.5

1). A retailer sells a certain item, and finds that 8.2 customers are lost for that

item per Rupee cost of the item. The price y (in rupees) of the item at the

end of x months is given by

2

27

4.0xy where x = 1,2,3,..10.


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How many customers are lost per month at the end of 9 months?

2) The supply equation for a certain kind of pencil is

xxy 23 2

where x paise is the price per pencil when 1000 y pencils are supplied. Find

the rate of change of supply per 1 paise change in price when the price is

80 paise

3) Water is being drained from a swimming pool and the volume of water in the

pool x minutes after the draining starts is given by

)801600(250 2xxy

(y being measured in litres). How fast is the water flowing out of the pool 5

minutes after the draining starts?

4) The annual earnings of ABC and Co Ltd. x years after January 1, 1984 is y

lakhs of rupees where

y = 1025

2 2xx

find

(a) the rate at which the earnings were growing as on January 1, 1984;

(b) the rate at which the earnings should be growing as on January 1, 1989,

1.6 MORE FORMULAS ON RATE-OF-CHANGE

In this section, we give three more formulas, which are very useful in

applications.

1.6.1: suppose u and v are functions of x. then

FORMULA6dx

dvu.v.

dx

du(uv)

dx

d

This is called the PRODUCT RULE. We illustrate its use by an example.


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Example: suppose ),2()1( 232 xxxy find dy/dx.

)23()1()2(2

)23(.2

23

)2()()(

)2(

2

)1()()1(Now

..therefore

.then

2and1write

2223

2

2

23

23

22

232

xxxxxx

xxuxudx

dy

xx

dx

dx

dx

dx

dx

d

xxdx

d

dx

dv

x

dx

dx

dx

dx

dx

d

dx

du

dx

dvuv

dx

du

dx

dy

uvy

xxvxu

1.6.2 : Again, suppose u and v are functions of x. Then

FORMULA7 )(1

)(2 dx

dvu

dx

duvvv

u

dx

d

This is called the QUOTIENT RULE. The following example illustrates its use.

)(1

1

)1()()1(

1)(

andthen

1andwrite

.find,

1

2 dx

dvu

dx

duvvdx

dy

dx

dx

dx

dx

dx

d

dx

dv

xdx

d

dx

du

v

uy

xvxu

dx

dy

x

xy


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2

2

2

)1(1

)1()1(

1

)1.1).1(()1(

1

x

xxx

xxx

In the above two examples, we made use of the formulas 6 and 7. What other

formulas we made use of ? Study these examples carefully and find them out.

1.6.3. CHAIN RULE

Suppose y is a function of u and u is a function of x, then we have

FORMULA8dx

du

du

dy

dx

dy.

This is called the CHAIN RULE. It is useful on many occasions.

Example:

Supposedx

dyfindxy

,2)1(3

Write 22 1 uythatsoxu

uudx

d

du

dy2)( 2

and )1()()1( 22

dx

dx

dx

dx

dx

d

dx

du

x2

So finally we have

dx

du

du

dy

dx

dy

.

= 2u.2x from above

= 4x(x2 +1)


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EXERCISE 1.6

1. If y = (x+2) (x-5)8, finddx

dy(Use Product Rule)

2. If y = (3x-2)3 (4x+1)4, finddx

dy(Use Product Rule)

3. If y =,63

12

x

x, find

dx

dy(Use Quotient Rule)

4. If y =,)1(

)1(4

3

x

x, find

dx

dy(Use Quotient Rule and Chain Rule)

(HINT : put u = (x-1)3 and v = (x+1)4.

Also put f = (x1) and g = x+1

Then y =v

u, u = f 3 and v = g 4

Also ).

dx

dg

dg

dv

dx

dvand

dx

df

df

du

dx

du

5. If y = (x3 + 1)2 (x2 + 1)3, finddx

dy.

(HINT : Use Product Rule and chain Rule. Put

u = (x3 + 1)2 and v = (x2 + 1)3 Also put

f = x3 +1 and g = x2 + 1.

Then y = uv, u = f 2 and v = g 3

Further ).dx

dg

dg

dv

dx

dvand

dx

df

df

du

dx

du


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6 . If ,)2

3( 2

x

xy find

dx

dy

7. If ,)

3

1( 2

2

2

x

xy find

dx

dy.

8. If y = (x2-1)3 (2x 2+ 3x +2)2 finddx

dy

9. Ifdx

dyfind

xxy ,

)2(

132

10.If11

122

2

xx

xwand

xx

xy

Verify that.dx

dw

dx

dy

1.7 SECOND-ORDER DERIVATIVE : ACCELERATION

In section 1.4 we became familiar with rate-of-change or derivative. On many

occasions we can also find the rate-of-rate-of-change or what is called the

second-order derivative.

We introduce this by means of an example. After this we will introduce the

notation for it and another name for it.

1.7.1: EXAMPLE:

Consider .58 24 xxy Then its derivative is given by

xxdx

dy1032 3 .

Notice that the right-hand-side expression is again a function of x. so writing

z = 32x3 +10x


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we get

1096 2xdx

dz

Thus we have

dx

dzx 1096 2

= derivative of z

= derivative of 32x3 + 10x

= derivative ofdx

dy

= derivative of (derivative of y)

1.7.2: This derivative of (derivative of y) is called the second derivative of yor second-order derivative of y. we denote it by the symbol

.2

2

dx

yd

Thus, for our example 1.7.1, we have

.1096 22

2

xdx

yd

1.7.3 : Recall from section 1.4 that

dx

dy

is also called the velocity (when we deal with problems involving motion), and

dx

dyis written as v (the first letter for VELOCITY)

i.e.,dx

dy= v

In this context, the second derivative


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2

2

dx

yd

is called the ACCELERATION and is denoted by a. (first letter for

acceleration). Thus we have

Acceleration = a

2

2

dx

yd

Note also that a =.dx

dv

1.7.4 EXAMPLE

A boat is moving along a straight river and its distance (from its starting point ) infeet is given by

542

3

3

1

12

1 234 xxxxy

where x is time measured in seconds. If v ft/sec. Is the velocity and a ft / sec2. is

the acceleration of the boat at x sec., find x, y and v when a = 0.

Now 433

123 xxx

dx

dyv

And .322 xxdx

dva

When ;032,0 2 xxgetwea

Or (x + 1)2 = 4 or x + 1= 2 (since time cannot be negative) or x = 1 sec.

So substituting in y and v we get


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ft/sec.3

12

43x111x31v

feet12

117

514xx12

3x1

3

1x1

12

1y

23

234

EXERCISE 1.7

i) Let m be the slope of the tangent to the curve

y = x32x2 + x.

Find the rate of change of m at the point (2,2).

ii) Find2

2

dx

ydwhere y = 7 x38x2

iii) Find the slope of the tangent at each point of the curve y= x 4 + x33x2

where the rate-of-change of the slope is zero.

iv) A stone is thrown vertically upwards. At time x (in seconds), its distance above

the ground is given by y (in feet) where

y = 128x 16x2

a) Find the velocity and acceleration

b) Find the velocity when x = 2, x = 4 and x = 6.

c) Find the distance above the ground, when v = 0.

d) Verify that acceleration is always constant.


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1.8 HIGHER ORDER DERIVATIVES

1.8.1. In section 1.7 we learnt how to calculate the second derivative of a given

function. There is nothing to stop us form computing the derivative of the

second derivative (which is called the third derivative of f(x)) and so on

The third derivative of f (x) or y is denoted by

3

3

dx

yd

and the fourth derivative is denoted by

4

4

dx

yd

and etcetera. In general the k th derivative is denoted by

k

k

dx

yd

where k can be 1 or 2 or 3 or

1.8.2. Recall that the first derivative or dx

dy

has several names:

1) FIRST DERIVATIVE (of y with respect to x )

2) VELOCITY (if y is the distance covered in time x)

3) SLOPE (of the tangent line to y = f (x))

4) RATE OF CHANGE.

Similarly, the second derivative or2

2

dx

ydhas several names:

1) SECOND DERIVATIVE2) ACCELERATION

3) CURVATURE (of the curve y = f(x))

4) RATE OF RATE-OF-CHANGE


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NOTE: We have not introduced the name (3) given above i.e. CURVATURE. We

dont have any intention of introducing it ever. It has been included for

completeness sake.

Thus the first and second derivatives have several names. But the third

derivative onwards, we dont have several names associated with them. THIS

SHOWS THAT THE FIRST AND-SECOND DERIVATIVES ARE VERY

IMPORTANT FOR APPLICATION. This is the reason for the various names for

these two.

1.8.3. It is high time that we illustrated the concept of higher order derivatives by

means of some examples.

1.8.4. EXAMPLE: Find all the derivatives of

y = x3 + x2 + x + 1

Here

123

)1()()()(

2

23

xx

dx

dx

dx

dx

dx

dx

dx

d

dx

dy

0

)6(

6

)2()6(

)26(

26

2)2(3

)1()2()3(

)123(

4

4

3

3

2

2

2

2

dx

d

dx

yd

dx

dx

dx

d

xdx

d

dx

yd

x

x

dx

dx

dx

dx

dx

d

xxdx

d

dx

yd


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28

0

)0(5

5

dx

d

dx

yd

This is because zero is also a constant and the result follows by formula 3.

Similarly we have

on.soand0,07

7

6

6

dx

yd

dx

yd

Thus we can write that

,.......6,5,40 kifdx

ydk

k

1.8.5: EXAMPLE

Find the second derivative of

1x

xy

Notice that our function has something (i.e., x) in the numerator and something

(i.e. x + 1) in the denominator.

Thus, if we take

).(now

wirtecanwe

1and

v

u

dx

d

dx

dy

v

uy

xv

xu


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30/89

30

So we get

2

4

22

22

2

2

)1()1(

1

))1(10)1((])1([

1

)(

xdx

d

x

xdx

dx

x

v

u

dx

d

dx

yd

Our next job is to calculate

2)1(xdx

d

This is very similar to the Example of section 1.6.3

So we proceed as therein.

Put .1)1( 21 xuandxy

)1().(

.

)1(

2

1

12

2

1

xdx

du

du

d

dx

du

du

dy

dx

dyx

dx

d

anduyThen

)1(2

2

12

x

u

u


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31

Thus, we have, finally

.)1(

2

)1(2)1(

1

)1()1(

1

3

4

2

42

2

x

xx

xdx

d

xdx

yd

Dont get discouraged by the above example. It looks formidable because it is

lengthy and it looks lengthy because we have exhibited all the steps. With some

experience, you can also do such problems. Not only that ! you can skip some of

the steps and arrive at the answer quickly!

So dont give up ! try and try again ! Here are some problems to give you enough

experience.

EXERCISE 1.8

Find all derivatives of

1) 2

2) x

3) x2

4) x2 + x + 2

5) if y = x6 + x 5+ x4 + x3 + x 2+ x + 1, verify that the seventh derivative of y

is zero.

6) if y = x6, verify that the seventh derivative of y is zero

7) what do you learn from the above two problems ?

8) Find the second derivative of

a)x

1

b)x

x 12


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32

1.9. TAYLORS FORMULA AND APPROXIMATIONS

1.9.1. POLYNOMIALS

We have already come across functions of the form

x2 + x + 1

x6 + 5x

.9

17

5

4

3

2 2xx

and so on. These functions, which are very important, are called

POLYNOMIALS. Thus, a polynomial has a finite number of terms, separated by +

or sign, where one of the terms may be a number only and all other terms are

of the form a number multiplied by a power of x.

For example, consider

2

9

17

5

4

3

2xx

(Number called constant term) ( xbymultiplied5

4 ) 2bymultiplied9

17 x

We said above that polynomials are important. There are several reasons for

this. One of them is that their values can be computed easily.

For example, consider the polynomial

X 2+ X +1

Let us write 12 xxy as we have done so far. Then, when x = 0, y = 1 when x

= 1, y = 3 and so on. Some of the values of y for various values of x are exhibited

in the table given below:

X -3 -2 -1 0 1 2 3

Y 7 3 1 1 3 7 13


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You can not only calculate the value of y for integer values of x, but also for any

value of x. look at the following table:

12 xxy

X 0.1 0.2 0.12 0.19 ?

Y 1.11 1.24 1.334

Now go ahead and compute the value of y when x = 0.19. The blank space in the

above table is for you to complete. Dont stop ! Calculate the values of y for

further values of x.

CAN YOU SEE THAT YOU CAN COMPUTE THE VALUE OF Y FOR ANY

VALUE OF X WHEN THE VALUE OF X CONTAINS AT MOST FOURDECIMALS ?

Now compute the value of y when

X = 0.123456789 !!!

Can you do it ? it will take a long time. Here is where a computer can be of use.

The computer will calculate the value of y within a second providing you have the

correct program. Such programs are available. look it up and do it.

It is great fun !

So we see that if we are satisfied with three or four decimal accuracy, then we

can compute the values of the given polynomial either by hand or by a calculator.

And, If we require higher level of accuracy or the given value of x contains more

than four decimals, then we can resort to the use of computer.

1.9.2: TAYLORS FORMULA

As explained above values of polynomials can be found by performing a finite

number of additions and multiplications.

However, there are functions, such as exponential and logarithm functions (about

which you will study in chapter 3), that cannot be evaluated easily.


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34

Then how to compute their values? One way is to go to a computer. A second

way is to look into a table giving the values of such functions. A third way is to

approximate such functions by polynomials (as closely as we can) and calculate

the value of these polynomials.

Of course the third method gives only an approximate value or so it appears. In

reality, all the above listed three methods give only approximate values never

the exact value to various levels. So, all is not lost by adopting the third

method.

This idea of approximating functions with polynomials was conceived by the

English mathematician Brook Taylor and the formula given below is called

Taylors Formula.

_______________________________________________

TAYLORS FORMULA

)()(!

)(

.....)(!2

)()(

!1

)()()(

)(

2

xRaxn

af

axaf

axaf

afxf

n

nn

where

Remainder.(x)R

n)times......times3times2times1(.....3.2.1!

)(

)()(

)()(

)()(

n

)(

2

2

nn

xfy

dx

ydaf

dx

fdaf

dx

dyaf

axn

nn

ax

ax

NOTE: The above formula holds under certain conditions. But most of our

functions will satisfy these conditions automatically.


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The remainder Rn (x) is a function of x, which in the case of a polynomial is

always zero. In all other cases (i.e. when the given function is NOT a polynomial)

Rn (x) is not zero.

We illustrate this idea with some examples.

1.9.2.1. EXAMPLE

Consider f(x) = x2 + x + 1. Taking a = 0 in Taylors Formula we get the following:

y = x2 + x + 1 = f (x)

a = 0

?)WHY(0.....

2]2[)()0(

1]12[)()0(

4

4

3

3

02

2

0

dx

yd

dx

yd

dx

ydf

xdx

dyf

xox

oxx

Thus we take n = 2, and get

).(1

)()0(!2

2

)0(!1

1

)()(2

2

xRxx

xRxxafxf

n

n

Thus Rn (x) = 0!

So, in this case, we see that

1) Rn (x) = 0 for n = 2

2)Taylors Formula gives the polynomial itself, if we take a = 0!

1.9.2.2: These two facts are true in the case of any polynomial.

So remember the following: -

If f (x) is a polynomial, then


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1. Rn (x) = 0 for some value of n. This value of n will vary from

polynomial to polynomial. In fact n is equal to the highest power of x

in f(x)( )

2. Taylors Formula gives the polynomial itself, if we take a = 0

Point (2) above tells an important fact : That is, if we consider only the first

n+1 terms on the righthandside of Taylors formula.

nn

axn

afax

afax

afaf )(

!

)(..)(

!2

)()(

!1

)()(

)(2

than what we get is a polynomial. This polynomial is known as TAYLORS

POLYNOMIAL. So we note the following: -

TAYLORS POLYNOMIAL

(B) )(!

)(...)(

!2

)()(

!1

)()(

)(2 ax

n

afax

afax

afaf

n

1.9.2.3: in point (2) of (A) we saw that Taylors formula gives the polynomial itself,

when a = 0. What happens if a is different from zero? The following example

illustrates this idea.

( ) For example, in the case of our polynomial

f (x) = x2 + x+ 1

the term x2 has the highest power, namely 2. Thus the highest power of x

in x2 + x + 1 is 2.

1.9.2.4. EXAMPLE:

Consider again our old friend f (x) = x

2

+ x + 1. Now we take a = 1. (There is nosanctity about 1. we can take any value like a = - 1; ,

2

1a a = - 0.04, a = 1.12

etc. some of these are given as exercises). Then

(A)


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37

2]2[)()1()(

3]12[)()1()(

.3)1()(

112

2

11

xx

xx

dx

ydfaf

xdx

dyfaf

faf

and as in Example 1.9.3, we have

.0...4

4

3

3

dx

yd

dx

yd

Thus we take n = 2 and get (Recall (A))

!)1()1(331.,.

)1()1(33

)1(!2

2)1(

11

33

)(!2

)(

)(!1

)(

)()(

22

2

2

2

xxxxei

xx

xx

ax

af

ax

af

afxf

Does this look surprising to you? Dont worry! We will now clarify it. Take the right

handside. It is equal to

3 + 3 (x-1) + (x-1)2

= 3 +3x3 + x22x + 1 (Recall the basic algebra)

= x2 + (3x2x) + (1+3-3)

= x2 + x +1

which is the left-hand-side expression !

Thus, all that we have done is to express the given polynomial as a Taylors

polynomial.

1.9.2.5: Note that our polynomial x2 + x + 1 is simpler than Taylors polynomial 3

+ 3(x1) + (x1)2.

[simpler in the sense that we can compute the values of x2 + x +1 more easily

than the values of 3 + 3(x- 1) + (x-1)2 ]. Then what is the use of Taylors Formula

? well, in the case of polynomials this formula is not of much use. But in the case

of functions, other than polynomials, it is of much use. This point will become

clear only after we study the exponential and logarithm functions.


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38

For the present, we give below an example to show that Rn(x) need not be zero.

1.9.2.6: EXAMPLE

Consider the function f(x) = .1

1

xNote that this is not a polynomial. We will take

a = 0 and calculate a few terms:

.6

)4(])1(

3.2[

0]1

1[)0()(

2

)4(])1(

2[

0]1

1[)0()(

1

)4(]

)1(

1[

]1

1[)0()(

11

1)0()(

04

3

3

03

2

2

02

0

problemseex

xxdx

dfaf

problemseex

xxdx

dfaf

problemsee

x

xdx

dfaf

faf

x

x

x

x

Thus we see that

)(1

)(!3

3.2)(

2

21

)()0(!3

)0()0(

!2

)0()0(

!1

)0()0()(

332

3

32

3

32

xRxxx

xRxxx

xRxf

xf

xf

fxf

We write this as

321)( xxxxf


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39

where means approximately equal to

Thus

32

11

1

xxxx

so we have approximated the functionsx1

1(which is not a polynomial ) by a

polynomial.

This idea of approximation is very important. We take it up in the next sec tion.

1.9.3 : LINEAR APPROXIMATIONS

We write Taylors Formula as an approximation formula

nn

axn

afax

afax

afafxf )(

!

)(...)(

!2

)()(

!1

)()()(

)(2

This is what we meant when we said that Taylors Formula gives polynomial

approximation to a general function. Go back to section 1.9.2 and read the five

paragraphs (before Taylors Formula) again. If we take 1n in the above

approximation we get:

FORMULA9 ).(.)()()( axafafxf

This is known as LINEAR APPROXIMATION FORMULA. It is very useful, when

xa is very small. This idea is illustrated by the following examples:

1.9.3.1: EXAMPLE

Suppose f (x) = x2 + x + 1. Find the value of f(x) when x =1.01. Of course, we can

find this value by direct substitution. But, if we want only an approximate value of

f(1.01), we can use the above formula as follows

Take x = 1.01 and a = 1. Then


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40

xa = 0.01 which is very small value.

Now 12)( xdx

dyxf

Thus 3]12[)1()( 1xxfaf

)1(01.0)1()01.1( fxff

03.3

301.03

so f(1.01) 3.03.

The actual value of f (1.01) is 3.0301. So our approximation is very good indeed!

Now, are you able to perceive the power of formula 9?

Before going to the next example, note the following:

FORMULA2 IS TRUE EVEN WHEN n IS A FRACTION. i.e., WE HAVE

1)( nn xnxdx

d

WHERE n CAN BE A FRACTION LIKE

.

3

7,

2

5,

3

2etc

1.9.3.2: EXAMPLE

Find an approximate value of

(1.1)1/3

So we consider the function

)1.1()1.1(

)(

31

31

fthatso

xxf

Thus we have to find an approximation for f (1.1).

Take x = 1.1 and a = 1.


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Also

3

1)1()(

.1

3

1

3

1

formulaaboveby3

1

)()(

32

32

1

3

1

31

fafThus

x

x

x

xdx

dxf

Hence, we have

03.1

03.01

1.03

11

1.0)1()1(

)().()(

)1.1()1.1(31

xff

axafaf

f

Where we have used the fact that

.1)1()1( 3/1f

1.9.3.3: EXAMPLE

Find an approximation for

8.73

Now 33131 )(takeweso.)8.7(8.7 xxf . Also we take x =7.8 and a = 8.


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Thus,

121

]1

3

1[)(

2)8()(

832

31

ax

af

af

9833.1

0167.02

2.012

12

)88.7(12

12

)().()(

)()8.7(8.73So 31

axafaf

xf

EXERCISE 1.9

1. Compute the values of x2 + x + 1 corresponding to the following values of

x.

- 0.18, - 2.01, 0.1234, 0.1121.

2. Compute the values of 3!, 4! and 5!.

3. What is the highest power of x in

a) x6 + x10 + 2?

b) 2 + x4 + x5 ?

c) x11 ?

4. If y =x1

1: verify that

.)1(

3.2,

)1(

2,

)1(

143

3

32

2

2 xdx

yd

xdx

yd

xdx

dy


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43

5. Let f (x) = x2 + x+1. Take n = 2 in Taylors Formula and find the Taylors

Formula for

a) a = -1 andb) a = - .

2

1

6. Let f (x) = x2 + x + 1. Use formula 9 to compute an approximate value for

a) f (0.18) Take a = 0

b) f (1.123) Take a=1c) f (-0.18) Take a = -1d) f (0.123456789) Take a = - 0.

7. Find an approximate value of

a) 283 Take x = 28 and a = 27b) 5.37 Take x = 37.5 and a = 36

c) 824 Take x = 82 and a = 81.

8. Find an approximation forf (2.01)

Where f (x) =2

1

x

(HINT: Take a = 2)


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Chapter 2

OBJECTIVE

In this block you will be introduced to the following ideas Related Ratesand Maxima Minima mainly from the point of view of applications to real life

situations.

Related Rates, as the name indicates, concern with rates (more than one)

related to one another by means of a relation. If the values of all, except one of

them are known, the value of the remaining rate can be found. This simple idea

is so very useful that many complex problems arising out of real-life situations

can be solved with surprising ease.

The theory of maxima-minima is an extensive one, our aim is to presentthe basics of this great idea and show how this nodding acquaintance can be

exploited to solving sufficiently complex problems.


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45

2.1 INTRODUCTION

In computer 1 you learnt about RATE and some applications of this idea to

simple real-life situations. In this chapter you will learn about related rates and

maxima-minima and how all these ideas can be used in more complex situations.

2.2 RELATED RATES

When two variable x and y are related by an equation, then their rates with

respect to time tdt

dyand

dt

dxwill also be related by an equation.

If the value of one of them is known at an instant of time, then the value of the

other can be calculated (for the same instant of time). Problems of this kind come

under the heading Related Rates.

The following examples will illustrate this idea by means of applications to real-

life situations.

2.2.1 EXAMPLE

A ladder 25 meters long is leaning against a vertical wall. If the bottom of the

ladder is pulled horizontally away from the wall at 3 meters per second, how fast

is the top of the ladder sliding sown the wall, when the bottom is 15 meters form

the wall?

Let

t = the time that has elapsed since the ladder started to slide down the

wall;

y = the distance form the ground to the top of the ladder at t seconds;

x = the distance from the bottom of the ladder to the wall at t seconds.

See the figure below.


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46

Because the bottom of the ladder is pulled horizontally away from the wall at 3

meters per second,

)1.....(3dt

dx

We want to find dy/dt when x = 15.

Note: x will be 15 meters at some instant of time. This instant of time is not givenexplicitly because it is not required for this (and most of the) problem (s). This

fact should be borne in mind.

From the figure we have

X2 + y2 = 252

= 625

i.e., y2 = 625x2 .. (2)

because x and y are functions of t, we differentiate on both sides of (1) withrespect to t and obtain (by chain rule)

2ydt

dxx

dt

dy2


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47

)4(.....20

15,

)3(.....

y

getwexwhenSo

dt

dx

y

x

dt

dy

from (2); and hence

[dt

dy]x=15 = - 3

20

15

because of (1), (3) and (4).

Thus .

4

12][ 15x

dt

dy

Hence, the top of the ladder is sliding down the wall at the rate of4

12 m / sec

when the bottom is 15 m from the wall. (The significance of the minus sign is that

y decreases as t increases).

2.2.2 EXAMPLE

Two cars, one going due east at the rate of 37.5 km/hr and the other going due

south at the rate of 30 km/hr, are traveling toward an intersection of the two

roads. At what rate are the two cars approaching each other at the instant whenthe first car is 40 km and the second car is 30 km from the intersection?

Let P be the intersection of the two roads. See the figure below:


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48

Let x = the distance of the first car form P at t seconds;y = the distance of the second car from P at t seconds;

z = the distance between the two cars at t seconds.

As t increases (i.e. as time passes) x decreases because the first car approaches

P. thus

.5.37dt

dx

Similarly, we have

.30dt

dy

we want to find dz/dt when x = 40 and y = 30.


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49

From the figure, we obtain

z2 = x2 + y2 .. (5)

Differentiating with respect to t (apply chain rule), we obtain

2zdt

dyy

dt

dxx

dt

dz22

and so

)6(.....dt

dy

z

y

dt

dx

z

x

dt

dz

When x = 40 and y = 30, we have z = 50 from (5). Thus, we have

.48

)30(50

30)5.37(

50

40

3040

xxdt

dz

yx

Therefore, at the instant under consideration, the cars are approaching each

other at the rate of 48 km/hr. (Again the minus sign shows that z decreases as t

increases. This fact is also clear from the situation because the two cars move to

wards P and has to decrease).

2.2.3 EXAMPLE

A pebble is dropped into a pond causing a circular ripple. A measuring device

indicates that at the time the radius of the circular ripple is 5 inches, the radius is

changing of at the rate of 3 inches per second. How fast is the area changing at

this instant of time?

Let A be the area of the circular ripple, when r is its radius. Then we know that

A = r2 . (7)

We are given that, when r = 5, .2dt

drThat is

25rdt

dr. . (8)


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50

From (7) we have

..2.dt

drr

dt

dA

So when r =5 we get

20

8210

52 55

by

dt

dr

dt

dAxr

Thus, when r =5 inches, the area is changing at the rate of 20 square inchesper second. (here the plus sign shows that the area is expanding at the above

rate. This is also clear from the context. Perform an experiment to get a better

understanding).

EXERCISE 2.2

1. Two trucks have a common starting point, but one moves southward at 30

km/hr and the other westward at 40 km/hr. at the end of 1 hr, how fast are

the trucks moving away from each other?

2. The marginal revenue corresponding to a quantity of 2 units is known to

be Rs. 10 per unit. If the price corresponding to 2 units is Rs. 8 per unit,

how fast is the price changing (with respect to quantity) when the quantity

is 2 units?

(HINT : Total Revenue = number of units x price per unit

Marginal Revenue = rate of change of total revenue

with respect to the quantity.

Quantity = number of units.

Let x = number of units at some particular instant

p = price per unit

R= Total revenue.


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51

Then

R = px

Marginal Revenue =dx

dpxp

dx

dR

We are given that

.28

102

xwhenp

dx

dR

x

3. a manufacturer finds that when 100 units of a commodity are sold, profit is

increasing at the rate of Rs. 4 per unit and price is decreasing at the rate

of Rs. 1 per unit. The price for 100 units is Rs. 60 per unit. How fast is the

total cost changing per unit at this level of sales?

(HINT: Again, let

x = number of units sold

P = profit

c = cost

p = price

Then P = xpc.)

4. A boy flies a kite at a height of 300 feet. The wind is carrying the kite

horizontally away from the boy at a rate of 2 feet per second. How fastmust the boy pay out the string when the kite is 500 feet away from him?

5. A man on a dock is pulling in a boat at the rate of 50 feet per minute by

means of a rope attached to the boat at water level. If the mans hands are

16 feet above the water level, how fast is the boat approaching the dock

when the length of the rope is 20 feet?

2.3 MAXIMA AND MINIMA

In many day-to-day problems we are concerned with maximum or minimum

values for example, what is the maximum profit one can obtain, what is the

maximum number of employees that can be used for a specific purpose, what

is the minimum loss one has to suffer in a venture etc.


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52

To give a specific illustration, assume that a manufacturer knows that he will

earn p(x) rupees if he produces x items of some commodity. If he produces

too few items, he will be unable to meet demand, and if he produces to many,

he will be overstocked, thus, he expects p to be small if x is too small or too

large. His problem is to determine the number x for which p is maximum. If we

call this number x0, then p(x0) will be the maximum earnings possible. (Note

that p is a function of x).

The following algorithm tells you how to solve a given problem involving

maxima or minima.

ALGORITHM

I. Determine the function y = f (x) involved in the problem.

II. Find dx

dy.

III. Solve the equation 0dx

dy.

This will give values for x,

Call the value a.

IV. Find

ax

dx

yd2

2

V. Evaluateax

dx

yd2

2

VI. (i) If above value is negative, x = a gives maximum; i.e. f(a) is the

maximum value required.

(ii) If the above value is positive, x = a gives minimum; i.e., f(a) is the

minimum value required.

CAUTION: A given problem will involve only one: either maximum or minimum.

No problem can involve both maximum and minimum.


53/89


54/89

54

2.3.2 EXAMPLE: A photographer has a thin piece of wood 16 inches long.

How should he cut the wood to make a rectangular picture frame that

encloses the maximum area?

Let x = the length of the rectangular frame

y = the width of the rectangular frame.

Then the area of the frame is given by

A = x y. . (9)

Note that we are restricted here by the total length of 16 inches of wood to make

up all the four sides. The total length of all the four sides of the frame is

x + y + x + y = 2x + 2y

And this must be 16. so we have

2x + 2y = 16

or

x + y = 8.

So we get y = 8x

And using this in (9) we get

A = x (8x)

= 8xx2. . (10)

Note that there are two sides of length x and their sum 2x should be less than 16.

That is

0 < 2x < 16 or

0 < x < 8. .. (11)

Now, from (10) we see that

. xdx

dA28 .


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So 0dx

dA

gives 82x = 0 or x = 4

Observe that the value 4 lies in the interval given by (11). Also we have

22

2

dx

Ad

and hence 242

2

xdx

Ad

which is negative. Thus x =4 gives the maximum.

So the maximum area is 16 from (10) and this occurs when the frame is a square

having sides equal to 4 inches. [Because y = 84 = 4].

2.3.3 EXAMPLE: The crime-rate index of a locality, denoted by y, and the

weekly contact-hours (denoted by x) that social workers spend in direct

contact with persons living in that area are related as follows:

Y = 0.001x20.2x + 12

where x can be any number between o and 150. What is the optimum number of

contact hours per week for this area?

Observe that the optimum number of contact hours per week is that value of x for

which y is a minimum; i.e. the value of y is as small as possible.

We have

.2.0002.0 x

dx

dy

so 0dx

dy

gives 0.002x0.2 = 0

Or

x = 100.


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Note that this value lies between 0 and 150. Further

002.02

2

dx

yd

And hence .002.01002

2

xdx

yd

Since 0.002 is a positive quantity, we see that the value x = 100 gives the

required minimum value for y.

Hence, minimum value of y = 0.001 (100)20.2 (100, + 12

= 2

Thus, the optimum number of weekly contact hours should be 100 and thecorresponding crime-rate index is 2.

EXERCISE 2.3

1. For a firm, the total cost and total revenue are given by the following

relations:

C = x 35x2 + 9x

and R = 34x3x 2

where x denotes the level of output and x can be any value between 0 and 7.

(a)Find the value of the output for which total revenue is maximum.

(b)Find the value of the output for which total profit is maximum.

2. When a person coughs, the diameter of the trachea decreases. The

velocity y of air in the trachea during a cough is related to the radius (x) of

the trachea by the equation:

Y = 5x2 (10x)

where x can take any value between 0 and 10. Find the value of x for

which the velocity is greatest and also find the maximum velocity of the

air.

3. A manufacturer produces x articles per day where x can be any value

between 0 and 50. The total cost per article is given by

10012

16 x

xy


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How many articles should be produced each day in order to minimize the cost?

4. A small amount of water is treated to control bacterial growth. After x days,

the concentration of bacteria per cubic centimeter is given by

y = 20x2200x + 640

over a 9-day period. How many days after treatment will the bacterial

concentration be lowest? What is the lowest concentration?

5. The sum of one number x, and three times a second number y is 80. Find

the numbers if the product of x by the square of y is maximum.


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Chapter 3

Objective

In this block you will be introduced to the ideas of exponential functions andintegrations.

Exponential functions occur very frequently in applications growth, decay and

waiting time distribution to name a view topics. It is the single most important

function which is ubiquitous and useful. Here you will learn its properties and its

inverse function called the natural logarithm function.

Closely related with differentiation is the idea of integration. Infact, integration is

the reverse process of differentiation. At the same time it is something more than

that. You will learn about antiderivation of simple functions, and how tomanipulate antiderivatives: further the idea of definite integral, which is again an

important idea, is introduced by means of wellmotivated examples and you are

shown how the two ideas- antiderivatives and definite integrals are related. In

this way one can compute the values of definite integrals.


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Exponential Function

3.1 Introduction

If any sufficiently large segment of population is measured in any reasonable

way, the normal mode of growth is exponential. In other words population grows

like compound interest, multiplied by a constant factor in equal periods of time.

What it means mathematically is that at any point of time the rate of growth is

proportional to the size of the population, is proportional to the total magnitude

already achieved. Thus bigger a population, faster it grows. The same is true of

the human population of the world or any particular country. The exponential

growth is thus of great significance. Let us try to build a mathematical model.

3.2 The Exponential Function

As a population following exponential growth gets multiplied by a constant factor

in equal intervals of time, it will double itself in certain equal intervals of time. If

we take such an interval of time to be one unit, we may say that the rate of

change of the population at any point of time will be equal to itself. In other words

if the population at any point of time t is denoted by the function f(t), we get.

.tfdt

tdf(1)

We may further assume that the population at any point of time where we startour measurement of time, to be one unit, i.e., f(t) = 1 at t = 0, or f(0) = 1.

If we write y = f(t) and 00 fy , then we have

.1, 0yydt

dy(2)

We have thus obtained a differential equation satisfied by y = f(t). We would like

to find a solution of the differential equation (2), that is we want to find a function,

which will satisfy (2). Let us call this function, exponential function and write the

same as exp(t).

3.3 Properties

The question arises what are the other properties of this function. At any time t,

the value of the concerned population will be exp(t) and at equal intervals of time


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it gets multiplied by a constant factor. Thus after an interval of time equal to a

units, the population will be some constant multiple of exp(t). In other words.

,expexp tCat (3)

Where C is some constant. To get the value of C, we consider what happens to

(3) when t = 0, We obtain from (3) by putting t = 0.

.0expexp Ca

But according to our assumption exp (0) = 1. Thus C = exp (a) and we obtain.

.expexpexp taat (4)

As the measurement of time t, could be started for any point of time, (4) will betrue for every value of t and a, positive as well as negative. Thus taking a = -t in

(4), we get.

Exp (0) = exp (-t) exp (t)

i.e.1

expexp

1exp t

tt (5)

Further for any positive integer p, we have

exp(pt) = exp (t + t + ...p times) = exp(t) exp (t).... expt(t) = ((exp(t))p

and .expexp

1

exp

1exp

P

Pt

tptpt

Hence for any integer ,0q we have

ttq

q

exp1exp

Consequently qttq

1)exp(

1exp


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Also ifq

pr is a rational number (p a positive integer and q an integer not equal

to zero), then it follows that

.expexpexpexp rqp

tttqPrt (6)

According to the definition of exp(t), its value at t = 0 is 1 and for any subsequent

value of t(>0) the value of exp(t) will be more than 1. Thus

.01exp tfort (7)

also for t = -a, where a is positive, we have

aat

exp

1expexp

Thus 0 < exp (t) < 1 for t < 0. (8)

It follows from (7) and (8) that

exp (t) > 0. (9)

for all values of t in R. Further from (5) and (6) we conclude that exp(t) takesvalues larger than any given positive number for sufficiently large positive values

of t and very small positive values for sufficiently large negative values of t. Thus

for the function exp(t).

Domain = R and Range = {x / x > 0}

and for large negative values of x, the x-axis is an asymptote to the graph of

exp(x).

The value of exp(1) is 2.718281828 (correct upto 9 places after the decimal) andis denoted by e. We can calculate this value using computer.

It follows from (6), by taking t = 1, that for any rational number r.

rr er 1expexp


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using this as the motivation we define

.exp Rxeveryforex x

In view of the above discussion and the values of exp(0) and exp(1) the graph of

exp (x) may be drawn as shown below:

3.4 Natural Logarithm

We recall that the domain of the exponential function is the set of all real

numbers R and the range is the set of all positive numbers. Further the property

exp(x+a) = exp(a) exp(x) and the result (4) tells us that exp (x+a) > exp(x) for

a>0. In other words we have exp (x1) < exp(x2) whenever x1 < x2. Thus the

exponential function takes distinct values at distinct points. It is thus a one-one

function. Consequently the function.

Exp : R R+

Is one-one and onto and therefore it possesses the inverse. The inverse of the

exponential function has domain R+ and range R and is called natural logarithmic

function. It is denoted by ln x. Thus

Ln : R+ R


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Is denoted as y = lnx wherever x = exp (y).

We now draw the graph of lnx by taking the reflection of the graph of y = exp x in

the line y = x. We note that ln(1) = 0 because exp (0) = 1 and ln e = 1 becauseexp(1) = e.

3.5 Properties of the function ln x

(i) 2121 lnlnln xxxx

(ii) xx

ln1

ln

(iii) 212

1 lnlnln xx

x

x

(iv) xrx r lnln r any rational number.

(v)x

xdx

d 1ln


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To establish (i) suppose

.lnln 2211 yxandyx

then .expexp2211

yxandyx

hence 2121 expexp yyxx

,exp 21 yy by (4)

Thus we obtain

212121 lnlnln xxyyxx

Taking ,1

1

2x

x we get

1

1

1lnln1lnx

x

which proves (ii) because ln 1 = 0.

(iii) is proved by writing2

1

2

1

1

xxx

x

and using (i) and (ii). To prove (iv) assume

,ln yx r where r is a rational number

Then yx r exp

Or yr

yx r1

expexp1

, by (6).

Hence xyr

ln1

Or y = r ln x

Thus ln xr = r ln x.


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To prove (v), assume y = ln x. Then x = exp(y). This on differentiation with

respect to x, gives

dx

dyyexp1

Thus .1

exp

1

xydx

dy

3.6 The Value of e

We have defined e = exp (1). In order to find the value of e, we once again look

at the derivative of ln x. Then

.1xxf

Hence .11f By definition of derivative we have

h

fhfLimitfh

1111

0

h

hith

1ln1lnlim

0

hh

Limith

1ln1

0

h

hhLimit

1

01ln

Using the fact that ln x is the inverse of exp x, we note that

h

hhLimit

1

011exp

i.e. hhhite

1

01lim


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Let us now evaluate this limit by using a calculator. We calculate the values of

hh1

1 for values of h = .1, .01, .001, etc. We find

hhh1

1

.1 2.59374246

.01 2.704813829

.001 2.716922932

.0001 2.718145927

.00001 2.718268237

.000001 2.718280469

.0000001 2.718281692

.00000001 2.718281815

This table gives us the value of e, correct upto six places after the decimal. Forstill better accuracy one can take values of h smaller than those considered in

this table.

However as suggested earlier, one may use a computer to find the value of e

correct to any places after the decimal.

3.7 Graphs

In any problem of growth of a certain type of population let us determine the

population ,......,, 210 xxx at equal intervals of time, say at the instances of time

,....,, 210 ttt


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and plot the graph of the points ,....,,,,, 221100 xtxtxt We now try to join the

points by drawing a smooth curve through them. If the curve resembles the graph

of exp(t), we conclude that the growth of population is exponential.

There is another way of looking at the problem. Instead of plotting the points

,,,,,, 332211 xtxtxt as above, we plot the points (t1, ln x1), ,..ln,,ln, 3322 xtxt

In the case of exponential growth we shall have the relation.

332211 exp,exp,exp txtxtx ,

or

,....exp,exp,exp 332211 txtxtx

for some constant . Hence we have

......ln,ln,ln

....ln,ln,ln

333222111

333222111

txztxztxzor

txztxztxz

This shows that the points ,....,,,,, 332211 ztztzt

will lie on a straight line as shown in the adjoining figure. As it is easy to visualize

a straight line in comparison to the graph of the exponential function, many a

time it is preferred to draw the curve using the natural logarithm of the observed

values of the population under consideration. If the points lie on a straight line we

conclude that the mode of growth is exponential.


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Chapter 4

4.1 Introduction

In elementary arithmetic there are certain simple, direct processes, like addition,

multiplication, and squaring a number. These can always be carried out within

the framework of the natural numbers: 0,1,2,3,4, what is 3 added to 4?

Answer is 7. What is 3 multiplied by 4? Answer is 12. What is the square of 3?

Answer is 9.

Later on, we learnt to reverse these operations. We learnt to subtract by

reversing additions: 3 and what makes 7? Answer is 4. We divided by reversing

multiplication: 3 times what is 12? Answer is 4. We reversed squaring to extract

square root: Which number squared gives 9? Answer is 3.

These reverse operations lead us to extend our ideas and discover new ones. In

fact, the idea of negative numbers, the idea of fractions, the idea of irrational

numbers and the even more striking idea of complex numbers were discovered

by this process of reversing.

In calculus also exactly the same kind of growth occurs. In Chapter 1 we learnt

how to compute derivatives of functions, and how this concept is used. For

example : What is the slope of the curve y = f(x) at any point? Answer is .dx

dy

Given a law y = f(x) which tells you where an object is at any time, what is its

velocity? Answer is .dx

dy.

We can easily reverse this and ask: Given the slope of a curve at any point, find

the curve or, given the law for the velocity, find the law for the position. In

symbols, given that

)(xgdx

dy

(where g(x) is a known function) find y as a function of x. This kind of problems

goes by the name: ANTIDERIVATION or INTEGRATION.


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Recall that when we learnt about rates, we studied two notions:

1. Derivative at a point give by

axdx

dy

and

2. Derivative (at any point) given by

.dx

dy

In the same way, in this Chapter you are going to study about two notions:

1. General (Integration) which solves the problem of finding the (unknown)

function y when xgdx

dy, (Here g(x) is the given function). And

2. Definite Integral which is the difference of the values of the anti-derivative

at two points.

(More will be said on this at the appropriate place).

4.2 Antiderivation

As explained above, anti-derivation or (general) integration solves the problem of

finding the (unknown) function y given that xgdx

dy(g(x) being the given

function).

For example, given that

.2xdx

dy(*)

We know that the answer is: x2 (Look up the formula 2) But, surprisingly, this is

not the only answer! any one of the following functions will also satisfy the above

given equation(*)


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2

3

2

2

2

2

2

x

x

x

and, in general, x2 + K, where K is any constant.

Note : Do you see why the above statements are true? Go back to Formula 3.

Thus, the general answer for the problem

.2xdx

dy

is given by .2 kxy

This is generally written as Cxy 2 (where the capital C is used to denote the

general constant).

This function

X2 + C

Is called (general) Antiderivatives of the function 2x, which appears on the right-

hand-side of (*). The process of finding antiderivatives is called

ANTIDERIVATION.

Remember: Anti derivative of a given function is not unique; i.e., given a

function g(x), the equation

xgdx

dy

has many answers. Any two of these answers differ by a constant. Thus, if we

know one answer, F(x) , then all the answers are known and they are given by

F(x) + C, for suitable values of C.


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4.2.2 Example

Find an anti-derivative of

8

55

2

3

1

xx

Using the above formulae, we get

Antder

13

1

13

1

3

1x

x

3

4

4

3x

Antder

15

2

15

2

5

2x

x

5

3

3

5x

Antder x8

5

8

5

Thus required anti-derivative is

Cxxx8

5

3

5

4

35

3

3

4

Before giving further formulae, we note the following: To every formula of

differentiation there is a corresponding formula of antiderivation.

Now commit the following formulae to memory.

Formula 3

Antder (f(x)+ g (x)) = Antder (f(x)) + Antder (g(x))

Note that we have already made use of this formula in examples given above.


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Formula 4

Antder (kf(x)) = k Antder (f(x))

Where k is constant

For example, Antder (3x2) = 3 Antder (x2)

,3

3 33

CxCx

where C, is some constant.

4.2.3 Example

Consider the following modification of example 2.2.

8

534 5

2

3

1

xx

Using the results of example 4.2. and the formulae 3 and 4 we get

Antder8

534 5

2

3

1

xx

=8

534 5

2

3

1

AntderxAntderxAntder

.8

553

8

5

3

53

4

34

5

3

3

4

5

3

3

4

Cxxx

Cxxx


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Formula 5

CeeAntderxx

Formula 6

0

ln1

xwhere

Cxx

Antder

Note that this formula fills up the lacuna in formula-1. In formula-1 the case n = -1

was excluded. The above formula shows what happens when n = -1.

Example: Find the antiderivatives of

xexx

1021

2

Note that 22

1x

x. We have

x

x

ex

xAntder

e

xx

Antder

102

1021

2

2

Cexx

Cexx

eAntderx

AntderxAntder

x

x

x

10ln21

10ln212

101

2

12

2


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Formula 7

Cea

eAntderaxax 1

where a is a non-zero constant

Formula 8

Integration by Parts

Antder xgxfdx

xdgxf - Antder xg

dx

xdf

Formula 8 goes by the name integration by parts. This is a very useful formulaand must be committed to memory firmly! The following example illustrates its

use:

4.2.4 Example

Find antiderivative of kxxe . So we have to find Antder kxxe . First of all we try to

express kxe in the form xgdx

dfor some suitable function g(x). Since

Cek

eAntderkxkx 1

by formula 7, we get

kx

kx

kxkx

e

kek

edx

d

kekdx

d

1

11

Thus kxkx ekdx

de

1


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76

And so we take .1 kxek

xg Thus we have

xgdx

dekx

where kxek

xg1

Next we take f(x) = x and apply formula8. We get

Antder xgdx

dxfAntderex

kx

1k

eAntderk

ex

xfdxdxgAntderxgxf

kxkx

kxkxek

Antderek

11

.11

111

11

2Ce

kek

Cekk

ek

eAntderk

ek

kxkx

kxkx

kxkx

In particular, taking k = -1. We get

Antder .Cexexe xxx


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Exercise 4.2

Find the antiderivatives of the following:

1. 35 32

4 xxx

2. 32

2

3

xx

3.3

2 13

x

xx(Hint: Divide each term by x3)

4.22 32x (Hint: Expand first)

5. x2 ex (Hint : Apply Formula8 twice)

6. x ln x (Hint: Recall 2

2

1x

dx

dx

7. ln x (Hint: Write lnx = 1.ln x and use )(1 xdx

d)

4.3 Definite Integral

4.3.1 In this section, we will introduce the important concept- Definite integral.

Roughly speaking, definite integral is an infinite sum; i.e. a sum containing a

limitless or unending number of terms. To put it in proper focus, we start by

considering sum of finitely many terms.

Consider the following sums.

1 + 2

1 + 2 + 31+ 2 + 3 + 4

1 + 2 + 3 + . + 100

These are all finite sums because we are adding a finite (may be two or three

or. > number of terms.


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Before proceeding further, we use a short-hand notation for representing sums of

above forms and infinite sums. This is the notation. ( is read as sigma. It

is Greek alphabet corresponding to the English alphabet S). Using this notation,

we can write

1 + 2 + 3 + + 100

in a compact form:

100

1n

n

For example,

1239

1

2

n

n

is a short form for

2222 1239.........321

Do you understand how useful this notation is?

4.3.2 Now let us move on to infinite sums. For example,

........3

1

2

1

1

1222

is an infinite sum. The pattern is : if you know a term, say, ,1

2n

then the next term

is2

1

1

n. Thus we can generate as many terms as we want and there will be no

end to this process! That is why, the above sum is called an infinite sum. (Theword infinite is the opposite or antonym of the word finite) Note that the above

sum can be written compactly as

21

1lim

k

n

kn


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This needs explanation:

n

k k12

1

is short for

2222

1.....

3

1

2

1

1

1

n

The symbolnlim stands for generate terms without end.

We will express the above sum in a slightly different way: let

2

1

xxf

Then

kfk

2

1

and the above sum is given by

kfn

kn 1

lim

Also note that in f(x), x takes only the values

1,2,3,., n.

We picture this as follows:

Now it is clear that the distance between any two consecutive x-values is 1(one).

So we write the above sum as


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1.lim1

kfn

kn

(*)

Now, we will see that the definite integral is a generalization of (*). Thus, we can

think of the definite integral as the generalized sum or generalized aggregate.

4.3.3 For the process of definite integration, we need the following data:

1. An interval [a,b] where a < b

2. A nice function f(x) defined on [a,b]. Given these, we divide the

interval [a,b] into n equal parts, as shown below:

Each part will have the same lengthn

ab. In each part we take a point:

x1 in the first part,

x2 in the second part

--------------------------

xk in the kth part;

--------------------------

xn in the nth part.

We form the sum

n

k

knk xfxfxfxfxf1

21 .........

Multiply this sum by the length of each part;

n

ab

we getn

k

kxfn

ab

1

Now consider


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n

k

k

n

xfn

abLim

1

......(**)

This value is called the definite integral of f(x) from a to b, and is denoted by

dxxf

b

a

The symbol is elongated S.

Note the close resemblance between (*) and (**)

Remember

1.b

adxxf is called the definite integral of f(x) from a to b.

2.b

adxxf is a generalized sum.

3. The meaning ofb

a

dxxf is given by (**)

4.3.4 We have seen above that the definite integral can be thought of as ageneralized sum. Since this concept is an important one, we will provide two

more motivations: One of these is given here and the other is given in the next

section.

Suppose you are traveling in a car going along a straight road with velocity

(speed) given by f(x) where x represents time. If you travel from time x = a to time

x = b, how much distance has been traveled?

We cannot find this distance by multiplying the velocity f(x) and the total time

traveled b-a; because the velocity is not a constant! It varies with time!

So we feel that, if we divide the total time b-a into several small intervals then an

argument similar to above will work-take sum of velocity in each interval

multiplied by the length of the interval.


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Also we feel that the smaller each interval is, the better will be our approximation!

THIS IS PRECISELY THE IDEA BEHIND DEFINITE INTEGRATION!!!

So proceeding as in 4.3.3., we see that the total distance traveled is given by

b

a

dxxf

Do you feel comfortable about this concept now?

4.3.5 Definite Integral as Area:

We now give another motivation for definite integral from another angle- from

geometric viewpoint.

Now go back to 4.3.3. We are given a nice function (i.e., a continuous function

which lies above the x-axis y = f(x) defined on [a,b]. See figure 1.

As before, we divide the interval [a,b] into n equal parts each of length

n

ab


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Where the points of division are

baaaaaaaa nnkk ,,......,,.....,, 11210

See Figure 2.

As before (4.3.3) we take points

nk xxxx ,.....,.....,, 21

in each part, as shown in figure 3.

We form the sum

n

k

kxfn

ab

1

n

k

kxfn

ab

1

.

Now the term kxf

n

ab. can be interpreted as the area of the rectangle whose

sides are kxfandn

ab. See figure 4.


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Thus the sum

n

k

kxfn

ab

1

.

can be interpreted as the sum of areas of rectangles each with base

n

ab

and heights nxfxfxf ,....,, 21 , see figure 5.


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Thus the sum can be taken as an approximation for the area under the curve

xfy from x = a to x = b.

Hence, when n becomes larger, the base of the rectangles become smaller and

the sum

n

k

kxfn

ab

1

.

gives closer approximation for the area under the curve.

So the value (**) gives the EXACT VALUE of the area under the curve y = f(x)

from x = a to x = b.

We thus have the important interpretation:

b

a

dxxf

represents the area under the curve y = f(x) from x = a to x=b! (Remember that

f(x) is nice)

4.4 Fundamental Theorem of Integration

As the title indicates this result is a basic one. It exhibits a remarkable relation

between the antiderivatives and the definite integral of a function. This relation is

quite useful in evaluating the definite integrals.

We state this result in an informal way:

Fundamental Theorem: If a function f(x) is nice over the interval [a,b], then

b

a

b

a

xfAntderdxxf

Note: If F(x) is any function, the symbol

b

axF

stands for the value F(b)F(a).


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aFbFxFb

a

The Fundamental Theorem tells you that in order to evaluateb

adxxf , it is

enough to

1. Find an antiderivatives of f(x);

2. Evaluateb

axfAntder .

Then 3. This is the required value ofb

adxxf .

The following example illustrates this idea.

4.4.1 Example:

Evaluate2

1

2dxx

Now Antder .3

32 xx

There is no need to add the constant C here, because we need only oneantiderivative.

Thus

2

1

2

1

32

3

xdxx

3

1

3

2 33

31

38

3

7


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87

4.4.2 Example

Find the area under the curve y = ex between x = 0 and x = 1.

The required area =

1

0

dxex

1

0

xe

= e1

4.4.3 Example

Evaluate0

dxexkkx

This integral is important.

Now Antder kxxke

kxkx ek

ex1

(Go back to example 4.2.5). Thus

0 0

1 kxkxkx e

k

exdxxke

k

1

Because 0lim&0lim kxx

kx

x

eex


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88

Exercise 4.4

Evaluate

1.

2

1

4

dxx

2.0

1

dxex

3. dxx1

0

2

1

4. dxxx3

2

3 4

5.2

0

dxex

6.2

1

1dxt

7. Find the area under y = x from x = 0 to x = 1. Sketch a figure. What do you see?

8. Find the area under y = 4x2 between x = -1 and x = 2.

9. Find1

02 1

dxx

x

(Hint : Find 12xdx

d

10. Find dxx 322

2


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