Single Integral Ver1

8/11/2019 Single Integral Ver1

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SINGLE INTEGRAL/STQM1124 CALCULUS II/SEM2 20122013

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TOPIC: SINGLE INTEGRAL

5.1 The area problem

The Area Problem: Given a function f that is continuous and nonnegative on

an interval [ ],a b , find the areaA,between the graph of fand the interval [ ],a b on the x axis.

A


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Divide the interval [ ],a b into n equal subintervals

The subintervals are given by [0, 1], [1, 2],,[1, ]. In general eachsubinterval is represented by [1 , ]where = 1,2,3, , .

Each subinterval has length 1 =

=where = 1,2,3, ,. For each subinterval [1 , ], construct a rectangle from the interval[1 , ] to any point on the curve and let denotes the area of therectangle on that interval.


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Formulas offor each approximation:

Left endpoint approximation : =(1) =(1) Right endpoint approximation : =() =()

Midpoint approximation : =

=

Thus, the areaAis approximated by

1 +2 ++ = =1

.

LE : =(0) +(1) ++(1) = (1)=1 RE : =(1) +(2) ++() = ()=1 MD : = + ++ = +2 =1


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Example 1: Using the left endpoints approximation, approximate the areabetween the graph() = + 2and the interval [0,4] for =2. Next, try for = 4.


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Note that the above approximation appears to become much betteras thenumber of rectangles increases or equivalently when

.


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Definition 1: The areaAthat lies under the graph of the nonnegative continuousfunction fis given by

= lim

= lim

(

1+

2+

+

)

=1

Note that,

Left endpoint

= lim

=1= lim(1)

=1

Right endpoint

= lim

=1= lim

(

)

=1

Midpoint

= lim

=1= lim

+2

=1


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In general,

we can also choose any point [1 , ]called sample or evaluationpoints

take the height of the rectangle on interval [1 , ]as()

and now a more general expression for the area is now given by

= lim

=1= lim()

=1.


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Example 2: Find the area (from Example 1) using the limit of the right endpointtechnique. (Answer = 16 unit

2)

-need to know Sigma Notation and Summation Formulas!

-refer to the page 373 (Soo T. Tan Calculus 2010)


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5.2 The definite integral

We now give the above limit a special name and notation.

Riemann sums and the definite integral

A partition, P of [, ] is a collection of points { =0,1,2, , =}where

=0 < 1 < 2


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Definition 2: Letis a continuous function defined for and letP= { =0, 1, 2, , =}

be a regular partition of the interval [,].We divide the interval into n subintervals of equal width = .We let

1,

2, ,

be n sample points where

is chosen arbitrarily from

[1 , ].

a) The Riemann sumfor this partition P and set of sample points is given by

()

=1.


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b) The definite integraloffrom a to b is given by

() =

lim()

=1

whenever the limit exists and is the same for any choice of sample points.

When the limit exists, we say that

is integrableon [

,

].


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The symbol was introduced by Leibniz and is called an integral sign.

It is an elongated Sand was chosen because an integral is a limit of sums.

In the notation

()

()is called the integrand,

The number ais called a lower limitand bis called an upper limitand bothare called limits of integration.

The procedure of computing an integral is called integration.


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Example 3: Use Riemann sums and a limit to compute the exact area under the

curve =2on [1, 3].


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If takes on both positive and negative values then the Riemann sum is thesum of the areas of the rectangles that lie above the x-axisand the negative ofthe areas of the rectangles that lie below the x-axis. We take the limit of suchRiemann sums, we will get a definite integral that can be interpreted as a netarea, that is, a difference of areas:

( 1) =1 22

0

=.


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If fis continuous on [ ],a b then the total area between the curve ( )=y f x andthe interval [ ],a b is given by

|()|

.


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Example 4: Find the net and total area of the curve = 3 on [1, 5].


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Distance problem:Find the distance traveled by an object during a certain period of time if thevelocity of the object is known at all times.

1. If the velocity remains constant, then

distance, d= velocity time

2. If the velocity of the object varies we cannot use the above formula.

Now suppose an object moves with velocity =() where .Assume that() 0, so the object always moves in the positive direction.The process goes as below:

a. Take velocity readings at times

=

0,

1,

2, ,

1,

=

so that the

velocity is approximately constant on each subinterval.

b. If these times are equally spaced then the time is

= .


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c. For first time interval, the velocity is(0)and1 =(0).

d. For second time interval, the velocity is(1)and2 =(1)and so on

e. Thus the total distance traveled during the time interval [, ]is given by

(1)

=1.

f. Finally the exact distance traveled is

= =

lim(1)

=1.


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Example 5: An object moving along a straight line has velocity function = 2 + 1. If the object starts at position 0, determine the total distancetraveled at time = 5.


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Example 6: Determine the integral

5 250

in terms of area.


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We have defined the definite integral in terms of a limit but under what conditionthe limit actually exists?

Theorem 1: If

is continuous on the closed interval [

,

], then

is integrable

on [, ].

Properties of the definite integral

Theorem 2:

a) If a is in domain fthen ( ) = 0a

a

f x d x

b) If fis integrable on [ ],a b then ( ) ( )= b a

a b

f x dx f x dx

Proof: Do it on your own!


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Theorem 3: If f and g are integrable on [ ],a b and if c is a constant then,cf f g are also integrable on [ ],a b and

a) =( )

b) ( ) ( )= b b

a a

cf x dx c f x dx

c) ( ) ( ) ( ) ( ) = b b b

a a a

f x g x dx f x dx g x dx

d) ( ) ( ) ( )= + f e f

d d e

f x dx f x dx f x dx for any three points , ,d e fin [ ],a b .


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Theorem 4: Let functions fand g be integrable on [ ],a b .

a) If ( ) [ ]

0, ,f x x a b then ( )

0

b

a

f x d x .

b) If ( ) ( ) [ ] , ,f x g x x a b then ( ) ( ) b b

a a

f x dx g x dx .

c) If () for then ( )() ( ).


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Discontinuities and integrability

Definition 3: A function fon an interval L is said to be bounded on that intervalif there is a positive number Msuch that

( ) ,M f x M x L .


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Theorem5: Let fbe a function that is defined on the finite closed interval [ ],a b .

a) If fhas finitely many discontinuities in [ ],a b but is bounded on [ ],a b then f

is integrable on [ ],a b .

b) If fis not bounded on [ ],a b then fis not integrable on [ ],a b .


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Antiderivative

Definition4: A function F is an antiderivative of a function g on an interval L if

( ) ( ) =F x g x for all x L .

Theorem 6: If F is an antiderivative of hthen ( ) +F x C is also an antiderivativeof hfor any constant .C

Proof:

Note that,

(

(

) +

) =

(

) +

=

(

) + 0 =

(

) =

.

Thus ( ) +F x C is also an antiderivative of h for any constant .C


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Theorem 7: (Mean Value Theorem) If f is continuous and differentiable on a

closed interval [ ],a b then there is at least one point (, )such that

() =() () .

Proof: Read from any calculus book.

Example 7: Validate the above theorem for() = 32on [1, 5].


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We next give a simple theorem without proof and that has some significantapplications.

Theorem 8: (Integral Mean Value Theorem) If f is continuous on

[ ],a b then

there is at least one point (, )such that

()

=()( ).


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Example 8: Verify the above theorem if () = 2 + 1 for [3,4] for = 1/2.

Single Integral Ver1

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