the definite integral part i tuesday, february

8/14/2019 The Definite Integral Part I Tuesday, February

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DEFINITE INTEGRAL

If fis a function defined for a x b,

we divide the interval [a, b] into nsubintervals

of equal width x= (b a)/n.

We let x0(= a), x1, x2, , xn(= b) be the endpointsof these subintervals.

We let x1*, x2*,., xn* be any sample points inthese subintervals, so xi* lies in the ith subinterval.

Definition 2


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DEFINITE INTEGRAL

Then, the definite integral of ffrom ato bis

provided that this limit exists.

If it does exist, we say fis integrable on [a, b].

1

( ) lim ( *)n

b

ia n

i

f x dx f x x

Definition 2


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INTEGRAL SIGN

The symbol was introduced by Leibniz

and is called an integral sign.

It is an elongated S.

It was chosen because an integral isa limit of sums.

Note 1


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In the notation ,

f(x) is called the integrand.

aand bare called the limits of integration;ais the lower limit and bis the upper limit.

For now, the symbol dxhas no meaning by itself;

is all one symbol. The dxsimply indicatesthat the independent variable is x.

( )

b

a f x dx

( )b

a f x dx

Note 1( )b

a f x dxNOTATION


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INTEGRATION

The procedure ofcalculating an integral

is called integration.

Note 1


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DEFINITE INTEGRAL

The definite integral is a number.

It does not depend on x.

In fact, we could use any letter in place of x

without changing the value of the integral:

( )b

a

f x dx

( ) ( ) ( )b b b

a a a f x dx f t dt f r dr

Note 2( )b

a f x dx


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RIEMANN SUM

The sum

that occurs in Definition 2 is called

a Riemann sum.

It is named after the German mathematicianBernhard Riemann (18261866).

1

( *)

n

i

i

f x x

Note 3


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RIEMANN SUM

So, Definition 2 says that the definite integral

of an integrable function can be approximated

to within any desired degree of accuracy by

a Riemann sum.

Note 3


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RIEMANN SUM

We know that, if fhappens to be positive,

the Riemann sum can be interpreted as:

A sum of areas of approximating rectangles

Note 3


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RIEMANN SUM

Comparing Definition 2 with the definition

of area in Section 5.1, we see that the definite

integral can be interpreted as:

The area under the curve y= f(x) from ato b

( )b

a f x dx

Note 3


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RIEMANN SUM

If ftakes on both positive and negative values,

then the Riemann sum is:

The sum of the areas of the rectangles that lieabove the x-axis and the negatives of the areas

of the rectangles that lie below the x-axis

That is, the areas ofthe gold rectanglesminus the areas of

the blue rectangles

Note 3


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RIEMANN SUM

When we take the limit of such

Riemann sums, we get the situation

illustrated here.

Note 3

Thomson Higher Education


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NET AREA

A definite integral can be interpreted as

a net area, that is, a difference of areas:

A1 is the area of the regionabove the x-axis and below the graph of f.

A2 is the area ofthe region below

the x-axis andabovethe graph of f.

1 2( )

b

a f x dx A A

Note 3

Thomson Higher Education


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INTEGRABLE FUNCTIONS

To simplify the calculation of the integral,

we often take the sample points to be right

endpoints.

Then, xi* = xiand the definition of an integralsimplifies as follows.


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INTEGRABLE FUNCTIONS

If fis integrable on [a, b], then

where

1

( ) lim ( )i

nb

ia n

i

f x dx f x x

andi

b a x x a i x

n

Theorem 4


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DEFINITE INTEGRAL

Express

as an integral on the interval [0, ].

Comparing the given limit with the limitin Theorem 4, we see that they will beidentical if we choose f(x) = x3 + xsin x.

3

1lim ( sin )

n

i i i in

i x x x x

Example 1


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DEFINITE INTEGRAL

In general, when we write

we replace:

lim by

xi* by x

xby dx

1

lim ( *) ( )n b

ian

i

f x x f x dx


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EVALUATING INTEGRALS

a.Evaluate the Riemann sum for f(x) = x3 6x

taking the sample points to be right

endpoints and a =0, b =3, and n =6.

b.Evaluate .3

3

0( 6 ) x x dx

Example 2


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With n= 6,

The interval width is:

The right endpoints are:x1 = 0.5, x2 = 1.0, x3 = 1.5,

x4 = 2.0, x5 = 2.5, x6 = 3.0

3 0 1

6 2

b ax

n

Example 2 a


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So, the Riemann sum is:

6

6

1

1

2

( )

(0.5) (1.0) (1.5)

(2.0) (2.5) (3.0)

( 2.875 5 5.625 4 0.625 9)3.9375

i

i

R f x x

f x f x f x

f x f x f x

Example 2 a


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Notice that fis not a positive function.

So, the Riemann sum does not

represent a sum of areas of rectangles.

Example 2 a


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With nsubintervals, we have:

Thus, x0 = 0, x1 = 3/n, x2 = 6/n, x3 = 9/n.

In general, xi= 3i/n.

3b ax

n n

Example 2 b


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asthe difference of areas A1 A2,whereA1 and ALUATING INTEGRALS

A2

are as shown.

Example 2 bHowever, itcan be

interpretedasthe


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This figure illustrates the calculation by showing the

positive and negative terms

in the right Riemann sum Rnfor n= 40.

Example 2 b


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A much simpler method for

evaluating the integral in Example 2

will be given in Section 5.3


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a.Set up an expression for as

a limit of sums.

b.Use a computer algebra system (CAS)

to evaluate the expression.

54

2

x dx

Example 3


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Here, we have f(x) = x4, a= 2, b= 5,

and

So, x0 = 2, x1 = 2 + 3/n, x2 = 2 + 6/n,

x3 = 2 + 9/n, andxi= 2 + 3i/n

3

b ax

n n

Example 3 a


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From Theorem 4, we get:

54

21

1

4

1

lim ( )

3 3lim 2

3 3lim 2

n

in

i

n

ni

n

ni

x dx f x x

if

n n

in n

Example 3 a


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We will learn a much easier

method for the evaluation of integrals

in the next section.

Example 3 b


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Evaluate the following integrals by interpreting

each in terms of areas.

a.

b.

12

01 x dx

3

0 ( 1) x dx

Example 4


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However, since y2 = 1

- x2, we get:

x2

+ y2

= 1

This shows thatthe graph of fis

the quarter-circlewith radius 1.

Example 4 a


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Therefore,

In Section 8.3, we will be able to prove that

the area of a circle of radius ris r2

.

12 21

40

1 (1)4

x dx

Example 4 a


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The graph of y= x 1

is the line with slope 1

shown here.

We compute the integralas the difference of theareas of the two triangles:

3 1 11 2 2 2

0( 1) (2 2) (1 1) 1.5 x dx A A

Example 4 b

the definite integral part i tuesday, february

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