The Definite Integral Part 2 Thursday, February

8/14/2019 The Definite Integral Part 2 Thursday, February

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5.2The Definite Integral Part 2

Thursday, February 4, 2010

In this section, we will learn about:

Integrals with limits that represent

a definite quantity.

INTEGRALS


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EVALUATING INTEGRALS

Since ,

we can interpret this integral as

the area under the curve

from 0 to 1.

Example 4 a

2( ) 1 0 f x x

21y x


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However, since y2 = 1

- x2, we get:

x2 + y2 = 1

This shows that

the 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 20

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

Example 4 b


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MIDPOINT RULE

We often choose the sample point xi*

to be the right endpoint of the ith subinterval

because it is convenient for computing

the limit.


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MIDPOINT RULE

However, if the purpose is to find

an approximation to an integral, it is usually

better to choose xi* to be the midpoint of

the interval.

We denote this by .i

x


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MIDPOINT RULE

Any Riemann sum is an approximation

to an integral.

However, if we use midpoints, we get

the following approximation.


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THE MIDPOINT RULE

1

1

11 12

( ) ( )

( ) ... ( )

where

and ( ) midpoint of ,

nb

iai

n

i i i i i

f x dx f x x

x f x f x

b ax

n

x x x x x


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MIDPOINT RULE

Use the Midpoint Rule with n= 5

to approximate

The endpoints of the five subintervalsare: 1, 1.2, 1.4, 1.6, 1.8, 2.0

So, the midpoints are: 1.1, 1.3, 1.5, 1.7, 1.9

2

1

1dx

x

Example 5


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MIDPOINT RULE

The width of the subintervals is:x= (2 - 1)/5 = 1/5

So, the Midpoint Rule gives:

2

1

1(1.1) (1.3) (1.5) (1.7) (1.9)

1 1 1 1 1 1

5 1.1 1.3 1.5 1.7 1.9

0.691908

dx x f f f f f x

Example 5


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MIDPOINT RULE

The approximation

M40 -6.7563

is much closer to

the true value -6.75 than

the right endpoint

approximation,

R40 -6.3998,

in the earlier figure.


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Notice that, if we reverse aand b, then x

changes from (b a)/nto (a b)/n.

Therefore,

If a= b, then x= 0, and so

( ) ( )a bb a

f x dx f x dx

( ) 0a

b f x dx

PROPERTIES OF DEFINITE INTEGRAL


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PROPERTIES OF THE INTEGRAL

We assume fand gare continuous functions.

1. ( ), where c is any constant

2. ( ) ( ) ( ) ( )

3. ( ) ( ) , where c is any constant

4. ( ) ( ) ( ) ( )

b

a

b b b

a a a

b b

a a

b b b

a a a

c dx c b a

f x g x dx f x dx g x dx

c f x dx c f x dx

f x g x dx f x dx g x dx


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PROPERTIES OF INTEGRALS

Use the properties of integrals to

evaluate

Using Properties 2 and 3 of integrals,we have:

12

0(4 3 ) x dx

1 1 12 2

0 0 0

1 1 2

0 0

(4 3 ) 4 3

4 3

x dx dx x dx

dx x dx

Example 6


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We know from Property 1 that:

We found in Example 2 in Section 5.1that:

1

04 4(1 0) 4dx

12

130 x dx

Example 6


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Thus,

1 1 12 2

0 0 013

(4 3 ) 4 3

4 3 5

x dx dx x dx

Example 6


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PROPERTY 5

Property 5 tells us how to combine

integrals of the same function over

adjacent intervals:

( ) ( ) ( )c b b

a c a f x dx f x dx f x dx


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If it is known that

find:

10 8

0 0( ) 17 and ( ) 12 f x dx f x dx

Example 7

10

8( ) f x dx


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By Property 5, we have:

So,

8 10 10

0 8 0( ) ( ) ( ) f x dx f x dx f x dx

10 10 8

8 0 0( ) ( ) ( )

17 125

f x dx f x dx f x dx

Example 7


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COMPARISON PROPERTIES OF THE INTEGRAL

These properties, in which we compare sizes

of functions and sizes of integrals, are true

only if a b.

6. If ( ) 0 for , then ( ) 0

7. If ( ) ( ) for , then ( ) ( )

8. If ( ) for , then

( ) ( ) ( )

b

a

b b

a a

b

a

f x a x b f x dx

f x g x a x b f x dx g x dx

m f x M a x b

m b a f x dx M b a

The Definite Integral Part 2 Thursday, February

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