Areas and Distances Friday, January 29, 2010

8/14/2019 Areas and Distances Friday, January 29, 2010

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5.1

Areas and Distances

Friday, January 29, 2010

INTEGRALS

In this section, we will learn that:

We get the same special type of limit in trying to find

the area under a curve or a distance traveled.


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AREA PROBLEM

We begin by attempting to solve

the area problem:

Find the area of the region Sthat lies

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


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RECTANGLES

For a rectangle, the

area is defined as:

The product of the

length and the width


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TRIANGLES

The area of a

triangle is:

Half the base times theheight


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POLYGONS

The area of a polygon

is found by:

Dividing it into triangles

and adding the areasof the triangles


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AREA PROBLEM

However, it isnt so easy to find the area

of a region with curved sides.

We all have an intuitive idea of what the areaof a region is.

Part of the area problem, though, is to make thisintuitive idea precise by giving an exact definitionof area.


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AREA PROBLEM

We first approximate the region Sby

rectangles and then we take the limit of

the areas of these rectangles as we increase

the number of rectangles.

The following example illustrates the procedure.


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AREA PROBLEM

Use rectangles to

estimate the area

under the parabola

y= x2 from 0 to 1,the parabolic

region Sillustrated

here.

Example 1


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AREA PROBLEM

We first notice that the area of Smust be

somewhere between 0 and 1, because S

is contained in a square with side length 1.

However, we cancertainly do betterthan that.

Example 1


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AREA PROBLEM

Suppose we divide S

into four strips

S1, S2, S3, and S4 by

drawing the verticallines x= , x= ,

and x= .

Example 1


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AREA PROBLEM

We can approximate

each strip by a

rectangle whose base

is the same as thestrip and whose

height is the same as

the right edge

of the strip.

Example 1


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AREA PROBLEM

In other words, the heights of these rectangles

are the values of the function f(x) = x2 at the

right endpoints of the subintervals

[0, ],[, ], [, ],and [, 1].

Example 1


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AREA PROBLEM

Each rectangle has

width and

the heights are ()2,

()2

, ()2

, and 12

.

Example 1


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AREA PROBLEM

If we let R4 be the sum of the areas

of these approximating rectangles,

we get:

22 2 231 1 1 1 1 1

4 4 4 4 2 4 4 4

15

32

1

0.46875

R

Example 1


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AREA PROBLEM

We see the area A of

Sis less than R4.

So, A


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AREA PROBLEM

Instead of using the

rectangles in this

figure, we could use

the smaller rectanglesin the next figure.

Example 1


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AREA PROBLEM

Here, the heights are

the values of fat

the left endpoints of

the subintervals.

The leftmost rectanglehas collapsed because

its height is 0.

Example 1


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AREA PROBLEM

The sum of the areas of these approximating

rectangles is:

22 22 31 1 1 1 1 1

4 4 4 4 4 2 4 4

7

32

0

0.21875

L

Example 1


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AREA PROBLEM

We can repeat this

procedure with a larger

number of strips.

Example 1


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AREA PROBLEM

The figure shows what happens when

we divide the region Sinto eight strips of

equal width.

Example 1


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AREA PROBLEM

By computing the sum of the areas of

the smaller rectangles (L8) and the sum of

the areas of the larger rectangles (R8),

we obtain better lower and upper estimatesfor A:

0.2734375 < A < 0.3984375

Example 1


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AREA PROBLEM

So, one possible answer to the

question is to say that:

The true area of Slies somewherebetween 0.2734375 and 0.3984375

Example 1


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AREA PROBLEM

We could obtain better

estimates by increasing

the number of strips.

Example 1


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AREA PROBLEM

The table shows the results of similar

calculations (with a computer) using n

rectangles, whose heights are found with

left endpoints (Ln)or right endpoints

(Rn).

Example 1


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AREA PROBLEM

In particular, we see

that by using:

50 strips, the area

lies between 0.3234and 0.3434

1000 strips, wenarrow it down

even moreA liesbetween 0.3328335and 0.3338335

Example 1


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AREA PROBLEM

From the values in the

table, it looks as if Rn

is approaching 1/3 as

nincreases.

We confirm this in

the next example.


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AREA PROBLEM

From this figure, it appears that, as n

increases, Rnbecomes a better and better

approximation to the area of S.


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AREA PROBLEM

From this figure too, it appears that, as n

increases, Lnbecomes a better and better

approximations to the area of S.

Thomson Higher Education


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AREA PROBLEM

Lets apply the idea of Examples 1 and 2

to the more general region Sof the earlier

figure.


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AREA PROBLEM

We start by

subdividing Sinto n

strips

S1, S2, ., Snof equalwidth.


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AREA PROBLEM

The width of the interval [a, b] is b a.

So, the width of each of the nstrips is:

b ax

n


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AREA PROBLEM

These strips divide the interval [a, b] into n

subintervals

[x0, x1], [x1, x2], [x2, x3], . . . , [xn-1, xn]

where x0 = aand xn= b.


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AREA PROBLEM

Lets approximate the ith strip Siby

a rectangle with width xand height f(xi),

which is the value of fat the right endpoint.

Then, the area of theith rectangle is f(xi)x.


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AREA PROBLEM

What we think of intuitively as the area of S

is approximated by the sum of the areas of

these rectangles:

Rn= f(x1) x + f(x2) x + + f(xn) x


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AREA PROBLEM

Here, we show this approximation for

n= 2, 4, 8, and 12.


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AREA PROBLEM

Notice that this approximation appears to

become better and better as the number of

strips increases, that is, as n .


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AREA PROBLEM

The area A of the region Sthat lies

under the graph of the continuous function f

is the limit of the sum of the areas of

approximating rectangles:

1 2

lim

lim[ ( ) ( ) ... ( ) ]

nn

nn

A R

f x x f x x f x x

Definition 2


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AREA PROBLEM

It can also be shown that we get the same

value if we use left endpoints:

0 1 1

lim

lim[ ( ) ( ) ... ( ) ]

nn

nn

A L

f x x f x x f x x

Equation 3


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SAMPLE POINTS

In fact, instead of using left endpoints or right

endpoints, we could take the height of the ith

rectangle to be the value of fat anynumber xi*

in the ith

subinterval [xi- 1, xi].

We call the numbers xi*, x2*, . . ., xn*the sample points.


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AREA PROBLEM

The figure shows approximating rectangles

when the sample points are not chosen to be

endpoints.


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AREA PROBLEM

Thus, a more general expression for

the area of Sis:

1 2lim[ ( *) ( *) ... ( *) ]nnA f x x f x x f x x

Equation 4


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SIGMA NOTATION

We often use sigma notation to write sums

with many terms more compactly.

For instance,

1 2

1

( ) ( ) ( ) ... ( )n

i n

i

f x x f x x f x x f x x


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AREA PROBLEM

Hence, the expressions for area

in Equations 2, 3, and 4 can be written

as follows:

1

1

1

1

lim ( )

lim ( )

lim ( *)

n

i

n i

n

in

i

n

in

i

A f x x

A f x x

A f x x


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AREA PROBLEM

Let A be the area of the region that lies under

the graph of f(x) = cos xbetween x= 0 and

x= b, where 0 b /2.

a. Using right endpoints, find an expression for Aas a limit. Do not evaluate the limit.

b. Estimate the area for the case b=/2 by takingthe sample points to be midpoints and using four

subintervals.

Example 3


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AREA PROBLEM

Since a= 0, the width of a subinterval

is:

So, x1

= b/n, x2

= 2b/n, x3

= 3b/n, xi

= ib/n,xn = nb/n.

0

b bx

n n

Example 3 a


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AREA PROBLEM

The sum of the areas of the approximating

rectangles is:

1 2

1 2( ) ( ) ... ( )(cos ) (cos ) ... (cos )

2

cos cos ... cos

n

n nR f x x f x x f x x x x x x x x

b b b b nb b

n n n n n n

Example 3 a


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AREA PROBLEM

According to Definition 2, the area is:

Using sigma notation, we could write:

lim

2 3lim (cos cos cos ... cos )

nn

n

A R

b b b b nb

n n n n n

1

lim cos

n

ni

b ibAn n

Example 3 a


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AREA PROBLEM

With n= 4 and b= /2, we have:

x= (/2)/4 = /8

So, the subintervals are:

[0, /8], [/8, /4], [/4, 3/8], [3/8, /2]

The midpoints of these subintervals are:

x1* = /16 x2* = 3/16x3* = 5/16 x4* = 7/16

Example 3 b


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AREA PROBLEM

The sum of the areas

of the four rectangles is:

4

4

1

( *)

( /16) (3 /16)

(5 /16) (7 /16)

3 5 7cos cos cos cos16 8 16 8 16 8 16 8

3 5 7cos cos cos cos 1.006

8 16 16 16 16

ii

M f x x

f x f x

f x f x

Example 3 b



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AREA PROBLEM

So, an estimate

for the area is:

A 1.006

Example 3 b



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DISTANCE PROBLEM

Now, lets consider the distance problem:

Find the distance traveled by an object during

a certain time period if the velocity of the

object is known at all times.

In a sense, this is the inverse problem of the velocityproblem that we discussed in Section 2.1


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CONSTANT VELOCITY

If the velocity remains constant, then

the distance problem is easy to solve

by means of the formula

distance = velocity x time


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DISTANCE PROBLEM

Suppose the odometer on our car is

broken and we want to estimate the

distance driven over a 30-second time

interval.

Example 4


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DISTANCE PROBLEM

We take speedometer

readings every five

seconds and record

them in this table.

Example 4


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DISTANCE PROBLEM

If we add similar estimates for the other time

intervals, we obtain an estimate for the total

distance traveled:

(25 x 5) + (31 x 5) + (35 x 5)+ (43 x 5) + (47 x 5) + (46 x 5)

= 1135 ft

Example 4


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DISTANCE PROBLEM

We could just as well have used the velocity

at the end of each time period instead of

the velocity at the beginning as our assumed

constant velocity.

Then, our estimate becomes:(31 x 5) + (35 x 5) + (43 x 5)

+ (47 x 5) + (46 x 5) + (41 x 5)= 1215 ft

Example 4


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DISTANCE PROBLEM

The similarity is explained when we sketch

a graph of the velocity function of the car

and draw rectangles whose heights are

the initial velocities foreach time interval.


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DISTANCE PROBLEM

The area of the first rectangle is 25 x 5 = 125,

which is also our estimate for the distance

traveled in the first five seconds.

In fact, the area of eachrectangle can beinterpreted as a distance,because the height

represents velocity andthe width represents time.


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DISTANCE PROBLEM

The sum of the areas of the rectangles is

L6 = 1135, which is our initial estimate for the

total distance traveled.


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Areas and Distances Friday, January 29, 2010

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