5.1 Areas and Distances 1 Dr. Erickson. 5.1 Areas and Distances2 time velocity After 4 seconds, the...

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Chapter 5 – Integrals 5.1 Areas and Distances 1 5.1 Areas and Distances Dr. Erickson

Transcript of 5.1 Areas and Distances 1 Dr. Erickson. 5.1 Areas and Distances2 time velocity After 4 seconds, the...

Page 1: 5.1 Areas and Distances 1 Dr. Erickson. 5.1 Areas and Distances2 time velocity After 4 seconds, the object has gone 12 feet. Since rate. time = distance:

1

Chapter 5 – Integrals

5.1 Areas and Distances

5.1 Areas and Distances Dr. Erickson

Page 2: 5.1 Areas and Distances 1 Dr. Erickson. 5.1 Areas and Distances2 time velocity After 4 seconds, the object has gone 12 feet. Since rate. time = distance:

5.1 Areas and Distances2

Consider an object moving at a constant rate of 3 ft/sec.

time

velocity

After 4 seconds, the object has gone 12 feet.

Since rate . time = distance:

If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.

ft3 4 sec 12 ft

sec

3t d

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Page 3: 5.1 Areas and Distances 1 Dr. Erickson. 5.1 Areas and Distances2 time velocity After 4 seconds, the object has gone 12 feet. Since rate. time = distance:

5.1 Areas and Distances3

Velocity is not Constant

If the velocity is not constant,we might guess that the distance traveled is still equalto the area under the curve.

(The units work out.)

211

8V t Example:

We could estimate the area under the curve by drawing rectangles touching at their left corners.

This is called the Left-hand Rectangular Approximation Method (L).

1 11

8

11

2

12

8t v

10

1 11

8

2 11

2

3 12

8

Approximate area: 1 1 1 31 1 1 2 5 5.75

8 2 8 4

4 0

41

b ax

n

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Page 4: 5.1 Areas and Distances 1 Dr. Erickson. 5.1 Areas and Distances2 time velocity After 4 seconds, the object has gone 12 feet. Since rate. time = distance:

5.1 Areas and Distances4

Right-hand Rectangular Approximation Method (R).

11

8

11

2

12

8

Approximate area: 1 1 1 31 1 2 3 7 7.75

8 2 8 4

3

211

8V t

4 0

41

b ax

n

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Page 5: 5.1 Areas and Distances 1 Dr. Erickson. 5.1 Areas and Distances2 time velocity After 4 seconds, the object has gone 12 feet. Since rate. time = distance:

5.1 Areas and Distances5

The area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles:

Definition #2 – Using Right End Points

1 2

1 2

lim

lim ( ) ( ) ... ( )

lim ( ) ( ) ... ( )

nn

nn

nn

A R

f x x f x x f x x

f x f x f x x

b ax

n

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Page 6: 5.1 Areas and Distances 1 Dr. Erickson. 5.1 Areas and Distances2 time velocity After 4 seconds, the object has gone 12 feet. Since rate. time = distance:

5.1 Areas and Distances6

We get the same values if we use left endpoints.

Definition Using Left Endpoints

0 1 1

0 1 1

lim

lim ( ) ( ) ... ( )

lim ( ) ( ) ... ( )

nn

nn

nn

A L

f x x f x x f x x

f x f x f x x

b ax

n

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Page 7: 5.1 Areas and Distances 1 Dr. Erickson. 5.1 Areas and Distances2 time velocity After 4 seconds, the object has gone 12 feet. Since rate. time = distance:

5.1 Areas and Distances7

Midpoint Rectangular Approximation Method (M)

Another approach would be to use rectangles that touch at the midpoint. This is the Midpoint Rectangular Approximation Method.

1.031251.28125

1.78125

Approximate area:

6.625

2.53125

t v

1.031250.5

1.5 1.28125

2.5 1.78125

3.5 2.53125

In this example there are four subintervals.As the number of subintervals increases, so does the accuracy.

211

8V t

4 0

41

b ax

n

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Page 8: 5.1 Areas and Distances 1 Dr. Erickson. 5.1 Areas and Distances2 time velocity After 4 seconds, the object has gone 12 feet. Since rate. time = distance:

5.1 Areas and Distances8

211

8V t

Approximate area:

6.65624

t v

1.007810.25

0.75 1.07031

1.25 1.19531

1.382811.75

2.25

2.75

3.25

3.75

1.63281

1.94531

2.32031

2.75781

13.31248 0.5 6.65624

width of subinterval

With 8 subintervals:

The exact answer for thisproblem is .6.6

4 0

80.5

b ax

n

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Page 9: 5.1 Areas and Distances 1 Dr. Erickson. 5.1 Areas and Distances2 time velocity After 4 seconds, the object has gone 12 feet. Since rate. time = distance:

5.1 Areas and Distances9

We can consider the midpoint to be sample points ( ) so we have the formula:

Definition Using Midpoints

1 2

1 2

* * *

* * *

lim

lim ( ) ( ) ... ( )

lim ( ) ( ) ... ( )

n

n

nn

n

n

A M

f x x f x x f x x

f x f x f x x

1

* * *2, ,..., nx x x

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Page 10: 5.1 Areas and Distances 1 Dr. Erickson. 5.1 Areas and Distances2 time velocity After 4 seconds, the object has gone 12 feet. Since rate. time = distance:

5.1 Areas and Distances10

We often use sigma notation to write the sums with many terms more compactly. For instance,

Sigma Notation

1 1 21

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

nn

i

f x x f x f x f x x

1

11

*

1

lim lim ( )

lim lim ( )

lim lim ( )i

n

n in n

i

n

n in n

i

n

nn n

i

A R f x x

A L f x x

A M f x x

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Page 11: 5.1 Areas and Distances 1 Dr. Erickson. 5.1 Areas and Distances2 time velocity After 4 seconds, the object has gone 12 feet. Since rate. time = distance:

5.1 Areas and Distances11

a) Estimate the area under the graph from x=0 to x=4 using four approximating rectangles and right endpoints. Sketch the graph and the rectangles. Is your estimate an underestimate or overestimate?

b) Repeat part (a) using left endpoints.

Example 1

( )f x x

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Page 12: 5.1 Areas and Distances 1 Dr. Erickson. 5.1 Areas and Distances2 time velocity After 4 seconds, the object has gone 12 feet. Since rate. time = distance:

5.1 Areas and Distances12

a) Graph the function.

b) Estimate the area under the graph from x=-2 to x=2 using four approximating rectangles and taking sample points to be (i) right endpoints and (ii) midpoints. In each case, sketch the graph and the rectangles.

c) Improve your estimates in part (b) using 8 rectangles.

Example 2

2

( ) , 2 2xf x e x

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Page 13: 5.1 Areas and Distances 1 Dr. Erickson. 5.1 Areas and Distances2 time velocity After 4 seconds, the object has gone 12 feet. Since rate. time = distance:

5.1 Areas and Distances13

Oil leaked from a tank at a rate of r(t) liters per hour. The rate decreased as time passed and values of the rate at two-hour time intervals are shown in the table. Find the lower and upper estimates for the total amount of oil that leaked out.

Example 3

t(h) 0 2 4 6 8 10

r(t) (L/h) 8.7 7.6 6.8 6.2 5.7 5.3

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Page 14: 5.1 Areas and Distances 1 Dr. Erickson. 5.1 Areas and Distances2 time velocity After 4 seconds, the object has gone 12 feet. Since rate. time = distance:

5.1 Areas and Distances14

Use the definition (#2 according to your book) to find an expression for the area under the graph of f as a limit. Do not evaluate the limit.

a)

b)

Example 4

4( ) , 1 16f x x x

( ) cos , 0 2f x x x x

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Page 15: 5.1 Areas and Distances 1 Dr. Erickson. 5.1 Areas and Distances2 time velocity After 4 seconds, the object has gone 12 feet. Since rate. time = distance:

5.1 Areas and Distances15

Determine a region whose area is equal to the given limit. Do not evaluate the limit.

Example 5

1

lim tan4 4

n

ni

i

n n

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