Area and the Definite Integral Tidewater Community College Mr. Joyner, Dr. Julia Arnold and Ms. Shirley Brown using Tans 5th edition Applied Calculus for

Area and the Definite Integral

Tidewater Community CollegeMr. Joyner, Dr. Julia Arnold and

Ms. Shirley Brownusing Tan’s 5th edition Applied Calculus for the managerial , life, and social sciences text


Suppose a state’s annual rate of oil consumption each year over a 4-year period is a constant function, such as f(t) = 2, where t is measured in years and f(t) is measured in millions of barrels. We would graph such a function as:

3

2

1

-1

2 4 6 t

t = 4

years

f(t)

f(t) = 2


How could you use the graph to answer the question, “How much oil was used during the four years”?

Answer:2(4-0) = 8 million barrels

It seems reasonable that the answer is 8 million barrels and this is the same as the area of the rectangle…the area under the “curve.” (Yes, even a straight line is considered a “curve” in mathematics!)


How would you find the area above the x-axis and under the curve f(x) = x from x = 0 to x = 4?

Does it seem reasonable to find the area of the triangle… the area under the “curve.”

6

5

4

3

2

1

-1

-2 2 4 6 8 10

x

f(x)

f x = x


Imagine now that our function, f(x) = x, represents the oil consumption of the same state over the same four years.

6

5

4

3

2

1

-1

-2 2 4 6 8 10

x

f(x)

f x = x

What was the oil consumption over the four years?

Let’s explore.

The average rate of change over the first year is (1 – 0)/2 = 0.5 million barrels per year.


For the first year, the state started off consuming no oil (zero barrels per year), and ended the year consuming oil at the rate of 1 million barrels per year.

Since the rate of increased consumption is linear, we can use the average rate of change over the year to represent the entire first year.

For one year, (the first year) the amount of oil consumed is: 1(0.5) = 0.5 million barrels

The average rate of change over the 2-year period is (2 – 0)/2 = 1 million barrels per year.


For the first two years, the state started off consuming no oil and ended the two-year year period consuming oil at the rate of 2 million barrels per year.

Again, since the rate of increased consumption is linear, we can use the average rate of change over the 2 years to represent the entire two-year period.

For two years. the amount of oil consumed is: 2(1) = 2 million barrels.


6

5

4

3

2

1

-1

-2 2 4 6 8 10

x

f(x)

f x = x

In a similar fashion we can reason that the total oil consumption is as follows:

Total Number of Years

Average Rate of Consumption

Total oil consumption

1 .5 1(.5) = .5

2 1 2(1) = 2

3 1.5 3(1.5) = 4.5

4 2 4(2) = 8

Lets compare these values to…the areas of the right triangles under the curve f(x).


Triangle areas:

Base(total number of years)

Height(average rate of consumption)

Area = ½(bh)(total oil consumption)

1 1 ½(1)(1) = 0.5

2 2 ½(2)(2) = 2

3 3 ½(3)(3) = 4.5

4 4 ½(4)(4) = 8

We get the same values!


How would we find the area under the following curve and above the x-axis?

The work is a little more complex, but the approach is the same…

area under the curve!

f(x)

x

2 on [0,2]( ) 1f x x

We could use rectangles to approximate the area.

First, divide the horizontal base with a partition. We will start with a partition that has two equal subintervals: [0,1] and [1,2]

2 on [0,2]( ) 1f x x f(x)

x


In this slide, the subinterval rectangles have been drawn using the right-hand endpoints of each subinterval. We could have used the left endpoints, the midpoints of the intervals, or any point in each subinterval.2 on [0,2]( ) 1f x x

0 1 21unit 1unit

22 +1 = 5

12 + 1 = 2


2 on [0,2]( ) 1f x x

0 1 21

5

2

1

The area of the first rectangle is 2 and is greater than the area under the curve. The area of the

second rectangle is 5 and is also greater than the area under the curve.

The total area of the rectangles is 2 + 5 = 7 which is greater than the desired area under the curve.


In this slide, let’s draw the rectangles using the left-hand endpoints.

0 1 21unit 1unit

12 +1 = 2

02 + 1 = 1

2 on [0,2]( ) 1f x x


The area of the first rectangle is 1 x 1 = 1 and is less than the area under the curve.

0 1 21unit 1unit

2 1

The area of the second rectangle is 1 x 2 = 2 and is also less than the area under the curve.

The total area of the rectangles using the left-hand endpoints is 1 + 2 = 3 which is less than the desired area under the curve.

Now, let’s calculate the areas of these new rectangles.

2 on [0,2]( ) 1f x x


Now we know that the area under the curve must be between 3 and 7. Can we get a more accurate answer? Yes! And to do so, we need to construct more subintervals using a finer partition.

2 on [0,2]( ) 1f x x


Area Under A Curve

The area under a curve on an interval can be approximated by summing the areas of individual rectangles on the interval.

If the rectangles are circumscribed above the curve, the area will be greater than the desired area.

(In our example, this area was 7.)

If the rectangles are inscribed under the curve, the approximated area will be less than the desired area.

(In our example, this area was 3.)


By using a finer partition (one with more subintervals), we make each rectangle narrower (increasing their number), and we get an area value that is closer to the true area under the curve.

By taking the limit as n (the number of rectangles) approaches infinity, the actual area is approached.

Area Under A Curve


Definition: A sum such as the one below is called a Riemann sum:

1 2 3( ) ( ) ( ) ... ( ) where nf x f x f x f x x b axn

Area Under A Curve


Let f(x) be a nonnegative, continuous function on the closed interval [a, b]. Then, the area of the region under the graph of f(x) is given by

where x1, x2, x3, … xn are arbitrary points in the n

subintervals of [a,b] of equal width .

1 2 3lim ( ) ( ) ( ) ... ( )nnA f x f x f x f x x GGGGGGGGGGGGGG

where b ax xn

Area Under A Curve

The area under a curve can be approached by taking an infinite Riemann sum.


On the following six slides, you will see the number of rectangles increasing for the graph of y = x3 on the interval [-1, 2] . Observe the way that the area of rectangles more closely approximates the area under the curve as the number of rectangles increases.


When there are 10 rectangles, Area 2.4675

20 rectangles, Area 3.091875





As you can see, the sum of the areas of the rectangles increases as you increase the number of rectangles.

As the number of rectangles increases, they begin to better “fit” the curve giving a closer and closer approximation of the area (3.75) under the curve and above the x-axis.


We are almost done with this lesson in that we can now turn our attention to the definition and discussion of the definite integral for all functions, not just nonnegative functions.


Let f(x) be defined on [a,b]. If

exists for all choices of representative points x1, x2, x3, …

xn in the n subintervals of [a,b] of equal

width , then this limit is called the definite

integral of f(x) from a to b and is denoted by

.

xxfxfxfxf nn

)(...)()()(lim 321

b axn

b

a

dxxf )(

The number a is the lower limit of integration, and the number b is the upper limit of integration.

Definition: The Definite Integral


1 2 3( ) lim ( ) ( ) ( ) ... ( )b

n na

f x dx f x f x f x f x x GGGGGGGGGGGGGG

Definition: The Definite Integral

Thus

As long as f(x) is a continuous function on a closed interval, it has a definite integral on that interval. f(x) is said to be integrable when its integral exists.


The definite integral of a function is a number.

example:

The indefinite integral of a function is another

function. example:

The Definite Integral

Please make this important distinction between the indefinite integral of a function and the definite integral of a function:

2

2

xxdx C

3

1

4x dx You will learn how to calculate this number in the next section of your textbook.


If f(x) is a nonnegative, continuous function on [a,

b], then is equal to the area of the

region under the graph of f(x) on [a, b].


Let’s take a quick look at the geometric interpretation of the definite integral, and we’ll be done.

( )b

af x dx


If f(x) is a nonnegative, continuous function on [a, b],

then is equal to the area of the region

under the graph of f(x) on [a, b].


( )b

af x dx

f(x)

y

x

( )b

aA f x dx

a b

What happens if f(x) is

not always nonnegative?


If f(x) is simply a continuous function on [a, b],

then is equal to the area of the region below

the graph of f(x) and above the x-axis minus the area of

the region above the graph of f(x) and below the x-axis on

[a, b].


( )b

af x dx

( )b

af x dx f(x)

y

xa b

1R

2R

3R

Area of region 1 –

Area of region 2 +

Area of region 3


Let’s look at a problem similar to #3 of the section 6.3 Exercises (page 475) in the textbook. We shall use right endpoints.

Let f(x) = 3x.

Part a: Sketch the region R under the graph of f(x) on the interval [0,2] and find its exact area using

geometry.




Solution - Part a: Since the area under the curve from [0,2] forms a right triangle, we can calculate the exact area using the geometry formula A= ½ bh ; b = 2 and h = 6, so ½ (2)(6) = 6

Area = 6

f(x) = 3x

0 21

6

4

2


Let f(x) = 3x.

Part b: Use a Riemann sum with four subintervals (n = 4) of

equal length to approximate the area of region R.

Choose the representative points to be the right

endpoints of the subintervals.



1 21/2 3/2

f(x) = 3x

Solution – Part b: Divide the length of the

interval (2) by the number of subintervals

(4). Each subinterval has a length equal to

½ unit;

. This produces the 4

subintervals:

[0, 1/2], [1/2, 1], [1, 3/2], and [3/2,

2] .

21

x

The Riemann Sum isA = [f(1/2) + f(1) + f(3/2) + f(2)](1/2)

1 3 13 3(1) 3 3(2)

2 2 2

3 9 1 12 1 153 6 9

2 2 2 2 2 2

A

A


Let f(x) = 3x.

Part c: Repeat part (b) with eight subintervals (n = 8) ofequal length.



Solution – Part c: Divide the length of the

interval (2) by the number of subintervals

(8). Each subinterval has a length equal to

1/4 unit;

. This produces the 8

subintervals:

The Riemann Sum isA = [f(1/4) + f(1/2) + f(3/4) + f(1) + f(5/4) + f(3/2) + f(7/4) + f(2)](1/4) 1 1 3 5 3 7 1

3 3( ) 3 3(1) 3 3 3 3(2)4 2 4 4 2 4 4

3 3 9 15 9 21 1 72 1 108 273 6 9

4 2 4 4 2 4 4 4 4 16 4

A

A

1

4x

[0, 1/4], [1/4, 1/2], [1/2, 3/4], [3/4, 1] .

[1, 5/4], [5/4, 3/2], [3/2, 7/4], and [7/4, 2]


Let f(x) = 3x.

Part d: Compare the approximations obtained in parts (b) and

(c) with the exact area found in part (a). Do the

approximations improve with larger n ?


Solution – Part d: Compare the values. What is you

conclusion? Hopefully, you see that the approximation

does improve with a finer partition (and thus, larger

values of n).


Documents

Area and the Definite Integral Tidewater Community College Mr. Joyner, Dr. Julia Arnold and Ms. Shirley Brown using Tans 5th edition Applied Calculus for