Area of a Plane Region

• We know how to find the area inside many geometric shapes, like rectangles and triangles.

• We will now consider finding the area under a curve.

• The following example will be instructive in developing a general procedure.

Area of a Plane Region

• Find the approximate area of the region bounded by the graphs of:

2

0

( )

1

3

y

f x x

x

x

0y 10

8

6

4

2

2

321

0y

2( )f x x

1x

3x

10

8

6

4

2

2

321

10

8

6

4

2

2

321

• The area we are trying to find is enclosed by the four curves.

10

8

6

4

2

321

• We begin by dividing the interval from x = 1 to x = 3 into 4 equal subintervals.

• Each of these subintervals are 0.5 units wide.

Length of Interval

Number of Subintervals

3 1

4

20.5

4

• In general, if there are n equal subintervals from x = a to x = b, the width of each will be

b ax

n

• In general, where i is the ith subinterval, the left endpoint of each interval can be determined by:

( 1)im a x i

• Using the formula in the current example, the left end points are:

( 1)im a x i

1 1 0.5(1 1) 1 0.5(0) 1m

2 1 0.5(2 1) 1 0.5(1) 1.5m

3 1 0.5(3 1) 1 0.5(2) 2m

4 1 0.5(4 1) 1 0.5(3) 2.5m

4 equal subintervals

10

8

6

4

2

21.5 2.51 3

10

8

6

4

2

21.5 2.51 3

• Draw a rectangle in each subinterval, with the left side of the rectangle touching the curve.

• Then find the height of each rectangle.

10

8

6

4

2

21.5 2.51 3

• To do this, use the left endpoint of each interval in the function

2(1) 1 1f

1,1

2(1.5) 1.5 2.25f 1.5,2.25

2(2) 2 4f

2,4

2(2.5) 2.5 6.25f

2.5,6.252( )f x x

10

8

6

4

2

21.5 2.51 3

• Find the area of each rectangle.

1,1

1.5,2.25

2,4

2.5,6.25

• This will be accomplished by multiplying the height (function value) times the width (always 0.5 in this example).

10

8

6

4

2

21.5 2.51 3

1,1

1.5,2.25

2,4

2.5,6.25 1 0.5 0.5

2.25 0.5 1.125

4 0.5 2

6.25 0.5 3.125

• In general, to find the area of the ith rectangle with left endpoints, use the following:

( )i iA f m x

10

8

6

4

2

21.5 2.51 3

• Find the total area of all the rectangles:

1,1

1.5,2.25

2,4

2.5,6.250.5 1.125 2 3.125

6.75 sq units

10

8

6

4

2

21.5 2.51 3

• Using sigma notation, the sum can be written as:

1,1

1.5,2.25

2,4

2.5,6.25

6.75 sq units

4

1

(4) ( )ii

s f m x

10

8

6

4

2

21.5 2.51 3

• The sum that we just found is called a Lower Sum since the rectangles are inscribed rectangles (all of them were below the curve).

1,1

1.5,2.25

2,4

2.5,6.25

• In general, to find the sum of the areas of all the rectangles using left endpoints, use the following:

1

( ) ( )n

ii

s n f m x

( 1)im a x i

b ax

n

• Width of Intervals:

b ax

n

Summary

3 1

4

0.5

Summary

• Left endpoint of the ith subinterval:

( 1)im a x i

1 0.5( 1)i

Summary

• Area of the ith rectangle using left endpoints:

( )i iA f m x

( ( 1))f a x i x

(1 0.5( 1))0.5f i

Summary

• Total area of inscribed rectangles using left endpoints:

1

( ) ( )n

ii

s n f m x

4

1

1 0.5 1 (0.5)i

f i

10

8

6

4

2

21.5 2.51 3

10

8

6

4

2

321

• Note that the area found using the rectangles is just an approximation of the actual area we wanted.

10

8

6

4

2

21.5 2.51 3

10

8

6

4

2

321

• Since the area found is less than the actual area, let’s repeat the process, only this time using the right endpoints.

• The width of each subinterval will be the same as before:

• Each of the subintervals are 0.5 units wide.

3 10.5

4

b ax

n

• In general, where i is the ith subinterval, the right endpoint of each interval can be determined by:

iM a x i

• Using the formula in the current example, the right end points are:

iM a x i

1 1 0.5(1) 1.5M

2 1 0.5(2) 2M

3 1 0.5(3) 2.5M

4 1 0.5(4) 3M

• Draw a rectangle in each subinterval, with the right side of the rectangle touching the curve.

• Then find the height of each rectangle.

10

8

6

4

2

21.5 2.51 3

10

8

6

4

2

21.5 2.51 3

• To do this, use the right endpoint of each interval in the function

2(1.5) 1.5 2.25f 1.5,2.25

2(2) 2 4f

2,4

2(2.5) 2.5 6.25f

2.5,6.252( )f x x

2(3) 3 9f

3,9

10

8

6

4

2

21.5 2.51 3

1.5,2.25

2,4

2.5,6.25

3,9• Find the area of each rectangle.

• This will be accomplished by multiplying the height (function value) times the width (always 0.5 in this example).

10

8

6

4

2

21.5 2.51 3

1.5,2.25

2,4

2.5,6.25

3,9

2.25 0.5 1.125

4 0.5 2

6.25 0.5 3.125

9 0.5 4.5

• In general, to find the area of the ith rectangle with right endpoints, use the following:

( )i iA f M x

10

8

6

4

2

21.5 2.51 3

1.5,2.25

2,4

2.5,6.25

3,9• Find the total area of all the rectangles:

1.125 2 3.125 4.5

10.75 sq units

10

8

6

4

2

21.5 2.51 3

1.5,2.25

2,4

2.5,6.25

3,9• Using sigma notation, the sum can be written as:

10.75 sq units

4

1

(4) ( )ii

S f M x

10

8

6

4

2

21.5 2.51 3

1.5,2.25

2,4

2.5,6.25

3,9• The sum that we just found is called a Upper Sum since the rectangles are circumscribed rectangles (all of the tops of the rectangles are above the curve).

• In general, to find the sum of the areas of all the rectangles using right endpoints, use the following:

1

( ) ( )n

ii

S n f M x

iM a x i

b ax

n

• Width of Intervals:

b ax

n

Summary

3 1

4

0.5

Summary

• Right endpoint of the ith subinterval:

iM a x i

1 0.5 i

Summary

• Area of the ith rectangle using right endpoints:

( )i iA f M x

( )f a x i x

(1 0.5 )0.5f i

Summary

• Total area of circumscribed rectangles using right endpoints:

1

( ) ( )n

ii

S n f M x

4

1

1 0.5 (0.5)i

f i

• Rather than calculating the area of each rectangle and finding the sum, we can use the formulas.

1

( ) ( )n

ii

S n f M x

4

1

1 0.5 (0.5)i

f i

4

2

1

1 0.5 (0.5)i

i

4

2

1

1 0.5 (0.5)i

i

4

2

1

0.5 1 0.25i

i i

4 5 4 5 90.5 4 0.25

2 6

0.5 4 10 7.5

0.5 4 10 7.5

0.5 21.5

10.75

• Note that this is the same value found earlier in calculating the sum of the areas of the circumscribed rectangles.

10

8

6

4

2

321

10

8

6

4

2

21.5 2.51 3

• Once again the area found using the rectangles is just an approximation of the actual area we wanted.

10

8

6

4

2

321

10

8

6

4

2

21.5 2.51 3

• In this case the approximation turns out to be larger than the actual area.

Area

• Left Endpoints • Right Endpoints

1

( ) ( )

10.75

n

ii

S n f M x

1

( ) ( )

6.75

n

ii

s n f m x

6.75 Actual Area 10.75

Conclusion• We know the actual area is between 6.75 sq

units and 10.75 sq units. This isn’t very “close”. How do we get a better estimate?

• There are two possibilities:

1. Use rectangles that are closer to estimating the area. In the current example, using the midpoint of the interval would give a better estimate.

2. Use more rectangles. It can be shown that as the number of rectangles approaches infinity, the area will be exact.

Definition of the Area of a Region in the Plane

• Let function f be continuous and nonnegative on the interval [a,b]

• The area of the region bounded by the graph of f, the x-axis, and the vertical lines x = a and x = b is given by:

1

Area lim ( )n

in

i

f c x

1where and i i i

b ax c x x

n

Definition of the Area of a Region in the Plane

1 i i ix c x

b a

xn

• This is stating that ci can be any point in the interval, including the left or right endpoints.

• This is the width of each interval.

Definition of the Area of a Region in the Plane

1

Area lim ( )n

in

i

f c x

Area of Rectangle

(height times width)

Add up all the areas of all the rectangles

Let the number

of rectangles approach infinity