MAT 21B – Lecture 1 – Area and Estimating with Finite Sums PDF

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MAT 21B – Lecture 1 – Area and Estimating with Finite Sums 

Previously from geometry, we know that the formula for the area of a rectangle is

b∗h , the area of a triangle is 



1 (b∗h) , and the area of a circle is 2

πr

2

.

There is a systematic way to estimate the area under a function by using finite sums of rectangles. Three ways to do this is by using the left endpoint rule, the right endpoint rule, and the midpoint rule. 2 Example: Estimate the area under the function, f ( x )=1−x on [0,1] by using two rectangles, or n=2 . o 1) Left endpoint rule: We observe that from the interval, 1−0=1 . To obtain rectangles of equal length, we can divide 1 by 2 to get two rectangles. Thus, the length for each rectangle is first rectangle that has height of height of

()

f 1 2

1 . Then we draw the 2

f ( 0) and the second rectangle has

. The sum of the areas of both rectangles is,

( ( ))

() ( ) ()

1 1 1 1 1 3 1 3 1 7 7 area ≈ ( f ( 0) ) + f = (1 )+ = 1+ = = =0.875 . 2 2 2 2 4 2 4 2 4 8 2

o

2) Right endpoint rule: The length of each rectangle is

1 . This time, 2

the rectangles are drawn from the right end of the interval such that the first rectangle will have a height of f ( 1 ) . The second rectangle has a

( 12 ) . Then 1 area ≈ f ( ) + ( f ( 1) ) = ( ) + ( 0 )= =0.375 (2 12 ) 12 21 34 12 38

height of

o

f

.

3) Midpoint rule: What remains the same is that each rectangle has a length of and

1 . We have two intervals within [0,1] which are: 2

1 [ 0, ] 2

1 [ , 1] . Then take the average of the endpoints for each interval 2

and divide that by 2. On

1 [ 0, ] , we have 2

1 1 −0 2 1 . This the 2 = = 2 4 2

value of x that will determine the height of that first rectangle. On

1 3 1− 1 [ , 1] , 2 = 2 3 . This x-value will be used in finding the height = 2 2 2 4 of second rectangle. Then

area ≈ .

( ( )) ( ( )) ( ) ( ) (

) ( )

1 1 1 15 1 7 1 15 7 1 22 11 1 3 + f = + = = = =0.6875 + f 4 4 2 2 2 16 2 16 2 16 16 2 16 16



When we have several intervals of equal length, we use the notation the length. For each of the examples above, we had

Using the left-endpoint rule, the



Or we can try

∆ x=

1 8

1 . We can try the 2

1 . 4 1 1 3 1 area ≈ (f ( 0 )+ f +f +f ) . 4 2 4 4

same rules for the same function, but with 

∆ x=

∆ x for

∆ x=

() () ()

in which case

() () () () () () ()

1 1 3 1 5 3 7 1 +f +f +f +f +f +f ) . area ≈ (f ( 0 ) +f 8 8 4 8 2 8 4 8

 

Using an upper sum always gives an overestimate of the area while using a lower sum gives an underestimate of the area. Example: Using an upper sum and a lower sum, estimate the area underneath the function, f ( x )=sin (πx) with four rectangles. o

1) First, we find that

∆ x=

1 . To obtain an upper sum, find the x-value 4

on the endpoint that yields the highest value for height within each of the four intervals representing the length of the rectangle. The length of first rectangle can be represented with interval

f

( 14 )> f (0)

rectangle. On

and so we use

f

( 14 )

()

1 1 1 1 >f ( ) [ , ] , f 4 2 2 4

of the second rectangle. Similarly, on

1 [ 0, ] . Here, 4

as the height of that first and thus,

1 3 [ , ] , 2 4

( 12 ) is the height 1 3 f ( ) > f ( ) therefore 4 2 f

( 12 ) is the height for the third rectangle. Lastly, on [ 43 , 1] , 3 3 f ( )> f (1) and so f ( ) is used as the height of the fourth 4 4 1 1 1 1 3 rectangle. Then, area ≈ (f ( ) +f ( )+f ( ) +f ( ) ) . 4 4 2 2 4 f

o

2) To obtain a lower sum, find the x-value on the endpoint that yields the lowest value for height within each of the four intervals. On interval

()

1 1 [ 0, ] , f ( 0)...