Drill
• A train travels at 80 mph for 5
hours. How far does it travel?
• A truck travels for an average of
48 mph for 3 hours. How far does
the truck travel?
• Beginning at a standstill, a car
maintains a constant acceleration
of 10 ft/s2 for 10 seconds. What is
its velocity after 10 seconds? Give
your answer in ft/sec and then
convert to mi/hr.
• 80(5) = 400 miles
• 48 (3) = 144 miles
• a(t) = 10
– v(t) = 10t + C, v(0) =
0, so v(t) = 10t
– v(10) = 100 ft/s
100
ft
sec

60 sec
1 min
68 . 18 mph

60 min
1hour

1mile
5280 ft

Estimating Finite Sums
Lesson 5.1
Objectives
• Students will be able to
– approximate the area under the graph of a
nonnegative continuous function by using
rectangle approximation methods.
– interpret the area under a graph as a net
accumulation of a rate of change.
Distance Traveled
• Figure 1: A train moves along a
track at a steady rate of 75 mph
from 7 AM to 9 AM. What is
the total distance traveled by
the train?
• Can be represented by
determining the area of the
rectangle above.
• The distance traveled
would be 75 mph (2
hours ) = 150 miles
• Looking at the
rectangle, A = length x
width = 150 miles
A car accelerates from 0 to 88 feet in 11 seconds. If the
acceleration is CONSTANT, how far will the car have traveled in
that time?
• Solution: Graph and
find the area of the
shape.
• Area = ½ base (height)
• = ½ (11)(88)
• = 484 feet
v(t) = 800 – 32t. Find the distance traveled
between 8 and 20 seconds.
• Solution: Graph and
find the area of the
shape.
• The shape is a
trapezoid.
• A = ½ h (b1 + b1 )
• h = 20 -8 = 12
• b1 = v(8) =544
• b2 = v(20) = 160
• A = ½ (12)(544 + 160)
• = 4224 units
Distance Traveled
• If the velocity varies
over the timer interval
[a,b], does the shaded
region give the distance
traveled?
• Yes, but because it is an
irregular region, we would
need to partition it into
narrow rectangles and add
the areas of each rectangle.
Please note: the smaller the
width of the rectangle, the
more accurate the area.
Example 1 Finding Distance Traveled
when Velocity Varies
A particle starts at x = 0 and moves along the x-axis with velocity
v(t) = 3t2 for time t > 0. Where is the particle at t = 6?
Solution: Graph the function of v and partition the time interval into
subintervals of length Δt.
We will define Δt to be 1, which means the base of each smaller
rectangle will be 1.
The height of each rectangle will be v(t), where t is represented by
some point within each interval along the x-axis.
Our intervals will be broken down as follows:
0
1
2
3
4
5
6
Choosing values of t:
• When we use the left-handed endpoints, the estimate is usually less than
the actual area under the curve
– In this case, the left-hand endpoints would be 0, 1, 2, 3, 4, 5, and to
find the height, we would find v(0), v(1), v(2), etc.
• When we use the right-handed endpoints, the estimate is usually more
than the actual area under the curve
– In this case, the left-hand endpoints would be 1, 2, 3, 4, 5, 6 and to
find the height, we would find v(1), v(2), v(3), etc.
• When we use the mid-points of the intervals, the estimate is usually most
accurate.
– In this case, the mid-points would be .5, 1.5, 2.5, 3.5, 4.5 and 5.5, and
to find the height, we would find v(.5), v(1.5), v(2.5), etc
Example 1 Finding Distance Traveled
when Velocity Varies
A particle starts at x = 0 and moves along the x-axis with
velocity v(t) = 3t2 for time t > 0. Where is the particle at
t = 6?
Left-Side Rectangles
1   v 0   v 1   v  2   v 3   v  4   v 5 
1   0  3  12  27  48  75 
Velocity
1  v 0   1  v 1   1  v  2   1  v 3   1  v  4   1  v 5 
165
Time
Example 1 Finding Distance Traveled
when Velocity Varies
A particle starts at x = 0 and moves along the x-axis with
velocity v(t) = 3t2 for time t > 0. Where is the particle at
t = 6?
Right-Side Rectangles
1   v 1   v  2   v 3   v  4   v 5   v 6 
1  3  12  27  48  75  108

Velocity
1  v 1   1  v  2   1  v 3   1  v  4   1  v 5   1  v 6 
273
Time
Example 1 Finding Distance Traveled
when Velocity Varies
A particle starts at x = 0 and moves along the x-axis with
velocity v(t) = 3t2 for time t > 0. Where is the particle at
t = 6?
Midpoint Rectangles
Velocity
1  v  0 . 5   1  v 1 . 5   1  v  2 . 5   1  v 3 . 5   1  v  4 . 5   1  v 5 . 5 
1  v  0 . 5   v 1 . 5   v  2 . 5   v 3 . 5   v  4 . 5   v 5 . 5 
1  0 . 75  6 . 75  18 . 75  36 . 75  60 . 75  90 . 75 
214 . 5
Time
Homework, day 1
• Page 270: 1-6
– Please note the following for 5 and 6
– LRAM: use the left endpoints
– RRAM: use the right endpoints
– MRAM: use the mid-points
Drill:
Find the area under the curve listed below, [0, 4].
Δx = 1 Determine using LRAM, RRAM and MRAM
V 
1
8
t 1
2
3
We could estimate the
area under the curve
by drawing rectangles
touching at their left
corners.
2
1
0
1
1
2
1
1
8
Approximate area:
11
1
8
1
1
2
2
1
8
5
3
4
 5.75
3
1
4
1
2
2
8
t
v
0
1
1
1
8
1
1
2
1
2
3
1
2
1
8

V 
1
t 1
2
3
8
2
1
0
1
1
1
2
1
8
1
3
2
1
8
2
4
3
We could also use a Right-hand Rectangular Approximation Method (RRAM).
Approximate area:
1
1
8
1
1
2
2
1
8
37
3
 7.75
4

t
V 
v
0 .5
1 .0 3 1 2 5
1 .5
1 .2 8 1 2 5
2 .5
1 .7 8 1 2 5
3 .5
2 .5 3 1 2 5
1
t 1
2
3
8
2
1
0
1
2
1 .0 3 1 2 5
1 .7 8 1 2 5
1 .2 8 1 2 5
3
4
2 .5 3 1 2 5
Another approach would be to use rectangles that touch at the midpoint. This is
the Midpoint Rectangular Approximation Method (MRAM).
Approximate area:
In this example there are four subintervals.
As the number of subintervals increases, so
does the accuracy.
6 .6 2 5

Example 2 Finding Information
Given Data
People who sail ships at sea use dead reckoning to calculate
the distance a ship has gone. (The term dead reckoning comes
from “ded-reckoning,” which is short for “deduced
reckoning.”) Suppose that a ship is maneuvering by changing
speed rapidly. The table shows its speeds at 2-min intervals.
(A knot – abbreviated kn – is a nautical mile per hour. A
nautical mile is about 2000 yd.)
Use this information to find the distance the ship traveled in
the 12-min time interval.
Time
(min)
0
2
4
6
8
10
12
Speed
(kn)
33
25
27
13
21
5
9
Example 2 Finding Information
Given Data
Time
(min)
Speed
(kn)
0
2
4
6
8
10
12
33
25
27
13
21
5
9
 27  
2
Left-Side Rectangles
2
60
2
60
33  
2
60
 25  
2
60
60
13  
33  25  27  13  21  5 
4 . 13 nautical
miles
2
60
21  
2
60
5 
Example 2 Finding Information
Given Data
Time
(min)
Speed
(kn)
0
2
4
6
8
10
12
33
25
27
13
21
5
9
Right-Side Rectangles
2
60
2
60
25  
2
60
 27  
2
60
13  
2
60
 25  27  13  21  5  9 
3 . 33 nautical
miles
21  
2
60
5  
2
60
9 
Example 3
• Dye Concentration Data (graph on p. 269)
Secon
ds
after
inject
ion5
5
41
9
11
13
15
17
19
21
23
25
27
29
31
Dye
conce
ntrati
on
0
3.8
8
6.1
3.6
2.3
1.45
.91
.57
.36
.23
.14
.09
0
• Estimate the cardiac output of the patient
whose data appears above.
• Cardiac output = amount of dye/area under
the curve
• Assume our patient’s amount of dye = 5.6 mg
Cardiac Output
To compute MRAM, we would actually need to draw the
rectangles and determine the height of each rectangle.
LRAM
• 2( 0 + 3.8 + 8 + 6.1 + 3.6 +
2.3 + 1.45 + .91 + .57 + .36 +
.23 + .14 + .09)
• 2(27.55) = 55.1 mg/L · sec
RRAM
• 2(3.8 + 8 + 6.1 + 3.6 + 2.3 +
1.45 + .91 + .57 + .36 + .23 +
.14 + .09 + 0)
• 2(27.55) = 55.1 mg/L · sec
5.6 mg/ 55.1 mg/L · sec = .102 L/sec
.102 L/sec X 60 sec/1 min = 6.10 L/minute
Homework
• Page 270: 15-19
• On #15, use MRAM, meaning you will have to
graph the rectangles and determine heights.