Transcript Calculus - integrals as net change

Greg Kelly, Hanford High School, Richland, Washington
A honey bee makes several trips from the hive to a flower
garden. The velocity graph is shown below.
What is the total distance traveled by the bee?
200  200  200 100  700
700 feet
100
ft
min 50
0
200ft
200ft
2
4
6
8
10
minutes
-50
200ft
100ft
-100

What is the displacement of the bee?
200  200  200 100  100
100 feet towards the hive
100
ft
min 50
0
200ft
200ft
2
4
6
8
10
minutes
-50
-200ft
-100ft
-100

To find the displacement (position shift) from the velocity
function, we just integrate the function. The negative
areas below the x-axis subtract from the total
displacement.
Displacement   V  t  dt
b
a
To find distance traveled we have to use absolute value.
Distance Traveled   V  t  dt
b
a
Find the roots of the velocity equation and integrate in
pieces, just like when we found the area between a curve
and the x-axis. (Take the absolute value of each integral.)
Or you can use your calculator to integrate the absolute
value of the velocity function.

2
Displacement:
1
1
2
1
0
1
2
13
2
4
5
1 1
1    2  1
2 2
2
-1
Distance Traveled:
velocity graph
-2
1 1
1   2  4
2 2
2
1
0
1
2
3
4
-1
-2
position graph
5
Every AP exam I have seen
has had at least one
problem requiring students
to interpret velocity and
position graphs.

In the linear motion equation:
dS
 V t 
dt
V(t) is a function of time.
dS  V  t  dt
For a very small change in time, V(t) can be
considered a constant.
S  V  t  t
We add up all the small changes in S to get
the total distance.
S  V1  t  V2  t  V3  t  
S  V1  V2  V3   t

S  V  t  t
We add up all the small changes in S to get
the total distance.
S  V1  t  V2  t  V3  t  
S  V1  V2  V3   t
k
S   Vn  t
n 1

S   Vn  t
As the number of subintervals becomes
infinitely large (and the width becomes
infinitely small), we have integration.
n 1
S   V  t  dt

This same technique is used in many different real-life
problems.

Example 5:
National Potato Consumption
The rate of potato consumption
for a particular country was:
C  t   2.2  1.1t
where t is the number of years
since 1970 and C is in millions
of bushels per year.
For a small
t , the rate of consumption is constant.
The amount consumed during that short time is C  t   t .

Example 5:
National Potato Consumption
C  t   2.2  1.1t
The amount consumed during that short time is C  t   t .
We add up all these small
amounts to get the total
consumption:
total consumption   C  t  dt
From the beginning of 1972 to
the end of 1973:
1
t
2.2

1.1
dt
 2.2t 
1.1
2
ln1.1
4
4
 7.066
t
2
million
bushels

Work:
work  force  distance
Calculating the work is easy
when the force and distance are
constant.
When the amount of force
varies, we get to use calculus!

Hooke’s law for springs:
F  kx
k = spring
constant
x = distance that
the spring is
extended beyond
its natural length

Hooke’s law for springs:
F=10 N
x=2 M
F  kx
Example 7:
It takes 10 Newtons to stretch a
spring 2 meters beyond its natural
length.
10  k  2
5k
F  5 x
How much work is done stretching
the spring to 4 meters beyond its
natural length?

F(x)
How much work is done stretching
the spring to 4 meters beyond its
natural length?
x=4 M
For a very small change in x, the
force is constant.
dw  F  x  dx
dw  5x dx
 dw   5x dx
4
W   5 x dx
0
F  x   5x
4
5 2
W x
2 0
W  40 newton-meters
W  40 joules
p