8/7/2019 Lecture 4.3 Integral as Net Change
1/14
8/7/2019 Lecture 4.3 Integral as Net Change
2/14
Theimage cannotbe displayed.Your computer may nothaveenough memory toopen theimage, or theimagemay havebeen corrupted.Restartyour computer, and then open thefileagain.I fthe red x stillappears, you may havetodelete theimage and then insertit again.
ft
min
minutes
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?
200ft
200ft
200ft
100ft
200 200 200 100 700 ! 700 feet
p
8/7/2019 Lecture 4.3 Integral as Net Change
3/14
Theimage cannotbe displayed.Your computer may nothaveenough memory toopen theimage, or theimagemay havebeen corrupted.Restartyour computer, and then open thefileagain.I fthe red x stillappears, you may havetodelete theimage and then insertit again.
ft
min
minutes
What is the displacement of the bee?
200ft
-200ft
200ft
-100ft
200 200 200 100 100 !
100 feet towards the hive
p
8/7/2019 Lecture 4.3 Integral as Net Change
4/14
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 totaldisplacement.
Displacementb
a
V t dt !
Distance Traveledb
a
t dt! To find distance traveled we have to use absolute value.
Find the roots of the velocity equation and integrate in
pieces, just like when we found the area between a curveand 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. p
8/7/2019 Lecture 4.3 Integral as Net Change
5/14
Theimage cannotbed isplayed.Your computer may nothaveenough memory toopen theimage,or theimage may havebeen corrupted.Restart your computer,and then open thefile again.If thered x stillappears,you may havetodeletethe imageand then insertit again.Theimage cannotbed isplayed.Your computer may nothaveenough memory toopen theimage,or theimage may havebeen corrupted.Restart your computer,and then open thefile again.If thered x stillappears,you may havetodeletethe imageand then insertit again.Theimage cannotbed isplayed.Your computer may nothaveenough memory toopen theimage,or theimage may havebeen corrupted.Restart your computer,and then open thefile again.If thered x stillappears,you may havetodeletethe imageand then insertit again.Theimage cannotbed isplayed.Your computer may nothaveenough memory toopen theimage,or theimage may havebeen corrupted.Restart your computer,and then open thefile again.If thered x stillappears,you may havetodeletethe imageand then insertit again.
Theimage cannotbed isplayed.Your computer may nothaveenough memory toopen theimage,or theimage may havebeen corrupted.Restart your computer,and then open thefile again.If thered x stillappears,you may havetodeletethe imageand then insertit again.Theimage cannotbed isplayed.Your computer may nothaveenough memory toopen theimage,or theimage may havebeen corrupted.Restart your computer,and then open thefile again.If thered x stillappears,you may havetodeletethe imageand then insertit again.Theimage cannotbed isplayed.Your computer may nothaveenough memory toopen theimage,or theimage may havebeen corrupted.Restart your computer,and then open thefile again.If thered x stillappears,you may havetodeletethe imageand then insertit again.
velocity graphTheimage cannotbed isplayed.Your computer may nothaveenough memory toopen theimage,or theimage may havebeen corrupted.Restart your computer,and then open thefile again.If thered x stillappears,you may havetodeletethe imageand then insertit again.
position graph
1
2
1
2
1
2
Displacement:
1 11 2 1
2 2 !
Distance Traveled:
1 11 2 4
2 2 !
Every AP exam I have seenhas had at least one
problem requiring students
to interpret velocity and
position graphs.
p
8/7/2019 Lecture 4.3 Integral as Net Change
6/14
In the linear motion equation:
dS V tdt
! V(t) is a function of time.
For a very small change in time, V(t) can be
considered a constant. dS V t dt!
S V t t ( ! ( We add up all the small changes in S to getthe total distance.
1 2 3S V t V t V t ! ( ( (
1 2 3S V V V t! (
p
8/7/2019 Lecture 4.3 Integral as Net Change
7/14
S V t t( ! ( We add up all the small changes in S to getthe total distance.
1 2 3S V t V t V t ! ( ( (
1 2 3
S V V V t! (
1
k
n
n
S V t
!
! (
1
n
n
S V t
g
!
! (
S V t dt!
As the number of subintervals becomes
infinitely large (and the width becomes
infinitely small), we have integration.
p
8/7/2019 Lecture 4.3 Integral as Net Change
8/14
This same technique is used in many different real-life
problems.
p
8/7/2019 Lecture 4.3 Integral as Net Change
9/14
Example 5: National Potato Consumption
The rate of potato consumption
for a particular country was:
2.2 1.1tC t !
where t is the number of years
since 1970 and is in millions
of bushels per year.
For a small , the rate of consumption is constant.t(
The amount consumed during that short time is . C t t (
p
8/7/2019 Lecture 4.3 Integral as Net Change
10/14
Example 5: National Potato Consumption
2.2 1.1
tC t !
The amount consumed during that short time is . C t t (
We add up all these small
amounts to get the totalconsumption:
total consumption C t dt!
4
2
2.2 1.1tdt
4
2
12.2 1.1
ln1.1
tt!
From the beginning of 1972 to
the end of 1973:
7.066}million
bushels
p
8/7/2019 Lecture 4.3 Integral as Net Change
11/14
8/7/2019 Lecture 4.3 Integral as Net Change
12/14
Hookes law for springs: F kx!
x = distance that
the spring is
extended beyond
its natural length
k= spring
constant
p
8/7/2019 Lecture 4.3 Integral as Net Change
13/14
Hookes law for springs: F kx!
Example 7:
It takes 10 Newtons to stretch a
spring 2 meters beyond its natural
length.
F=10 N
x=2 M
10 2k!
5 k! 5F x!
How much work is done stretching
the spring to 4 meters beyond its
natural length?
p
8/7/2019 Lecture 4.3 Integral as Net Change
14/14
F(x)
x=4 M
How much work is done stretching
the spring to 4 meters beyond its
natural length?
For a very small change in x, the
force is constant.
dw F x dx!
5 dw x dx!
5 dw x dx!
4
0
5 W x dx!
4
2
0
5
2W x!
40W ! newton-meters
40W ! joules
5F x x!
T