force F(N)
Topic 2.2 Extension 05
Area and the definite integral
To follow the worksheet "Initial value problems"
1)
Perhaps you recall from Physics 1 that the impulse of a rocket
engine is given by the area under the force (thrust) vs. time curve.
Consider the force curve shown below.
Since it is curved it will be
difficult to get the exact area. However, we will try.
20
15
10
5
0
force F(N)
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
time t(sec)
First estimate of the total area.
We estimate by vertical rectangles
at 1-second intervals, as shown below. Sum up the four areas:
20
15
10
5
19 = 9
0
0.0
0.5
1.0
119 = 19
1.5
2.0
119 = 19
2.5
3.0
19 = 9
3.5
4.0
force F(N)
time t(sec)
56
Area = _____________Ns.
Area = 9 + 19 + 19 + 9 = 56 Ns
2)
One source of error is that the rectangles "miss" some of the
area.
With a yellow pencil color in the "missing" area:
Another
source of error is that the rectangles encompass some "extra" area.
With a blue pencil color in the "extra" area:
Yet another source of
error is that you have to estimate the height of each rectangle from
NO
the graph. Do you have to estimate the width? _____.
3)
Now it just so happens that the force
t(s)
F(N)
curve shown above has a known equation;
2
0.5
8.75
namely F(t) = 20t - 5t . Use this equation
1.5
to find the exact height of each of the
18.75
four rectangles:
Note that we pick the
2.5
18.75
height at the center of the interval. Now
3.5
8.75
recalculate your area from 1) using the
55
exact (formula) height for each rectangle.
Area = _____________
Ns.
Area = 8.75 + 18.75 + 18.759 + 8.75 = 55 Ns
1
4)
What was the size of t in 2) and 3)? _________
s.
What could
you do to t in order to get a more accurate estimate of the area?
Make the time interval smaller - use more and smaller rectangles.
5)
Second (better) estimate.
Now find the area using half-second
intervals. The rectangles are shown. Be sure to use the formula for
the rectangle heights and make a table similar to that in 3):
20
17.1875 19.6875 19.6875 17.1875
15
12.1875
12.1875
10
4.6875
5 4.6875
Area = 53.75 Ns
0
0.0
0.5
.54.8675
2.34375
1.0
1.5
.512.1875 .517.1875
6.09375
8.59375
2.0
2.5
time t(sec)
.519.6875 .519.6875
9.84375
9.84375
1
3.0
3.5
4.0
.517.1875 .512.1875 .54.8675
8.59375
6.09375
2.34375
6)
Now do the following integral:
∫(20t - 5t2) dt
= 20t2/2 - 5t3/3 + C
4 s. Now substitute this
7)
What is the ending t in the graph? t = __
53.3333 + C Ns.
value of t into your answer for 6). You get _________
0 s. Now substitute
8)
What is the beginning t in the graph? t = __
this value of t into your answer for 6). You get _________
+ C Ns.
0
9)
Now subtract the Ns answer of 7) from the Ns answer of 8).
53.3333 Ns.
do you get? _________
What
10)
Compare
compare?
they
this
answer
with
the
Ns
answer
of
5).
How
do
53.3333 is very similar to 53.75
11)
What happened to the arbitrary constant C?
53.3333 + C - (0 + C)  the C cancels in the subtraction.
12)
We have just done a process called finding the definite integral.
We write
4 upper limit of integration
53.3333 Ns.
(20t - 5t2) dt = _________
0 lower limit of integration
∫
13)
We call the “4” the upper limit of integration and we call the
“0” the lower limit of integration. Label them.
14)
In review, the definite integral gives the area under the curve
that is being integrated.
To find the definite integral we do the
following steps. First, we integrate:
4
4 square bracket
(20t - 5t2) dt =
20t2/2 - 5t3/3
notation
0
0
Note the square bracket notation which has the upper and lower limits
of integration shown.
Label the square bracket notation.
Why isn’t
the “C” shown?
C cancels in the subtraction.
∫
]
[
15)
Second, we substitute the upper and lower limits into the
integral, and subtract.
Note that we have not bothered to write the
arbitrary constant C, this time. Why can we ignore it?
[
4
20t2/2 - 5t3/3
]0
=
20·42/2 - 5·43/3
–
=
53.3333 Ns
_________
-
=
53.3333 Ns.
_________
20·02/2 - 5·03/3
0
_________
Ns
16)
Consider the area shown below, bounded by f(x) = x3 and x = 4.
Find its value, by using the appropriate definite integral. Be sure to
show each step of the process clearly, using the appropriate notation.
f(x) = x3
4
(x3) dx =
0
∫
x = 4
4
[ ]0
x4/4
2
= 44/4 - 04/4 = 64