More Applications of Definite Integrals 341
Distance Traveled=
∫ 10
0
∣∣
v(t)
∣∣
dt
∣∣
2 − 6 e−t
∣∣
=
{
−( 2 − 6 e−t)if 0≤t< 1. 09861
2 − 6 e−tift≥ 1. 09861
∫ 10
0
| 2 − 6 e−t|dt=
∫ 1. 09861
0
−(2− 6 e−t)dt
+
∫ 10
1. 09861
( 2 − 6 e−t)dt
= 1. 80278 + 15. 803 = 17. 606.
Alternatively, use the [nInt] function on
the calculator.
EnternInt(abs(2− 6 e∧(−x)),x,0,10)
and obtain the same result.
- Build a table of value for
dy
dx
=x−y.
y=− 2 y=− 1 y= 0 y= 1 y= 2
x=− 20 − 1 − 2 − 3 − 4
x=− 11 0 − 1 − 2 − 3
x= 02 1 0 − 1 − 2
x= 13 2 1 0 − 1
x= 24 3 2 1 0
Draw short lines at each intersection with
slopes equal to the value of
dy
dx
at that point.
22.
dP
dt
=. 65 P
(
1 −
P
50
)
can be separated
and integrated by partial fractions.∫
dP
P
+
∫
dP
( 50 −P)
=
∫
. 65 dt
produces ln|P|+ln
∣∣
50 −P
∣∣
=. 65 t+C 1 and
P
50 −P
=C 2 e.^65 t,so
P=
50 C 2 e.^65 t
1 +C 2 e.^65 t
. Since one person knows
the rumor on day zero, 1=
50 C 2
1 +C 2
and
C 2 =
1
49
. The model for the population
becomesP=
50 e.^65 t
/
49
1 +
(
e.^65 t
/
49
)=
50 e.^65 t
49 +e.^65 t
=
50
49 e−.^65 t+ 1
. Half the
population of the office would be 25
people, so solve fortin 25=
50 e.^65 t
49 +e.^65 t
.
Since t≈5.987, half of the office will have
heard the rumor by the sixth day.
- The logistic model becomes
dP
dt
=kP
(
1 −
P
200
)
since the carrying
capacity is 200. Separate the variables∫
200 dP
P( 200 −P)
=
∫
kdtand integrate by
partial fractions
∫
dP
P
+
∫
dP
200 −P
=
∫
kdt.
You find ln
∣∣
∣∣ P
200 −P
∣∣
∣∣=kt+C 1. Exponentiate
to get
P
200 −P
=ekt+C^1 =C 2 ekt. Solving
forPproducesP=
200 C 2 ekt
1 +C 2 ekt
. On day
zero, two students are infected, so
2 =
200 C 2
1 +C 2
andC 2 =
1
99
. On day five, 12
students are infected, so 12=
200 e^5 k
/
99
1 +
(
e^5 k
/
99
)
andk≈.369.