More Applications of Definite Integrals 341Distance Traveled=∫ 100∣∣
v(t)∣∣
dt∣∣
2 − 6 e−t∣∣
={
−( 2 − 6 e−t)if 0≤t< 1. 09861
2 − 6 e−tift≥ 1. 09861
∫ 100| 2 − 6 e−t|dt=∫ 1. 098610−(2− 6 e−t)dt+
∫ 101. 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 andP
50 −P
=C 2 e.^65 t,soP=
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
andC 2 =1
49
. The model for the population
becomesP=50 e.^65 t/
491 +(
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 bypartial fractions∫
dP
P+
∫
dP
200 −P=
∫
kdt.You find ln∣∣
∣∣ P
200 −P∣∣
∣∣=kt+C 1. Exponentiateto getP
200 −P
=ekt+C^1 =C 2 ekt. SolvingforPproducesP=
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.