Find the radius and interval of convergence of the following power series:
1
k ( x + 2)
1.
k
2.
k =1
3. f ( x) = 1 + x + 4 x 2 + 9 x 3 + 16 x 4 .
n =1
4.
1
2
n
xn
1
k + 2x
k
k =0
5.
(1)k
k =0
1
x 2 k +1
(2k + 1)!
6.
3k k
x
k =0 2k
n
n =0
7.
(1)
8.
(1
PROPORTION TYPE DIFFERENTIAL EQUATIONS
1. Due to a curse imposed by a neighboring town, the members of a particular town are driven away.
The rate at which the population is declining is approximately 2 P people per month when the
population is P. Suppose
In problems 1 8 find the general solution to the differential equation.
1.
dy
= kx
dx
2.
dy
= ky
dx
3.
dy
= x2 + k2
dx
4.
dy
= y2 + k 2
dx
5.
dy
= y + ky
dx
6.
dy
= y+k
dx
7.
dy
= kx x
dx
8.
dy
= ky ( x 1)
dx
In problems 9 10, solve the initial value prob
A particular disease can spread in one of four possible ways. Let P be the number of individuals
infected at time t, P (0) = Po , and the total population is 200. The four models are shown below.
Model A
dP
=k
dt
dP
= k (200 P )
dt
Model C
Model B
dP
= kP
Area/mass
Section 8.4
1. Find the area of the region bounded by y = x + 6 , y = 5 x and the x-axis.
A. Use slices perpendicular to the x-axis.
B. Use slices perpendicular to the y-axis
2 A. Find the area of the region bounded by the first arch of y = sin
INTEGRALS AND SERIES
[7.7]
Definition of convergence of improper integrals:
Suppose f(x) is positive for x a .
lim a f ( x) dx
b
If
b
is a finite number, we say that
a
a
f ( x) dx
converges and define
f ( x) dx = lim a f ( x) dx .
b
b
Otherwise, we say th
Specific Example
General
The problem:
The problem:
Find the area of the irregular shaped
region bounded by f ( x ) , the x-axis,
over the interval [a, b].
The quantity we want to find depends on something
that varies. Applications come from geometry
(leng
Power Series
1. Find the radius and interval of convergence of
2k k
x using the steps below:
k
k =0 5
(i) a k =
(ii) a k +1 =
(iii) Simplify
(iv) lim
a k +1
a k +1
k
ak
ak
=
=
(v) Radius
(vi) Interval
2. Repeat the process shown in problem 1 for the foll
Series and Improper Integrals
1. Show that the series
1
converges by comparing it to the improper integral
4
n =1 n
2. Show that the series
n =1 n
3. Show that the series
1
n
n =1
p
diverges by comparing it to the improper integral
1
x
4
dx .
1
1
Section