1. Compute the
following integrals:
Solution. First do long division to write
as 
.
The denominator factors as x2(x2 + 1) so we can use partial fractions to write
Comparing the coefficients of x 2, x2, x, 1 gives the
equations A +C = 3, B+D = 2, A = 1,
B = 1. Plugging the values for A and B back into the first two equations gives C
= 2 and
D = 1. The integral is now
Solution. Let x = 2 tanθ so dx = 2 sec2 θ dθ . Then the
integral becomes
Solution. Use integ ration by parts . We can let
and dv = dx, so v = x
and 
. Then the integral becomes
2. (a) Sketch the curve
r 2 = cos 2θ in polar coordinates and compute the area inside the
curve.
Solution. Since r2 is always ≥ 0, there will only
be
points on the curve when cos 2θ ≥ 0, which means
 and
 . For the area inside
the curve we use the formula  . By symmetry
we can integrate from 0 to π /4 and then multiply by
4. Thus we get |
 |
(b) Compute the length of the curve parametrized by x =
cos3 t, y = sin3 t, for 0 ≤ t ≤ 2π .
Solution. The arclength is
3. Consider the region
between the curve y = sin x and the x-axis for 0 ≤ x ≤2π .
Compute the volume of the solid obtained by rotating this region about the
y-axis.
Solution. Using cylindrical shells, the volume would be
.
However, the function sin x is negative in the interval [ π, 2π ] so the integral
would be
negative on this interval, which is not what we want. So we integrate from 0 to
π with a
plus sign and from π to 2π with a minus sign:
Using integration by parts with u = x and dv = sin x,
hence du = dx and v = −cos x, we
get
4. Determine whether
each of the following series converges:
. Use the integral
test: 
which is finite so
the series converges.
. The individual terms have absolute value
, so the terms do
not approach zero and the series diverges.
. Using the ratio test, we look at
.
To compute the limit as n → ∞ we take the logarithm:
.
This has the indeterminate form ∞ ยท 0 so we apply l’Hopital’ s rule , first
rewriting it as
. Then taking derivatives of numerator and
denominator gives
Thus 
. This is less
than 1 so the series converges.
(d) Determine the values of x > 0 for which the series
verges.
Solution. There are n+1 terms in the denominator so when x = 1 the series
becomes the
harmonic series 
which diverges. When n < 1 we
have 
n + 1 so 
and the series diverges by
comparison with the
harmonic series. When x > 1 we compare the series with the geometric series
. We
have 
so 
. Since x > 1 the
geometric series converges, hence the given
series also converges when x > 1.
5. (a) Determine the values of x for which
the series 
converges.
Solution. Ratio test: 
.
Thus the series converges if |x − 1|2 < 1/2, or equivalently
, or 
. At the endpoints of the interval we have
(x−1)2 = 1/2 so the series becomes
which converges, being
a p-series with p > 1. Thus the interval of convergence is
(b) Compute the Maclaurin series for the function
.
Solution. Let 
, so
. Multiplying this
by x2, we get 
.
(c) Find the Taylor series 
for ln x and
determine the interval of convergence.
Solution. First we will find a similar series for 1/x, then integrate this to
get a series for
ln x. We have
Integrating this, we get
We determine C by setting x = 2, which makes the series
have the value 0 so C = ln 2. To
determine the interval of convergence we use the ratio test:
Thus the series converges when |x − 2|/2 < 1, or |x − 2| <
2, which means 0 < x < 4.
At the endpoint x = 0 we have the series
which diverges. At x = 4 we have
which converges by the alternating series
test. The interval of convergence
is therefore 0 < x ≤ 4.
(d) Suppose the ratio test yields the information that a certain series
has
radius of convergence equal to R. Show that the derivative series
has
the same radius of convergence R.
Solution. To apply the ratio test to the series
we compute the limit
. For the derivative series
the
corresponding limit is 
.
This is the same
limit as before, so the radius of convergence is the same.
6. (a) Find the general solution of the
differential equation x2y′ − y = 1.
Solution. There are two methods that work : separation of variables , and finding
an
integrating factor μ(x). Let’s use the latter method. First rewrite the equation
in the
form y′ − x -2y = x -2, so P(x) = −x -2 and 
.
After
multiplying the equation y′−x -2y = x -2 by μ(x) we get
. Integrating
both sides, we get 
. Solving this for y gives
.
(b) Find the solution of y′′ + 2y′ + 2y = 0 satisfying y(0) = 1 and y′(0) = 0.
Solution. The auxiliary equation is r2+2r+2 = 0 with roots
so the general solution is y = Ae-x cos x + Be-x sin x. Plugging in the
condition y(0) = 1
gives A = 1. For the condition y′(0) = 0 we first compute
. Then y′(0) = −A + B = 0 so B = A = 1. The
final answer is
y = e-x(cos x + sin x).
(c) Find a differential equation satisfied by all of the hyperbolas xy = C, for
C an arbitrary
constant.
Solution. Differentiating xy = C implicitly gives x(dy/dx) + y = 0, so dy/dx =
−y/x.
(d) Find a family of curves, each of which intersects each hyperbola in part (c)
at right
angles.
Solution. The slope of these curves will be the negative reciprocal of the
slopes −y/x in
part (c). Thus we have the differential equation dy/dx = x/y. This can be
rewritten as
y dy = x dx. Integrating this, we get 
. This
can also
be written as y2 − x2 = C for a new constant C.
