Spring 2016 Math 152
8.9 Improper Integrals:
3. Improper Integrals:
Overview of Material for Test II
courtesy: Amy Austin
Trig Integrals
Case I: Integrals where one (or both) of the limits is infinite: Your goal is to determine whether the improper integral converges (finite value) or diverges (infinite value).
Z ∞
Z t
a.)
f (x) dx = lim
f (x) dx
t→∞
a
8.3 Integration by Trig Substitution
b.)
a2 + x2 : Substitute x = a tan θ ,− π2 < θ <
a
Z
∞
f (x) dx = lim
π
2
c.)
f (x) dx =
−∞
Z
Z
a
f (x) dx
t
a
f (x) dx +
π
2
√
Note: If in the form ax2 + bx + c, you must complete
the square, and then do the correct trig substitution.
8.4 Integration by Partial Fractions
2. Partial Fractions:
g(x)
where g(x) and h(x) are polynomih(x)
als and the degree of h(x) is BIGGER than the degree of
g(x) (if this is not true, you must first do long division).
In order to integrate a partial fraction problem, you must
first find the Partial Fraction Decomposition:
Suppose f (x) =
Case I: h(x) is a product of linear factors, none repeating, then the Partial Fraction Decomposition has the
form:
A
B
x+1
=
+
(x − 2)(2x − 11)
x − 2 2x − 11
Case II: h(x) is a product of linear factors, some repeating, then the Partial Fraction Decomposition has the
form:
A
C
B
C
x+1
=
+
+
+
3
2
(x − 1)(x − 2)
x − 1 x − 2 (x − 2)
(x − 2)3
Case III: h(x) contains irreducible quadratic factors,
none repeating, then the Partial Fraction Decomposition
has the form:
A
Bx + C
x+1
=
+ 2
(x − 2)(x2 + 1)
x−2
x +1
Once you find A, B, etc., then you integrate the result.
Z
∞
f (x) dx, then try
a
−∞
to evaluate both integrals.
π
2
x2 − a2 : Substitute x = a sec θ , 0 ≤ θ ≤ π, θ 6=
a
t→−∞
−∞
1. Look for the form. There are 3 cases:
a2 − x2 : Substitute x = a sin θ , − π2 ≤ θ ≤
Z
Case II: Integrals where there is a discontinuity on the
interval [a, b]:
a.) Suppose f (x) is discontinuous at x = a: Then
Z b
Z b
f (x) dx = lim
f (x) dx
a
t→a+
t
b.) Suppose f (x) is discontinuous at x = b: Then
Z b
Z t
f (x) dx = lim
f (x) dx
a
t→b−
a
c.) If f (x) is discontinuous at some c where a < c < b,
then
Z b
Z c
Z b
f (x) dx, then try to evalf (x) dx +
f (x) dx =
a
a
uate both integrals.
c
• Comparison Theorem for Improper Integrals:
a.) Suppose f (x) and g(x) are continuous, positive
functions on the interval [a, ∞). Also, suppose that
f (x) ≥ g(x) on the interval [a, ∞). Then:
Z ∞
Z ∞
(i) If
f (x) dx converges, so does
g(x) dx.
a
a
Z ∞
f (x) dx diverges, no conclusion can be
(Note: If
aZ
∞
drawn about
g(x) dx).
a
Z ∞
Z ∞
(ii) If
g(x) dx diverges, so does
f (x) dx.
a
a
Z ∞
(Note: If
g(x) dx converges, no conclusion can be
a Z
∞
f (x) dx).
drawn about
a
Note: The way you choose the comparison function: You
take the largest part of the numerator over the largest
part of the denominator on the interval [a, ∞). Once you
find the comparison function, you must determine the
direction of the inequality, then integrate the comparison
function and draw the correct conclusion.
9.3 Arc Length:
4. There are three possible formulas which gives the length
of a curve:
a.) If y = f (x), a ≤ x ≤ b, then
of the curve
sthe length
2
Z b
dy
dx
from x = a to x = b is L =
1+
dx
a
b.) If x = g(y), c ≤ y ≤ d, then
of the curve
sthe length
2
Z d
dx
1+
from y = c to y = d is L =
dy
dy
c
c.) If x = f (t) and y = g(t), α ≤ t ≤ β, then the length
of the curve from t = α to t = β is
s
Z β 2 2
dy
dx
+
dt
L=
dt
dt
α
9.4 Surface Area of Revolution:
5. Revolution around the x axis:
a.) If the curve y = f (x), a ≤ x ≤ b is revolved around
the x axis, then s
the resulting surface area is given by
2
Z b
dy
dx
f (x) 1 +
SA = 2π
dx
a
b.) If the curve x = g(y), c ≤ y ≤ d is revolved around
the x axis, then
surface area is given by
s the resulting
2
Z d
dx
dy
SA = 2π
y 1+
dy
c
c.) If the curve x = f (t) and y = g(t), α ≤ t ≤ β,
is revolved around the xsaxis, then the resulting surface
2 2
Z β
dy
dx
area is SA = 2π
g(t)
+
dt
dt
dt
α
6. Revolution around the y axis:
a.) If the curve y = f (x), a ≤ x ≤ b is revolved around
the y axis, then
surface area is given by
s the resulting
2
Z b
dy
dx
SA = 2π
x 1+
dx
a
b.) If the curve x = g(y), c ≤ y ≤ d is revolved around
the y axis, then s
the resulting surface area is given by
2
Z d
dx
SA = 2π
g(y) 1 +
dy
dy
c
c.) If the curve x = f (t) and y = g(t), α ≤ t ≤ β,
is revolved around the y axis, then the resulting surface
area is given by
s 2
Z β
2
dx
dy
+
dt
SA = 2π
f (t)
dt
dt
α
10.1 Sequences:
7. A sequence is an infinite, ordered list of numbers. There
are three main points emphasized in this section:
a.) Limit of a sequence: If lim an = L, then we say the
n→∞
sequence {an } converges to L. If lim an = ∞ or does
n→∞
not exist, then we say the sequence {an } diverges.
b.) Bounded sequence: We say a sequence is bounded
below if there is a number N so that an ≥ N for all
n. We say a sequence is bounded above if there is a
number M so that an ≤ M for all n. If an is bounded
both above and below, then we say the sequence
is bounded.
c.) Increasing or decreasing sequence: We say a sequence
an is increasing if an < an+1 for all n ≥ 1. We say a sequence an is decreasing if an > an+1 for all n ≥ 1. Note:
For a sequence to be increasing or decreasing, it need not
be true for ALL n, it just must be eventually true, from
some value m to ∞. If a sequence is either increasing or
decreasing, then we say the sequence is monotonic.
d.) Recursive sequences: A recursive sequence is a sequence where a1 is given, and an+1 = f (an ). First,
find the first few terms of the sequence to get a feel for
whether the sequence converges. Then, to find the limit,
take the limit of both sides of the recursive definition.
10.2 Series
8. Def: Let {an } = {a1 , a2 , a3 , ..., ...} be a sequence. We
∞
P
define the infinite series to be
an = a1 + a2 + ... +
n=1
an + ... + .... In other words, a series is the sum of a
sequence. The main focus of chapter 10 is to determine
when the sum is finite.
9. Def: Let
∞
P
an be a series. We will construct the se-
n=1
quence of partial sums
{sn } = {s1 , s2 , s3 , ..., ...} as follows:
s1 = a1
s2 = a1 + a2
s3 = a1 + a2 + a3
Therefore a general formula for sn is
n
P
ai = a1 + a2 + ... + an .
sn =
i=1
If lim sn = s, where s is finite, then we say the series
n→∞
∞
P
an converges with sum s. If lim sn is infinite or
n=1
n→∞
∞
P
does not exist, then we say the series
n=1
an diverges.
10. Finding the sum of a series:
a.) Geometric Series: The geometric series
∞
P
arn−1
n=1
converges if |r| < 1 and diverges if |r| ≥ 1. If |r| < 1,
∞
P
a
then the sum is
arn−1 =
.
1
−
r
n=1
b.) Telescoping series: A telescoping series is a series of
∞
P
the form
(an+i − an ) for some integer i ≥ 1. The
n=1
series ’collapses’. To find the sum of a telescoping series,
find a formula for sn and then find lim sn .
n→∞
c.) If an explicit formula is given for the nth partial sum
∞
P
of the series, sn , then
an = lim sn .
n=1
n→∞
The Test for Divergence: If lim an 6= 0, then
n→∞
∞
P
an
n=1
diverges. NOTE: The converse is not necessarily true: If
∞
P
lim an = 0, then the series
an does not necessarily
n→∞
n=1
converge. Therefore if you find that lim an = 0, then
n→∞
the divergence test fails.