Math 113 Exam 2-Review
October 14, 2016
Exam 1 covers 7.4-8.5 in the textbook. It is advisable to also review the
material from the first midterms as this will be helpful in solving some of the
problems.
7.4 Integration by Partial Fractions
For a function of the form
f (x) =
P (x)
Q(x)
where P and Q are polynomials one can use partial fractions to solve the
integral. The idea is to breakdown the polynomial into the factors to simplify
the integral. Before doing this make sure that the degree of P is less than
the degree of Q. If not, then you need to use long division first.
Now factor the polynomial Q(x). This will already be done for you, or Q
will be a quadratic that easily factors, or a higher degree polynomial of a nice
form so you can easily factor. The result should be linear factors (possibly
repeated) or quadratic factors with complex roots.
For a linear factor (ax + b) separate these off with a constant on top that
will be determined later. If the factor is repeated than one must repeat the
term. So if (2x + 3)3 is on the denominator we need
B
C
A
+
+
.
2
2x + 3 (2x + 3)
(2x + 3)3
The same is true for the quadratic factors with complex roots, but in this
case on the numerator we always need an expression of the form Ax + B.
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Now recombine the expression with the constants of the form A or Ax+B
into a long expression on the numerator and solve using the original equation.
If all of the roots are unique then instead of solving a system of equations
you can substitute in the factors to solve for the constants. If there are
repeated roots or complex roots (possibly also repeated) then you need to
solve the resulting system of equations to find these.
For the midterm you should be able to express the general form given a
factored polynomial (good multiple choice question) and solve the integral
for the free response.
To integrate the linear terms are easy as they are of the form
A
(ax + b)k
that integrates easily.
For quadratic factors you will usually divide into 2 parts. The first is the
derivative of the bottom so of the form
2ax + b
C 2
ax + bx + c
and the second is of the form
ax2
C
.
+ bx + c
The second expression will usually be a tan θ substitution, possibly after
completing the square. So you will have two components to the solution a
component that is of the form constant times ln |ax2 + bx + c| and a term
involving tan−1 .
7.5 Strategy for Integration
This section overlaps with the first half of the chapter that was tested
on midterm 1 so you should at least review the first half of the chapter for
this section. Before beginning these integrals you usually need to do a usubstitution or simplify the expression first. After this the expression should
be similar to problems from the first half of the chapter. Usually, these are
simpler expressions than the ones in sections 7.1-7.3 after the first step.
If you aren’t sure where to begin, then just try something. It may not
be correct, but often will lead to the correct method.
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7.7 Approximate Integration
If you remember the formulas this should be relatively easy questions on
the exam, but watch algebraic errors (especially for multiple choice problems). You will need to have the formulas for the left point approximations,
right point approximations, midpoint approximations, trapezoidal rule, and
Simpson’s rule memorized for the exam. These can be found on pages 515518 in the text book. For this you need to simply be able to compute with
low values of n and compute the approximate integral.
You do not need to have the error formulas memorized, but you should
be able to use them to see how accurate a given approximation is guaranteed
to be.
7.8 Improper Integrals
For this section is is good to review some properties of limits. In general
we have two kinds of integrals. The first is an expression where the integral
goes to ∞ or −∞. The second is where the integral either includes an
asymptote or approaches an asymptote.
Remark 0.1 Watch out for integrals that are discontinuous between a and b
(usually in the form of an asymptote). Do not try to solve by the Fundamental
Theorem as it does not apply!
First kind of integral is of the form
Z
Z ∞
f (x)dx or
a
f (x)dx.
−∞
a
In these you replace the value infinity and take a limit. So for example
Z t
Z ∞
lim
f (x)dx =
f (x)dx
t→∞
a
a
provided the limit exists. If the limit does not exist (so could
R ∞ be infinite),
then we say the integral diverges. For integrals of the form −∞ f (x)dx you
split this up into two expressions such as
Z 0
Z ∞
f (x)dx +
f (x)dx.
−∞
0
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If both integrals converge, then the value of the integral is the sum of the
limits. If either diverges, then the entire term diverges.
The second type is where the integral approaches an asymptote. In this
case one takes one-sided limits depending on how the integral approaches an
asymptote. So if there is an asymptote at b, then
Z b
Z t
f (x)dx = lim−
f (x)dx
t→b
a
a
provided the limit exists. If the asymptote lies between a and b, then split
this into two components. One will have a left handed limit, the other a
right handed limit. The integral will exist provided each each of the onesided limits exist and then the integral is the sum of the values. If either
integral diverges, then the entire integral diverges. R
∞
We also could have an integral of a mixed form a f (x)dx where there
is an asymptote at a. In this case we split up the integral and again look at
each one separately. So
Z ∞
Z c
f (x)dx
f (x)dx +
c
a
where c is some arbitrary point between a and ∞.
The other part of this section involves the Comparison Theorem.
Theorem 0.2 Suppose that f and g are continuous functions with f (x) ≥
g(x) ≥ 0 for x ≥ a.
R∞
R∞
1. If a f (x)dx is convergent, then a g(x)dx is convergent.
R∞
R∞
2. If a g(x)dx is divergent, then a f (x)dx is divergent.
A similar statement can be made if there is an asymptote and the same
result applies. The next result is useful when there is an asymptote at a.
Theorem 0.3 Suppose that f and g are continuous functions with f (x) ≥
g(x) ≥ 0 for x > a.
Rb
Rb
1. If a f (x)dx is convergent, then a g(x)dx is convergent.
Rb
Rb
2. If a g(x)dx is divergent, then a f (x)dx is divergent.
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In applying both theorems we need to keep track of the following two
ideas.
Z ∞
1
dx
xn
1
converges for n > 1 and diverges for 0 < n ≤ 1. On the other hand, we have
Z 1
1
dx
n
0 x
converges for 0 < n < 1 and diverges for n ≥ 1.
For applications of the theorem then we want to look at powers of x and
see which case we fall into before we begin (look at leading powers of x).
After we know that then we can make the necessary approximations as in
the homework.
8.1 Arc Length
The formulas we need here are
Z
Z bp
1 + [f 0 (x)]2 dx and L =
L=
d
p
1 + [g 0 (y)]2 dy.
c
a
The main points to remember are
1. make sure and differentiate f or g,
2. be careful with arithmetic, and
3. look for ways to simplify the integrand.
For multiple choice you may just be asked to set it up. For free response
after the simplification it should be quite easy to integrate.
The arc length function is
Z xp
1 + [f 0 (t)]2 dt.
s(x) =
a
8.2 Area of a Surface of Revolution
When revolving a function about the x-axis we either have
Z b
Z d
p
p
0
2
2πf (x) 1 + [f (x)] dx or
2πy 1 + [g 0 (y)2 dy.
a
c
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When revolving a function about the y-axis we either have
Z b
Z d
p
p
0
2
2πx 1 + [f (x)] dx or
2πg(y) 1 + [g 0 (y)2 dy.
a
c
Sometimes you can use either function. Often one is easier and usually
one cannot be found (you cannot solve for one of the variables in place of the
other). For the exam you will either be asked to set up the integral (such as
in the multiple choice portion) or solve completely.
As in the previous section any small mistakes in the setup or computing
derivatives will most often make the integral virtually unsolvable. If this
happens go back and check your work. After setting up the integral remember to do a lot of simplification before trying to evaluate.
8.3 Applications to Physics and Engineering
The problems involve hydrostatic pressure and moments. For hydrostatic
pressure the integrals always take the form
Z b
Constant · Depth · Length dx.
a
The “Constant” will be given to you. For metric units this will be the density
and you will need to multiply by 9.8. For standard units the constant does
not need to be multiplied by anything. For depth you will have a term of
the form x + c or c − x depending on your choice of coordinates. For the
length this will be a function in terms of x. The function will depend on
your coordinates and on the geometry (shape) of the object.
For moments you need to simply compute
Z
Z
1 b1
1 b
x[f (x) − g(x)]dx and ȳ =
([f (x)]2 − [g(x)]2 )dx.
x̄ =
A a
A a 2
The term A is just area or in other words
Z b
f (x) − g(x)dx.
a
The other term you may need is moment about the y-axis that is My =
Rb
ρ a x(f (x) − g(x))dx. where ρ is the density. Don’t be confused because of
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the My and used to compute the x̄ term. Likewise, the moment about the
Rb
x-axis is Mx = ρ a 12 ([f (x)]2 − [g(x)]2 )dx.
8.5 Probability
R ∞ A function f is a probability density function if f (x) ≥ 0 for all x and
f (x)dx = 1. If we want the probability that we will between values a
−∞
and b then we have
Z b
P (a ≤ X ≤ b) =
f (x)dx.
a
The average value is
Z
∞
xf (x)dx.
µ = x̄ =
−∞
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