Download 1 Super-Brief Calculus I Review.

CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
For the purposes of this class, we will regard calculus as the study of limits and limit processes.
Without yet formally recalling the definition of a limit, let’s recall the main ways in which one applies
the concept in an introductory Calculus I course (which the student should have recently taken).
1
Super-Brief Calculus I Review.
The Derivative. Given a function f and a point x, we wish to compute the instantaneous rate of
change of f at x. Geometrically, we may identify this instantaneous rate of change with the slope of
the line tangent to the graph of f at the point x. How can we compute such a thing? The process
is achieved in just two major steps: a “geometric approximation” step and then a “limit” step.
y2 − y1
to compute the average rate of change between x and any
x2 − x1
other nearby point, which we denote x + h. Since the function f determines the y-values at x
and x + h, the formula we obtain is the famous difference quotient below:
(1) Use the slope formula
f (x + h) − f (x)
h
If we take the distance h to be a very small number, then this slope should be a very close
approximation to the slope we hope to compute. Indeed, in general a smaller choice of h should
lead to a closer approximation.
(2) Since step (1) gives us an infinite family of arbitrarily close approximations to the slope of the
tangent line, the natural final step is to let h shrink to 0, i.e. compute the limit as h → 0 of our
approximations. So we obtain the formula
f 0 (x) = lim
h→0
f (x + h) − f (x)
h
We call this value f 0 (x) the derivative of f at x, and it is precisely the slope of the tangent line
we seek.
The Definite Integral. Now, given a function f and two points a and b with a < b, we wish to
compute the area under the curve of f , bounded by the lines x = a and x = b. Again we proceed by
a “geometric approximation” step, and then a “limit” step. The student should certainly observe the
clear analogy between this process and the one above for computing the derivative!
(1) First we consider any positive integer n, and we subdivide the interval [a, b] into n equal pieces,
indexed by the endpoints a = x0 < x1 < x2 < ... < xk < ... < xn = b. The width of each piece
[xk−1 , xk ] is ∆x = xk − xk−1 = b−a
n . From each piece [xk−1 , xk ], we select any test point xk , and
we imagine drawing a rectangle with base [xk−1 , xk ], and height determined by the test point
f (xk ). The area of this rectangle is obviously (height)·(width)= f (xk ) · ∆x. Adding together
the area of each of the n rectangles, we obtain the Riemann sum:
n
X
k=1
1
f (xk ) · ∆x
2
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
This Riemann sum gives the approximate area of the region bounded by the curve of f . It is easy
to see that, in most cases, choosing a finer mesh, i.e. subdividing [a, b] into more, smaller pieces,
will yield a closer approximation to the actual area. In other words, choosing larger values of n
(equivalently, smaller values of ∆x) should lead to a closer approximation to the area we hope
to compute.
(2) Now, again, step (1 yields for us an infinite family of arbitrarily close approximations to the true
area under the curve. So again we take a limit, this time as n → ∞, or equivalently, as ∆x → 0.
We get the formula for the definite integral of f from a to b:
Rb
a
f (x)dx = lim
n→∞
n
X
f (xk ) · ∆x
k=1
This integral, when it exists, is the area under the curve of f bounded by x = a and x = b, as
desired.
The two processes above should have been the main focus of the reader’s Calculus I course, together
with the following remarkable theorem:
Theorem 1.1 (The Fundamental Theorem of Calculus). Let f be any continuous function and let a be
any point. Define the area function A of f (centered at a) by the rule
A(x) =
Rb
a
f (t)dt
Stated informally, A is the function which takes a real number x for input, and for output computes
the definite integral of f between the (non-variable) point a and the input x, i.e. gives the area under
the curve of f between a and x. Then the following two statements hold:
(1) A0 (x) = f (x); and
(2)
Rb
a
f (x)dx = F (b) − F (a), where F is any antiderivative of f .
So, again informally, the FTC (Fundamental Theorem of Calculus) may be read as: (1) The definite
integral function of f is an antiderivative of f ; and (2) any antiderivative of f also computes the definite
integral of f . In other words, integration and antidifferentiation are essentially the same thing. So the
(perhaps) seemingly unrelated geometric processes of taking the derivative and computing the definite
integral are inverse processes of one another.
One of our major goals for the course will be to develop some new fundamental applications of the
“limit process,” especially the development of infinite sequences and series. Before we turn to this, we’ll
warm up our existing Calculus I integration skills by pursuing some applications.
2
Area Between Two Curves
Example 2.1. Find the area of the region between the functions f (x) = x2 −4x+10 and g(x) = 4x−x2 ,
for 1 ≤ x ≤ 3.
Example 2.2. Find the area of the region bounded by the graphs of f (x) = x2 −5x−7 and g(x) = x−12.
Example 2.3. Find the area of the region bounded by the graphs of y =
3
8
x2 ,
y = 8x, and y = x.
Volume of a Three-Dimensional Solid
Consider a 3-dimensional solid that extends in the x-direction from x = a to x = b. To estimate the
volume of this solid, we could take a ”3-dimensional Riemann sum:” first we subdivide the interval [a, b]
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
3
into n equal subintervals, thereby slicing the solid into n different 3-dimensional chunks, each of width
∆x = b−a
n . Then we estimate the volume of each of the n chunks by taking the area of one cross-section
(call it A(xk ), for 1 ≤ k ≤ n), and multiplying it by the width
The sum of these volume estimates
P∆x.
n
gives an estimate for the volume of the entire solid, i.e. V ≈ k=1 A(xk )∆x.
Then, as in the case with integration, to find an exact volume all we need to do is take the limit of
the above sum as n runs to infinity. So we get
V = lim
n
X
n→∞
Z
b
A(x)dx
A(xk )∆x =
a
k=1
where A(x) is a function which takes a point x in [a, b] for input and returns the area of the cross section
of the solid at x.
Example 3.1. Compute the volume of the solid whose base is bounded by the curve x = 4 − y 2 and the
y-axis in the xy-plane, and whose vertical cross sections (perpendicular to the x-axis) are half-circles.
Example 3.2. Calculate the volume V of a pyramid of height 12 m whose base is a square of side 4 m.
4
Solids of Revolution - Washer Method and Shell Method
Example 4.1. Let R be the region bounded by the curve f (x) = (x + 1)2 , the x-axis, and the lines
x = 0 and x = 2. Find the volume of the solid of revolution obtained by revolving R about the x-axis.
The previous example suggests a general method for computing the volumes of solids of revolution.
Disk Method about the x-axis. Let f be continuous with f (x) ≥ 0 on the interval [a, b]. If the
region R bounded by the graph of f , the x-axis, and the lines x = a and x = b is revolved about the
x-axis, the volume of the resulting solid of revolution is
V =
Rb
a
π(f (x))2 dx.
√
Example 4.2. The region R is bounded by the graphs of f (x) = x and g(x) = x2 between x = 0 and
x = 1. What is the volume of the solid that results when R is revolved about the x-axis?
Washer Method about the x-axis. Let f and g be continuous functions with f (x) ≥ g(x) ≥ 0 on
[a, b]. Let R by the region bounded by the curves y = f (x) and y = g(x), and the lines x = a and x = b.
When R is revolved about the x-axis, the volume of the resulting solid of revolution is
V =
Rb
a
π(f (x)2 − g(x)2 )dx.
Example 4.3. The region R is bounded by the graphs of f (x) = x2 + 2 and g(x) = 4 − x2 . Find the
volume of the solid that results when R is revolved about the line y = −3.
Now let f be a continuous function, and let R be the region bounded by the graph of f , the x-axis,
and the lines x = a and x = b. The washer method gave us an easy way to compute the volume of the
solid generated when R is revolved around the x-axis, but what about when R is revolved about the
y-axis instead? We wish to have a convenient method for measuring this kind of solid of revolution.
Let’s consider again a type of 3-dimensional ”Riemann sum.” Divide the interval [a, b] up into n equal
subintervals, which divides R up into n chunks. When one of these chunks is revolved about the y-axis,
instead of a prism, we get a a cylindrical shell. To estimate the volume of this shell, imagine cutting it
in one place and stretching it out to look like a rectangular prism. Then its volume will be given by its
4
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
width, which is the length of the subinterval ∆X, times its height f (xk ) (where xk is some point in the
subinterval), times its length, which is the circumference of the circle of radius 2πxk . So the volume of
each chunk is approximately 2πxk · f (xk ) · ∆x.
When we add
Pnup the volumes of each of the n chunks, we get an estimate for the volume of the solid,
given by V ≈ k=1 2πxk f (xk )∆x. Then to obtain the exact volume, we again take the limit is n → ∞,
Rb
to find that V = a 2πxf (x)dx.
The above gives a computation of the simplest case, where R is bounded by just one function f , and
the x-axis. The shell method given below will be true in the more general case, where R is bounded
above by a function f and bounded below by a function g.
Shell Method about the y-axis. Let f and g be continuous functions with f (x) ≥ g(x) on [a, b]. If
R is the region bounded by the curves y = f (x) and y = g(x) between the lines x = a and x = b, the
volume of the solid generated when R is revolved about the y-axis is
V =
Rb
a
2πx(f (x) − g(x))dx.
Example
Let R be the region bounded by the graph of f (x) = sin x2 , the x-axis, and the vertical
p4.4.
π
line x = 2 . Find the volume of the solid generated when R is revolved about the y-axis.
Example 4.5. Let R be the region bounded by the graphs of f (x) = x(5 − x) and g(x) = 8 − x(5 − x).
Find the volume of the solid obtained by rotating R about the y-axis.
5
Integration by parts.
At this point in our development of the calculus, the student should recognize that our differentiation
techniques (chain rule, product rule, quotient rule, etc.) give us a tremendous amount of power to compute derivatives, but that our integration techniques are comparatively lacking. So far the only major
integration tool we have developed is the substitution rule, which one should think of as a “reverse chain
rule.” We now wish to expand the number of integration tools at our disposal by introducing a “reverse
product rule.”
So consider a pair of differentiable functions u(x) and v(x). The product rule tells us that
d
dx u(x)v(x)
= u0 (x)v(x) + u(x)v 0 (x)
and hence
u(x)v 0 (x) =
d
dx u(x)v(x)
− v(x)u0 (x)
Now we antidifferentiate both sides (and invoke the Fundamental Theorem of Calculus) to obtain an
integration rule, which we call integration by parts:
R
u(x)v 0 (x)dx = u(x)v(x) =
R
v(x)u0 (x)dx
Note that in the above expression we need not include any +C term, since there is an indefinite integral on both sides of the equation. We may also write the rule for integration by parts in the following
memorable compact fashion, by using just a slight abuse of notation:
R
udv = uv −
R
vdu
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
5
Example 5.1. Compute the following antiderivatives using integration by parts.
R
(1) R xex dx
(2) x sin xdx
It may be necessary to use integration by parts more than one time, as the following example illustrates.
Example 5.2. Compute the following antiderivatives.
R
(1) R x2 ex dx
(2) e2x sin xdx
The Fundamental Theorem of Calculus also implies that we may use integration by parts to compute
definite integrals, using the following rule.
Rb
a
u(x)v 0 (x)dx = [u(x)v(x)]ba −
Rb
a
v(x)u0 (x)dx
Example 5.3. Compute the following definite integrals.
R2
(1) 1 ln xdx
R π/2
(2) 0 x cos 2xdx
6
Trigonometric Integrals.
In the next few examples we will make use of the following well-known trigonometric identities, called
respectively the Pythagorean identity and the half-angle formulas.
sin2 x + cos2 x = 1
sin2 x = 21 (1 − cos 2x)
cos2 x = 12 (1 + cos 2x)
Example 6.1. Evaluate the following integrals.
R
(1) R cos5 xdx
(2) R sin3 x cos4 xdx.
(3) sin4 xdx
R
General Strategy for Evaluating Integrals of the Form sinn cosm xdx: If n is odd use substitution rule with u = cos x and if m is odd use substitution rule with u = sin x. If both n and m are even,
resort to the half-angle formulas (this is usually a bit longer).
Example 6.2. Evaluate the following integrals.
R
(1) R sin4 x cos2 xdx
(2) sin3 x cos−2 xdx
The student can see from the above examples that evaluating integrals involving powers of trigonometric functions often involves using identities to reduce the exponents involved to lower, more manageable
levels. Of course, this process can tend to become tedious when dealing with large exponents. For this
purpose, we introduce the following reduction formulas which the student may use at will from now on:
R
sinn xdx = −
R
cosn xdx =
sinn−1 x cos x n − 1 R
+
sinn−2 xdx
n
n
cosn−1 x sin x n − 1 R
+
cosn−2 xdx
n
n
6
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
R
R
tann xdx =
secn xdx =
tann−1 x R
− tann−2 xdx
n−1
secn−2 x tan x n − 2 R
secn−2 xdx
+
n−1
n−1
The above will more than suffice for our purposes. The student can consult Rogawski Section 7.2 for
a much more complete list of reduction formulas.
Example 6.3. Evaluate sec6 xdx.
Our last two problems are substitution rule problems.
R
Example 6.4. Find tan xdx.
R
Example 6.5. Find sec xdx.
Fact 6.6. The following hold:
R
R
7
tan xdx = ln | sec x| + C
sec xdx = ln | sec x + tan x| + C
Trigonometric Substitutions
The main trick we will develop in this section is based√on the following idea: Suppose we are trying to
evaluate an integral which involves a term of the form a2 − x2 , where a is some positive real number.
For this term to make sense, we must have x2 ≤ a2 and hence −a ≤ x ≤ a. So −1 ≤ xa ≤ 1, and hence
we may set θ = arcsin xa . In other words, we may write x = a sin θ for some angle θ. Then if we make
p
p
√
a change of variable, we may write a2 − x2 = a2 − a2 sin2 θ = a 1 − sin2 θ = a| cos θ|. In other
words, by substituting a trigonometric function into some expression, it may become possible to simplify
the original expression using our familiar trigonometric identities. We’ll use this idea for the next few
examples.
R
Example 7.1. Evaluate (16−x12 )3/2 dx.
Example 7.2. Use integration to compute the area of a perfect circle of radius r.
Fact 7.3. The following are all equivalent forms of the Pythagorean trigonometric identity (Why?):
sin2 x = 1 − cos2 x
cos2 x = 1 − sin2 x
sec2 x = tan2 x + 1
tan2 x = sec2 x − 1
General Strategy for Evaluating Integrals Using a Trigonometric Substitution: First identify
whether or not a straightforward u-substitution is a better idea. If not, then
(1) if you see a term of the form a2 − x2 for some number a, try x = a sin θ.
(2) if you see a term of the form x2 + a2 for some number a, try x = a tan θ.
(3) if you see a term of the form a2 − x2 for some number a, try x = a sec θ.
Example 7.4. Evaluate
R
x2
dx.
(4−x)5/2
Example 7.5. Evaluate
R
1
(1+x2 )2 dx.
Example 7.6. Evaluate
R
x2
√1
dx.
x2 −9
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
8
7
Integration of Rational Functions by the Method of Partial Fractions.
Z
Example 8.1. Evaluate
3x
dx.
x2 + 2x − 8
Solution. The key to evaluating this integral is knowing that every rational function may be rewritten
as a partial fraction expansion. Note that the denominator x2 + 2x − 8 in the integrand above factors
into (x + 4)(x − 2). Then it is possible to find two real numbers A and B for which
3x
A
B
3x
=
=
+
.
x2 + 2x − 8
(x + 4)(x − 2)
x+4 x−2
The only trick is to find just what the values of A and B should be. To find the values of A and B,
multiply both sides of the equation above by (x + 4)(x − 2) to clear all denominators. Then combine
like terms:
3x = A(x − 2) + B(x + 4)
= Ax − 2A + Bx + 4B
= (A + B)x + (−2A + 4B)
Now we have obtained the equality 3x = (A + B)x + (−2A + 4B); it follows that we must have
3 = A + B and 0 = −2A + 4B. So we have a system of two equations in two variables; now we may
resort to our algebra techniques! If we carefully solve the given system, we should get A = 2 and B = 1.
Thus we may convert our original integral into the following, much easier problem:
Z
Z
Z
3x
2
1
dx
=
dx
+
dx
2
x + 2x − 8
x+4
x−2
= 2 ln |x + 4| + ln |x − 2| + C.
3x + 7x − 2
dx.
x3 − x2 − 2x
Z
5x2 − 3x + 2
dx.
x3 − 2x2
Example 8.2. Evaluate
Example 8.3. Evaluate
2
Z
Note: The same techniques we have used above will work for the last example above, if we keep track
of one important detail: Notice that the denominator x3 − 2x2 factors into x2 (x − 2). Since x appears
as a factor with multiplicity 2; we must represent x twice in the partial fraction expansion: once as an
x term, and once as an x2 term. In other words, we will be able to find a partial fraction expansion of
the form
A
B
5x2 − 3x + 2
C
= + 2+
,
2
x (x − 2)
x
x
x−2
for some real numbers A, B, and C. Otherwise we may proceed as usual.
Z
7x2 − 13x + 13
Example 8.4. Evaluate
dx.
(x − 2)(x2 − 2x + 3)
Note: To solve the above, we must again keep track of an important detail. Notice that the polynomial
x2 − 2x + 3 which appears in the denominator is an irreducible quadratic, i.e. it cannot be factored
into binomials with real coefficients. We may still find a partial fraction expansion for the integrand,
8
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
but it will take the form
7x2 − 13x + 13
A
Bx + C
=
+ 2
,
2
(x − 2)(x − 2x + 3)
x − 2 x − 2x + 3
where A, B, and C are some real numbers. A similar principle holds whenever an irreducible quadratic
appears in the denominator of a rational expression. These integrals can in general be challenging to
compute, and may involve both the method of completing the square (from the student’s previous algebra
course) and the use of the arctangent function.
Z
z+1
dz.
Example 8.5. Evaluate
z(z 2 + 4)
9
Improper Integrals.
Example 9.1. Let b be any real number greater than 1. Compute the following integrals.
Rb
(1) 1 1dx
(2)
Rb
1
dx
1 x2
Solution. We have
Z
b
1dx = [x]b1
1
=b−1
and
Z
1
b
1
1
dx = [− ]b1
x2
x
1
=1− .
b
Let us consider the geometric meanings of the expressions we have computed above. If we sketch the
Rb
picture, the first integral 1 1dx may be interpreted as the area of the box with the x-axis for a base,
the line x = 1 for a left side, the line y = 1 for a top side, and the line x = b for a right side. If we
allow b to grow large, i.e. we allow the right side of the box to slide further right, the integral confirms
our intuition that the area of the box becomes larger and larger. It grows unboundedly, i.e. we have
Z b
lim
1dx = lim (b − 1) = ∞.
b→∞
1
b→∞
Rb
On the other hand, the second integral 1 x12 dx = 1− 1b behaves remarkably differently. If we compute
the area bounded above by the graph of x12 between, say, 1 and 2, we get 1 − 12 = 12 . If we compute from
1 to 3 we get 32 ; if we compute from 1 to 4 we get 34 , and so on. In general the area is increasing, which
confirms our intuition. But what happens as b becomes very large? Notice that no matter how large b
is, the area 1− 1b will never exceed 1, i.e. the area does not grow unboundedly as in the previous example.
Thus the region bounded by the curve x12 , the x-axis, and the line x = 1 is infinitely long but has
Z b
1
1
finite area. Since lim
dx = lim 1 − = 1, we say that the region has area 1.
b→∞ 1 x2
b→∞
b
This is our first glimpse at a phenomenon that we will consider extensively during our sequences/series
unit later in the semester: that of an infinite “process” which yields a finite result.
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
9
Definition 9.2. Let f be a continuous function. We define improper integrals with infinite bounds
of integration as below, provided the given limits exist:
Z ∞
Z b
f (x)dx = lim
f (x)dx for any real number a;
b→∞
a
Z
b
a
Z
b
f (x)dx for any real number b;
f (x)dx = lim
a→−∞
−∞
Z
∞
Z
c
a→−∞
Z
b
f (x)dx + lim
f (x)dx = lim
−∞
a
b→∞
a
f (x)dx for any real number c.
c
If the limit exists and is finite, then we say the integral converges. If the limit does not exist, or is
equal to ∞ or −∞, then we say the integral diverges.
Example 9.3. Evaluate each integral.
R∞
(1) 0 e−3x dx
(2)
R∞
(3)
R∞
0
1
1
1+x2 dx
x−1 dx
Example 9.4. Let R be the region bounded by the graph of y = x−1 and the x-axis, to the right of the
line x = 1.
(1) What is the volume of the solid generated when R is revolved about the x-axis?
(2) What is the volume of the solid generated when R is revolved about the y-axis?
R∞
Example 9.5. Compute 0 xe−x dx.
We may also define improper integrals for functions with vertical asymptotes, as we see in the next
example.
R1
Example 9.6. What should the value of 0 √1x dx be?
Definition 9.7. Let f be a function continuous at x 6= p and with a vertical asymptote at x = p. Define
the improper integrals (with unbounded integrand) as follows:
Z p
Z c
f (x)dx = lim
f (x)dx for a < p;
c→p−
a
b
Z
p
Z
a
a
b
Z
f (x)dx = lim+
b
d→p
Z
f (x)dx = lim−
c→p
a
f (x)dx for p < b;
d
c
Z
f (x)dx + lim+
d→p
a
f (x)dx for a < p < b.
d
Again, we say the integral converges if the limit exists and is finite; otherwise we say the integral
diverges.
R3
1
Example 9.8. Compute −3 √9−x
dx.
2
Example 9.9. State whether the following integrals converge or diverge. If they converge, give the
value of the integral. (Be careful!)
R1 1
(1) −1 x1/3
dx
(2)
R1
1
dx
−1 x3
Example 9.10.
(1) Determine for what values of p the integral
R∞
1
1
xp dx
converges or diverges.
10
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
(2) Determine for what values of p the integral
R1
1
dx
0 xp
converges or diverges.
Theorem 9.11. Let p > 0. Then
(1) if p > 1,
R∞
(2) if p < 1,
R∞
1
1
(3) if p = 1, both
10
=
1
xp dx
diverges but
R∞
1
1
p−1
R1
1
xp dx
1
xp dx
but
and
1
dx
0 xp
diverges.
R1
1
dx
0 xp
R1
1
dx
0 xp
p
1−p .
=
diverge.
Probability and Integration
Definition 10.1. A probability density function p(x) is a non-negative function (i.e. p(x) ≥ 0 for
all x) with the property that
R∞
−∞
p(x)dx = 1.
A random variable is an unknown quantity X that takes numerical values, subject to some probability distribution. For real numbers a < b, the probability that X is between a and b is denoted
P (a ≤ X ≤ b).
We say that X is a continuous random variable if there exists a continuous probability density
function p(x) with the property that
P (a ≤ X ≤ b) =
Example 10.2. Let p(x) =
Rb
a
p(x)dx, for all a < b.
C
x2 +1 .
(1) Assume p(x) is a probability density function. Find C.
(2) Suppose X is a continuous random variable with density function p. Find the probability that
X is between 1 and 4.
Example 10.3. Let p(x) be any probability density function, and X an associated random variable.
Find the probability that X = a, where a is any real number.
Definition 10.4. An exponential probability density is a function of the form
p(t) = 1r e−t/r , where r is some positive real number.
The number r in the above expression is the mean or average value of a random variable X with
density function p.
Example 10.5. The waiting time T between customer arrivals in a drive-through fast food restaurant
is a random variable with exponential probability density. If the average waiting time is 60 seconds,
what is the probability that a customer will arrive within 30 to 45 seconds after another customer?
Definition 10.6. The standard normal density is the function
p(x) =
2
√1 e−x /2 .
2π
The above has average value 0 and standard deviation 1.
More generally, the normal density with mean µ and standard deviation σ is the function
p(x) =
2
2
√1 e−(x−µ) /(2σ ) .
σ 2π
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
11
If a random variable has a normal density function then we say it has a normal distribution or a
Gaussian distribution.
11
Numerical Integration
The last integration tool we will develop is not really a method for integrating at all, but rather a
technique for closely estimating the value of a definite integral when finding an explicit antiderivative
is impossible. (For example, we have no explicit antiderivative for the standard normal density p(x) =
2
√1 e−x /2 .)
2π
The following two methods give area estimates using Riemann sums.
Trapezoidal Rule. Let f (x) be an integrable function on [a, b], and let N be a positive integer. Setting
∆x = b−a
, x0 = a, and xj = x0 + j∆x for each integer j ≤ N , the N -th trapezoidal approximation
Rb N
to a f (x)dx is
TN = 12 ∆x[f (x0 ) + 2f (x1 ) + 2f (x2 ) + ... + 2f (xN −1 ) + f (xN )].
Midpoint Rule. Let f (x) be an integrable function on [a, b], and let N be a positive integer. Setting
∆x = b−a
letting cj denote the midpoint of
N , x0 = a, and xj = x0 + j∆x for each integer j ≤ N , and
Rb
each interval [xj−1 , xj ], the N -th midpoint approximation to a f (x)dx is
MN = ∆x[f (c1 ) + f (c1 ) + ... + f (cN )].
Definition 11.1. Define the numbers Error(TN ) and Error(Mn ) by
Error(TN ) = |
Rb
Error(MN ) = |
Rb
a
a
f (x)dx − TN |;
f (x)dx − MN |.
Theorem 11.2 (Error Bounds). Let f (x) be integrable on [a, b]. Assume f 00 (x) exists and is continuous.
Let K2 be any number for which |f 00 (x)| ≤ K2 for all x in [a, b]. Then
Error(TN ) ≤
Example 11.3. Let f (x) =
√
K2 (b − a)3
K2 (b − a)3
and Error(Mn ) ≤
.
2
12N
24N 2
x.
R4√
(1) Calculate the estimates T6 and M6 for 1 xdx. (4.6615 and 4.6692 respectively.)
(2) Calculate the error bounds for T6 and M6 . (0.0156 and 0.0078 respectively.)
(3) Calculate the integral exactly. How close were we?
R3
2
Example 11.4. Find N such that Tn approximates 0 e−x dx with an error of at most 10−4 = 0.0001.
(N = 213 ≥ 212.1.)
Simpson’s Rule. Let f (x) be an integrable function on [a, b], and let N be a positive even integer.
Setting ∆x = b−a
x0 = a, and xj = x0 + j∆x for each integer j ≤ N , the N -th Simpson’s rule
N , R
b
approximation to a f (x)dx is
SN = 13 TN/2 + 23 MN/2 = 13 ∆x[f (x0 ) + 4f (x2 ) + 2f (x2 ) + ... + 4f (xN −3 ) + 2f (xN −2 ) + 4f (xN −1 ) + f (xN )].
12
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
Theorem 11.5 (Error Bound for Simpson’s Rule). Let f (x) be integrable on [a, b]. Assume the fourth
derivative f (4) (x) exists and is continuous on [a, b]. Let K4 be any number for which |f (4) (x) ≤ K4 | for
all x in [a, b]. Then
Error(SN ) ≤
K4 (b − a)5
.
180N 4
Example 11.6. Find N such that the approximation SN to
0.000001. (N = 46 ≥ 45.45.)
12
R3
1
dx
1 x
has an error of at most 10−6 =
Arc Length and Surface Area
Fact 12.1. Let f (x) be a differentiable function, and assume f 0 (x) is continuous on [a, b]. The the arc
length s of the graph of f (x) from x = a to x = b is
s=
Rbp
a
1 + [f 0 (x)]2 dx.
Example 12.2. Find the arc length s of the graph of f (x) =
1 3
12 x
+ x−1 over [−1, 3].
Example 12.3. Estimate the arc length s of the graph of y = sin x over [0, π], using Simpson’s rule
with N = 6. (s ≈ 3.82.)
Fact 12.4. Let f (x) be a differentiable function, and assume f (x) ≥ 0 and f 0 (x) is continuous on [a, b].
The surface area S of the surface obtained by rotation the graph of f (x) about the x-axis for a ≤ x ≤ b
is
S = 2π
Rb
a
p
f (x) 1 + [f 0 (x)]2 dx.
Example 12.5. Calculate the surface area of a sphere of radius R.
Example 12.6. Find the surface area S of the surface obtained by rotating the graph of f (x) =
x1/2 − 31 x3/2 about the x-axis for 1 ≤ x ≤ 3.
13
Fluid Pressure and Force
Fact 13.1. Let g be the gravity constant, i.e. g ≈ 9.8 meters or g ≈ 32 feet. The pressure p at depth
h in a fluid of mass density ρ is
p = ρgh .
The pressure acts at each point on an object in the direction perpendicular to the object’s surface at
that point. If the pressure is constant on a surface of area A, then the fluid force F exerted on the
surface is
F = pA.
Example 13.2. Calculate the fluid force on the top and bottom of a box of dimensions 2 × 2 × 5 meters,
submerged in a pool of water with its top 3 meters below the water surface. The density of water is
ρ = 103 kg/m3 .
Example 13.3. Calculate the fluid force on each side of the box in the example above.
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
14
13
Sequences
Definition 14.1. A sequence is an infinite list of numbers, enumerated by the natural numbers starting from 1. We use the following notations to refer to the sequence whose n-th term is the number an :
(a1 , a2 , a3 , ..., an , ...) = (an )∞
n=1 = (an ).
We should start with a few examples of sequences.
Example 14.2. For each sequence defined below, write out the first 5 terms of the sequence.
(1) Let a1 = 1, and for each integer n ≥ 1, let an+1 = 3 + an .
(2) Let f (x) = x3 − 1. For each integer n ≥ 1, let an = f (n).
(3) For each integer n ≥ 1, let an =
1
2n .
(4) For each integer n ≥ 1, let an be the n-th digit after the decimal point in the decimal expansion
of the number π − 3.
Each of the examples above clearly defines a sequence, but they are all defined by different means.
For the first example, we say that the sequence (an ) is defined by a recurrence relation, or we say that
it is defined implicitly. For the second and third examples, the sequence (an ) is defined explicitly, i.e.
each term an may be computed exactly as a well-defined function f of n.
We have not defined the fourth example via a recurrence relation nor by an explicit formula, but it is
still a valid sequence nonetheless.
Example 14.3. Let (an ) = (−2, 5, 12, 19, ...) and let (bn ) = (3, 6, 12, 24, 48, ...).
(1) Find recurrence relations that could generate the terms of the sequences (an ) and (bn ).
(2) Using the relations you found, compute what the next two terms could be in each sequence.
(3) For each sequence, find an explicit formula for the n-th term generated by your recurrence
relation.
Definition 14.4. If the terms of a sequence (an )∞
n=1 become arbitrarily close to some fixed number L
as n grows sufficiently large, then we say that L is the limit of (an ), or that (an ) converges to L. We
also write
lim an = L, or (an ) → L,
n→∞
so long as the latter notation does not introduce confusion about what limit we are taking.
If there is no such number L then we say the sequence diverges.
Example 14.5. Write the first four terms of each sequence. Do you believe the limit converges or
diverges? If you believe it converges, then to what value?
2
∞
2n − 1
(1)
2n2
∞n=1
(−1)n
(2)
n2 + 1 n=1
(3) (cos(πn))∞
n=1
(4) (cos(2πn))∞
n=1
14
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
(5) (n)∞
n=1
(6) (n!)∞
n=1
(7) (an )∞
n=1 , where a1 = 1 and an+1 = −2an for all integers n ≥ 1.
(8) (an )∞
n=1 , where an is the n-th digit after the decimal point in the decimal expansion of π − 3
The next theorem makes it easier to compute limits of explicitly-defined sequences.
Theorem 14.6. Suppose f is a function for which f (n) = an for all positive integers n. If lim f (x) =
x→∞
L, then lim an = L.
n→∞
Example 14.7. Does the converse of the above theorem hold? That is, is it true that if f (n) = an for
some function f , and lim an = L, then lim f (x) = L?
n→∞
x→∞
In addition to the above theorem, the following nice properties all hold for any convergent sequences
(an ), (bn ) with lim an = A and lim bn = B..
n→∞
n→∞
(1) lim (an ± bn ) = A ± B;
n→∞
(2) lim can = cA, where c is any constant;
n→∞
(3) lim an bn = AB; and
n→∞
(4) lim
n→∞
an
A
= , provided B 6= 0.
bn
B
Example 14.8. Compute the limits of the following sequences, if they exist.
3n3
(1) (an ) where an = 3
.
n +1
n + ln n
(2) (bn ) where bn =
.
n3/2
(3) (cn ) where cn = e−n n10 .
Theorem 14.9 (Squeeze Theorem for Sequences). Let (an ), (bn ), and (cn ) be sequences such that for
some integer M ,
bn ≤ an ≤ cn for all n > M .
If lim bn = lim cn = L, then so too does lim an = L.
n→∞
n→∞
n→∞
n
Example 14.10. Prove that lim
n→∞
R
= 0 for all real numbers R > 0.
n!
Definition 14.11. If (an ) is a sequence for which an+1 ≥ an for every n, then we say (an ) is nondecreasing. If (an ) satisfies an+1 ≤ an for every n, then we say (an ) is nonincreasing. If (an ) is either
nonincreasing or nondecreasing, we say (an ) is monotone. If there exists any real number M for which
|an | ≤ M for every integer n, then we say the sequence (an ) is bounded.
Example 14.12. Are the following sequences monotone? Bounded?
(1) (1 − n1 )∞
n=1
(2) (15 − 3n)∞
n=1
Theorem 14.13. Every bounded monotone sequence converges to some limit.
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
15
15
Relative Growth Rates
Definition 15.1. Let (an ) and (bn ) be two sequences for which (an ) → ∞ and (bn ) → ∞. Let
bn
M = lim
, if this limit exists.
n→∞ an
(1) If 0 < M < ∞, we say that (an ) and (bn ) have comparable growth rates.
(2) If M = 0, we say that (an ) grows faster than (bn ).
(3) If M = ∞, we say that (bn ) grows faster than (an ).
Fact 15.2. The following sequences are ranked in order, from the slowest growth rate to the fastest
growth rate.
(1) (ln n)∞
(Logarithmic Growth)
n=1 .
(2) ((ln n)r )∞
(Logarithmic Growth)
n=1 where r > 1.
(3) (nq )∞
(Polynomial Growth)
n=1 where 0 < q < 1.
(4) (n)∞
(Special Case of Polynomial Growth: Linear Growth)
n=1 .
(5) (np )∞
(Polynomial Growth)
n=1 where p > 1.
(6) (bn )∞
(Exponential Growth)
n=1 where b > 1.
(7) (n!)∞
(Factorial Growth)
n=1 .
(8) (nn )∞
(Superexponential Growth)
n=1 .
Example 15.3. For any pair of sequences (an ) and (bn ) in the list above, verify the order of growth
bn
. (L’Hospital’s rule is a helpful tool here.)
rates by computing lim
n→∞ an
16
Infinite Series
Now we arrive at one of the major developments of the course- the extension of the basic arithmetical
operation of addition, which up until now in our education has formally applied only to finite collections
of numbers, to encompass infinite collections of numbers. We have already seen during our study of
improper integrals (see, for example, the comments following Example 9.1) that it is possible and natural
for an “infinite process” to yield only a ”finite result.” We involve limits to formalize this idea.
Let’s look at some explicit examples.
Example 16.1. Find a geometric way to visualize the “infinite sum”
1
1
1
1
1
2 + 4 + 8 + 16 + ... + 2n + ...
What should the value of this sum be? Consider the area of a unit square in the xy-plane.
Example 16.2. Does it make sense to compute the infinite sum 1 + 1 + 1 + 1 + ... ?
Now we will formalize the concept of “infinite addition” by applying the technology of sequences
which we developed in the previous section.
Definition 16.3. Let (an )∞
n=1 be an infinite sequence. We refer to the infinite sum
a1 + a2 + a3 + ...
as an infinite series. We denote this series by
∞
X
an
n=1
which is a natural extension of our sigma notation for finite sums.
Note here that sequences and series are intimately connected, but that they represent different concepts. A sequence is always just an infinite list of numbers without any additional structure; a series
always refers to an infinite addition of numbers.
16
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
Definition 16.4. Let
∞
X
an be an infinite series. Define a new sequence (Sn )∞
n=1 as follows:
n=0
S1 = a1
S2 = a1 + a2
S3 = a1 + a2 + a3
S4 = a1 + a2 + a3 + a4
...
In general, define Sn by Sn = a1 + a2 + ... + an for every integer n ≥ 1.
We call this sequence (Sn ) the sequence of partial sums of
L, then we say that the series
∞
X
∞
X
an . If (Sn ) converges to some limit
n=0
an converges to L, and we write
n=0
∞
X
an = L.
n=0
If the sequence (Sn ) diverges, then we say that the series
∞
X
an diverges.
n=1
Example 16.5.
(2) Compute
(1) Compute
∞
X
∞
X
1
, if the series converges.
2n
n=1
1, if the series converges.
n=1
Example 16.6. Consider the infinite series
∞
X
1
.
n(n
+ 1)
n=1
(1) Find the first four terms in the sequence of partial sums.
(2) Do you think the series converges? If so, to what sum?
Example 16.7. Compute
∞
X
(−1)n n, if the sum converges. (Be careful!)
k=1
17
Geometric Series and Telescoping Series
Definition 17.1. A geometric series is a series of the form
∞
X
arn = a + ar + ar2 + ar3 + ...
n=0
where a and r are some fixed real numbers.
Example 17.2. Consider the series
3+
3
2
+
3
4
+
3
8
+
3
16
+ ...
Is the series geometric? Does it converge, and if so to what value?
It is not hard to see that if
∞
X
arn is a geometric series where |r| ≥ 1, then the series diverges. On
n=0
the other hand, if we take |r| < 1, then we can prove the following theorem.
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
Theorem 17.3 (Convergence of Geometric Series). Let
∞
X
17
arn be a geometric series with |r| < 1. Then
n=0
the series always converges, and moreover we can compute the sum by the following formula:
∞
X
arn =
n=0
a
.
1−r
Proof. For each n set
Sn = a + ar + ... + arn−1 + arn
So (Sn ) is the sequence of partial sums. We must check if (Sn ) converges.
We can do this by making the following clever observation. Multiply both sides of the equation above
by r. Then we get
rSn = r(a + ar + ... + arn )
= ar + ar2 + ... + arn + arn+1 .
Subtracting the two equalities we’ve obtained, and then solving for Sn , we get the following.
Sn − rSn = (a + ar + ... + arn ) − (ar + ar2 + ... + arn + arn+1 )
Sn − rSn = a − arn+1
(1 − r)Sn = a(1 − rn+1 )
Sn = a ·
1 − rn+1
.
1−r
So we have an explicit formula for the value of each partial sum Sn = a ·
do is take the limit. Since |r| < 1, we have lim rn+1 = 0, and hence
1−r n+1
1−r .
All that remains to
n→∞
∞
X
n=0
an = lim Sn
n→∞
= lim a ·
n→∞
1 − rn+1
1−r
1
1−r
a
=
.
1−r
=a·
So the series converges to
a
as we hoped.
1−r
Example 17.4. Compute the following geometric sums, if they converge.
18
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
(1)
(2)
∞
X
n=0
∞
X
1.1n
e−k
n=0
∞
X
n
3
(3)
3
4
n=0
n
∞
X
3
(4)
3
4
n=2
n
∞
X
3
(5)
3 −
4
n=0
Example 17.5. Suppose (an ) is a P
sequence for which limn→∞ an = L, where L > 0 is some positive
∞
number. Is it possible for the series n=1 an to converge?
Theorem 17.6 (Divergence Test). If
converge to 0, then
∞
X
∞
n=1 an
converges, then lim an = 0. Equivalently, if (an ) does not
n→∞
an diverges.
n=1
Example 17.7 (Telescoping Series). Evaluate the following series.
∞ X
1
1
−
(1)
3k
3k+1
k=1
∞
X
1
(2)
(k + 1)(k + 2)
k=1
18
The Harmonic Series
Consider the infinite series:
∞
X
1
1 1 1 1
= 1 + + + + + ...
n
2 3 4 5
n=1
The series above is called the harmonic series. Does the harmonic series converge or diverge?
Since the sequence ( n1 )∞
n=1 converges to 0, it seems plausible that the harmonic series might converge.
Computing the first few terms in the sequence of partial sums, we may obtain:
S1 = 1
3
S2 =
2
11
S3 =
6
25
S4 =
12
...
The trend of convergence or divergence may not be clear from the first few terms. What about if we
consider very large n-th partial sums? It can be shown that the sum of the first 1, 000, 000 terms of the
harmonic series is less than 15. The number of atoms comprising the planet Earth is (roughly) 1050 ;
the sum of the first 1050 terms of the harmonic series is approximately 115.6. The number of atoms in
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
19
the observable universe is (very roughly) 1080 ; the sum of the first 1080 terms of the harmonic series is
approximately 184.7.
It may be unclear from this information whether the series in fact converges or diverges. In fact, we
can show through an elementary argument that the harmonic series diverges. Simply make the following
observation:
∞
X
1
1 1 1 1 1 1
1
+ + + + + + +...
= |{z}
1 +
n
2
|{z} |3 {z 4} |5 6 {z 7 8}
n=1
≥1/2
≥1/2
≥1/2
≥1/2
∞
X
1
must involve adding terms whose sum is ≥
n
n=1
times, the sum must diverge to infinity. It just diverges very, very slowly!
Since computing the infinite sum
1
2
infinitely many
Example 18.1. Prove that lim ln x = ∞.
x→∞
Theorem 18.2 (Integral Test). Suppose f is a continuous, positive, decreasing function for x ≥ 1 and
∞
X
an and the improper
suppose an = f (n) for every positive integer n ≥ 1. Then the infinite sum
n=1
R∞
integral 1 f (x)dx either both converge, or else both diverge. However, if they both converge, their
values need not in general be the same.
Example 18.3. Determine whether the following series converge or diverge.
∞
X
n
(1)
n
+
1
n=0
∞
X k
(2)
k2 + 1
(3)
(4)
k=1
∞
X
√
k=3
∞
X
1
2k − 5
1
, where p > 1 is some fixed real number.
p
n
n=1
Theorem 18.4 (Convergence of p-series).
∞
X
1
always converges if p > 1, and diverges otherwise.
np
n=1
Proof. This theorem is the direct consequence of Theorem 9.11 and Theorem 18.2.
In the next few sections we will develop many methods for determining whether a given series converges
or diverges. In general it will not always be completely clear which test, if any, will be the appropriate
one to apply when faced with any particular series. It is up to the student to develop the intuition and
problem-solving skills for attacking such problems.
19
The Direct Comparison Test
Theorem 19.1 (Direct Comparison Test). Let
∞
X
n=1
that 0 < an ≤ bn for every integer n ≥ 1.
∞
∞
X
X
(1) If
bn converges, then
an converges.
n=1
n=1
an and
∞
X
n=1
bn be series with positive terms, and such
20
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
(2) If
∞
X
an diverges, then
n=1
∞
X
bn diverges.
n=1
Example 19.2. Determine whether the following series converge.
(1)
(2)
(3)
∞
X
n=1
∞
X
k=1
∞
X
k=1
20
√
1
n3n
k3
−1
2k 4
ln k
k3
The Limit Comparison Test
∞
X
Theorem 20.1 (Limit Comparison Test). Let
an and
n=1
lim
n→∞
(1) If 0 < L < ∞, then
(2) If L = 0 and
∞
X
∞
X
an and
n=1
∞
X
n=1
bn converges, then
n=1
∞
X
(3) If L = ∞ and
bn diverges, then
n=1
∞
X
bn be series with positive terms, and let
n=1
an
= L.
bn
bn either both converge or both diverge.
∞
X
n=1
∞
X
an converges.
an diverges.
n=1
Example 20.2. Determine whether the following series converge.
(1)
(2)
∞
X
k 4 − 2k 2 + 3
k=1
∞
X
k=1
21
2k 6 − k + 5
ln k
k2
Alternating Series
Example 21.1. Determine if the series
∞
X
(−1)k+1
k=1
k
converges or diverges. (We call this series the
alternating harmonic series.)
k
X
1 1 1
(−1)k+1
(−1)k+1
= 1 − + − + ... +
, so Sk is the sequence
Solution. For each k ≥ 1 let Sk =
k
2 3 4
k
n=1
of partial sums of the given series.
First consider the sequence which consists only of the even terms (S2 , S4 , S6 , ..., S2k , ...). Notice that
since 13 > 41 , we have 13 − 14 > 0, and hence S4 = S2 + 13 − 14 > S2 . In fact for any even number 2k, we
1
1
1
1
have 2k+1
− 2k+2
> 0 and hence S2k+2 = S2k + 2k+1
− 2k+2
> S2k . It follows that the sequence of even
terms (S2 , S4 , ...) is strictly increasing.
In addition, the sequence (S2 , S4 , ...) is bounded above by 1 = S1 . So the sequence (S2 , S4 , ...) is a
bounded monotone sequence, and hence it converges to some limit L by our theorem from earlier.
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
21
If we consider the sequence (S1 , S3 , S5 , ..., S2k−1 , ...) of odd partial sums, then arguments which are
very similar to the above will show that the sequence is strictly decreasing and bounded below by S2 ;
so the sequence of odd partial sums also converges to some limit, say M , by our same theorem from earlier.
If we can check that L = M , then we will have shown the series converges; otherwise we will have
1
shown it diverges. To see the relationship between the two limits, simply observe that S2k = S2k−1 − 2k
for every positive integer k. Hence we have
L = lim S2k
k→∞
1
= lim S2k−1 −
k→∞
2k
1
k→∞ 2k
= lim S2k−1 − lim
k→∞
=M −0
= M.
So the series converges!
Question 21.2. To what value does the alternating harmonic series converge? We don’t have the tools
to compute this right now, but we will see the answer in the future.
Notice that the argument above depends only on three facts: that the sequence of terms in the series
alternates signs; that the terms of the (non-alternating) harmonic series are a decreasing sequence; and
that the terms converge to 0. So the solution above may be easily modified to prove the following
theorem (which we leave to the reader):
Theorem 21.3 (Alternating Series Test). Suppose (an )∞
n=1 is a sequence of positive terms for which
(1) an+1 ≤ an for every positive integer n, and
(2) lim an = 0.
n→∞
Then the alternating series
∞
X
(−1)n+1 an converges.
n=1
Example 21.4. Determine whether the following series converge or diverge.
(1)
∞
X
(−1)k+1
k2
k=1
(2) 2 −
(3)
+
4
3
−
5
4
+
6
5
− ...
∞
X
(−1)k ln k
k=2
22
3
2
k
Absolute vs. Conditional Convergence
Definition 22.1. Let
∞
X
an be a series. We say that the series
n=1
(1) converges absolutely if
∞
X
n=1
|an | converges.
22
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
∞
X
(2) converges conditionally if
an converges but
n=1
Fact 22.2. If
∞
X
∞
X
|an | converges then
n=1
∞
X
|an | does not converge.
n=1
an converges.
n=1
Example 22.3. Do the following series converge absolutely? Conditionally?
P∞ (−1)n−1
(1)
n=1
n2
P∞ (−1)n−1
√
(2)
n=1
n
23
The Ratio Test
Theorem 23.1 (Ratio Test). Let
∞
X
an be a series. Let
n=1
an+1 .
ρ = lim n→∞
an (1) If 0 ≤ ρ < 1, the series converges absolutely.
(2) If ρ > 1 (including r = ∞), the series diverges.
(3) if ρ = 1, the test is inconclusive.
Example 23.2. Determine whether the following series converge.
(1)
(2)
∞
X
10n
n!
n=1
∞
X nn
n=1
24
n!
The Root Test
Theorem 24.1 (Root Test). Let
∞
X
an be a series. Let
n=1
L = lim
p
n
n→∞
|an |.
(1) If 0 ≤ L < 1, the series converges absolutely.
(2) If L > 1 (including p = ∞), the series diverges.
(3) If L = 1, the test is inconclusive.
Example 24.2. Determine whether the following series converge.
(1)
(2)
n
∞ X
4n2 − 3
n=1
∞
X
k=1
7n2 + 6
2k
k 10
General Strategy for Determining Convergence of a Series. There is no hard and fast rule for
determining whether a series converges or diverges, and in general the student should build up their
own intuition on the subject through practice. However, the following mental checklist can be a helpful
guideline.
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
(1) Divergence Test. Given a series
∞
X
23
an , the first thing one should always ask is, do the terms
n=1
(an ) → 0? If not, we are done and the series diverges by the DT. If the terms do shrink to zero,
however, we have to put more thought into the problem.
(2) Very Nice Series. Is the series a p-series with p > 1, or a geometric series, or a telescoping
series? If so we are done and the series converges by our general theorems. (Note that sometimes
∞
X
1
is a telescoping series in disguise.
a telescoping series can be “hidden”: for example
n(n
+ 1)
n=1
The method of partial fractions may reveal these.)
(3) Alternating Series Test. If the series is alternating and the terms go to zero, then the series
converges by the AST.
(4) Integral Test. Do the terms of the series look like a function we could probably integrate? (As
∞
X
2n
an example, the series
shouts “Substitution Rule” when we consider its improper
(1
+
n2 )2
R ∞ n=1
integral analogue 1 2x(1 + x2 )−2 dx. Remember the series and the improper integral are not
equal to one another, but they do either both converge or both diverge.)
(5) Direct Comparison Test and Limit Comparison Test. Do the terms an of the series behave
similarly to those of one of the Very Nice Series? If so, a comparison test may be appropriate.
∞
X
n
Usually the LCT is more powerful than the DCT. (For examples, the series
3
n + 5n2 − 17
n=1
∞
X 1
for very large n and so we think it probably converges; conversely
seems to behave like
n2
n=1
∞
X
3n7 + 100
the series
seems to behave like a constant multiple of the harmonic series for very
n8
n=1
large n, so we think it diverges. Usually the natural choices for a comparison test are a p-series
or a geometric series.)
(6) Ratio Test. Are we looking at a series of fractions where the question of convergence seems to
∞
X
n3
come down to comparing the growth rate of the numerator vs. the denominator? e.g.
or
3n
n=1
∞
∞
X
X
n
1000n
or
. Then Ratio Test has a good chance of working. (Ratio Test is almost
(ln n)4
n!
n=1
n=1
always the most natural choice if there is a factorial n! term appearing anywhere in the series.)
(7) Root Test. Are we looking at a series where “to the n-th power” occurs in a significant way?
n
−n2
∞ X
2
1
e.g.
or
1+
. Root Test might be easiest.
5n + 1
n
n=1
(8) Words of Warning. Your first attempt at a test may not be the right choice. Be aware of
when a test fails (for instance if the limit in Ratio/Root Test is exactly 1, or extremal cases of
Limit Comparison Test where limit is 0 or ∞, etc.) and be prepared to try a different strategy.
On that note, it is often very helpful to do some private scratchwork where you determine for
yourself if a particular test solves the problem, and then carefully write a solution for full credit
in a homework, quiz, or exam setting.
For practice, try working out each of the examples given in the strategy outline above!
24
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
25
A Note On Polynomials
Recall that a polynomial is a function of the form
p(x) = c0 + c1 x + c2 x2 + ... + cn xn =
n
X
ck xk ,
k=0
where c0 , c1 , ..., cn are some real numbers, which we call coefficients. The positive integer n in the
above is called the degree of the polynomial p(x).
Now let a be any real number, and consider the function q(x) defined below:
q(x) = c0 + c1 (x − a) + c2 (x − a)2 + ... + cn (x − a)n =
n
X
ck (x − a)k .
k=0
It is obvious that q(x) is another polynomial of degree n, and that q(x) = p(x − a). The reader should
recall from a previous course that if q(x) = p(x − a), then the graph of q(x) is just a shift of the graph
of p(x); in particular, a horizontal shift to the right, of magnitude a. If this is the case we say that the
polynomial q(x) is centered at a.
Now that we have defined infinite sums, and recovered some geometric intuition for finite polynomials,
it is not too much of a leap to our next topic: infinite polynomials.
26
Power Series
Let us start off with a concrete example. Recall from our section on geometric series that the following
equation holds:
∞
X
rk = 1 + r + r2 + r3 + ... =
k=0
1
1−r
whenever r is a real number with |r| < 1. Let us now define a function f by the following rule:
∞
X
xk = 1 + x + x2 + x3 + ...
f (x) =
k=0
Notice that f (x) only makes sense for numbers x which satisfy |x| < 1, since the series above only
converges for such x. So the domain of f is the open interval (−1, 1). Moreover, by our geometric series
1
formula, we have f (x) = 1−x
for all x in (−1, 1); so f is continuous and infinitely differentiable on this
restricted domain. This function f is the simplest example of an “infinite polynomial.” We refer to such
functions as power series.
Definition 26.1. A power series has the general form
∞
X
ck (x − a)k ,
k=0
where a is a real number called the center of the series, and c0 , c1 , c2 , ... are real numbers called the
coefficients of the series. The set of values x for which the series converges is called the interval of
convergence of the series. The radius of convergence of the series, denoted R, is the distance from
the center of the series to the boundary of the interval of convergence.
Example 26.2. Find the interval and radius of convergence for each power series.
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
(1)
(2)
(3)
25
∞
X
(−1)k (x − 2)k
k=0
∞
X
k=0
∞
X
4k
xk
k!
k!xk
k=1
Theorem 26.3. A power series
∞
X
ck (x − a)k centered at a converges in one of three ways:
k=0
(1) The series converges absolutely for all x, in which case the interval of convergence is (−∞, ∞)
and the radius of convergence is R = ∞.
(2) There is a real number R > 0 such that the series converges absolutely for |x − a| < R, and
diverges for |x − a| > R, in which case the radius of convergence is R.
(3) There series converges only at a, in which case the radius of convergence is R = 0.
Example 26.4. Use the Ratio Test to find the radius and interval of convergence of
∞
X
(x − 2)k
√
.
(1)
k
k=1
∞
X
(−1)n
(x − 5)n .
(2)
nn
4
n=1
Theorem 26.5. Suppose f (x) =
∞
X
ck xk and g(x) =
k=0
∞
X
dk xk are two power series with the same
k=0
interval of convergence I. Then the following hold.
∞
X
(1)
(ck + dk )xk converges to the function (f + g)(x) on I.
k=0
(2) xm
∞
X
ck x k =
∞
X
ck xk+m converges to xm f (x) on I whenever m is a non-negative integer.
k=0
k=0
m
(3) If h(x) = bx
where m is a positive integer and b is a real number, then
∞
X
ck (h(x))k converges
k=0
to the composite function f (h(x)) for all x such that h(x) is in I.
Example 26.6. Find the power series and interval of convergence for the following functions.
x5
(1)
1−x
1
(2)
1 − 2x
1
(3)
1 + x2
27
Differentiating and Integrating Power Series
Theorem 27.1. Let f (x) =
∞
X
ck xk be a power series with interval of convergence I.
k=0
(1) f is a continuous function on I.
(2) The power series may Rbe differentiated or integrated term by term, and the resulting power series
converges to f 0 (x) or f (x)dx, respectively, at all points in the interior of I.
26
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
∞
Example 27.2. Let f (x) =
X
1
=
xk for |x| < 1.
1−x
k=0
(1) Differentiate the series above to find a power series for f 0 (x), and identify the function it represents.
(2) Integrate the series above term by term and identify the function it represents.
(3) To what value does the alternating harmonic series converge?
Example 27.3. Find power series representations centered at 0 for the following functions and give
their intervals of convergence.
(1) arctan x
1+x
(2) ln
1−x
Fact 27.4 (Some Maclaurin Series (See Definition 28.3)). In the previous examples we have shown:
(1)
∞
X
1
=
xn on (−1, 1).
1 − x n=0
(2)
∞
X
1
=
(n + 1)xn on (−1, 1).
(1 − x)2
n=0
(3) ln(1 − x) =
(4) arctan x =
28
∞
X
xn
on [−1, 1).
n
n=1
∞
X
(−1)n x2n+1
on (−1, 1).
2n + 1
n=0
Taylor Series
We have shown that power series always represent continuous functions. Now we will consider the
converse: given an arbitrary continuous function, may it be represented as some convergent power series? We will attack this problem in the usual way: given a function f , we will attempt to find some
polynomials which closely approximate it; if we can do such a thing, then we will pass to a limit.
Example 28.1. Let f (x) = ln x.
(1) Find a linear function p1 which approximates f at x = 1, in the sense that p1 (1) = f (1) and
p01 (1) = f 0 (1).
(2) Find a quadratic function p2 which approximates f at x = 1, in the sense that p2 (1) = f (1),
p02 (1) = f 0 (1), and p002 (1) = f 0 (1).
(3) Find a cubic function p3 which approximates f at x = 1, in the sense that p3 (1) = f (1),
(3)
p03 (1) = f 0 (1), p003 (1) = f 00 (1), and p3 (1) = f (3) (1).
(4) Find an n-th degree polynomial pn which satisfies p(k) (1) = f (k) (1) for every non-negative integer
k ≤ n.
Definition 28.2. Let f be a function with at least n derivatives at a point a. The nth-order Taylor
polynomial for f centered at a, denoted pn , is given by
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
pn (x) = f (a) + f 0 (a)(x − a) +
27
f (n) (a)
f 0 (a)
(x − a)2 + ... +
(x − a)n
2
n!
or
pn (x) =
n
X
f (k) (a)
k=0
k!
(x − a)k
(k)
This function pn has the property that pn (a) = f (k) (a) for all non-negative integers k ≤ n.
Definition 28.3. Suppose f is a function with derivatives of all orders on an interval containing the
point a. The Taylor series for f centered at a is
∞
X
f (k) (a)
k=0
k!
(x − a)k
A Taylor series centered at 0 is called a Maclaurin series.
Example 28.4. Compute the Maclaurin series for the following functions, and give the interval of
convergence.
(1) f (x) = sin x
(2) f (x) = cos x
(3) f (x) =
1
1−x
(4) f (x) = ln(1 + x)
(5) f (x) = ex
(6) f (x) = e3x
(7) f (x) = x2 ex
Fact 28.5 (More Maclaurin Series). The following equalities hold.
(1) sin x =
∞
X
(−1)n x2n+1
for all x.
(2n + 1)!
n=0
∞
X
(−1)n x2n
(2) cos x =
for all x.
(2n)!
n=0
(3) ex =
∞
X
xn
for all x.
n!
n=0
∞
X
1
(4)
=
(−1)n xn on (−1, 1).
1 + x n=0
(5) ln(1 + x) =
∞
X
(−1)n−1 xn
on (−1, 1].
n
n=1
28
29
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
Parametric Equations
Definition 29.1. Suppose f (t) and g(t) are two continuous functions in one variable t. Let C be the
set of all points (x, y) in the Cartesian plane such that (x, y) = (f (t), g(t)) for some input t. Then C is
called a parametrized curve, and f (t), g(t) are called parametric equations. We denote by ~c(t) the
function
~c(t) = (f (t), g(t)),
so ~c is a function which takes a single real number t for input and returns a pair (x, y) as output. This
function ~c is called a parametrization of C.
Example 29.2. Sketch the curve C parametrized by the equations x = 2t − 4 and y = 3 + t2 .
Example 29.3. Let C be the curve parametrized by ~c(t) = (2t − 4, 3 + t2 ). Describe C as the graph of
an equation of the form y = f (x).
Example 29.4. A bullet follows the trajectory parametrized by ~c(t) = (80t, 200t − 4.9t2 ), where t is
measured in seconds and distance in meters.
(1) Find the bullet’s height at t = 5 seconds.
(2) Find its maximum height.
Fact 29.5 (Parametrization of a Line).
(1) The line of slope m passing through the point (a, b) is
parametrized by ~c(t) = (a + rt, b + st), where m = s/r.
(2) The line passing through the points (a, b) and (c, d) is parametrized by ~c(t) = (a + t(c − a), b +
t(d − b)).
Example 29.6. Find a parametrization of the line of slope 4 passing through (3, −1).
Fact 29.7 (Parametrization of Circles and Ellipses).
(1) The circle of radius R centered at (a, b) is
parametrized by ~c(t) = (a + R cos θ, b + R sin θ) of 0 ≤ θ ≤ 2π.
(2) The ellipse which is the graph of ( xa )2 + ( yb )2 = 1 is parametrized by ~c(t) = (a cos t, b sin t).
Example 29.8. Sketch the curve parametrized by ~c(t) = (t2 + 1, t3 − 4t. Label the points corresponding
to t = 0, ±1, ±2, ±2.5).
Fact 29.9. Let ~c(t) = (x(t), y(t)) where x and y are differentiable functions of t. Assume x0 (t) is
continuous and x0 (t) 6= 0. Then
dy
=
dx
y 0 (t)
x0 (t) .
Example 29.10. Let C be the curve parametrized by ~c(t) = (t2 + 1, t3 − 4t).
(1) Find an equation for the tangent line at t = 3.
(2) Find all points in C where the tangent line is horizontal.
30
Arc Length
Fact 30.1. Let C be the curve parametrized by ~c(t) = (x(t), y(t)) for a ≤ t ≤ b, where x0 (t) and y 0 (t)
exist and are continuous. Then the arc length s of C is
s=
Rbp
a
(x0 (t))2 + (y 0 (t))2 dt.
Example 30.2. Compute the circumference of a circle of radius R.
Example 30.3. Calculate the length of the curve C parametrized by ~c(t) = (2(t − sin t), 2(1 − cos t)) for
0 ≤ t ≤ 2π.
p
Definition 30.4. If ~c(t) is a parametrization, the speed of ~c is ds
(x0 (t))2 + (y 0 (t))2 .
dt =
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
29
Example 30.5. The motion of a particle in the plane is given by ~c(t) = (2t, 1 + t3/2 ). Find
(1) the particle’s speed at time t = 1.
(2) the distance traveled s and the displacement d during the interval 0 ≤ t ≤ 4.
31
Polar Coordinates
Definition 31.1. To a point P in the Cartesian plane, we associate the pair (r, θ), where r is the
distance from P to the origin, and θ is the angle formed between the line connecting the origin to P
and the x-axis (in the counter-clockwise direction). We call the pair (r, θ) the polar coordinates of P .
Often one will require that r ≥ 0, but we will not do so here.
Fact 31.2 (Polar-to-Cartesian and Cartesian-to-Polar Conversions). Suppose (x, y) are the Cartesian
coordinates and (r, θ) are the polar coordinates of some particular point. Then
x = r cos θ and y = r sin θ,
and
r2 = x2 + y 2 and tan θ = xy .
Example 31.3.
(1) Find the Cartesian coordinates of (r, θ) = (3, 5π
6 ).
√
(2) Find the polar coordinates of (x, y) = (4, −4 3).
Example 31.4. Show that there are infinitely many polar representations of (−1, 1) by exhibiting all
of them.
√
Example 31.5. Find the polar equation of the line through the origin of slope
3
3 .
Example 31.6. Show that r = d sec(θ − α) is the equation of the line L whose point closest to the
origin is (d, α).
Example 31.7. Sketch the graph of r = 5.
Example 31.8. Find the polar equation of the line L tangent to the circle r = 4 at the point (r, π3 ).
Example 31.9. Sketch the graph of r = 10 cos θ.
Example 31.10. Sketch the graph of r = 2 cos θ − 1.
32
Area and Arc Length in Polar Coordinates
Fact 32.1. If f (θ) is a continuous function, then the area bounded by the graph of r = f (θ) (in polar
coordinates) between the rays θ = α and θ = β (where α < β and β − α ≤ 2π) is
1
2
Rβ
α
r2 dθ =
1
2
Rβ
α
f (θ)2 dθ.
Example 32.2. Compute the area of the right semicircle with equation r = 4 sin θ.
Example 32.3. Sketch r = sin 3θ and compute the area of one “petal.”
Example 32.4. Find the area of the region inside the circle r = 2 cos θ but outside the circle r = 1.
Fact 32.5. If f (θ) is a differentiable function, then the arc length of the graph of the polar equation
r = f (θ) from θ = α to θ = β is given by
Rβp
s = α f (θ)2 + f 0 (θ)2 dθ.
Proof. We may parametrize the graph of r = f (θ) using the parametrization ~c(t) = (x, y) = (r cos θ, r sin θ) =
(f (θ) cos θ, f (θ) sin θ). Now compute the derivatives of x and y using the product rule:
30
CALCULUS MATH 166 FALL 2013 (COHEN) LECTURE NOTES
x0 (θ) = −f (θ) sin θ + f 0 (θ) cos θ and y 0 (θ) = f (θ) cos θ + f 0 (θ) sin θ.
If we square both derivatives, add them together, and take the square root, we get
p
p
x0 (t)2 + y 0 (t)2 = f (θ)2 + f 0 (θ)2 .
Now the conclusion follows from our old arc length formula.
Example 32.6. Find the total length of the circle r = 2a cos θ.