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MATH 1432 (CALCULUS II) LECTURE NOTES
INVITATION-ONLY SECTION
VAUGHNCLIMENHAGA
Contents
I Integration 1
1 Review of integration and the substitution rule 1
2 Integration by parts 3
3 Trigonometric integrals 6
4 More trigonometric integrals 10
5 Trigonometric substitutions 14
6 Complicated quadratics 18
7 Rational functions 19
8 General partial fraction decompositions 23
9 Numerical integration 29
10 Improper integrals 31
II Applications of integration 40
11 Arc length and the catenary 40
12 Surface area 44
13 Physical applications 49
14 Two- and three-dimensional objects 52
15 *Probability 55
III Differential equations 59
16 Ideas and examples 59
17 *Separable differential equations 63
18 *Other population models 67
19 *Linear differential equations 69
20 Coupled differential equations 73
IV Parametric curves and polar coordinates 77
21 Parametric curves 77
22 Calculus with parametrizations 80
23 Geometry of parametric curves 82
24 Polar coordinates 85
25 Calculus with polar coordinates 88
V Sequences and series 93
Date: July 15, 2020.
i
ii VAUGHNCLIMENHAGA
26 Sequences 93
27 Summing an infinite series 98
28 The integral test 101
29 Comparison tests and alternating series 105
30 Absolute convergence, ratio and root tests 107
31 Power series 110
32 Calculus with power series 113
33 Taylor and Maclaurin series 117
VI Conic sections, planetary motion 126
34 Parabolas 126
35 Ellipses (and hyperbolas) 131
36 Kepler and Newton 136
1
Part I. Integration
Lecture 1 Review of integration and the substitution rule
Stewart §5.5, Spivak Ch. 19
1.1. Definite and indefinite integrals
Last semester, we motivated the introduction of integrals by considering the question
of how to determine areas. This led us to two definitions:
(1) the definite integral Rb f(x)dx is a number obtained as a limit of Riemann sums,
a
which depends on the interval [a,b] and can be interpreted as an area;
(2) the indefinite integral R f(x)dx is a function whose derivative is f(x).
The two are related by the Fundamental Theorem of Calculus, which has two halves.
The first half says that definite integrals can be used to find indefinite integrals (an-
tiderivatives), since d Rx f(t)dt = f(x).
dx a
The second half goes in the opposite direction, and says that indefinite integrals can
be used to find definite integrals: if F(x) = R f(x)dx is an indefinite integral of f, so
that F′(x) = f(x) at every x, then Rbf(x)dx = F(b)−F(a).
a
Although the first half guarantees that every continuous function has an indefinite
integral, it does not give a general procedure for writing down an elementary formula
for R f(x)dx. Our emphasis for the next little while will be on this process, which is
essential if we are to use the second half of the FTC effectively.
By “elementary formula”, we mean a formula that can be written down in terms
of constants, polynomials, rational functions, exponentials, trigonometric functions,
and logarithms using addition, subtraction, multiplication, and division. For exam-
ple, F(x) = tan−1(x) is an elementary formula, but F(x) = Rx 1 2 dt is not elementary
0 1+t
because it involves an integral, even though it represents the same function.
Given an integral R f(x)dx, then, our goal will be to find an elementary formula for
it. Bear the following warning in mind, though: not every integral admits an elemen-
1 R 2
tary formula. For example, it is possible to show that sin(x )dx does not have an
elementary formula, and in fact there is a sense in which most indefinite integrals do
not have elementary formulas. Nevertheless, a great many of them do, including some
of the most important ones, and so we will turn our attention now to finding them.
1.2. Substitution rule
The first method of integration is by direct inspection: we have a list of functions
F(x) whose derivatives f(x) = F′(x) are known, and if f happens to appear on the
corresponding list of derivatives, then we can simply read off the indefinite integral
R f(x)dx = F(x)+C.
1
The proof involves tools that go beyond the scope of this course, and we will not discuss it.
2
The second method, which we encountered briefly last semester, is the substitution
rule. This is a consequence of the chain rule for differentiation, which says that if F,g are
′ ′ ′
differentiable functions, then F ◦g is differentiable and has (F ◦g) (x) = F (g(x))g (x).
In particular, if F′(x) = f(x) so that F gives the indefinite integral of f, then we have
(F ◦g)′ = (f ◦g)·(g′); this can be written in the form
Z f(g(x))g′(x)dx = F(g(x)).
It is usually easier to remember and apply this rule if we introduce a new variable
u=g(x), and observe that d F(u) = f(u), so that the above formula becomes
du
(1.1) Z f(g(x))g′(x)dx = Z f(u)du.
It is common to rewrite the formula g′(x) = du as du = g′(x)dx, in which case (1.1)
dx
appears to become almost trivial:
Z f(g(x))g′(x)dx = Z f(u)du.
|{z} | {z }
u du
We emphasize, though, that the formula du = g′(x)dx is purely a bookkeeping device
rather than a valid part of a proof, because we have not yet given du and dx any
independent meaning of their own. We will continue to use it because it simplifies the
appearance of various computation, but please remember the logical order of things:
(1.1) justifies this formula, rather than the other way round.
R √ 2 2
Example1.1. Wecancompute x 1+x dxbyputtingu=1+x sothatdu=2xdx,
and we obtain
Z √ 2 Z √ 2 Z 1 1/2 1 2 3/2 1 2 3/2
x 1+x dx= 1+x ·xdx= u du = · u +C= (1+x) +C.
| {z } |{z} 2 2 3 3
√ 1du
u 2
Example 1.2. To find R tanxdx, we can write tanx = sinx and notice that the deriva-
cosx
tive of cosx appears in the numerator (up to a negative sign), so putting u = cosx gives
du = −sinxdx and
Z tanxdx=Z sinx dx=Z −du =−ln|u|+C =−ln|cosx|+C =ln|1/cosx|+C
cosx u
=ln|secx|+C.
There is no universal procedure telling us how to make the change of variables u =
g(x), but these examples illustrate some guidelines that are helpful to keep in mind: it
is reasonable to try setting u as the input of some function in the integrand (the square
root function in Example 1.1), or as an expression whose derivative also appears in the
integrand (the cosine function in Example 1.2). Sometimes it even works to let u be
the entire integrand: for example, in R √2x+1dx we can take u = √2x+1 so that
u2 = 2x+1 and 2udu = 2dx, and we get
Z √ Z 1 3 1 3/2
2x+1 dx = u·udu= u +C= (2x+1) +C.
| {z } |{z} 3 3
u udu
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