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J I M M Y B R O O M F I E L D
A B R I E FG U I D E T OC A L C U L U S I I
T H E U N I V E R S I T Y O F M I N N E S O TA
Copyright © 2015 Jimmy Broomfield
http://moodle.umn.edu
License: Creative Commons CC BY-NC 4.0http://creativecommons.org/licenses/by-nc/4.0/
First printing, March 2015
Contents
Integration Techniques 5
Basic Integrals 5
Integration By Parts 6
Trigonometric Identities 9
Trigonometric Substitution 12
Partial Fraction 16
Improper Integrals 19
Applications of Integration 23
Arc Length 23
Area of a Surface of Revolution 25
Application to Physics and Engineering 27
Fluid Force and Fluid Pressure 28
Moments and Centers of Mass 30
Centroid of a Lamina 31
Differential Equations 35
Introduction 35
Slope Fields and Euler’s Method 38
Separable Equations 38
Models for Population Growth 38
Linear Equations 41
Predator-Prey Systems 42
Glossary of Terms 42
Integration Techniques
Throughout this guide, we will present results without proof. If youwould like a proof of the techniques and theorems that we use, pleasesee "Calculus Early Transcendentals", volume I, 7E by James Stewart.To begin this guide, we will review basic integration formula thatevery calc II student should know, and then move to new techniques.
Basic Integrals
The integrals below are essential formulas the should be memorized.If you struggle with a few of them, please practice until you havecommitted them to memory. Also note that the constants have beenleft out of the table below for convenience.
∫xn dx =
xn+1
n + 1, n 6= −1
∫ 1x
dx = ln |x|
∫ex dx = ex
∫ax dx =
1ln a
ax
∫sin x dx = − cos x
∫cos x dx = sin x
∫sec2 x dx = tan x
∫csc2 x dx = − cot x
∫sec x tan x dx = sec x
∫csc x cot x dx = − csc x
∫ 1a2 + x2 dx =
1a· tan−1
(xa
)
∫ 1√a2 − x2
dx = sin−1(
xa
)
6
Along with these integral formulas, you should also be comfort-able with the technique of uuu-substitution. The next two examplesillustrate how u-substitution works with respect to changing boundsof integration.
Example 1.1. Evaluate the following integral
∫ π/2
0
sin x1 + cos x
dx
Solution: Let u = cos x, then du = − sin x. Now we may rewrite theintegral above as:
−∫ 0
1
11 + u2 du
Notice that limits of integration have changed from 0 and π/2 to 1 and 0 This example should be straightforward, but if you feel youneed more practice,choose afew extra problems to work.
respectively. This is due to the fact that when the function u(x) = cos(x)is evaluated at 0 and π/2, we get
u(0) = 1 u(π/2) = 0
Therefore our integral becomes:∫ 1
0
11 + u2 du
= arctan(u)∣∣∣∣1
0
= arctan(1) − arctan(0)
= π/4
Integration By Parts
The first new technique of calculus II tat we will introduce is Inte-gration by Parts. This technique will give a possible way to integrateproducts of functions. Without further ado, the formulas for integra-tion by parts is
∫u dv = uv −
∫v du
∫ b
au dv = uv
∣∣∣∣ba−∫ b
av du
This equation is often remembered by the saying "the integral ofu dv is uv minus the integral of v du". We will now consider threeexamples of integration by parts that will illuminate the possibleways to use this formula.
7
Example 1.2. Evaluate∫ 2
1ln x dx.
As you read through thisguide, note that theseexamples are a minimal setof problems to study from,and it will do little goodto only read these problems.You should choose problemsto study in each section. Itis often said that "Math isnot a spectator sport", thismeans you must do the workby yourself on your own time.
Solution: In this example, it does not seem that we have any reasonablechoices for u and dv. When this occurs, it is often useful to choose dv to bedx. This leaves us with
u = ln x dv = dx
du =1x
dx v = x
Therefore when we use our formula for integration by parts we get:
∫ 2
1ln x dx = x ln x
∣∣∣∣21−∫
xdxx
= x ln x∣∣∣∣2
1−∫ 2
1dx
= x ln x∣∣∣∣2
1− x
∣∣∣∣21
= 2 ln 2 − 0 − 2 + 1
= 2 ln 2 − 1
Example 1.3. Evaluate the integral∫
eθ sin 2θ dθ.
Solution: In this example, the key idea is to try integration by parts untilyou get back to the original integral you wish to evaluate. Following thispath, we have
u1 = sin 2θ dv1 = eθ
du1 = 2 cos 2θ v1 = eθ
Under this choice of u and dv, the integral above becomes∫eθ sin 2θ dθ = sin 2θ eθ − 2
∫cos 2θ eθ dθ
Then if we use integration by parts again with
u2 = cos 2θ dv2 = eθ
du2 = −2 sin 2θ v2 = eθ ,
8
we get: ∫eθ sin 2θ dθ = sin 2θ eθ − 4 cos 2θ eθ − 4
∫sin 2θ dθ
Now if we subtract the far right integral from each side, we get
5∫
eθ sin 2θ dθ = sin 2θ eθ − 4 cos 2θ eθ
Hence the answer we get is
∫eθ sin 2θ dθ =
sin 2θ eθ − 4 cos 2θ eθ
5+ C
This final example of this section will show how to use integrationby parts multiple times. A type of integral that often arises in appli-cations is the integration of a function of the form xn g(x), whereg(x) can be integrated n times. When faced with a situation likethis, we may use a variation of integration by parts known as TabularIntegration.
Example 1.4. Evaluate the integral∫
x2e2x dxTry to understand why this works!
Solution: To use tabular integration in solving this problem, we will makea table in which we place the derivatives of x2 in the right column and theintegrals of e2x in the left column. After constructing this table, we will addthese terms as depicted in the following table.
D I
x2 e2x
2x12
e2x
214
e2x
018
e2x
=12
x2e2x − 12
xe2x +14
e2x
+
−
+
Therefore ∫x2 e2x dx =
12
x2e2x − 12
xe2x +14
e2x + C
9
Trigonometric Identities
In this section we will focus on the trigonometric identities that willhelp in simplifying certain types of integrals. Before we begin, it willbe important to introduce the identities that we will be using. Theseidentities will be essential to doing the problems and therefore mustbe memorized. This will help you in exams when you do not havetime to derive such formulas. The necessary identities are
Pythagorean Identities
sin2 x + cos2 x = 1
sec2 x = tan2 +1
csc2 x = cot2 +1
Half Angle Formulas
sin2 x =12(1− cos 2θ) cos2 x =
12(1 + cos 2θ)
Double Angle Formula
sin 2x = 2 sin x cos x
These identities are far a from complete, but they will suffice formost of the problem that we will encounter. To illustrate this, we willpresent the common strategies for integrating functions of the formf (x) = sinn x cosn x.
Cosine is odd
If the power of cosine is odd, then we will use the Pythagoreanidentity to convert all but one of the cosines to 1− sin2 x raise to somepower. This remaining cosine will serves as du in the substitutionu = sin x. This will give the following where n = 2k + 1.
∫sinm x cosn x dx =
=∫
sinm x(cos2 x)k cos x dx
=∫
sinm x(1− sin2 x)k cos x dx
=∫
um(1− u2)k du
10
Sine is odd
If the power of sine is odd, then we will use the Pythagorean identityto convert all but one of the sines to 1− cos2 x raise to some power.This remaining sine will serves as du in the substitution u = cos x.This will give the following where m = 2k + 1
∫sinm x cosn x dx
=∫(sin2 x)k cosn x sin x dx
=∫(1− cos2 x)k cosn x sin x dx
= −∫(1− u2)kun du
Both Sine and Cosine are even
If the powers of sine and cosine are both even, we use the half angleor double angle formulas to reduce powers until we are left with asimpler function that can be integrated.
We will now use these ideas to work through a few examples. We will see that these strategiescan also be applied to powersof secant and tangent.Example 1.5. Evaluate
∫sin3 θ cos4 θ dθ
Solution: We first notice that the power of sine is odd, therefore we will usethe identity sin2 x = 1− cos2 x.
∫sin3 θ cos4 θ dθ
=∫(1− cos2 θ) cos4 θ sin θ dθ
= −∫(1− u2)u4 du
= −∫(u4 − u6) du
= −u5
5− u7
7
=17
cos7 θ − 15
cos5 θ + C
11
Example 1.6. Evaluate∫
cos4 2x dx.
Solution: To evaluate this integral, we will have to use the half angle for-mula to reduce powers of cosine. This will give
∫cos4 2x dx
=14
∫ (1 + cos 2x
)2
dx
=14
∫ (1 + 2 cos 2x + cos2 2x
)dx
=x4+
sin 2x4
+14
∫cos2 2x dx
=x4+
sin 2x4
+18
∫1 + cos 4x dx
=x4+
sin 2x4
+x8+
sin 4x32
=3x8
+sin 2x
4+
sin 4x32
+ C
Example 1.7. Evaluate∫
sin2 x cos2 x dx
Solution: To evaluate this integral, notice that the function we are inte-grating can be rewritten as (sin x cos x)2. This can be rewritten using thedouble angle formula to become (sin 2x)2/4. Now we can use the half angleformula as in the last example to rewrite this. Therefore the integral can beevaluated as.
∫sin2 x cos2 x dx
=14
∫(sin 2x)2 dx
=18
∫(1− cos 4x) dx
=x8− sin 4x
32+ C
12
In this section, we will conclude by deriving a formula for integratingsecant.
Example 1.8. Evaluate∫
sec x dx.
Solution: For this integral, we will use a specific trick that shows howmultiplication by a special term can often help in simplifying a problem. Theterm that we will multiply this integral by is
sec x + tan xsec x + tan x
Therefore we have ∫sec x dx
=∫
sec xsec x + tan xsec x + tan x
dx
=∫ sec2 x + sec x tan x
sec x + tan xdx
Now we let u = sec x + tan x, and du = sec2 x + sec x tan x. Thus we get
∫ 1u
du
ln|u|+ C
= ln | sec x + tan x|+ C
Please note that the examples that were provided in this section donot show all of the possible difficulties that can arise from these typesof integrals.
Trigonometric Substitution
In this section we will explore a type of integral substitution knownas an inverse substitution. When we perform u-substitution, we havea way to determine u from x. That is, we may write u as functionu(x). An inverse substitution instead will give a way to determine When we perform an inverse substi-
tution we must make sure that such asubstitution is one-to-one.
x from u. That is, we will be able to write x as a function x(u). Inthis section, we will consider a specific set of inverse substitutionsinvolving trigonometric functions.
13
Table of Substitutions
Expression Substitution Identity√a2 − x2 x = a sin θ 1− sin2 θ = cos2 θ
√a2 + x2 x = a tan θ 1 + tan2 θ = sec2 θ
√x2 − a2 x = a sec θ sec2 θ − 1 = tan2 θ
As commented earlier, these substitutions are only valid if the sub-stitution is one-to-one. Since these trigonometric functions are notone to one, we must restrict θ to be within a valid domain. The tablebelow gives the valid domains for these substitutions.
Substitution Interval
x = a sin θ θ ∈[− π
2,
π
2
]
x = a tan θ θ ∈[− π
2,
π
2
]
x = a sec θ θ ∈[
0 ,π
2
]or θ ∈
[π ,
3π
2
]
Before working some example, we remark that the reason thesesubstitutions are called "inverse" substitutions is because in order torewrite θ in terms of x we must invert any final answer we get. To seethis in action, let’s work some examples.
Example 1.9. Evaluate∫
x3√
1− x2 dx
Solution: For this example, we will let x = sin θ, then dx = cos θ. Then
∫x3√
1− x2 dx
=∫
sin3 θ√
1− sin2 θ cos θ dθ
=∫
sin3 x cos2 x dx
=∫
sin2 x cos2 x sin x dx
=∫(1− cos2 x) cos2 x sin x dx
Now if we use a u-substitution with u = cos θ, we have the following:
14
∫u4 − u2 du
=u5
5− u3
3+ C
=cos5 θ
5+
cos3 θ
3+ C
Finally we must invert this equation to get back to x. In this case we useright triangle trigonometry to determine what cos θ is in terms of x. Sincesin θ =
x1
, we will represents the transformation with the following triangle.
x1
√1− x2
θ
Therefore cos θ =√
1− x2, and our answer becomes
(1− x2)5/2
5− (1− x2)3/2
3C
Example 1.10. Evaluate∫ 1√
4x2 + 1dx.
Solution: For this example, we will proceed as above and let x =12
tan θ.
Then dx =12
sec2 θ our problem becomes
∫ 1√4x2 + 1
dx
=12
∫ sec2 θ
sec θdθ
=12
∫sec θ dθ
=12
ln | sec θ + tan θ|+ C
Again, we will use right triangle trigonometry to invert this expression.The diagram below will help us do so.
15
2x
√4x2 + 1
1θ
Hence tan θ = 2x. Then our answer becomes
12
ln |√
4x2 + 1 + 2x|+ C
Our final example will be to show how the limits of integrationchange when using a trigonometric substitution.
Example 1.11. Evaluate∫ 2√
3
√x2 − 3
xdx.
Solution: For this problem we will use the trigonometric substitution
x =√
3 sec θ and dx =√
sec θ tan θ dθ
The key point in this example is in converting the limits of integration. Todo this, we consider x to be a function of θ and set
√3 =
√3 sec θ and
2 =√
3 sec θ. These equations become sec θ = 1 and sec θ = 2/√
3, andthey have the solutions θ = 0 and θ = π/6. Therefore the integral becomes
∫ 2√
3
√x2 − 3
xdx
∫ π/6
0
(√
3 tan θ)(√
3 sec θ tan θ)√3 sec θ
dθ
=∫ π/6
0
√3 tan2 θ dθ
√3∫ π/6
0(sec2 θ − 1) dθ
√3[
tan θ − θ
]π/6
0
= 1−√
3π
6
16
Partial Fraction
Another great technique that we can use for simplifying integralsis the method of partial fractions. The following few steps give ageneral method for decomposing a rational function P(x)/Q(x).
1. Polynomial division if improper: If the degree of the numeratoris greater than or equal to the degree of the denominator, preformpolynomial long division on the numerator to obtain
Q(x)P(x)
= f (x) +Q1(x)P(x)
Where deg(P(x)) is less than deg(Q(x)).
2. Factor the denominator: Factor the denominator into linear andquadratic factors of the following form Any polynomial with real number
coefficients can be completely factoredinto linear and quadratic factors.
(px + q)m and (ax2 + bx + c)n
where each factor is irreducible.
3. Linear factors: For each linear factor of the form (px + q)m, thepartial fraction decomposition of this term must be
A1
(px + q)+
A2
(px + q)2 + · · ·+ Am
(px + q)m
4. Quadratic factors: For each factor of the form (ax2 + bx + c)n, thepartial decomposition of this term must be
B1x + C1
(ax2 + bx + c)+
B2x + C2
(ax2 + bx + c)2 + · · ·+ Bnx + Cn
(ax2 + bx + c)m
In the steps above, the terms A1, ..., Am, B1, ..., Bn, and C1, ..., Cn areunknown coefficients that must be solved for. The steps above givethe general decomposition for the method of partial fractions, butwe must also know how to solve for these coefficients. The followingexamples will show how to solve for these coefficients in a fewdifferent situations.
Example 1.12. Evaluate∫ 5x2 + 20x + 6
x3 + 2x2 + xdx
Solution: First be must factor the denominator. Doing this we have
x3 + 2x2 + x = x(x + 1)2
Therefore we have the following partial fraction decomposition Notice that we must include a term for(x + 1) and (x + 1)2).
17
5x2 + 20x + 6x(x + 1)2 =
Ax+
Bx + 1
+C
(x + 1)2
Now in order to solve for A, B, and C, we notice that if we get a commondenominator on the right, we get a numerator of
A(x + 1)2 + Bx(x + 1) + Cx
Therefore since the numerators of these two fractions must be the same, wehave
5x2 + 20x + 6 = A(x + 1)2 + Bx(x + 1) + Cx
Now notice that the key point in finding A, B, and C will be to chooseconvenient values of x to substitute into the equation above. If we substitutex = 0, then we have
6 = A
Now we will let x = −1. This gives
C = 9
Finally we will let x = 1 and use the fact that A = 6 and C = 9. This willgive us the following
31 = 20 + 2B + 9
B = −1
Therefore coming back to our original problem, we have
∫ 5x2 + 20x + 6x3 + 2x2 + x
dx
=∫ A
x+
Bx + 1
+C
(x + 1)2 dx
= 6 ln |x| − ln |x + 1|+ −9x + 1
+ C
= ln∣∣∣∣ x6
x + 1
∣∣∣∣+ 9x + 1
+ C
During this next example we will see how to use partial fractionswhen dealing with repeated quadratic factor.
Example 1.13. Evaluate∫ 8x3 + 13x
(x2 + 2)2 dx.
18
Solution: Since the denominator is already factored and the degree of thenumerator is less than the degree of the denominator, we have the followingdecomposition
8x3 + 13x(x2 + 2)2 =
Ax + B(x2 + 2)
+Cx + D(x2 + 2)2
For This example, we will start to solve for these coefficients by the samemethod as we did before. That is, we will set the numerators of the left andright hand side of the equation. Therefore we have
8x3 + 13x = (Ax + B)(x2 + 2) + Cx + D
8x3 + 13x = Ax3 + 2Ax + Bx2 + 2B + Cx + D
8x3 + 0x2 + 13x + 0 = Ax3 + Bx2 + (2A + C)x + (2B + D)
At this point, we will set the coefficients of the right and left hand sides tobe equal. Therefore B = 0 and 2B + D = 0. This gives us D = 0 as well.Further, A = 8 and 2A + C = 13, giving C = −2. Thus our originalintegral becomes ∫ 8x3 + 13x
(x2 + 2)2 dx
=∫ ( 8x
x2 + 2+
−3(x2 + 2)2
)dx
= 4 ln(x2 + 2) +3
2(x2 + 2)+ C
Finally, we will show an example that seems to be a partial fractionsproblem, but as it turns out, partial fractions will not help.
Example 1.14. Evaluate∫ x + 4
x2 + 2x + 5dx.
Solution: The reason that partial fractions will not help with this problemis because the function above has already been decomposed. In this example,instead of trying to proceed by direct integration, we will “complete thesquare" in the denominator. Therefore we have We have split the constant part of the
numerator up so that we can use au-substitution on the first integral.
∫ x + 4(x + 1)2 + 4
dx
∫ x + 1(x + 1)2 + 4
dx +∫ 3
(x + 1)2 + 4dx
19
Notice that we can use a u-substitution for which the left integral willbecome a natural logarithm and the right one will be an arctangent integral.Therefore we get
∫ x + 1(x + 1)2 + 4
dx +∫ 3
(x + 1)2 + 4dx
=∫ u
u2 + 4du +
∫ 3u2 + 4
du
=12
ln |u2 + 9| − 23
arctan u/3 + C
=12
ln |(x + 1)2 + 9| − 23
arctan((x + 1)2/3) + C
Improper Integrals
The final topic in this chapter is the evaluation of improper integrals.The integrals that we plan to study will be split between Type-Iimproper integrals and Type-II improper integrals. To begin, let usdefine what we mean by the previous sentence.
Definition 1.15. If f (x) is continuous on the implied domain, then animproper integral of Type-1 is an infinite integral the evaluation of one ofthe following
∫ ∞
af (x) dx = lim
t→∞
∫ t
af (x) dx
∫ a
−∞f (x) dx = lim
t→∞
∫ b
tf (x) dx
∫ ∞
−∞f (x) dx = lim
t→∞
∫ t
af (x) dx + lim
t→∞
∫ a
−tf (x) dx
Definition 1.16. An improper integral of Type-II is an integral given inone of the following forms
1. If f is continuous on [a, b), but discontinuous at b, then the followingintegral is improper
∫ b
af (x) dx = lim
t→b−
∫ t
af (x) dx
20
2. If f is continuous on (a, b], but discontinuous at a, then the followingintegral is improper ∫ b
af (x) dx = lim
t→a+
∫ b
tf (x) dx
3. If f is continuous on [a, c) ∪ (c, b], but discontinuous at c, then thefollowing integral is improper∫ b
af (x) dx = lim
t→c−
∫ t
af (x) dx + lim
t→c+
∫ b
tf (x) dx
In order to talk about evaluating these integrals, we must considertwo more definitions. These are given below
Definition 1.17. An improper integral of type I or II converges if the limitsgiven in the definition exist and are finite.
If the corresponding limits are infinite or do not exist, then the integral issaid to diverge.
With this information, let us do some examples to illustrate howthese concepts can play out.
Example 1.18. Evaluate∫ ∞
−∞
ex
1 + e2x dx.
Solution: Notice that this function is continuous on the interval (−∞, ∞),therefore it is an improper integral of type I, and we can split the integralwith a = 0, giving us
∫ ∞
−∞
ex
1 + e2x dx
limt→−∞
∫ 0
t
ex
1 + e2x dx + limt→∞
∫ t
0
ex
1 + e2x dx
= limt→−∞
[arctan ex
]0
t+ lim
t→∞
[arctan ex
]t
0
= limt→−∞
(π
4− arctan et
)+ lim
t→∞
(arctan et − π
4
)
=π
4− 0 +
π
2− π
4=
π
2
Since both of the integrals above are convergent, the original integral isconvergent.
21
Example 1.19. Evaluate∫ 2
−1
1x3 dx.
Note that this integral can be definedby using a Cauchy principle value. Thisterm is named after Augustin LouisCauchy, a French mathematician, wholaid some of the foundation for therigorous approach to calculus known asreal analysis.
Solution: Notice that this integral is not a type I improper integral becausethe limits of integration are finite, however when x = 0 there is a discontinu-ity. This means that the integral is of type II. Therefore we must evaluate itas follows ∫ 2
−1
1x3 dx
= limt→0−
∫ t
−1
1x3 dx + lim
t→0+
∫ 2
t
1x3
= limt→0−
[− 1
2x2
]t
−1+ lim
t→0+
[− −1
2x2
]2
t
= −∞ + ∞
Since neither of these integrals converge, the original integral is convergent.
Finally we will conclude this section with the comparison test anda result that will give you a set of test functions for the comparisontest.
In logic, the reversal of the hypothesisand the conclusion of a statement isknown as the converse. For more infor-mation on this, please see wikipedia
Theorem 1.20. Suppose that f and g are continuous functions on [a, ∞)
with f (x) ≥ g(x) ≥ 0 for all x ≥ a. Then
1. If∫ ∞
af (x) dx is convergent, then
∫ ∞
ag(x) dx is convergent.
2. If∫ ∞
ag(x) dx is divergent, then
∫ ∞
af (x) dx is divergent.
In the two pieces of this theorem, the part of the sentence thatcomes before “then” is the hypothesis of the statement and the partthat comes after is the conclusion. It is very important to note that wecannot switch the hypothesis and conclusion of these two statements.For example, if we look at point 1, if the integral of f (x) diverges, itdoes not tell us if the integral of g(x) converges.
To finish this section, we have the following theorem.
Proposition 1.21.∫ ∞
1
1xp dx is convergent for p > 1 and divergent for p ≤ 1.
Please try to work through the proof of the result above noticing thedifference in the anti-derivative when p = 1.
Applications of Integration
In this section we will consider a few applications of integration. Westart with the calculating arc length of a function.
Arc Length
Arc length ApproximationTo begin this section, we will give a definition and the equations tocalculate the arc length of a function.
Definition 1.22. A curve f (x) in the x-y plane is called rectifiable if it hasfinite length.
Theorem 1.23. Let the function given by y = f (x) represent a smooth curveon the interval [a, b]. The arc length of f between a and b is
s =∫ b
a
√1 + [ f ′(x)]2 dx
Similarly, for a smooth curve given by x = g(y), the arc length of g
x
f (x)
LinearApproximation
(x1, y2)
x0
(x2, y2)
x0 + ∆x
∆y
∆xf (x1) + ∆y
∆s =√
∆x2 + ∆y2
between c and d is
s =∫ d
c
√1 + [g′(x)]2 dx
Proof. This will be a sketch of the proof of the above and should notbe considered rigorous. Notice that we could approximate the arclength of a curve by performing a series of linear approximations asindicated to the right. If we break up the interval [a, b] into n equallyspaced intervals of length ∆x, then our approximation would amountto the following sum
n
∑i
√∆x2
i + ∆y2i
where ∆yi = f (xi)− f (xi−1). Then notice that we can also rearrangethis to be
n
∑i
√1 +
[∆yi∆xi
]2
∆x.
Now if we take the limit of this summation, letting n go to infinity,we get the integral ∫ b
a
√1 +
[dydx
]2
dx
24
Now we will work a few examples to illustrate how to use theseequations.
Example 1.24. Find the arc length of f (x) = x3/6 + 1/2x on the interval[0.5, 2] Solution: To begin, we will find f ′(x). Differentiating this function,we have
f ′(x) =3x2
6− 1
2x2
Therefore the arc length is calculated as follows
s =
∫ 2
0.5
√1 +
[12
(x2 − 1
2
)]2
dx
=
∫ 2
0.5
√14
(x4 + 2 +
1x4
)dx
=
∫ 2
0.5
√1
4x4
(x8 + 2x4 + 1
)dx
Notice that in this example, we pulledout the denominator in the square rootterm. This allowed us to factor of thepolynomial
x8 + 2x4 + 1
as(x4 + 1)2
=
∫ 2
0.5
12x2
√(x4 + 1
)2
dx
=∫ 2
0.5
12x2 (x4 + 1) dx
=∫ 2
0.5
12
(x2 +
1x2
)dx
=12
(136− 47
24
)
=12
[x3
3− 1
x
]2
1/2
=3316
25
Example 1.25. Find the arc length of the graph (y − 1)3 = x2 on theinterval [0, 8].
Solution: We will begin by solving for x in terms of y. This gives
x = ±(y− 1)3/2
Since we are considering the graph of the function over [0, 8], we will takethe positive root of this. Now we will take the derivative of this function toget
dxdy
=32
√y− 1
Since y = 1 when x = 0 and y = 5 when x = 8, we will integrate√1 + ( dx/ dy2) over the interval y ∈ [1, 5]. Therefore
∫ 5
1
√1 +
((3/2)2(y− 1)
)dx
=∫ 5
1
√(9/4)y− (5/4) dx
=12
∫ 5
1
√9y− 5 dx
=118
[(9y− 5)3/2
3/2
]5
1dx
=127
(403/2 − 43/2) ≈ 9.073
This will conclude the examples of this section. The equation for arclength is not difficult to remember, but it will take many examples tobecome proficient. Make sure to work through homework until youcan work problems quickly.
Area of a Surface of Revolution
In this section we will give another application related to arc length.We will define the surface of revolution to be the surface resultingfrom revolving the graph of a function f (x) around a line. Thefollowing gives a way to calculate the area of a surface of revolution.
Theorem 1.26. Let y = f (x) have a continuous derivative on the interval[a, b]. The area S of the surface of revolution formed by revolving the graph
26
of f about a horizontal or vertical axis is given by
S = 2π∫ b
ar(x)
√1 + [ f ′(x)]2 dx
where r(x) is the distance between the graph of f and the axis of revolution.If x = g(y) on the interval [c, d], then the surface area is
S = 2π∫ d
cr(y)
√1 + [g′(x)]2 dy
where r(y) is the distance between the graph of g and the axis of revolution.y
x2
4
Rotation of f (x) = x2 about the y-axis.
Example 1.27. The arc of the parabola y = x2 from x = 0 to x = 2 isrotated about the y-axis. Find the resulting surface area.
Solution: To begin, notice that
dydx
= 2x.
Therefore
S =∫ 2
02πx
√1 +
(dydx
)2
dx
= 2π∫ 2
0x√
1 + 4x2 dx
Now we will let u = 1 + 4x2, and thus du = 8x dx. Then with the properlimits of integration, we have
S =π
4
∫ 17
1
√u du
=π
4
[(2/3)u3/2
]17
1
=π
6(17√
17− 1)
Example 1.28. The arc of y = cos x from x = 0 to x = 2π is rotated aboutthe y-axis. Find the resulting surface area.
y
Rotation of f (x) = cos x for x ∈ [0, 2π]about the y-axis.
Solution: To begin, notice that
dydx
= − sin x
27
Therefore
S =∫ 2π
02πx
√1 + sin2(x) dx
For this example, we concede that the integral above cannot be evaluated an-alytically by our methods. However if we evaluate this integral numerically,we have the following
S ≈ 150.82
Example 1.29. (Gabriel’s Horn) For our final example, we will calculatethe surface area of Gabriel’s Horn. This is the surface area of revolution ofy = 1/x about the x-axis from x = 1 to x = ∞.
Solution: To begin, notice that
dydx
=−1x2
Therefore
S =∫ ∞
12π f (x)
√1 + f ′(x)2 dx
= 2π∫ ∞
1
(1x
)√1 +
1x4 dx.
Now we will use the comparison test for integrals with the following obser-vation √
1 +√
1x4 > 1
on the interval [1, ∞). Therefore we have
S = 2π∫ ∞
1
(1x
)√1 +
1x4 dx
≥ 2π∫ ∞
1
1x
dx = ∞
This concludes our examples in this section. Make sure to workthrough several examples to see the difficulties and intricacies ofworking surface area problems.
Application to Physics and Engineering
In this section we will consider two applications. The first will beto study fluid pressure and force. The section application will be toconsider the center of mass of a uniform mass distribution.
28
Fluid Force and Fluid Pressure
In this section we will consider two principles that will help us tocalculate pressure and force on an object submerged in a fluid. Thefirst physical law that we will consider is Pascal’s Principle. Thisstates that the pressure exerted by a fluid on an object at depth ddd istransmitted equally in all directions.
The second principle that we need is that the fluid pressure in-creases the deeper that an object is submerged. This is seen in thefollowing formula
Definition 1.30. The pressure on an object at depth d is defined to be theforce per unit are:
P =FA
= ρgd
where ρ is the density of the fluid and g is the acceleration constant due togravity. The constant g is given by g = 9.8m/s2 and the density of water isgiven by ρ = 1000kg/m3. Likewise we can solve for pressure to give
PA = F =ρgdA
.
Remark 1.31. Notice that the principles above allow us to use integration tocalculate the force exerted by a fluid on a vertical plate. This is given in thefollowing definition.
Definition 1.32. The force F exerted by a fluid of density ρ on a verticalplane region from y = a to y = b is given by
ρg∫ b
ah(y)L(y) dy
where h(y0) gives the depth of the fluid at y = y0, and L(y0) gives thehorizontal length of the region at height y = y0.
Now let us consider an example using this formula.
Example 1.33. A trapezoidal plate having the dimensions 8m across on thetop, 6m across on the bottom, and a height of 5m, is submerged in a pool ofwater. Find the force on the plate if it is submerged vertically so that the topis 4m below the surface of the pool.
Solution: First we must find formulas for h(y) and L(y). For h(y), we willset the x-axis to sit on the top of the water, and we will center the y-axis tothe center of the trapezoid. This will lead to
h(y) = −y.
Next we will notice that L(x) is given by two times the x coordinate ofan edge of the plate. This can be seen in the diagram on the following page.
29
Next, observe that we have L(y) in terms of x. This means that we mustfind a relationship between the x coordinate of the right edge of the plate andy. To do this, we will use the point slope formula to find the equation for theline passing through the points (3,−9) and (4,−4). Therefore we have
y + 9 = 5(x− 3)
and this leads to the following expression for x
x =y + 24
5.
Therefore
L(y) =25(y + 24).
Finally, we must find the limits of integration. These are given by a =
−9 and b = −4. Therefore we have the following calculation
30
F = ρg∫ −4
−9h(y)L(y) dy
= 9800∫ −4
−9(−y)
(25
)(y + 24) dy
= 3920∫ −9
−4y2 + 24y dy
= 3920[
y3
3+ 12y2
]−9
−4
= 3920(
16753
)
≈ 2.19× 106N
For this section, we have only completed one example, but please checkmoodle for other examples.
Moments and Centers of Mass
In this next section, we will consider the problem of finding thecenter of mass of a two dimensional plate, and the moment of asystem of masses. We will begin this section by first considering aone dimensional problem.
To find the center of mass or balancing point of a one-dimensionalsystem of two point masses of the following system
is given by the following formula
x =M1x1 + M2x2
M1 + M2
31
This can be extended to several point masses m1, ..., mn lying alongthe x-axis at positions x1, ..., xn. In this case, we have the following
x = (1/M)n
∑i=1
mixi
Where M =n
∑i=1
mi is the total mass of the system and the terms mixi
are referred to as moments about the point (0, 0).
This can be further extended to a system of masses in two dimen-sions. Consider a system of n particles with masses m1, ..., mn locatedat the points (x1, y1), ..., (xn, yn) in the xy-plane. Following the one-dimensional case, we define the moment of the system about theyyy-axis to be
My =n
∑i=1
mixi
and the moment of the system about the xxx-axis is
Mx =n
∑i=1
miyi.
Then the center of mass of this system is given by (x, y) where
x =My
m, y =
Mx
m,
and
m =n
∑i=1
mi
Centroid of a Lamina
Finally, we will extend our results to find the centroid of a flat plateor lamina in the xy-plane. The following gives us the result that weseek
Definition 1.34. Let f and g be continuous functions such that f (x) ≥g(x) on [a, b], and consider the lamina of uniform density ρ bounded by thegraphs of y = f (x) and y = g(x) and a ≤ x ≤ b.
1. The moments about the xxx-axis and yyy-axis are
Mx = ρ∫ b
a
[f (x) + g(x)
2
][ f (x)− g(x)] dx
My = ρ∫ b
ax[ f (x)− g(x)] dx.
32
2. The center of mass is given by (x, y), where
x =My
m,
y =Mx
m,
and
m = ρ∫ b
a[ f (x)− g(x)] dx
These formulas can be simplified to the following
1) x =1A
∫ b
ax[ f (x)− g(x)] dx
2) y =1A
∫ b
a
12
[f (x)2 − g(x)2
]dx
where A is the area between f (x) and g(x).
Now we will work through a few examples that show how such acalculation is performed.
Example 1.35. Find the center of mass of the region bounded by the curvesy = x2 and x = y2.
Solution: To begin, we notice that the region is bounded between x = 0 andx = 1 and
√x ≥ x2 on this interval. Therefore we will let f (x) =
√x and
g(x) = x2. Now we will calculate the area of this region.
A =∫ 1
0[√
x− x2] dx
=
[23
x3/2 − d f rac13x3]1
0
=13
33
Now we will calculate x and y.
x = 3∫ 1
0x[√
x− x2] dx
= 3∫ 1
0x3/2 − x3 dx
= 3[(
25
)x2/5 −
(14
)x4]1
0
= 3(
320
)= 9/20
y =32
∫ 1
0[x− x4] dx
=32
[(12
)x2 −
(15
)x5]1
0
=32
(3
10
)=
920
Therefore we have
(x, y) =(
320
,3
20
)
Example 1.36. Find the centroid of the region bounded by y = sin x andy = cos x between x = 0 and x = π/4.Solution: To begin, we calculate the area between these two curves. Sincecos x ≥ sin x on this interval, we have
A =∫ π/4
0[cos x− sin x] dx
=
[sin x + cos x
]π/4
0
=√
2− 1
Now we will calculate x and y. Notice that in the following calculations,
34
we will use integration by parts.
x =1√
2− 1
∫ π/4
0[x(cos x− sin x)] dx
=1√
2− 1
[x(
sin(x) + cos(x))∣∣∣∣π/4
0−∫ π/4
0(cos x + sin x) dx
]
=1√
2− 1
[(π/4)
(√2
2+
√2
2
)− 0 ·
(0 + 1
)+[− sin(x) + cos(x)
]π/40
]
=1√
2− 1
[(π/4)
(−√
2)+
(−√
22
+
√2
2− 0− 1
)]
=π√
2− 44√
2− 4
and
y =1√
2− 1
∫ π/4
0
12
[cos2 x− sin2 x
]dx
=1√
2− 1
∫ π/4
0
14(1 + cos(2x)− 1 + cos(2x)) dx
=1√
2− 1
∫ π/4
0
12
cos(2x) dx
=1
2(√
2− 1)
(sin x cos x
)∣∣∣∣π/4
0
=1
4(√
2− 1)
Therefore the centroid of this region is
(x, y) =(
π√
2− 44√
2− 4,
14(√
2− 1)
)This concludes our excursion into applications of calculus to physics.Make sure to work through several example in order to gain profi-ciency in these exercises.
Differential Equations
Introduction
In this chapter, we will discuss an important area of mathematicsknown as differential equations. Many subjects outside of mathuse models to predict future outcomes. Differential equations areparticularly important in modeling quantities that change with time.The following are important differential equations that are used tomodel phenomena in various fields.
• Schrödinger’s equation (physics)
• Navier-Stokes equation (physics)
• Rate equation (chemistry)
• Differential form of Gibbs Equation (chemistry)
• Solow-Swan Model (economics)
• Black-Scholes (economics)
• Hodgkin-Huxley model (biology)
• Lotka-Volterra (biology)
Let us now consider the vocabulary of this chapter. A differentialequation is an equation that contains a dependent variable, usu-ally "y", its derivatives y′, y′′,..., y(n), and possibly an independentvariable x. The following are examples of differential equations
y′′ − y = 1 y′ + y = sin x
y′′ − y = 4e−x y′ + exy = arctan(x)
The order of a differential equation is the order of the highest deriva-tive in the differential equation. The equations in the left column aresecond order differential equations and the right column containsfirst order differential equations.
36
A solution to a differential equation is a function f such that theequation is satisfied when y is replaced by f (x). For example f1(x) =sin(x) is a solution to the equation y′′ + y = 0 since
f ′′1 (x) = − sin(x).
Notice that in this example, f2(x) = cos x is another solution to thisdifferential equation. It turns out that A f1(x) and B f2(x) are alsosolutions, where A and B are constants. The following example givesa way to show whether or not a function is a solution to a differentialequation.
Example 1.37. Show that yg = A sin x + B cos x is a solution to thedifferential equation y′′ + y = 0.
Solution: Since yg = A sin x + B cos x and y′′g = −A sin x− B cos x, wehave
y′′g + yg = (−A sin x− B cos x) + (A sin x + B cos x) = 0.
Since yg satisfies the equation y′′ + y = 0, it follows that yg is a solution tothe differential equation.
In the example given above, the set F1 = {A sin x : A is a constant}is called a family of solutions. Similarly F2{B cos x : B is a constant}is also a family of solutions. It turns out that the differential equationy′′ + y = 0 only has two families of solutions and combinations offunctions from both of these families is also a solution. Now we willintroduce a definition that will allow us to discuss solutions.
Definition 1.38. If F1 and F2 are two sets of functions, then a linearcombination of these sets is a function of the form
h(x) = a · f (x) + b · g(x)
Where a and b are constants.
Example 1.39. Using the example above,
h(x) = 2 sin x + 3 cos x
is a linear combination of sin x and cos x.
Finally we can restate the result of example 1.37 succinctly as anylinear combination of sin x and cos x is a solution of y′′ + y = 0.
This leads to a natural question. If F1 and F2 are two families of solu-tions to a differential equation, is it true that any linear combination
37
from these two families is also a solution? In general, the answer tothis is question is no. If however, the differential equation is "linear"(to be defined later), then the answer is yes. Now let us consideranother definition
Definition 1.40. If F1,..., Fn are the families of solutions to a lineardifferential equation, then the general solution is defined to be
a1 f1(x) + a2 f2(x) + ... + an fn(x)
Where the ai’s are constants and fi ∈ Fi.
With all this talk about general solutions and family of solutions, onemight expect that differential equations have an infinite number ofsolutions. This is true for a general differential equation, but oftenthere is additional information that restricts us to have only onesolution. The piece of information that gives us a particular solutionis called an initial condition. An initial value is often given by one ofthe following conditions
y(x0) = y0 or y(t0) = y0
An initial value along with a differential equation is called an initialvalue problem. We will now give two examples of how to solve aninitial value problem.
Example 1.41. For the differential equation xy′ − 3y = 0 and initial valuey(−3) = 2, verify that y1 = Cx3 is a solution. Further, use the initial valuefind the particular solution.
Solution: Notice that y′1 = 3Cx2, therefore
xy′1 − 3y1 = 3Cx3 − 3Cx3 = 0
and this shows that y1 a solution to xy′ − 3y = 0. Now we may apply theinitial value to obtain the following
2 = y1(−3)
2 = C(−33)
− 227
= C
Therefore the particular solution is
y = − 227
x3
Example 1.42. Find the particular solution to the differential equationy′′ + y = 0 with the initial conditions y(0) = 1 and y′′(π/4) = 2
38
Solution: From a previous example, we know that the general solution tothis differential equation is y1 = a sin x + b cos x. Now we can apply theinitial value y(0) = 1 to obtain
A sin(0) + B cos(0) = 1
B = 1
Now we can apply the initial value y(π/4) = 2 and use the fact that B = 1to obtain
A sin(π/4) + cos(π/4) = 2
A(√
2/2) +√
2/2 = 2
A + 1 = 4/√
2
A = 2√
2− 1
Therefore the particular solution is
y = (2√
2− 1) sin x + cos x
We will conclude this section by conceding the fact it can be verydifficult to solve certain differential equations, and in some cases,it can be impossible. In the next section we will find a numericalmethod for solving first order differential equations.
Slope Fields and Euler’s Method
Separable Equations
Models for Population Growth
For this section, we will be brief. We will introduce the two modelsof population growth that are of interest to us and then we will workand example of each. To begin, let us define the two models.
Definition 1.43. The initial-value problem
dPdt
= kP P(0) = P0
is the differential equation that models exponential growth/decay. k is knownas the relative growth rate. If k > 0, this represent exponential growth.If k < 0, it represents exponential decay.
39
Definition 1.44. The initial-value problem
dPdt
= kP(
1− PM
)P(0) = P0
is the differential equation that models logistic growth.
The solution to the exponential growth/decay problem is found witha simple separation of variables. The solution is
P(t) = P0ekt
The solution to the logistic growth problem is found using a sepa-ration of variables along with partial fraction decomposition. Thesolution is
P(t) =M
1 + Ae−kt where A =M− P0
P0
To begin, let us do an example of exponential decay.
Example 1.45. (Newton’s Law of Cooling) Let T represent the temper-ature of an object in a room whose temperature is kept at a constant 60oF.Newton’s law states that the rate of change in the temperature of an objectis proportional to the difference between the temperature of the object andthat of the surrounding medium. If the object cools from 100oF to 90oF in10 minutes, how much longer will it take for the temperature to decrease to80oF? Solution: To begin, notice that our differential equation is
dTdt
= k(T − 60), 80 ≤ T ≤ 100
We will now solve this initial value problem with separation of variables.
dTdt
= k(T − 60)
dTT − 60
= k dt
∫ 1T − 60
dT =∫
k dt
ln |T − 60| = kt + C1
Since T > 0, |T − 60| = T − 60, and we have the following
T − 60 = ekt+C1 =⇒ T = 60 + Cekt
40
From the initial condition T(0) = 100, we have 100 = 60 + Ce0. ThusC = 40. Further, since T(10) = 90, we have
90 = 60 + e10k
30 = 40e10k
k = (1/10) ln 3/4
Therefore we have the following solution
80 = 60 + 40eln((3/4)1/10)t
20 = 40eln((3/4)1/10)t
1/2 = eln((3/4)1/10)t
ln(1/2) = ln((3/4)1/10)t
t =ln(1/2)
ln((3/4)1/10)≈ 24.09 minutes
This shows that it will take 24.09 minutes for the object to reach 80oF.
Example 1.46. The Pacific halibut fishery has been modeled by the differen-tial equation
dydt
= ky(
1− yM
)
where y(t) is the biomass (total mass of the members of the population) inkilograms at time t (measured in years), and k = 0.71
(a) If y(0) = 2× 107 kg, find the biomass a year later.
(b) How long will it take for the biomass to reach 4× 107 kg?
Solution: We will solve the problem using separation of variables along withpartial fraction decomposition. To begin, we have
41
dydt
=
(kM
)· y(M− y)
M dyy(M− y)
= k dt
∫ My(M− y)
dy =∫
k dt
∫ [1y+
1M− y
]dy = kt + C1
ln |y| − ln |M− y| = kt + C1
ln∣∣∣∣M− y
y
∣∣∣∣ = −kt + C1
M− yy
= e−kt+C1 = Ce−kt
M− y = Cye−kt
(Ce−kt + 1)y = M
y =M
1 + Ce−kt
Now we will use the initial value y(0) = 2.7× 107 and M = 8× 107 toobtain
C =8× 107 − 2× 107
2× 107 = 3
Therefore after one year, the biomass will be
y = 8× 107/(1 + 3e−0.71) ≈ 3.23× 107 kg
Linear Equations
A first-order linear differential equation is an equation of the follow-ing form
y′ + P(x)y = Q(x)
42
where P and Q are continuous function on a given interval. Linearfirst order equations show up in many applications to science. Tosolve such a differential equation, we will use the method of multi-plying the differential equation by an integrating factor. To see howthis will be achieved, consider the following example
Example 1.47. Using the product rule and the fundamental theorem ofcalculus, we have
ddx
[ye∫
P(x) dx]= y′e
∫P(x) dx + y
ddx
(e∫
P(x) dx
= y′e∫
P(x) dx + ye∫
P(x) dx[
ddx
∫P(x) dx
]
= e∫
P(x) dx[
y′ + P(x)y]
Therefore we have
e∫
P(x) dx[
y′ + P(x)y = Q(x)]
⇐⇒ ddx
[ye∫
P(x) dx]= e
∫P(x) dxQ(x)
=⇒ ddx
[ye∫
P(x) dx]= e
∫P(x) dxQ(x)ye
∫P(x) dx =
∫e∫
P(x) dxQ(x) dx
This leads to the following definition
Definition 1.48. The integrating factor of a first order linear differentialequation is given by
I(x) = e∫
P(x) dx
To solve a linear first order differential equation, we must multiply both sidesby I(x) and integrate.
Predator-Prey Systems
Glossary of Terms
• Autonomous Equation - An autonomous equation is a differentialequation that depends only upon the dependent variable. In
43
general, a first order autonomous equation is of the form
y′ = f (y)
• Carrying Capacity - The carrying capacity of the logistic growthmodel is the maximum population that the environment is capableof sustaining over a long period of time.
• Ordinary Differential Equation - A differential equation is amathematical equation involving a dependent variable, its deriva-tives, and possibly an independent variable. An ordinary differen-tial equation does not involve the derivatives of an independentvariable.
• Direction Field - A graphical representation of the solutions of afirst order differential equation. It is created by placing small linesegments, that represent slope, on several points in the xy-plane.
• Equilibrium Solution - An equilibrium solution is a solutionthat is constant with respect to the independent variable of adifferential equation.
• Euler’s Method - Euler’s method is a numerical method for solv-ing first order differential equations. (Pronounced as "Oil-ers" not"Yew-lers")
• Explicit Solution - A solution that depends only upon the inde-pendent variable.
• Exponential Growth - A model for population growth that as-sumes that the change in population is proportional to a growthconstant times the current population.
• Family of Solutions - A set of solutions to a differential equationthat differ by multiplication or addition of a constant. It mayturn out that a differential equation has more than one family ofsolutions.
• General Solution - A general solution to a differential equation isof the form
y1 + y2 + ... + yn
where each yi are from a family of solutions.
• Implicit Solution - A solution that depends on both the indepen-dent and dependent variable of the differential equation.
44
• Initial Condition - An initial condition is the value of the indepen-dent variable at a particular value of the dependent variable. Thisis often given in the following form
y(t0) = y0 or y(x0) = y0
• Initial Value Problem - An initial value problem is a differentialequation along with an initial value. An initial value is needed toproduce a unique solution to a differential equation.
• Integrating factor - A function that is chosen to aid in solving alinear differential equation. For a linear differential equation of theform
y′ + P(x)y = Q(x)
the integrating factor is
I(x) = e∫
P(x) dx
• Linear First Order Equation - A differential equation of the form
y′ + P(x)y = Q(x)
where P(x) and Q(x) are continuous function depending onlyupon x.
• Logistic Model - A model for population growth that assumesexponential growth with the exception that population must bebounded by a constant known as the carrying capacity.
• Lotka-Volterra Equations - A system of two equations that de-scribe the relationship between two populations. One being apredator population, and the other being the prey population.
• Numerical Solution - A method/algorithm designed to approx-imate the solution of a differential equation. For example seeEuler’s method.
• Order of a Differential Equation - The order of a differentialequation is the highest derivative in the equation.
• Orthogonal Trajectory - An orthogonal trajectory of a family ofcurves is a curve that intersects each curve of the family of curvesat a ninety degree angle.
• Particular Solution - A particular solution is a solution curve thatis determined by an initial condition.
• Phase Plane - Given a system of differential equations with depen-dent variables R and W, the RW-plane is called
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• Phase Portrait - A geometric representation of the trajectories of asystem of differential equations. If the dependent variables of thesystem are R and W, then this consists of equilibrium points, theslope field determined by dW/ dR, and typical trajectories in thephase plane.
• Phase Trajectory - A solution curve drawn in the phase plane iscalled a phase trajectory.
• Predator-Prey Equations - See Lotka Volterra Equations
• Relative Growth Rate - The constant given in the exponentialgrowth model.
• Slope Field - See direction field.
