Dynneson Calculus

7/23/2019 Dynneson Calculus

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UNIVERSITY OFTEXAS ATELPAS O

Behind the Curtains; Calculus Demystified

Andrew Dynneson, M.A.

Last Edit: April 7, 2015


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Behind the Curtains; Calculus Demystified.

2014 by Andrew Dynneson. All rights reserved.

Contact Author: [email protected] || +1 831.apachi-1

Alpha Edition- 2014 - University of Texas at El Paso.

Andrew Dynneson [email protected] i


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CONTENTS

0 Introduction 1

1 Various Equation Derivations 2

1.1 Exponential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 DeMoivre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2.1 Polygons in the complex-plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 Unit Circle in the Complex Plane Continued (Eulers Formula) . . . . . . . . . . . . . . . . . . . . . . . 4

1.4 Hyperbola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Trigonometric Identities 6

2.1 Sum and Product Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.1.1 Asin x+Bcos x=

A2 + B2 sin(x+) or

A2 +B2 cos(x+) . . . . . . . . . . . . . . . . . . . . 6

1 Limits 8

1.1 Formal Definition of Limits and Proofs of Limit Properties. . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.1.1 - Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2 Exponential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 Differentiation in Theory 11

2.1 Inverse Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.2 Exponential and Logarithmic Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 Differentiation in Practice 123.1 Extreme Values of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1 Integration in Theory 14

1.1 Area Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.1.1 Summation Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.1.2 Sigma. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.1.3 Summation Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.1.4 Riemann Summations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.2 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.2.1 Simpsons Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2 Integration in Practice 19

2.1 Logarithmic Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.1.1 A few Trigonometric Integrals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3 Volume 20

Appendix: Hass Integration Tables 22

Appendix: Derivations that do not use Calculus 29

3.1 Geometric Derivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.1.1 Area of a slice of pizza and arc-length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.1.2 Surface Area of a frustum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Andrew Dynneson [email protected] ii


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CHAPTER0

INTRODUCTION

Book I Precalculus

Andrew Dynneson [email protected] 1


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CHAPTER1

VARIOUSEQUATIONDERIVATIONS

1.1 EXP ON ENT IAL

The constantebecomes approached as the number of compoundingsnbecome large. Here, the rate is 100%, and

the time intervalt= 1. Then, we let ngrown larger and see that the compounding equation begins to approach theideal of continuous compounding by way of computation:

In Calculus, we will be able to get-at this number more exactly. Next, once we see that (1 + 1/n)n e, we canadapt this approximation to see that the compounding equation approximates continuous compounding for n

large enough.

Letrbe the desired rate. Replacingnwithn/ris okay, because if our rate is less than 100%, thenn/rbecomes

larger sincer< 1, and our calculation converges even faster. On the other hand, ifr> 1, then it is true thatn/rbecomes smaller, but I claim that we only need to takenout further in that case to get our calculation to converge

to the desired level of accuracy. Thus the claim is that:

1+ 1

n/r

n/r e

The nextstepis to divide1/(n/r)

=r/n. Also, takingthe rth powerof both sides yields: 1+

rnn = 1+

1n/rn/r

r

erNext, take the tth power of both sides to introduce time-increments into the equation:

1+ rn

nt er t. Fi-nally, multiply both sides of the equation by the initial value P, and we have successfully derived the continuous-

compounding equation:

P

1+ rn

nt Per t



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1.2 DEMOIVRE

1.2.1 POLYGONS IN THE COMPLEX-PLANE

One of the many many reasons that DeMoivres theorem is so useful is that it gives us a really cool way to program

polygons.

Consider the example of an heptagon. Dividing the unit-circle into seven equal arcs reveals each to be = 2/7

One of the vertices should be (1,0). The next vertex, we label , and the special math word for it is primitive.

Notice that = cos+ isin.Next, squaring this complex number reveals the next vertex my DeMoivres Theorem: 2 = cos2 = isin2, so

that themth vertex is m = cosm+ isinm, and 7 = (1,0) = 0.One could just as easily use this process to build the regular n-gon, since = 2/n, and then= cos+ isin,

the primitive vertex, and the subsequent vertices:m = cosm+ isin m.*Notice that as the number of vertices of our regular n-gon becomes larger and larger, the polygon will begin to

approximate a circle, the edges becoming finer and finer.

Image:



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1.3 UNI TCIRCLE IN THECOMPLEXPLANECONTINUED(EULERSFORMULA)

"one of the most remarkable, almost astounding, formulas in all of mathematics." Richard Feynman

We have already seen that the unit-circle in the complex plane is mapped-out by cos+ isin.Beforewe cancontinue, I will need to introduce some important approximations, namely, as the angle becomes

small, approaching zero, let us see what happens to our elementary trig-functions:

Notice that for a desired level of accuracy, whenever 0, we see that cos 1 and sin . These are impor-tant approximations which are used often in physics.

In particular, for this section, for any angle, notice that/ngoes to zero as nbecomes larger and larger, so

that fornlarge enough, /n

0, and: cos n+ isin n 1+ in().Next, recall that negative and rational exponents made perfect sense when they were explored, and irrational

exponents almost made sense from an approximation stand-point. However, at this point we are going to jump

into the deep-end of abstraction, and attempt to discuss a complex-exponent! Even the existence ofi is highly

questionable, to attempt to take a number to the power ofiis highly questionable. However, the result is of such

fundamental importance and beauty that one wonders.

DeMoivre allows us to take the nth power of both sides of () to get cos+ isin1+ in

n. Now, whatever ei

is, it should be approximated by the same as er. Therefore, fornlarge enough, we have:

cos+ isin

1+ in

n ei

Notice that the left-most and right-most sides have no reference to n, only that nneeds to be large enough to

make the approximation accurate enough. Now, we will do a Calculus-thing, lettingnactually go to infinity will

cause the approximations to become exact! And: cos+ isin = ei , and the complex exponential actually mapsout the unit-circle, in much the same way as DeMoivre maps out a polygon!

1.4 HYP ER BOL A

Figure and proof skeleton:



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For a specific distance c, let the focus F1 be the point whose distance is calong the x-axis in the negative

direction. Similarly, let the focus F2be the point whose distance iscalong thex-axis in the positive direction. The

distance between the fociF1andF2is 2c. Also, fix a value for a

The hyperbola is defined as the collection of points Pwhose difference of the distance from F2to Pand F1to P

is fixed at 2a. More explicitly, |distance(F2, P) distance(F1, P)| = 2a.We will show that the equation of the points along the hyperbola is given by the equation:

x2

a2 y

2

c2 a2= 1

The vast majority of students skip this derivation altogether. The more astute will derive it with the use of a

computer algebra software. Only the truly mad would attempt to derive this by hand. So, here we go:

First, notice that along the x-axis, this points (a,0) are the appropriate distance from our foci, because theirdistance is (c+ a) (c a) = 2a, showing that these two points are on the hyperbola.

We derive the left-hand branch, the right-hand branch should be similar. LetP(x,y) be a point on the left-

branch. By the distance formula, distance(F2, P) =

(xc)2 +y2 and distance(F1, P) =

(x+c)2 +y2, so that bydefinition:

2a=

(xc)2 +y2

(x+c)2 +y2

It is from consideration of this equation that the result will come. Left-hand-side and right-hand-side of the

equation will be abbreviatedL.H.S.andR.H.S., respectively. Squaring both sides yields:

4a2 = (

(x c)2 +y2

(x+ c)2 +y2)2 = (x c)2 +y2 2

(xc)2 +y2

(x+ c)2 +y2 + (x+ c)2 +y2

= x2 2xc+ c2 +y2 2

(x c)2 +y2

(x+ c)2 +y2 + x2 +2xc+ c2 +y2

= 2x2 + 2y2 +2c2 2

(xc)2 +y2

(x+c)2 +y2 = 2x2 + 2y2 + 2c2 2[((xc)2 +y2)((x+c)2 +y2)]1/2

4a2 = 2x2 + 2y2 +2c2 2[(x2 c2)2 +y2[(xc)2 + (x+ c)2]+y4]1/2 2[(x2 c2)2 +y2[(x c)2 + (x+c)2]+y4]1/2 = 2x2 +2y2 +2c2 4a2

[(x2 c2)2 +y2[(xc)2 + (x+ c)2] +y4]1/2 = x2 +y2 + c2 2a2()(L.H.S.)

=[x4

2c2x2

+c4

+y2(x2

2xc

+c2

+x2

+2xc

+c2)

+y4]1/2

= [c4 2c2x2 + x4 + 2c2y2 + 2x2y2 +y4]1/2 = x2 +y2 + c2 2a2() ()c4 2c2x2 + x4 + 2c2y2 + 2x2y2 +y4 = (x2 +y2 +c2 2a2)2

(R.H.S.) = (x4 + x2y2 +c2x2 2a2x2)+ (x2y2 +y4 +c2y2 2a y2)+ (c2x2 + c2y2 + c4 2a2c2) 2a2(x2 +y2 +c2 2a2)=x4 +y4 +c4 +4a4 +

2x2y2 +2c2x2 +

2c2y2 4a2x2 4a2y2 4a2c2

= () =c4 2c2x2 +x4 +2c2y2 +

2x2y2 +y4

0 = 4a4 + 4c2x2 4a2x2 4a2y2 4a2c2 0 = a4 +c2x2 a2x2 a2y2 a2c2 = x2(c2 a2) a2y2 a2(c2 a2)

x2(c2 a2) a2y2 = a2(c2 a2) x2

a2 y

2

c2 a2= 1

As desired



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CHAPTER2

TRIGONOMETRICIDENTITIES

2.1 SUM ANDPRODUCTFORMULAS

2.1.1 Asin x+Bcos x=

A2 +B2 sin(x+)O R

A2 +B2 cos(x+)

Considering the figure:

There is an angle such that sin= Bkand cos= Ak, where k=

A2 + B2. This is taken care of even whenAorBis negative. Then:

Asin x+ Bcos x= k

A

ksin x+ B

kcos x

= k(cossin x+ sincos x)

By the sum-angle identity, this is equal toksin(x+), and tan= BA.Source:

By using the phase-shifting identity,

A2 + B2 sin(x+) =

A2 +B2 cos(x+ 2 )



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Book II Differentiation and Limits



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CHAPTER1

LIMITS

1.1 FORMALDEFINITION OFL IMITS ANDPROOFS OFL IMITPROPERTIES

Note to Instructor: Many instructors choose to skip this section because it is too advanced for most first-year

students of Calculus. If it is required as part of the curriculum; make sure to calm the students down (had one

student walk out of the class muttering obscentities, and several others drop as a result of this lecture, and the

topic did not even show up on the exams). Nevertheless, exposure to formal \limits is invaluable, even if not all

of its subtleties are grasped the first-time-around.

The limit properties appear as Theorem 1.2 in Larsons Calculus book[3], section 1.3.

First assume that we are given that limxcf(x) = L, also we are given an > 0.

1. To show that for any real-number b, limxcbf(x) = bL, we know that there is a > 0 such that for anyxin the

interval |xc| < , then we can say that f(x) is in the interval |f(x) L| < |b| .From this we know that for anyxin |x c| < , then |bf(x) b L| = |b(f(x) L)| = |b||f(x) L| < |b| |b|=.Next, suppose that lim

xcg(x) = K.2. Then, we want to show that lim

xcf(x)+ g(x) = L+ K.To show this, we first need the Triangle Inequality, that is, for any two numbersxandy, we know that |x+y|

|x| + |y|. There is a way to visualize this using a triangle, however I usually find it easier to say that introducingabsolute values on the inside will only make the quantity larger, because off-set signs will detract from eachother,

whereas if both of the signs ofxand yare the same, then the sum will be larger than if they are not the same:

|x+y| | |x|+ |y||, and then because ||x|+ |y|| is positive, can drop the outer-most absolute value:|x+y| | |x|+ |y| |= |x|+ |y|.Now, we can return to showing that lim

xcf(x) + g(x) = L+ K. We know that there is a1 such that|x c| < 1implies that |f(x) L| < /2. Similarly, there is a 2such that |x c| < 2implies that |g(x) K| < /2.

Then, let= min{1,2} (Takethe smaller of the two deltas). Then, |xc| < implies that |f(x)+g(x)(L+K)| =|f(x) L+ g(x) K| |f(x) L|+ |g(x) K| < 2 + 2= (Where the is because of the Triangle Inequality).

3. Showing that limxcf(x) g(x) = LK is slightly more difficult, and may not be what you expect.

First, There are 1,2 > 0, and letting= min{1,2}, then |xc| < implies that |f(x)L| < and |g(x)K| 0 such that |g(x) K|


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Next, the hat-trick: 1g(x) 1K=

Kg(x)g(x)K= 1|K g(x)| |Kg(x)| = 1|K| 1|g(x)| |Kg(x)| < 1|K| 2|K| |g(x) K| < 2|K|2 |K|

2

2 =

This shows that limxc

1g(x)= 1K.

To finish, we only need to use 3: lim

xc

f(x)g(x)

=lim

xcf(x) 1g(x)

=L1K

=L/K.

5. For integer exponents, it is easy to showthat limxc[f(x)]

n = Ln, because we can use rules 3 and 4. For example,limxc[f(x)]

4 = limxc[f(x) f(x) f(x) f(x)] = LLLL= L

4. Similarly, limxc[f(x)]

3 = limxc

1[f(x)]3

= 1L1L1L.For rational exponents, it will be easier to show a stronger rule first, namely that the limit can be evaluated

inside of a continuous function.

Source:

This next Theorem is modified slightly from Larsons [3] Theorem 1.5. its proof is also modified from[3], Ap-

pendix A.

This deals with the composition of two functions, and shows that their limits also compose. Suppose that we

know that limxcg(x) = Land that limxLf(x) = K. Then, we want to show that The function composition, limxcf(g(x)) =

K.

Proof. Given > 0, we know that there exists a > 0 such that |u L| < implies that |f(u) K| < .Furthermore, we know that there exists a > 0 such that |x c| < implies that |g(x) L| < .Simply letu= g(x), then |xc| < implies that |g(x) L| < , which then implies that |f(g(x)) K| <

1.1.1 - LIMITSThe infinite case is similar, and I claim that infinite case depends on the finite case provided you first show the

composition rule. The argument is very similar to the finite case. Let > 0 be given.Suppose that lim

xf(x) = Land limxLg(x) = K. Then, we want to show that the composition, limxg(f(x)) = K.Based on the given information, for any > 0, there exists a such thatx> implies that |f(x) L| < . Also

from the given, there exists a > 0 such that|y L| < implies that|g(y) K| < . Composing the two, lettingy=f(x) yields that |g(f(x)) K| <

Next, suppose that limxcf(x) = , and limxg(x) = K, then want to show that the composition limxcg(f(x)) = K.

From the given information, for any, there is a > 0 such that |xc| < implies that f(x) >.Also, from the given, there is a specific such thaty> implies that |g(y) K| < .We can now compose, lettingy=f(x), and so |g(f(x)) K| < , as desiredThe other composition rules should be similar in their justifictation.

Next, I show the other operations, let = +,, ,. This uses primarily the composition rule and the finite rules:

limxf(x) g(x) = limx0+ f

1

x

g

1

x

= lim

x0+f

1

x

lim

x0+g

1

x

= lim

xf(x) limxg(x)

We have now established the rule for the operations = ,+,, ,.

1.2 EXP ON ENT IAL

Lety= limx

1+ 1x

x, we want to show that y= e.

Since the natural logarithm is continuous, taking the logarithm of both sides yields:

lny= limx ln

1+ 1

x

x= lim

xxln

1+ 1x

= lim

xln1+ 1x

1/x

()

Next, since 1/x 0 as x , and does so in a positive/decreasing way, (*) is equal to limt0+

ln(1+t)t . Since ln1 = 0,

can subtract it from the numerator, and since ln is differentiable, by definition we have:

http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspxhttp://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspxhttp://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx


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lny= limt0+

ln(1+ t) ln1t

= dd x

ln x

x=1

= 1x

x=1

= 1

Since lny= 1, this implies thaty= e, as desired. Source: Appendix of[3]



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CHAPTER2

DIFFERENTIATION INTHEORY

2.1 INVERSEDERIVATIVES

First Apply Axlers treatment of inverse functions[1] (p. 204). Then, switch to Hass[2] p.104. This is a visual for

b=f(a) f1(b) = a, supposingfis invertible. Then:

(f1)(b) = 1f(a)

= 1f(f1(b))

.

Then, a notation change:

d f1

d x

x=b

= 1d fd x

x=b

In addition to the graph, this can also be shown using the chain rule:

f(f1(x)) = x 1 = dd x

f(f1(x)) =f(f1(x)) dd x

f1(x) dd x

f1(x) = 1f(f1(x))

.

2.2 EXP ONE NT IA L AN DLOGARITHMICDERIVATIVES

Sinceax = eln ax = exln a, then dd xax = dd xexln a = exln a ln a= ax ln aApplying the conversion formula[1], logax= ln xln a. Then, dd xlogax= 1xln a.



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CHAPTER3

DIFFERENTI ATION INPRACTICE

3.1 EXT RE ME VALUE S OF FUNCTIONS

Theorem 1. Extreme values of continuous functions occur at critical values, or at end-points.

The intuition is as follows. Suppose thatcis a maximum or a minumum for a continuous function f. Iff(c) is

undefined, i.e. f is not differentiable atc, thencis a critical for f by definition. So, suppose thatf(c) exists. Then,it is either positive, negative, or zero. We want to show that it only makes sense for f(c) to be zero. If it is positve ornegative, then whenxis sufficiently close toc, the graph off looks like one of these two graphs:

Clearly, in either case,ccan be neither a max nor a min, because you need only to go a little to the left or right

to get a value on f that is more or less thanf(c).

Therefore, by process of elimination, f(c) = 0, andcis a critical value for f.Theorem 2. RollesTheorem: If f is continuous on[a, b] and differentiable on(a, b), and suppose f(a) =f(b). Then,there exists a c between a and b such that f(c) = 0.

Iffis a constant function, then f = 0. If fis not a constant function, then there is anxbetweenaandbsuchthat f(x) = f(a) nor f(b). By Extreme Value Theorem, continuity of f implies that fattains a maximum and aminimum on (a, b), and one of these is not one of the endpoints because f(x) is more-than or less-than f(a) and

f(b). Label this extreme value c. Finally, previous theorem3.1implies that cmust be a critical, and differentiability

on (a, b) implies thatf(c) = 0.



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Book III Integration and Series



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CHAPTER1

INTEGRATION INTHEORY

1.1 A RE AAPPROXIMATION

1 .1 .1 SUMMATIONNOTATION

Fibonacci SequenceFirst, an example of a famous sequence of numbers. This is commonly attributed to the

mathematician Fibonacci of Pisa, although it is believed that he didnot invent it. The sequence beginswith 0 and1,then the next numberis found by addingthe previous two. Thefirst several terms: 0,1, 1,2, 3,5,8, 13,21,34,55,89,144,233,37

if you letA(n) be the nth element in the sequence, then A(n) = A(n1)+A(n2) [i.e., the sum of the previoustwo elements. This is an example of a recursively defined sequence. It requires that you know A(0)=0 andA(1) = 1.

This sequence is very interesting, and involves the golden-ratio, which unfortunately goes beyond the scope of

this course, except to say that the sequence has a closed-form definition in addition to the recursive definition.

The golden-ratio:

= 1+

5

2 1.6180339887

Believe it or not, it can be shown that A(n) = n (1)n

5, test it on your calculators!

1.1.2 SIGMA

When you must add many terms, it is often convenient/necessary to use summation notation. We use the Greek

letter sigma () to indicate that we are taking a summation. In the previous example, we could abbreviate adding

any number of terms in the Fibonacci sequence, lets say the first 16 terms:

0+ 1+ 1+2+ 3+ 5+8+ 13+ 21+34+55+ 89+ 144+ 233+ 377+ 610+ 987 =A(0)+A(1)+A(2)+A(3)+A(4)+A(5)+A(6)+A(7) +A(8)+A(9)+A(10)+A(11)+A(12)+A(13)+A(14)+A(15)

=15

n=0A(n)

Since A(0) = 0, why not just call it

15n=1A(n). We will study summations in intricate detail in this course.

Figure: Fibonacci Spiral

1.1 .3 SUMMATIONF ORMULAS

The fact thatn

i=1C= Cnis clear, pretty much from the definitionof multiplication, for example 2+2+2+2+2 = 25.



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To see thatn

i=1i= n(n+1)

2 , visualize this as dot-counting:

You want to count the number of red-dots. So, you make a congruent triangle of green dots inverted, and placed

on top of the red-triangle. Then, theresult is a rectangle, the base of which has n+1 dots, and the height has ndots.Then, there are a total ofn(n+ 1) dots in the rectangle, exactly half of which are red.

Source:

Next, to see thatn

i=1i2 = n(n+ 1)(2n+ 1)

6 , see the diagram below:

One of the pyramids pictured in the top-left has the desired volume. Thus, when you put the three together,

slicing half of the top piece, and reflecting it, it forms a perfect rectangular prism. Then, the following formula is

achieved:

3 n

i=1i2 = n(n+ 1)(n+ 1/2)

Simplification yields the desired formula.[4]

The Evidence thatn

i=1i3 =

ni=1

i

2can be seen in the following diagram[4]:



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Then substitute the previous formulan

i=1i= n(n+1)

2 , to achieve the desired result:

ni=1

i3 =

ni=1

i

2=

n(n+1)2

2= n

2(n+1)24

.

1.1 .4 RIEMANN SUMMATIONS



20/34

1.2 THE FUNDAMENTAL THEOREM OFCAL CU LU S

Now that you have completed these activities, you are ready to derive one of the most powerful results in Calculus!

What you have shown essentially is that the derivative of the area under a curve function is equal to the function of

the curve itself. In other words, if you know the anti-derivative, you can find the area.

The first part of the Fundamental Theorem is stated precisely: Let fbe a continuous function on the interval

[a, b], andF(x) =x

a f(t)d t. We know thatF(x) is continuous on [a, b] and differentiable on (a, b).

F(x) = dd x

xa

f(t)d t=f(x).

The second part ofThe Fundamental Theoremstates that if fis a continuous function on the interval [a, b],

andFis an anti-derivative off on [a, b], then we can calculate the area exactly:

ba

f(x)d x= F(b) F(a)

The proof of this is based on what we have already shown, namely that F(x) = dd xx

a f(t)d t. This is incredible,

since it implies that there is a constant Csuch that:

F(x)+

C=

x

a

f(t)d t

Then, substitutex= ainto this function to get:

F(a)+C=a

af(t)d t= 0

Since there is no area within a vertical line atx= a. But this implies that C=F(a), which yields:

F(x) F(a) =x

af(t)d t

Now, simply substitutex= binto this function to get the desired result:

ba f(t)d t= F(b) F(a)

1.2.1 SIMPSONSMETHOD

Small parabolas can be used to approximate a curve even more accurately than rectangles or trapazoids.

Divide the interval [a, b] into an even number of sub-intervals. Consider a single cross-section, as though it

were centered at the origin (for the area will be the same no matter where it is horizontally located.



21/34

TheAreaof this cross-section can be found by integrating under the curve:

hh

Ax2+B x+C d x= Ax3

3 +B x

2

2 +C x

hh

= Ah3

3 + Bh

2

2 +C h

A(h)3

3 + B(h)

2

2 C h

= 2Ah

3

3 +2C h= h

3(2Ah2+6C)

Now, from the points on the curve, (h,y0),(0,y1), (h,y2), we achieve the system of equations;

y0=

Ah2

Bh

+C

y1 = Cy2 = Ah2 + Bh+C

Next, by subtractingy1 = Cfrom both sides, we achieve:

y0 y1 = Ah2 Bhy2 y1 = Ah2 + Bh

Finally, adding them together reveals:y0 +y2 2y1 = 2Ah2, substituting them intoAreaof the section,Area= h3 (y0 +4y1 +y2). And, adding all of the areas of each section together:

ba

f(x)d xh

3 (y0+4y1+y2)+

h

3 (y2+4y3+y4)+. . .+

h

3 (yn2+4yn1+yn) =

h

3 (y0+4y1+2y2+4y3+2y4+. . .+2yn2+4yn1+yn).[



22/34

CHAPTER2

INTEGRATION INPRACTICE

2.1 LOGARITHMICI NTEGRATION

2 .1 .1 A F EWTRIGONOMETRIC I NTEGRALS

sec xd x=sec xsec x+ tan xsec x+ tan x

d x=

sec2 x+sec xtan xsec x+ tan x

d x

Substitution ofu= sec x+ tan xyields d u= sec xtan x+ sec2 xd x, and the integral becomes

d uu = ln |u| =

ln |sec x+ tan x|+C.In order to integrate csc x, we must first find a few more derivatives. For example, dd xcsc x= dd x 1sin x= cos xsin2 x=csc xcot x[By quotient rule].Similarly, the dd xcot x= dd x cos xsin x= sin

2 xcos2 xsin2 x

= csc2 x. With these derivatives in mind, we can integrate:

csc xd x=

csc xcsc x+cot xcsc x+cot xd x=

csc2 x+csc xcot x

csc x+ cot x d x

Then, substitution ofu= csc x+cot ximplies (1)du= csc xcot x+csc2 xd x, andthe integral becomes (1)

d uu=

ln |u| = ln |csc x+cot x|+C.



23/34

CHAPTER3

VOLUME

Exercise 1: Volume of a thickened cylinder[5]. Let Abe the surface area of a toilet-paper roll of height and radius

r. Recall that the Surface Area of such a cylinder is equal to the circumference of the circle times height:

A= 2r. Also consider a solid shell around this cylinder of thicknessh(see figure).

Letk= 1/r. In terms of Aand k, what is the Volume of this shell? Change your answer into one of thefollowing forms:

(a) h+kh2

(b) A(h+ 12 kh2)(c) Ah+ 12 kh(d) A(h2 + kh2)

Solution:

(b)V= 2r+h

rrd r =[(r+h)2 r2] = (2r)h+h2 = Ah+ 1

2rAh2 = A(h+ 1

2kh2)

Exercise 2: Sphere. Derive formulas for the Volume of a sphere, given the radius. Be sure to cite your sources.

Solution:Note to TA: If someone submitted a solution that does not use Integration, please email it to me.

Volume:The radius of the sphere isr.



24/34

radius of a cross-section:x=

r2 z2Area of a cross-section:A=(r2 z2)

V= rr

r2 z2d z= 43r3

Source:

Exercise 3: Spherical Shell[5]. LetAbethe surface areaof a sphereof radius R, andsurround the spherewith a sphericalshell of thicknessh(See figure)1.Recall that the surface area of a sphere is A= 4R2

Find the volume of this shell, in terms ofAandR, and convert your answer into one of the following forms:

(a) A[ hR+ h3

3R2]

(b) 1+ hR+ h2

2R2

(c) Ah[1 + hR+ h2

3R2]

(d) h[1+ hR+ h2

3R2]

Solution:

(c)V= 43[(R+h)3 R3] = 43 (3R2h+3Rh2 +h3) = 4R2h+4Rh2 + 43h3 = Ah+ Ah2

R + Ah3

3R2= Ah[1+ hR + h

2

3R2]

1Figure: http://ned.ipac.caltech.edu/level5/March06/Overduin/Figures/figure2.jpg



25/34

APPENDIX: HASSINTEGRATIONTABLES

[2]



26/34

Basic Forms

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

Forms Involving

21.

22.

23. 24.

25. 26.

27. 28. L2ax + b

x dx = 22ax + b + bL dx

x2ax + bL A2ax + b Bn

dx = 2

a

A2ax + b Bn + 2n + 2

+ C, n Z -2

L dx

xsax + bd =

1bln` x

ax + b + CLxsax + bd

-2dx =

1

a2cln ax + b + b

ax + bd + C

Lxsax + bd-1

dx = x

a - b

a2 ln ax + b + CLsax + bd

-1dx =

1aln ax + b + C

Lxsax + bdn

dx = sax + bdn + 1

a2 cax + bn + 2 - bn + 1d + C, n Z -1, - 2

Lsax + bdn

dx = sax + bdn + 1

asn + 1d + C, n Z -1

ax b

L dx

2x2 - a2 = cosh-1 x

a + C sx 7 a 7 0dL dx

2a2 + x2 = sinh-1 x

a + C sa 7 0d

L dx

x2x2 - a2 = 1

asec-1 ` xa` + CL

dx

a2 + x2

= 1

atan-1 x

a + C

L dx

2a2 - x2 = sin-1 x

a + CLcoshx dx = sinhx + C

L sinhx dx = coshx + CL cotx dx = ln sinx + CL tanx dx = ln secx + CL cscxcotx dx = -cscx + CL secxtanx dx = secx + CL csc

2x dx = -cotx + C

L sec2

x dx = tanx + CL cosx dx = sinx + CL sinx dx = -cosx + CLa

xdx =

ax

lna + C sa 7 0, a Z 1d

Lex

dx = ex + CLdxx = ln x + C

L xn

dx = x

n + 1

n + 1 + C sn Z -1dLk dx = kx + C sany number kd

T-

A BRIEF TABLE OF INTEGRALS


27/34

29. (a) (b)

30. 31.

Forms Involving

32. 33.

34.

35.

36.

37.

38.

39.

40. 41.

Forms Involving

42. 43.

44. 45.

46.

47. 48.

49. 50.

51.

Forms Involving

52.

53. L2x2 - a2 dx =

x

22x2 - a2 - a2

2 lnx +2x2 - a2 + C

2x2 - a2 + C= lnx +L dx

2x2 - a2

x2 a2

L dx

x22a2 - x2 = -

2a2 - x2a

2x

+ CL

dx

x

2a2 - x2

= -1aln

à +2a2 - x2

x

` + C

L

x2

2a2 - x2

dx = a

2

2

sin-1xa -

1

2

x2a2 - x2 + CL2a2 - x2

x2

dx = -sin-1xa -

2a2 - x2x + CL

2a2 - x2x dx =2a2 - x2 - alnà +2a

2 - x2

x ` + CLx

22a2 - x2 dx = a48

sin-1xa -

18

x2a2 - x2 sa2 - 2x2d + CL2a

2 - x2 dx = x

22a2 - x2 + a2

2sin-1

xa + CL

dx

2a2 - x2 = sin-1 x

a + C

L dx

sa2 - x2d2 =

x

2a2sa2 - x2d +

1

4a3ln`x + ax - a + CL

dx

a2 - x2

= 12a

ln`x + ax - a + Ca2 x2

L dx

x22a2 + x2 = -

2a2 + x2a

2x

+ CL dx

x2a2 + x2 = -1alnà +2a

2 + x2

x ` + CL

x2

2a2 + x2 dx = - a

2

2 lnAx +2a2 + x2 B + x2a2 + x2

2 + C

-2a2 + x2

x + CL2a2 + x2

x2

dx = lnAx +2a2 + x2 BL2a2 + x2

x dx =

2a

2

+ x2

- alnà +2a2 + x2

x ` + C

Lx22a2 + x2 dx = x

8sa2 + 2x2d2a2 + x2 - a4

8 lnAx +2a2 + x2 B + C

+ a

2

2 lnAx +2a2 + x2 B + CL2a

2 + x2 dx = x

22a2 + x2

L dx

2a2 + x2 = sinh-1 x

a + C = lnAx +2a2 + x2 B + CL

dx

sa2 + x2d2 =

x

2a2sa2 + x2d +

1

2a3tan-1

xa + CL

dx

a2 + x2

= 1

atan-1 x

a + C

a2 x2

L dx

x22ax + b = -

2ax + bbx

- a

2bL dx

x2ax + b + CL2ax + b

x2

dx = -2ax + b

x + a

2L dx

x2ax + b + CL

dx

x2ax - b = 2

2b tan-1A

ax - bb

+ CL dx

x2ax + b = 1

2b ln2ax + b -2b2ax + b +2b ` + C

T-2 A Brief Table of Integrals


28/34

54.

55.

56.

57.

58.

59.

60.

61. 62.

Trigonometric Forms

63. 64.

65. 66.

67.

68.

69. (a)

(b)

(c)

70. 71.

72. 73.

74.

75.

76. m - 1

m + nL sinn

axcosm - 2 ax dx, m Z - n sreduces cosm axdL sinn

axcosm ax dx = sinn + 1 axcosm - 1 ax

asm + nd +

n Z - m sreduces sinn axdL sinn

axcosm ax dx = -sinn - 1 axcosm + 1 ax

asm + nd +

n - 1m + nL sin

n - 2axcosm ax dx,

Lsinaxcosax dx = -

1aln cosax + C

L cosn

axsinax dx = -cosn + 1 ax

sn + 1da + C, n Z -1L

cosaxsinax

dx = 1

aln sinax + CL

sinn axcosax dx = sinn + 1 ax

sn + 1da + C, n Z -1

Lsinaxcosax dx = -

cos 2ax

4a + C

Lcosaxcosbx dx = sinsa - bdx

2sa - bd +

sinsa + bdx

2sa + bd + C, a2 Z b 2

Lsinaxsinbx dx = sinsa - bdx

2sa - bd -

sinsa + bdx

2sa + bd + C, a2 Z b 2

Lsinaxcosbx dx = -cossa + bdx

2sa + bd -

cossa - bdx

2sa - bd + C, a2 Z b 2

L cosn

ax dx = cosn - 1 axsinax

na + n - 1

n L cosn - 2

ax dxL

sinn ax dx = -sinn - 1 axcosax

na + n - 1

n

L

sinn - 2 ax dx

L cos2

ax dx = x

2 +

sin 2ax4a

+ CL sin2

ax dx = x

2 -

sin 2ax4a

+ C

Lcosax dx = 1

asinax + CLsinax dx = -1acosax + C

L dx

x22x2 - a2 =

2x2 - a2a

2x

+ CL dx

x2x2 - a2 = 1

asec-1 ` xa + C = 1acos-1 ` ax + CL

x2

2x2 - a2

dx = a

2

2 lnx +2x2 - a2 + x

22x2 - a2 + C

L2x2 - a2

x2

dx = lnx +2x2 - a2 -2x2 - a2x + CL2x2 - a2

x dx =2x2 - a2 - asec-1 ` xa + CLx

22x2 - a2 dx = x8s2x2 - a2d2x2 - a2 - a4

8 lnx +2x2 - a2 + C

Lx A2x2 - a2 B

n

dx =A2

x2 - a2 B n + 2

n + 2 + C, n Z -2

L dx

A2x2 - a2 B n =x A2x2 - a2 B2 - n

s2 - nda2 -

n - 3

sn - 2da2L dx

A2x2 - a2 B n - 2, n Z 2L A2x

2 - a2 Bn

dx =x A2x2 - a2 B n

n + 1 -

na2

n + 1L A2x2 - a2 B

n - 2dx, n Z -1

A Brief Table of Integrals T-


29/34

77.

78.

79. 80.

81.

82.

83. 84.

85. 86.

87. 88.

89. 90.

91. 92.

93. 94.

95. 96.

97. 98.

99.

100.

101. 102.

Inverse Trigonometric Forms

103. 104.

105.

106.

107.

108.Lxn tan-1 ax dx =

xn + 1

n + 1tan-1 ax -

a

n + 1L x

n + 1dx

1 + a2x2, n Z -1

Lxn cos-1 ax dx =

xn + 1

n + 1cos-1 ax +

a

n + 1L x

n + 1dx

21 - a2x2, n Z -1Lx

n sin-1 ax dx = x

n + 1

n + 1sin-1 ax -

a

n + 1L x

n + 1dx

21 - a2x2, n Z -1L tan

-1ax dx = xtan-1 ax -

12a

lns1 + a2x2d + CL

cos-

1 ax dx = xcos-

1 ax - 1

a

21 - a2x2 + C

Lsin

-1 ax dx = xsin

-1 ax +

1a

21 - a2x2 + C

L cscn

axcotax dx = -cscn ax

na + C, n Z 0L secn

axtanax dx = secn ax

na + C, n Z 0

L cscn

ax dx = -cscn - 2 axcotax

asn - 1d +

n - 2n - 1L csc

n - 2ax dx, n Z 1

L secn

ax dx = secn - 2 axtanax

asn - 1d +

n - 2n - 1L sec

n - 2ax dx, n Z 1

Lcsc2 ax dx = -1acotax + C

Lsec2 ax dx = 1atanax + C

Lcscax dx = -1aln cscax + cotax + CLsecax dx =

1aln secax + tanax + C

L cotn

ax dx = -cotn - 1 ax

asn - 1d - L cot

n - 2ax dx, n Z 1L tan

nax dx =

tann - 1 ax

asn - 1d - L tan

n - 2ax dx, n Z 1

L cot2

ax dx = -1acotax - x + CL tan

2ax dx =

1atanax - x + C

Lcotax dx = 1

aln sinax + CL tanax dx = 1

aln secax + C

Lx n cosax dx = x

n

a sinax - na

Lx n - 1 sinax dx

Lx n sinax dx = - x

n

a cosax + na

Lx n - 1 cosax dx

Lxcosax dx = 1

a2cosax +

xasinax + CLxsinax dx =

1

a2sinax -

xacosax + C

L dx

1 - cosax = -

1acot

ax

2 + CL

dx

1 + cosax =

1atan

ax

2 + C

b2 6 c2L

dx

b + ccosax =

1

a2c2 - b 2 ln`c + bcosax +2c2 - b 2 sinax

b + ccosax ` + C,

L dx

b + ccosax =

2

a2b 2 - c2 tan-1 cA

b - cb + c

tanax

2d + C, b 2 7 c2

L dx

1 - sinax = 1atanap

4 + ax

2b + CL dx

1 + sinax = -1atanap

4 - ax

2b + C

b2 6 c2L

dx

b + csinax =

-1

a2c2 - b 2 ln`c + bsinax +2c2 - b 2 cosax

b + csinax ` + C,

L dx

b + csinax =

-2

a2b 2 - c2 tan-1 cA

b - cb + c

tanap4

- ax

2b d + C, b 2 7 c2



30/34

Exponential and Logarithmic Forms

109. 110.

111. 112.

113.

114.

115. 116.

117.

118. 119.

Forms Involving

120.

121.

122.

123.

124.

125.

126.

127. 128.

Hyperbolic Forms

129. 130.

131. 132.

133. L sinhn

ax dx = sinhn - 1 axcoshax

na - n - 1

n L sinhn - 2

ax dx, n Z 0

L cosh2

ax dx =sinh 2ax

4a +

x

2 + CL sinh

2ax dx =

sinh 2ax4a

- x

2 + C

Lcoshax dx = 1

asinhax + CLsinhax dx = 1

acoshax + C

L dx

x

22ax - x

2= -

1a A

2a - xx + CL

x dx

22ax - x

2= asin-1 ax - aa b -22ax - x2 + C

L22ax - x2

x2

dx = -2A2a - x

x - sin-1 ax - aa b + C

L22ax - x2

x dx =22ax - x2 + asin-1 ax - aa b + CLx22ax - x

2dx =

sx + ads2x - 3ad22ax - x26

+ a

3

2sin-1 ax - aa b + C

L dx

A22ax - x2 Bn =sx - ad A

22ax - x2 B

2 - n

sn - 2da2 +

n - 3

sn - 2da2L dx

A22ax - x2 B n - 2L A22ax - x

2 Bn

dx =sx - ad A22ax - x2 Bn

n + 1 +

na2

n + 1L A22ax - x2 B

n - 2dx

L22ax - x2

dx = x - a

2 22ax - x2 + a2

2sin-1 ax - aa b + C

L dx

22ax - x2 = sin-1 ax - aa b + C

22ax x2, a>0L

dx

xlnax

= ln ln ax + C

Lx

-1slnaxdm dx = slnaxdm + 1

m + 1

+ C, m Z -1

Lxnslnaxdm dx =

xn + 1slnaxdm

n + 1 -

m

n + 1Lxnslnaxdm - 1 dx, n Z -1

L lnax dx = xlnax - x + CLeax cosbx dx =

eax

a2 + b 2

sacosbx + bsinbxd + C

Leax sinbx dx =

eax

a2 + b 2

sasinbx - bcosbxd + C

Lxn

bax

dx = x

nb

ax

alnb -

n

alnb Lxn - 1

bax

dx, b 7 0, b Z 1 L

xn

eax

dx = 1

axn

eax -

na

L

xn - 1

eax

dx

L

xeax

dx = e

ax

a2 sax - 1d + C

Lbax

dx =1

ab

ax

lnb + C, b 7 0, b Z 1Le

axdx =

1ae

ax + C

A Brief Table of Integrals T-


31/34

134.

135. 136.

137. 138.

139. 140.

141. 142.

143.

144.

145. 146.

147. 148.

149.

150.

151. 152.

153.

154.

Some Definite Integrals

155. 156.

157.

L

p>2

0

sinn x dx =

L

p>2

0

cosn x dx =

d1 # 3 # 5 # # sn - 1d

2 # 4 # 6 # # n# p

2, if nis an even integer 2

2 # 4 # 6 # # sn - 1d3 # 5 # 7 # # n , if nis an odd integer 3

Lq

0

e-ax2

dx = 12A

p

a , a 7 0Lq

0

xn - 1

e-x

dx = snd = sn - 1d!, n 7 0

Leax coshbx dx =

eax

2 c e bx

a + b +

e-bx

a - bd + C, a2 Z b 2L

e ax sinhbx dx = e

ax

2 c e

bx

a + b - e

-bx

a - b d + C, a2 Z b 2L csch

naxcothax dx = -

cschn axna + C, n Z 0L sech

naxtanhax dx = -

sechn axna + C, n Z 0

L cschn

ax dx = -cschn - 2 axcothax

sn - 1da -

n - 2n - 1L csch

n - 2ax dx, n Z 1

L sechn

ax dx = sechn - 2 axtanhax

sn - 1da +

n - 2n - 1L sech

n - 2ax dx, n Z 1

L csch2

ax dx = -1acothax + CL sech

2ax dx =

1atanhax + C

Lcschax dx =

1aln` tanh

ax

2` + CLsechax dx =

1asin

-1

stanhaxd + C

L cothn

ax dx = -cothn - 1 ax

sn - 1da + L coth

n - 2ax dx, n Z 1

L tanhn

ax dx = -tanhn - 1 ax

sn - 1da + L tanh

n - 2ax dx, n Z 1

L coth2

ax dx = x - 1

acothax + CL tanh2

ax dx = x -1

atanhax + C

Lcothax dx = 1

aln sinhax + CL tanhax dx =1

alnscoshaxd + CL

xn

coshax dx = x

n

a sinhax - n

a

Lx

n - 1sinhax dx

Lx

n

sinhax dx = x

n

a coshax - n

a

Lx

n - 1coshax dx

Lxcoshax dx = x

asinhax - 1

a2coshax + CLxsinhax dx =

xacoshax -

1

a2sinhax + C

L coshn

ax dx = coshn - 1 axsinhax

na + n - 1

n L coshn - 2

ax dx, n Z 0



32/34

APPENDIX: DERIVATIONS THAT DO NOT USECALCULUS

3.1 GEOMETRICDERIVATIONS

3.1 .1 A REA OF A SLICE OF PIZZA AND ARC-LENGTH

The circle has arear2. Withmeasured in radians, you can imagine that a slice is a proportion, or fraction of

the overall circle. The ratio is : 2. And so, the area of the slice:

A=r2 2

= 12r2

.

Sometimes it is easier to measure by the length of the arc. by using a similar proportion as before, the circum-ference of the circle is 2r, and so the length of the arc on the slice: S= 2r

2= r .

Now, using circumference proportionS: 2r, the area of the slice is:

A=r2 S2r

= 12

S r

.

3 .1.2 SURFACEAREA OF A FRUSTUM

2Here r is the radius of the top-circle, Ris the radius of the bottom circle, and Lis the length of the lateral-edge (see

figure).

2Source: http://www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-lateral-area-of-a-right-circular-cone



33/34

Notice that an invisible cone-top is added to the figure, and two more measurement are made: L1is the total

length of the side of the cone, andL2is the length of the invisible cone that was added to the frustum.

Notice also that a ratio of hypotenuse to side lengths of two right-triangles can be made:

L1

R =L

R r.

From this, we see thatL1 =RL

R r. Also, note thatL2 = L1 L=RL

R r L.The circumferences of bottom and top circles are respectively: s1= 2Rand s2= 2r. Take scissors and cut

alongL, and unroll to make:

Notice thats1ands2now form arcs. So, from the pizza-slice area formula, the Surface Area of the frustum is:

S A= 12

s1L1 1

2s2L2 =R

RL

R r r r L

R r =(R2 r2)L

R r =(R+ r)L

If you know the length of the height h instead of the length of the side L, you can move halongr until r

vanishes. This will form a right-triangle whose legs are R r andh, and hypotenuse L. Then, by Pythagorean:L=(R r)

2 +h2.



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BIBLIOGRAPHY

[1] Sheldon Axler.Precalculus, a Prelude to Calculus. John Wiley and Sons, 2009.

[2] Hass, Weir, and Thomas.University Calculus. Addison-Wesley, 2007 and 2012.

[3] Larson and Edwards.Calculus. Brooks Cole Cengage Learning, 2010.

[4] Nelsen.Proofs without Words: Exercises in Visual Thinking. MAA, 1993.

[5] Shlomo Sternberg.Semi-Riemann Geometry and General Relativity. September 2003.

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Dynneson Calculus