For student 3 (Newton+ Optimization+integration).docx

OPTIMIZATION

Optimization Problems

An optimization problem seeks to find the largest ( or smallest) value of a quantity given certain limits or constraints. These problems arise in many areas of application. In solving these problems the greatest challenge is to convert the word problem into a mathematical optimization problem by setting up the function to be maximized or minimized.

Steps:

1. Understand the problem:

Read until you clearly understood.

- What is the unknown quantity to be optimized? - What are the given quantities?

- What are the given conditions unknown?

2. Draw a diagram:

Identity the given and required quantities on the diagram.

3. Introduce Notation:

Assign a symbol to the quantities to be maximized or minimized, Q or f, and the other unknowns quantities and level the diagram with these symbols.

4. Express Q as a function of the other variables from step 3. Also, express any relationships or conditions among the other variables with equations.

5. Rewrite, if necessary, Q as a function of just one of the variables and determine the domain of Q.

6. Use the methods of calculus to find the required optimum value.

Example:

2- Find the radius and height of the cylinder of the largest volume inside a cone with radius 6 and height 10.

4- Find the point on the curve y=x2_{that is closest to the point (18,0)}

Newton – Raphson Method(Newton’s Method)

- A technique to approximate the solution to an equation f x

 

0.

Let f be a continuous differentiable function on an open interval with a real root.

Steps:

1. Guess a first approximation to a solution of the equation f x

 

0. ( A graph of f x

 

0 may help)

2. Compute a second approximation using the formula,

 

 

 

1 , 0

n

n n n

n f x

x x f x

f x

   _   .

3. Repeat step 2 until it closes enough to the root.

Indeterminate Forms and L’Hospital’s Rule

Indeterminate Forms of Type 0

0 and

 

A limit of the form

 

 

lim

x a f x g x

 where xlimaf x

 

and xlimag x

 

are either both 0 or both  are called an indeterminate forms of type 0

0 and



 respectively.

Example: find 1 ln lim 1 x x x   . ???????

Theorem: (L’Hospital’s Rule)

Let f and g be differentiable functions with g x

 

0 on an open interval containing a, except possibly at a. Suppose

 

 

lim  x a

f x

g x produces an indeterminate form

0 0 or

  .

If

 

 

lim    x a f x

g x exists, or is + or -,then,

 

 

 

 

lim lim     

x a x a

f x f x

g x g x .

Moreover, this statement is also true in the case of a limit as _ 

x a , xa, x  ,

or as x  .

Applying L’Hospital’s Rule

Step 1: Check that the limit of f x

   

g x is indeterminate form of type 0

0 or

  .

Step 2: Differentiate f and g separately.

Step 3: Find the limit of f

   

x g x . If this limit is finite, +, or -, then it is equal

to the limit of f x

   

g x .

2-1

ln lim

1 x

x x

 

3-x x

x

2 sin lim

0



4- ₂

0

tan lim

x x

x 

5- _x

x _e

x

  lim

Other Indeterminate Forms

a) Indeterminate Products: 0. or 0.

If lim

 

0 and lim

 

 

x a f x  x a g x    , then x alim f x g x

   

   0



0 



To evaluate these limits,

i) rewrite f x g x

   

 as either

 

 

or

 

 

1 1

f x g x

g x f x .

ii) apply L’Hospital’s Rule

Example:

1- x x

xlim0 .ln

2- x x

x

2 sec ) tan 1 ( lim

4





b) Indeterminate Differences:

  

If lim

 

and lim

 

x a f x   x a g x  , then x alim f x

 

g x

 

   .

To evaluate these limits,

i) rewrite f x

 

g x

 

into a quotient by using a common denominator or rationalization or factoring out a common factor

Example:

1- 

  

  



 _{x x}

x _sin

1 1 lim

0

2-



x x x



x  

2

lim

3- lim



ln( 2 1)





 x x

c) Indeterminate Powers: _{0 ,}0 _0_,1

Several indeterminate forms arise from the lim

 

g x  x a f x 

1. lim

 

0 and lim

 

0

x a f x  x a g x  , type

0

0

2. lim

 

and lim

 

0

x a f x   x a g x  , type

0



3. lim

 

1 and lim

 

x a f x  x a g x  , type 1 

In these cases,

i) let y f x

 

g x 

ii) taking the natural logarithm:

 

 

lny g x lnf x iii) If lim

 

ln

 

x a g x f x exists and equals L,

then lim

 

g x  x a f x  =

L e .

Example: 1-





x

x

x

1 0

1 lim_ 



2-





x x x 1 0 sin 1 lim_ 

CHAPTER 4: INTEGRATION 4.2: THE INDEFINITE INTEGRAL

Antiderivatives:

Definition: A function F is called an antiderivative of f on an interval I if

F ’(x) = f (x) for all x I . Or

dx

d _{F (x) = f (x) for all x I .}

Theorem: If F is an antiderivative of f on an interval I, then the most general anti-derivative of f on I is

 

 

G x F x C , where C is an arbitrary constant.

Notation: antidifferentiation or integration is the process of finding anti-derivatives.

Here are some anti-differentiation formulas that you need to know well.

Function Antiderivative

a f x

 

a F x

 

+ C

 

 

f x g x F x

 

G x

 

+ C



1



n

x n   1

1 n x _C n   

sin x - cos x + C

cos x sin x + C 2

sec x tan x + C

sec x tan x sec x + C

Example: find the antiderivatives of the following functions;

- 5

2 ) (x x

f 

- f(x)5sinx

- f(x)sec2 xsecxtanx

- ( ) 1₃

x x

Indefinite Integrals

Notation:

_

f x dx

 

F x

 

C , where F (x ) is an antiderivative of f and C is an arbitrary constant.

Rules:

_

c f x dx

 

c f x dx

_

 

 

 

 

 

f x g x dx f x dx g x dx

    

 







 

 

 

 

a f x b g x dx a f x dx b g x dx







Formulas:

k dx kx C



where k is any constant 1

, 1

1 n

n x

x dx C n

n



   





1_{dx ln x C}

x  



x x

e dx e C



ln x

x a

a dx C

a

 



sinx dx  cosx C



cosx dx sinx C



2

sec x dx tanx C



2

csc x dx  cotx C



secx tanx dx secx C



csc cotx x dx  cscx C



1

2 2

1 _dx 1_tan x _C

x a a  a

   _{ }  _{ }



1 2 2

1 _{dx sin} x _C

Example:

-



xdx

-





3x6 2x2 7x1



dx

- dy y y y



            4 10 ₃ 4 3 - dt t t t

- dx x

x x



sec_2coscos

- dx

x



_1sin1

-



cot2 x dx

-



x dx

2 cos2

4.3: TECHNIQUES OF INTEGRATION

In this chapter we develop techniques for using the basic integration formulas to obtain indefinite integrals of more complicated functions.

1. The Substitution Rule

The method of substitution is used to evaluate integrals of the form

   

f g x g x dx_ _ 



.

The integral is rewritten in the form

_

f u du

 

, where u  g x

 

. The important step is to recognize a proper choice for u and the corresponding du g x dx

 

.

Example:

-





x2 1



50.2xdx

-

_

   



 _ 5

8 3 1

-



sin



2x9



dx

-



tanxdx

-



cotxdx

-



sin



sin



.cosd

-



dx

x x

cos

-



x2 x1dx

-



 

dx x

x

2

5 sin

-



sec43xdx

TECHNIQUES OF INTEGRATION

7.2. Integration by Parts (page 491)

In this section we develop a general method for attacking integrals of the form

   

f x g x dx



Integration by parts is a procedure based on reversing the product rule for differentiation. Recall that, if f and g are differentiable functions, then

   

   

   

d f x g x f x g x g x f x

dx      

   

   

   

f x g x 

_

_f x g x g x f x dx _

f x g x

   



_

f x g x dx

   

 

_

g x f x dx

   



_

f x g x dx f x g x

   

 

   



_

g x f x dx

   



Let u f x

 

andv g x

 

. Then du f x dx 

 

and dv g x dx

 

.

Thus,

_

u dv uv 

_

v du

The aim in using integration by parts is to obtain a new integral that is easier to evaluate than the original. A strategy that often works is to choose u and dv so that u become “simpler” when differentiated , while dv can be readily integrated to obtain v.

Notes:

1. Polynomial, exponential, trigonometric - functions can differentiate and integrate

2. logarithmic, inverse trigonometric

Example:

1-

_

xcosx dx

2-



x.ex dx

4-



xlnx dx

5-





dx x

1

tan

6-

_



dx x

2 cos 1

Repeated Integration By Parts

It is sometimes necessary to use integration by parts more than once in the same problem.

Example:

1-

_

x2cosx dx

2-



x3sinx dx

4-



x2. x1 dx

5-



ex.sinx dx

The Definite Integral

Definition: Let f x be a function defined on a close interval [a, b]. The definite

 

integral of f from a to b is

 

 

1 lim b n i n _i a

f x dx f x x

 _









When this limit exists, we say that f is integrable on the interval [a, b].

The sum

 

* 1

n i i

f x x







is called a Riemann Sum.

Riemann Sum

Let f x be a function defined on a closed interval [a, b].

 

Steps:

1. Partition the interval [a, b] into n subintervals of equal width x b a n



  .

Let x₀ a x x, ₁, ₂, , x_n1, x_n b be the endpoints of these subintervals.

2. Let x x₁, ₂, ,x_n _{be any sample points in these intervals. So}

i

x _{lies in the i th}

subinterval _x_i1,x_i_.

3. Form the sum.

 

1 1

 

2 2

 

n n n

R _f x _{ }x f x _x _{ }f x _x

 

1

n

k k

k

f x x







 .

Theorem ( The existence of Definite Integral)

If f is continuous on an interval [a, b], then f is integrable on [a, b].

Theorem: If f is integrable on [a, b], then

 

 

1 lim b n i n _i a

f x dx f x x

 _









where x b a n



  and x_i   a i x .

Properties of the Definite Integral

If f and g are integrable on [a, b], then

1.

 

0

a

a

f x dx 



2.

 

 

a b

b a

f x dx   f x dx





3.





b

a

c dx c b a 



, where c is any constant

4. b[

 

 

] b

 

b

 

a a a

f x g x dx  f x dx  g x dx







5.

 

 

b b

a a

c f x dx c f x dx





, where c is any constant

6.

 

c

a

f x dx



+

 

b

c

f x dx



=

 

b

a

f x dx



, where a c b

7. If f x

 

0 for a x b, then

 

b

a

f x dx





0.

8. If f x

 

g x

 

for a x b, then

 

b

a

f x dx





b

 

a

g x dx



.

9. If m f x

 

M for a x b, then





 





b

a

The Fundamental Theorem of Calculus

Part 1: If f is continuous on [a, b] and F is any antiderivative of f , then

 

 

 

b

a

f x dx F b F a



Example:

Evaluate the following:

1-

_

9

4 2

dx x x

2-



2

0 5

sin 

3-





4

4

tan sec





xdx x

4-

_



2

6

2 )

sin 2 (





dx x x

5- 2



2



3 0

1 x  x dx



or

ii) change the limit to fit u:

Example:

Use the inverse of property (6) to evaluate the following:

1-



3

0

) (x dx

f if

  

 

 

2 2

3

2 )

(

2

x x

x x x

2-



2

0

) (x dx

f where

        2 1 1 0 ) ( 2 x x x x x f 3-2 2

.

x dx





4-



4

Theorem: ( The total area between the continuous curve y=f(x) over the interval [a,b] and the x-axis)

Total area=



b

a

dx x f( )

Example:

Find the total area between the curve y=f(x) and the x-axis over the interval [a,b]:

1- 2

1 x

y  over [0,2]

2- y3sinx over [0,]