ESSON 6.1 NTIDERIVATIVES AND INDEFINITE …teachers.sduhsd.net/tnguyen/documents/Notes6Key_001.pdf4. Find the summation of the heights: 1 n i i f c 5. Multiply the Summation of the

Mrs. Nguyen – Honors PreCalculus – Chapter 6 Notes – Page 1

NOTES 6: INTEGRATION Name:______________________________ Date:________________Period:_________ Mrs. Nguyen’s Initial:_________________

LESSON 6.1 – ANTIDERIVATIVES AND INDEFINITE INTEGRATION Review Problems: Find '( )f x for the following problems 1. 5( )f x x

2. 2( ) 3f x x

3. ( ) 5f x

4. ( ) 5f x x

5. 2

1( )

2f x

x

6. ( )f x x

7. 5 31 2( ) 5

5 3f x x x x

8. 22 3

( ) 23

xf x

x

Note: If you are given '( )f x , to find ( )f x we need to use the process called finding the “ANTIDERIVATIVE.” Definition of an Antiderivative

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

'( ) ( ) .F x f x x in I

Note: F is NOT unique!


Theorem 6.1: Representation of Antiderivatives

If F is an antiderivative of f on an interval I, then G is an antiderivative of f on the interval iff G is of the form

( ) ( ) ,G x F x C x in I where C is a constant

Notation for Antiderivatives

Let ( )F x be the function, then

'( ) ( )dy

F x f xdx

.

or

( ) ( )dy

f x dy f x dxdx

( ) ( )y f x dx F x C

Note:

( )f x is the integrand dx is the variable of integration C is the constant of integration

Example 1: a. Given 3

1( )G x x and 3

2 ( ) 2G x x , then

1 '( )G x

2 '( )G x b. Now given 2'( ) 3G x x , then

( )G x ___________ or ___________ or __________

C is called the Constant of integration

( ) ( )G x F x C is called the General Antiderivative

3( )G x x C is called the General Solution of the differential equation

2'( ) 3G x x is called the Differential Equation

Examples: Find the antiderivative of the following functions

2. Given ( ) 5, ( )f x find F x ( )F x ______

An Antiderivative of 5

( )F x ______+C General Solution

3. Given ( ) 5 , ( )f x x find F x ( )F x ______

An Antiderivative of 5x

( )F x ______+C General Solution

Basic Integration Rules

'( ) ( )F x dx F x C

Integration is the “inverse” of differentiation.

( ) ( )d

f x dx f xdx

Differentiation is the “inverse” of integration.


Differentiation Formulas Integration Formulas

1. 0d

Cdx

1. 0dx C

2. dkx k

dx 2. kdx kx C

3. ( ) '( )d

kf x kf xdx

3. ( ) ( )kf x dx k f x dx

4. ( ) ( ) '( ) '( )d

f x g x f x g xdx

4. ( ) ( ) ( ) ( )f x g x dx f x dx g x dx

5. 1n ndx nx

dx 5.

1

; 11

nn x

x dx C nn

Note: Antidifferentiation is also called “indefinite integration” Practice Problems: Evaluate the indefinite integral. 1. k dx

2. 10xdx

3. udu

4. 22 1t dx

5. 22 1t dt

6. 2m dm

7. 2

3/ 2

1xdx

x

8. 2

1dx

x


x

y

Initial Conditions and Particular Solutions

Given ( ) ,F x a we can determine what C is. Example:

Find ( )F x given 2

1'( ) , 0F x x

x

and the initial condition (1) 0F .

Graph:

Example 7 on pg 393


LESSON 6.2 – AREA Sigma Notation

The sum of n terms 1 2 3, , ,..., na a a a is written as

1 2 31

...n

i ni

a a a a a

where: i index of summation,

thia i term of the sum,

n Upper bound of summation, 1 Lower bound of summation

Practice Problem 1: Evaluate the following sums.

a. 6

1i

i

b. 10

1

5i

c. 7

2

1

11

j

jn

d. 5

2

1i

i

e. 6

3

1i

i

f. 7

3

1k

k

g. 1

( )n

ii

f x x


1. 1

n

i

c cn

Example 1: 10

1

5i

2. 1

( 1)

2

n

i

n ni

Example 2: 6

1i

i

3. 2

2

1

( 1)(2 1) (2 3 1)

6 6

n

i

n n n n n ni

Example 3: 5

2

1i

i

Theorem 6.2: The Summation Formulas 4.

2 23

1

( 1)

4

n

i

n ni

Example 4: 6

3

1i

i

Practice Problems: Evaluate the following sums.

2. 15

3

1

2i

i i

Simplified answer in terms of n: ______________________ Answer: ______________________


3. 2

1

1n

i

i

n

for 10, 100, 1000, 10000,n and

Simplified answer in terms of n: ______________________ As ,n the limit is: ______________________

4. 2

41

4 ( 1)n

i

i i

n

for 10, 100, 1000, 10000,n and


n 2

1

1n

i

i

n

10

100

1000

10000

n

2

41

4 ( 1)n

i

i i

n

10

100

1000

10000


6. Find the limit of 2

8 4 4( )

3 3s n

n n as n .

Answer: ______________________

7. Find the limit of

2

118( )

2

n ns n

n

as n .

Answer: ______________________

Practice Problems: Find a formula for the sum of n terms. Use the formula to find the limit as n .

8. 2

1

16lim

n

ni

i

n



9. 2

1

1lim

n

ni

i

n n


10. 3

1

2 2lim 1

n

ni

i

n n



Understanding the Upper and Lower sums

Review: EXTREME VALUE THEOREM: If f is continuous on a closed interval ,a b , then f has

both a minimum and a maximum on the interval.

Consider the area of the region bounded by the curve ( ), , , .f x x axis x a and x b

Like before, we would want to divide the region into small rectangles with equal width. The number of rectangles is denoted by “n.” As n the area of the region becomes more accurate.

We can begin by subdividing the interval ,a b

into n subintervals (or smaller rectangles). The

width of each rectangle is b a

width xn

.

The endpoints of the intervals are as follows: 0 1 2 ...a x a x a x a n x

Because f is continuous, the EXTREME VALUE THEOREM pg 314 guarantees the existence of a minimum and a maximum value of ( )f x in each interval.

Minimum Value of ( )f x is denoted by ( )if m Maximum Value of ( )f x is denoted by ( )if M

Next, we can consider INSCRIBED rectangles and CIRCUMSCRIBED rectangles

Area of inscribed rectangles is the lower sum

Area of circumscribed rectangles is the upper sum


Formulas for the lower sum and upper sum Note: We know that

( ) ( )s n Area of region S n

Lower Sum:

1

( ) ( )n

ii

s n f m x

Upper Sum:

1

( ) ( )n

ii

S n f M x

Left Endpoint and Right Endpoint of each Rectangle

LEP: ( 1)ic a i x

REP: ic a i x

Theorem 6.3: Limits of the Lower and Upper Sums

Let f be continuous and nonnegative on the interval ,a b . The limit as n of both the lower and the

upper sums exist and are equal to each other. That is

1 1

lim ( ) lim ( ) lim ( ) lim ( )n n

i in n n ni i

s n f m x f M x S n

Definition of the Area of a Region in the Plane

Let f be continuous and nonnegative on the interval ,a b . The area of the region bounded by the graph of

f, the x-axis, the vertical lines x a and x b is

1

lim ( )n

ini

Area f c x

Guidelines to find the area of a region 1. Find the width:

b ax

n

2. Find the expression for the endpoint: ic a i x Note: As n , ( ) ( )s n Area of region S n , therefore, we

can use either the LEP or the REP expression to find the heights of the

rectangles. ic a i x is a simpler expression to use.

3. Find the expression for the heights: ( )if c

4. Find the summation of the heights: 1

( )n

ii

f c

5. Multiply the Summation of the heights and the

width: 1

( )n

ii

x f c

6. Take the limit as n : 1

lim ( )n

ini

f c x


Example: Find the area of the region bounded by 2( ) , , 0, 2f x x x axis x and x by using the Limit Definition.

1. b a

xn

1.

2. ic a i x

2.

3. ( )if c

3.

4. 1

( )n

ii

f c

4.

5. 1

( )n

ii

x f c

5. Simplified answer in terms of n:

6. 1

lim ( )n

ini

f c x

6.


Practice Problem 11: Find the area of the region bounded by 2( ) 4 , , 1, 2f x x x axis x and x by using the Limit Definition.

1. b a

xn

1.

2. ic a i x

2.

3. ( )if c

3.

4. 1

( )n

ii

f c

4.

5. 1

( )n

ii

x f c


6. 1

lim ( )n

ini

f c x

6.


Practice Problem 12: Find the area of the region bounded by 3( ) 1, 1 2h y y y by using the Limit Definition.

1. 2. 3. 4. 5. Simplified answer in terms of n: 6.


Practice Problem 13: Find the area of the region using upper and lower sums

Practice Problem 14: Use the upper and lower sums to approximate the area of the region using the indicated number of subintervals for 2y x .

Lower Upper 1st rectangle

2nd rectangle

3rd rectangle

4th rectangle

5th rectangle

6th rectangle

7th rectangle

8th rectangle

Area


LESSON 6.3 – RIEMANN SUMS AND DEFINITE INTEGRALS Definition of a Riemann Sum

Let f be defined on the closed interval ,a b , and let be a partition of

,a b given by

0 1 2 1... n na x x x x x b

where ix is the width of the thi

subinterval. If ic is any point in the thi subinterval, then the sum

1

( )n

i ii

f c x

, 1i i ix c x is called a

Riemann Sum of f for the partition .

In other words: we can divide the area of the region into rectangles with different widths to find the area.

Definition of a Definite Integral

Let f be defined on the closed interval

,a b and the limit 0

1

lim ( )n

i ixi

f c x

exists, then f is integrable on ,a b

and the limit is denoted by

0

1

lim ( ) ( )n b

i i axi

f c x f x dx

.

The limit is called the definite integral of f from a to b. The number a is the lower limit of inegration, and the number b is the upper limit of integration.

Note: x is the width of

the largest subinterval of a partition. Important Note:

A definite integral is a number (area under the curve).

An indefinite integral is a family of functions.

Theorem 6.4: Continuity Implies Integrability

If a function f is continuous on the closed interval ,a b , then f is

integrable on ,a b .

Theorem 6.5: The Definite Integral as the Area of a Region

Let f be continuous and nonnegative on the interval ,a b . Then

the area of the region bounded by the graph of f, the x-axis, the

vertical lines x a and x b is given by ( )b

aArea f x dx .


Example 1: Evaluate 3

2xdx

by using the Limit Definition.

1. i

b ax

n

1.

2. i ic a i x

2.

3. ( )if c

3.

4. 1

( )n

ii

f c

4.

5. 1

( )n

i ii

x f c


6.

01

lim ( )

( )

n

i ixi

b

a

Area f c x

f x dx

6.


Practice Problem 1: Evaluate 2 2

13 2x dx

by using the Limit Definition.

Simplified answer in terms of n: ______________________ Answer: ______________________

Practice Problems: Evaluate areas of common geometric figures

1. 3

14dx

2. 3

02x dx

3. 2 2

24 x dx


Definitions of Two Special Definite Integrals

1. If f is defined at ,x a

then ( ) 0a

af x dx

2. If f is integrable on ,a b ,

then ( ) ( )a b

b af x dx f x dx

Examples:

1. 2

2x dx

2.

Theorem 6.6: Additive Interval Property

If f is integrable on the three closed intervals determined by a, b, and c, then

( ) ( ) ( )b c b

a a cf x dx f x dx f x dx

Example: 1

1x dx

Theorem 6.7: Properties of Definite Integrals

If f and g are integrable on ,a b and k is a constant, then the

functions of kf and f g are integrable on ,a b , and

1. ( ) ( )b b

a akf x dx k f x dx

2. ( ) ( ) ( ) ( )b b b

a a af x g x dx f x dx g x dx

Examples for Theorem 6.7: Given:

4 3

2

4

2

4

2

60

6

2

x dx

xdx

dx

1. 4 3

26 2x x dx 2.

2 3

2x dx

Theorem 6.8: Preservation of Inequality

1. If f is integrable and nonnegative on the closed interval ,a b , then

0 ( )b

af x dx .

2. If f and g are integrable on the closed interval ,a b and ( ) ( )f x g x for every

x in ,a b , then ( ) ( )b b

a af x dx g x dx

Figure on pg 416


Practice Problem 4: Express the limit as a definite integral on 0, 3 , where ic is

any point in the thi subinterval.

2

01

lim 4n

i ixi

c x

Practice Problem 5: Set up a definite integral that yields the area of the region. a. 2( ) 4f x x

b. 2

1( )

1f x

x

Practice Problem 6: Sketch the region whose area is given by the definite integral. Then evaluate its area.

1

11 x dx

Practice Problem 7: Given 1

1

1

0

( ) 0

( ) 5

f x dx

f x dx

, Evaluate:

a. 0

1( )f x dx

b. 1 0

0 1( ) ( )f x dx f x dx

c. 1

13 ( )f x dx

d. 1

03 ( )f x dx


LESSON 6.4 – THE FUNDAMENTAL THEOREM OF CALCULUS You have been introduced to two branches of calculus:

1. Differential Calculus The Tangent Problem 0

( ) ( )'( ) lim

x

f x x f xf x

x

2. Integral Calculus The Area Problem 0

1

lim ( ) ( )n b

i i axi

f c x f x dx

These two problems are connected by a theorem called: THE FUNDAMENTAL THEOREM OF CALCULUS. Theorem 6.9: The Fundamental Theorem of Calculus

If a function f is continuous on the closed interval ,a b and F

is an antiderivative of f on the interval ,a b , then

( ) ( ) ( )b

aArea f x dx F b F a

Example 1: Evaluating a Definite Integral

4

1

2udu

u

Practice Problems: Evaluate the following definite integral

1. 1

0 3

x xdx

2. 3 2

12 3x x dx


3. 3

02 3x dx

With an absolute value function, we need to find the area by breaking up the function at its vertex. The vertex is: 2 3 0x x

32 3 ,

22 33

2 3 ,2

x xx

x x

We can rewrite the integral as:

3 3/ 2 3

0 0 3/ 22 3 2 3 2 3x dx x dx x dx

4. 4 2

04 3x x dx

The vertex is: 2 4 3 0x x x

2

2 2

2

4 3,

4 3 4 3,

4 3,

x x x

x x x x x

x x x

The integral can be re written as:


Theorem 6.10: Mean Value Theorem for Integrals

If f is continuous on the closed interval ,a b , then there exists

a number c in the closed interval ,a b such that

( ) ( ) ( )b

af x dx f c b a

Note: Area of the inscribed rectangles Area of the region Area of the circumscribed rectangles. Theorem 6.10 means that somewhere between the inscribed and circumscribed rectangles, there exists a rectangle whose area is precisely equal to the area under the curve.

Figures on pg 423

Review: Theorem 5.4 – Let f be continuous on the closed interval ,a b and differentiable on the

interval ,a b , then there exists a

number c in ,a b such that

( ) ( )'( )

f b f af c

b a

.

Definition of the Average Value of a Function on an Interval

If f is integrable on ,a b , then

the average value of f on the

interval is 1

( )( )

b

af x dx

b a

Note: ( )f c in the Mean Value Theorem for Integral is the average value of f on ,a b .

( ) ( ) ( )

1( ) ( )

( )

b

a

b

a

f x dx f c b a

f x dx f c Average Valueb a


Practice Problems: Find the value(s) of c guaranteed by the Mean Value Theorem for Integrals 5. 3( ) , 0, 2f x x

Area ( )( )f c b a Area Solve for c

6. ( ) 2 , 0, 2f x x x


Practice Problems: Find the average value of the function over the interval and all values of x in the interval for which the function equals its average value. 7. ( ) 2 , 0, 4f x x x

8.

2

1( ) , 0, 2

3f x

x


If f is continuous on an open interval I containing a, then, for every x in the interval,

( ) ( )x

a

df t dt f x

dx

Note: This theorem is only be used when asked to find '( )F x and the lower bound is a constant.

Theorem 6.11: The Second Fundamental Theorem of Calculus

The Definite Integral as a function of x.

( ) ( )

tan

x

a

F is a functionof x

F x f t dt

Cons t f is a functionof t

The Definite Integral as a number.

tan

( ) ( )

tan

b

a

Cons t

F x f x dx

Cons t f is a functionof x

Example 2: Evaluate the function 2

0( ) 3 3

xF x t dt at

1 1 30, , , , 1.

4 2 4x

2

0( ) 3 3

xF x t dt

Area

(0)F

1

4F

1

2F

3

4F

1F

Note: ( )F x is the antiderivative of ( )f x ( ) '( )f x F x

( ) '( ) ( )F x F x dx f x dx

( ) ( ) '( )d

F x f x F xdx


Practice Problem 9: Evaluate the function 1

( )x

F x y y dy at 2, 5, 8.x

Round to 4 decimal places.

1

( )x

F x y y dy

(2)F

(5)F

(8)F

Example 3: Use the 2nd Fundamental Theorem of Calculus to evaluate '( )F x given

2

0( ) 1

xF x t dt

Example 4: Find '( )F x given 3

2

0( )

xF x t dt .

Method 1 2nd Fundamental Theorem of Calculus

'( ) ( )dF du d du

F x F xdu dx du dx

=3

2 2

0 0

x ud du d dut dt t dt

du dx du dx

=


Practice Problem 10: Find '( )F x given 2

31

1( )

xF x dt

t .

Method 1 2nd Fundamental Theorem of Calculus

Practice Problem 11: Find '( )F x given 2

21( )

1

x tF x dt

t

.

Practice Problem 12: Find '( )F x given

3 3

0( ) 1

xF x t dt .


LESSON 6.5 – INTEGRATION BY SUBSTITUTION Review Problems: Find '( )f x

1. 42( ) 1f x x

2. 3( ) 1f x x

3. ( ) 5 1f x x

Pattern Recognition If ( ( )) '( ( )) '( )

dF g x F g x g x

dx , then

'( ( )) '( ) ( ( ))F g x g x dx F g x C

Example: Analyze Review problems 1-3

1.

2.

3.

Theorem 6.12: Antidifferentiation of a Composite Function

Let g be a function whose range is an interval I, and let f be a function that is continuous on I. If g is differentiable on its domain and F is an antiderivative of f on I, then

( ( )) '( ) ( ( ))f g x g x dx F g x C

If ( )u g x , then '( )du g x dx and

( ) ( )f u du F u C

Example: Evaluate

22 1 2x x dx


Practice Problems: Evaluate the following infinite integrals using pattern recognition. 1. 3 3 1x dx

2. 22 1x x dx

3. 3 25 1x x dx

4.

3

2416

xdx

x

5. 2 1x dx

6. 3 1x dr

Change of Variables

Let ( )u g x , then ( ( )) ( )f g x f u

( ( )) '( ) ( ) ( ) ( ( ))f g x g x dx f u du F u C F g x C

1. 2 1x dx

Examples: Evaluate using the change of variables method (or substitution) 2. 2 1x x dx

Note: Pattern recognition does not work. Must use substitution!


Practice Problems: Evaluate 3 4 5t t dt using both methods.

Pattern recognition Change of variable (or Substitution)

Theorem 6.13: The General Power Rule for Integration

If g is a differentiable function of x, then

1( )

( ) '( ) , 11

nn g x

g x g x dx C nn

Equivalently, if ( )u g x , then

1

, 11

nn u

u du C nn

Example:

Theorem 6.14: Change of Variables for Definite Integrals

If the function ( )u g x has a continuous derivative on the closed interval ,a b and f is continuous on the

range of g, then ( )

( )( ( )) '( ) ( )

b g b

a g af g x g x dx f u du

Example:


Method 1

Method 2: Using Theorem 6.14

Example: Evaluate

1 32

01x x dx

Practice Problem 8: Evaluate

4 22 3

28x x dx

Practice Problem 9: Evaluate

2

11 2x x dx


Theorem 6.15: Integration of Even and Odd Functions Let f be integrable on the closed interval ,a a .

If f is an even function, then

0( ) 2 ( )

a a

af x dx f x dx

If f is an odd function, then

( ) 0a

af x dx

Examples: Evaluate the definite integrals

2 2 2

2( 1)x x dx

Determine is the function is odd, even, or neither:

2 5 3

24 6x x x dx

Determine is the function is odd, even, or neither:

Documents

ESSON 6.1 NTIDERIVATIVES AND INDEFINITE …teachers.sduhsd.net/tnguyen/documents/Notes6Key_001.pdf4. Find the summation of the heights: 1 n i i f c 5. Multiply the Summation of the