Calculus Date: 3/7/2014 ID Check Obj: SWBAT connect Differential and Integral CalculusDo Now: pg 307 #37B #23HW Requests: SM pg 156; pg 295 #11-17 odds, 31-35 odds SM 162In class pg 316 #2, 8, 14, 18 group 319 #1-4HW: Read pg 305 Ex 8 Read pg 314 Error Analysis pg 316 #1-9 odds, 13-19 oddsAnnouncements:Saturday Tutoring 11-1 (Derivatives)Mock AP Exam during ACT Testing“There is something in every one of you that waits and listens for the sound of the genuine in yourself. It is the only true guide you will ever have. And if you cannot hear it, you will all of your life spend your days on the ends of strings that somebody else pulls.” ― Howard Thurman

MaximizeAcademicPotential

Trapezoidal RuleTo approximate , use

T = (y0 + 2y1 + 2y2 + …. 2yn-1 + yn)

where [a,b] is partitioned into n subintervals of equal length h = (b-a)/n.

( )b

af x dx

2

h

Trapezoidal RuleTo approximate , use

T = (y0 + 2y1 + 2y2 + …. 2yn-1 + yn)

where [a,b] is partitioned into n subintervals of equal length h = (b-a)/n.

( )b

af x dx

2

h

Equivalently,T = LRAMn + RRAMn

2where LRAMn and RRAMn are the Riemann sums using the left and right endpoints, respectively, for f for the partition.

Using the trapezoidal rule

Use the trapezoidal rule with n = 4 to estimate

h = (2-1)/4 or ¼, so

T = 1/8( 1+2(25/16)+2(36/16)+2(49/16)+4) = 75/32 or about 2.344

2 2

1x dx

EX 2: Trapezoidal RuleT = (y0 + 2y1 + 2y2 + …. 2yn-1 + yn)

where [a,b] is partitioned into n subintervals of equal length h = (b-a)/n.

2

h

Interval [0,1] [1,2] [2,3] [3,4] 4,

X 0 .5 1 1.5 2

Y = x2 0 .25 1 2.25 4

T = (y0 + 2y1 + 2y2 + 2y3 + y4)

T = ¼ (0 + 2(.25) + 2(1) + 2(2.25) + 4) = 11/4

2

h

Simpson’ Rule

To approximate , use

S = (y0 + 4y1 + 2y2 + 4y3…. 2yn-2 +4yn-1 + yn)

where [a,b] is partitioned into an even number n subintervals of equal length h =(b –a)/n.

Simpson’s Rule assumes that a figure with a parabolic arc is used to compute the area

( )b

af x dx

3

h

Using Simpson’s Rule

Use Simpson’s rule with n = 4 to estimate

h = (2 – 1)/4 = ¼, so

S = 1/12 (1 + 4(25/16) + 2(36/16) + 4(49/16) + 4)

= 7/3

2 2

1x dx

EX 2: Simpson’s Rule

Interval [0,1] [1,2] [2,3] [3,4] 4

X 0 .5 1 1.5 2

Y = x2 0 .25 1 2.25 4

The Definite Integral

When we find the area under a curve by adding rectangles, the answer is called a Rieman sum.

211

8V t

subinterval

partition

The width of a rectangle is called a subinterval.

The entire interval is called the partition.

Subintervals do not all have to be the same size.

211

8V t

subinterval

partition

If the partition is denoted by P, then the length of the longest subinterval is called the norm of P and is denoted by .P

As gets smaller, the approximation for the area gets better.

P

0

1

Area limn

k kP

k

f c x

if P is a partition of the interval ,a b

0

1

limn

k kP

k

f c x

is called the definite integral of

over .f ,a b

If we use subintervals of equal length, then the length of a

subinterval is:b a

xn

The definite integral is then given by:

1

limn

kn

k

f c x

1

limn

kn

k

f c x

Leibnitz introduced a simpler notation for the definite integral:

1

limn b

k ank

f c x f x dx

Note that the very small change in x becomes dx.

b

af x dx

IntegrationSymbol

lower limit of integration

upper limit of integration

integrandvariable of integration

(dummy variable)

It is called a dummy variable because the answer does not depend on the variable chosen.

b

af x dx

We have the notation for integration, but we still need to learn how to evaluate the integral.

time

velocity

After 4 seconds, the object has gone 12 feet.

In section 6.1, we considered an object moving at a constant rate of 3 ft/sec.

Since rate . time = distance: 3t d

If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.

ft3 4 sec 12 ft

sec

If the velocity varies:

11

2v t

Distance:21

4s t t

(C=0 since s=0 at t=0)

After 4 seconds:1

16 44

s

8s

1Area 1 3 4 8

2

The distance is still equal to the area under the curve!

Notice that the area is a trapezoid.

211

8v t What if:

We could split the area under the curve into a lot of thin trapezoids, and each trapezoid would behave like the large one in the previous example.

It seems reasonable that the distance will equal the area under the curve.

211

8

dsv t

dt

31

24s t t

314 4

24s

26

3s

The area under the curve2

63

We can use anti-derivatives to find the area under a curve!

Riemann Sums

• Sigma notation enables us to express a large sum in compact form

1 21

.....n

k nk

a a a a

Calculus Date: 2/18/2014 ID Check Objective: SWBAT apply properties of the definite integralDo Now: Set up two related rates problems from the HW Worksheet 6, 10HW Requests: pg 276 #23, 25, 26, Turn in #28 E.CIn class: Finish Sigma notation Continue Definite IntegralsHW:pg 286 #1,3,5,9, 13, 15, 17, 19, 21, Announcements:“There is something in every one of you that waits and listens for the sound of the genuine in yourself. It is the only true guide you will ever have. And if you cannot hear it, you will all of your life spend your days on the ends of strings that somebody else pulls.” ― Howard Thurman

MaximizeAcademicPotential

Turn UP! MAP

When we find the area under a curve by adding rectangles, the answer is called a Rieman sum.

211

8V t

subinterval

partition

The width of a rectangle is called a subinterval.

The entire interval is called the partition.

Subintervals do not all have to be the same size.

211

8V t

subinterval

partition

The width of a rectangle is called a subinterval.

The entire interval is called the partition.

Let’s divide partition into 8 subintervals.

Pg 274 #9 Write this as a Riemann sum. 6 subintervals

211

8V t

subinterval

partition

If the partition is denoted by P, then the length of the longest subinterval is called the norm of P and is denoted by .P

As gets smaller, the approximation for the area gets better.

P

0

1

Area limn

k kP

k

f c x

if P is a partition of the interval ,a b

0

1

limn

k kP

k

f c x

is called the definite integral of

over .f ,a b

If we use subintervals of equal length, then the length of a

subinterval is:b a

xn

The definite integral is then given by:

1

limn

kn

k

f c x

1

limn

kn

k

f c x

Leibnitz introduced a simpler notation for the definite integral:

1

limn b

k ank

f c x f x dx

Note that the very small change in x becomes dx.

Note as n gets larger and larger the definite integral approaches the actual value of the area.

b

af x dx

IntegrationSymbol

lower limit of integration

upper limit of integration

integrandvariable of integration

(dummy variable)

It is called a dummy variable because the answer does not depend on the variable chosen.

Calculus Date: 2/19/2014 ID Check Objective: SWBAT apply properties of the definite integralDo Now: Bell Ringer QuizHW Requests: pg 276 #25, 26, pg 286 1-15 odds In class: pg 276 #23, 28 Continue Definite IntegralsHW:pg 286 #17-35 odds Announcements:“There is something in every one of you that waits and listens for the sound of the genuine in yourself. It is the only true guide you will ever have. And if you cannot hear it, you will all of your life spend your days on the ends of strings that somebody else pulls.” ― Howard Thurman

MaximizeAcademicPotential

Turn UP! MAP

Bell Ringer Quiz (10 minutes)

Riemann Sums

• LRAM, MRAM,and RRAM are examples of Riemann sums

• Sn =

This sum, which depends on the partition P and the choice of the numbers ck,is a Riemann sum for f on the interval [a,b]

1

( )n

k kk

f c x

Definite Integral as a Limit of Riemann Sums

Let f be a function defined on a closed interval [a,b]. For any partition P of [a,b], let the numbers ck be chosen arbitrarily in the subintervals [xk-1,xk].

If there exists a number I such that

no matter how P and the ck’s are chosen, then f is integrable on [a,b] and I is the definite integral of f over [a,b].

01

lim ( )n

k kPk

f c x I

Definite Integral of a continuous function on [a,b]

Let f be continuous on [a,b], and let [a,b] be partitioned into n subintervals of equal length Δx = (b-a)/n. Then the definite integral of f over [a,b] is given by

where each ck is chosen arbitrarily in the kth subinterval.

1

lim ( )n

kn

k

f c x

Definite integral

This is read as “the integral from a to b of f of x dee x” or sometimes as “the integral from a to b of f of x with respect to x.”

( )b

af x dx

Using Definite integral notation

2

1

3 2

1

lim (3( ) 2 5)

(3 2 5)

n

k kn

k

m m x

x x dx

The function being integrated is f(x) = 3x2 – 2x + 5 over the interval [-1,3]

Definition: Area under a curve

If y = f(x) is nonnegative and integrable over a closed interval [a,b], then the area under the curve of y = f(x) from a to b is the integral of f from a to b,

( )

b

aA f x dx

We can use integrals to calculate areas and we can use areas to calculate integrals.

Nonpositive regions

If the graph is nonpositive from a to b then

( )b

aA f x dx

Area of any integrable function

= (area above the x-axis) –

(area below x-axis)

( )b

af x dx

MaximizeAcademicPotential

Turn UP! MAP

Integral of a Constant

If f(x) = c, where c is a constant, on the interval [a,b], then

( ) ( )b b

a af x dx cdx c b a

Evaluating Integrals using areas

We can use integrals to calculate areas and we can use areas to calculate integrals.

Using areas, evaluate the integrals:

1)

2)

3

2( 1)x dx

2 2

24 x dx

Evaluating Integrals using areas

Evaluate using areas:

3)

4) (a<b)

8

24dx

(2 1)b

ax dx

Evaluating integrals using areas

Evaluate the discontinuous function:

Since the function is discontinuous at x = 0, we must divide the areas into two pieces and find the sum of the areas

= -1 + 2 = 1

2

1

xdxx

Integrals on a Calculator

You can evaluate integrals numerically using the calculator. The book denotes this by using NINT. The calculator function fnInt is what you will use.

= fnInt(xsinx,x,-1,2) is approx. 2.04

2

1sinx xdx

Evaluate Integrals on calculator

• Evaluate the following integrals numerically:

1) = approx. 3.14

2) = approx. .89

1

20

4

1dx

x25

0

xe dx

Rules for Definite Integrals

1) Order of Integration:

( ) ( )a b

b af x dx f x dx

Rules for Definite Integrals

2) Zero: ( ) 0a

af x dx

Rules for Definite Integrals

3) Constant Multiple:

( ) ( )

( ) ( )

b b

a a

b b

a a

kf x dx k f x dx

f x dx f x dx

Any number k

k= -1

Rules for Definite Integrals

4) Sum and Difference:

( ( ) ( )) ( ) ( )b b b

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

Rules for Definite Integrals

5) Additivity:

( ) ( ) ( )b c c

a b af x dx f x dx f x dx

Rules for Definite Integrals

6) Max-Min Inequality: If max f and min f are the maximum and minimum values of f on [a,b] then:

min f ∙ (b – a) ≤ ≤ max f ∙ (b – a)( )b

af x dx

Rules for Definite Integrals

7) Domination: f(x) ≥ g(x) on [a,b]

f(x) ≥ 0 on [a,b] ≥ 0

( ) ( )b b

a af x dx g x dx

( )b

af x dx (g =0)

Using the rules for integration

Suppose:

Find each of the following integrals, if possible:a) b) c)

d) e) f)

1

1( ) 5f x dx

4

1( ) 2f x dx

1

1( ) 7h x dx

1

4( )f x dx

4

1( )f x dx

1

12 ( ) 3 ( )f x h x dx

1

0( )f x dx

2

2( )h x dx

4

1( ) ( )f x h x dx

Calculus Date: 2/27/2014 ID Check Obj: SWBAT connect Differential and Integral CalculusDo Now:

http://www.youtube.com/watch?v=mmMieLl-Jzs HW Requests: 145 #2-34 evens and 33HW: Complete SM pg 156, pg 306 #1-19 odds Announcements:Mid Chapter Test Fri. Sect. 6.1-6.3Careful of units, meaning of area, asymptotes, properties of integrals

Handout InversesSaturday Tutoring 10-1 (limits)“There is something in every one of you that waits and listens for the sound of the genuine in yourself. It is the only true guide you will ever have. And if you cannot hear it, you will all of your life spend your days on the ends of strings that somebody else pulls.” ― Howard Thurman

MaximizeAcademicPotential

The Fundamental Theorem of Calculus, Part I

( )x

af t dt

Antiderivative

Derivative

Applications of The Fundamental Theorem of Calculus, Part I

1.

2.

cos cosxd

tdt xdx

2 20

1 1

1 1

xddt

dx t x

Applications of The Fundamental Theorem of Calculus, Part I

22 2

1cos cos (2 ) 2 cos

xdtdt x x x x

dx

Applications of The Fundamental Theorem of Calculus, Part I

Applications of The Fundamental Theorem of Calculus, Part I

Find dy/dx.

y =

Since this has an x on both ends of the integral, it must be separated.

2

2

1

2

x

txdt

e

Applications of The Fundamental Theorem of Calculus, Part I

=

2 20

2 2 0

1 1 1

2 2 2

x x

t t tx xdt dt dt

e e e

22

0 0

1 1

2 2

x x

t tdt dt

e e

Applications of The Fundamental Theorem of Calculus, Part I

=

=

22

1 1(2) (2 )

2 2x x

xe e

2 2

2 2

22xx

x

ee

The Fundamental Theorem of Calculus, Part 2

If f is continuous at every point of [a,b], and if F is any antiderivative of f on [a,b], then

This part of the Fundamental Theorem is also called the Integral Evaluation Theorem.

( ) ( ) ( )b

af x dx F b F a

Applications of The Fundamental Theorem of Calculus, Part 2

End here

Calculus Date: 2/27/2014 ID Check Obj: SWBAT connect Differential and Integral CalculusDo Now:

http://www.youtube.com/watch?v=mmMieLl-Jzs HW Requests: 145 #2-34 evens and 33HW: SM pg 156 Announcements:Mid Chapter Test Fri. Sect. 6.1-6.3Careful of units, meaning of area, asymptotes, properties of integrals

Handout InversesSaturday Tutoring 10-1 (limits)“There is something in every one of you that waits and listens for the sound of the genuine in yourself. It is the only true guide you will ever have. And if you cannot hear it, you will all of your life spend your days on the ends of strings that somebody else pulls.” ― Howard Thurman

MaximizeAcademicPotential

The Fundamental Theorem of Calculus, Part I

( )x

af t dt

Antiderivative

Derivative

Applications of The Fundamental Theorem of Calculus, Part I

The Fundamental Theorem of Calculus, Part 2

If f is continuous at every point of [a,b], and if F is any antiderivative of f on [a,b], then

This part of the Fundamental Theorem is also called the Integral Evaluation Theorem.

( ) ( ) ( )b

af x dx F b F a

AntidifferentiationA function F(x) is an antiderivative of a function f(x) if

F’(x) = f(x) for all x in the domain of f. The process of finding an antiderivative is called antidifferentiation.

If F is any antiderivative of f then

= F(x) + C

If x = a, then 0 = F(a) + C C = -F(a)

= F(x) – F(a)

( )x

af t dt

( )x

af t dt

Calculus Date: 3/3/2014 ID Check Obj: SWBAT connect Differential and Integral CalculusDo Now: Put up your designate problem from the final exam.

HW Requests: SM pg 156; pg 306 #1-19 odds HW: pg 306 #1-19 odds if not completed#21-39 odds

Announcements:Handout Inverses sent via emailSaturday Tutoring 10-1 (Derivatives)Mock AP Exam during ACT Testing“There is something in every one of you that waits and listens for the sound of the genuine in yourself. It is the only true guide you will ever have. And if you cannot hear it, you will all of your life spend your days on the ends of strings that somebody else pulls.” ― Howard Thurman

MaximizeAcademicPotential

Applications of

The Fundamental

Theorem of Calculus,

Part I

Applications of

The Fundamental

Theorem of Calculus,

Part I

Pg 307 #22 Construct a function of the form

Calculus Date: 3/4/2014 ID Check Obj: SWBAT connect Differential and Integral CalculusDo Now: Put up your designate problem from the final exam. Pg 306 #32, 34

HW Requests: SM pg 156; pg 306 #1-19 , 21-39 odds if not completed oddsHW: pg 295 #11-17 odds, 31-35 oddsPg 307 #41-49 oddsAnnouncements:Handout Inverses sent via emailSaturday Tutoring 10-1 (Derivatives)Mock AP Exam during ACT Testing“There is something in every one of you that waits and listens for the sound of the genuine in yourself. It is the only true guide you will ever have. And if you cannot hear it, you will all of your life spend your days on the ends of strings that somebody else pulls.” ― Howard Thurman

MaximizeAcademicPotential

2. Find the total area of the regionNet Area: Area below the x axis is counted as negative

Total Area: Area below the x axis is counted positive.

Pg 307 #42Solve analytically and using fnint

Pg 307 #42

Average (Mean) Value

1( )

b

af x dx

b a

Find the average value of f(x) = 4 – x2 over the interval [0,3]. Does f take on this value at some point in the given interval?

Pg 295 #12

Mean Value Theorem for Definite Integrals

If f is continuous on [a,b], then at some point c in [a,b],

1( ) ( )

b

af c f x dx

b a

Applying the Mean Value

Av(f) = = 1/3 (3) = 1

f(x) = 4- x2

f(c) =14 – x2 = 1 when x = ± √3 but only √3 falls in

the interval from [0,3], so x = √3 is the place where the function assumes the average.

3 2

0

1(4 )

3 0x dx

Use fnInt

1( ) ( )

b

af c f x dx

b a

Using the rules for definite integrals

Show that the value of is less than 3/2

The Max-Min Inequality rule says the max f . (b – a) is an upper bound.The maximum value of √(1+cosx) on [0,1] is √2 so

the upper bound is: √2(1 – 0) = √2 , which is less than 3/2

1

01 cos xdx