Calculus Date: 3/12/2014 ID Check Obj: Objective: Properties of Definite Integrals Do Now: Pop Quiz...

Calculus Date: 3/12/2014 ID Check Obj: Objective: Properties of Definite Integrals Do Now: Pop Quiz See BelowHW Requests: SM 165, 166, pg 307-308 #57, 59HW: pg 331 #1-19 odds•Announcements:Saturday Tutoring 10-1 (Derivatives)Mock AP Exam during ACT Testing

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• Evaluate the integral analytically by using the Fundamental Theorem of Calculus.

• 1. 2. •

DIFFERENTIAL EQUATIONS

• A differential equation is an equation that contains a derivative. For example, this is a differential equation.

• From antidifferentiating skills from last chapter, we can solve this equation for y.

THE CONCEPT OF THE DIFFERENTIAL EQUATION

• The dy/dx = f(x) means that f(x) is a rate. To solve a differential equation means to solve for the general solution. By integrating. It is more involved than just integrating. Let’s look at an example:

EXAMPLE 1Find the general solution to the exact differential equation.

• GIVEN• Multiply both sides by dx to isolate

dy. Bring the dx with the x and dy with the y.

• Since you have the variable of integration attached, you are able to integrate both sides. Note: integral sign without limits means to merely find the antiderivative of that function

• Notice on the right, there is a C. Constant of integration.

• General Solution includes C the constant on integration.

dxxxdy

dxxxdx

C?? What is that?

• The derivative of a constant is 0. • when you integrate, you have to take into account that there

is a possible constant involved. • Theoretically, a differential equation has infinite solutions.• To solve for C, you will receive an initial value problem

which will give y(0) value. Then you can plug 0 in for x and the y(0) in for y.

• Continuing the previous problem, let’s say that y(0)=2.

Solving for c.

Initial value

Differential Equations

Differential EquationsExample 3 Solving an Initial Value Problem

Find the particular solution to the equations

dy/dx = ex -6x2 whose graph passes through

the point (1,0).

Exit Ticket:Pg 330 #1, 14, 20

SLOPE FIELDS***AP CALCULUS MATERIAL ONLY***

• We just solved for the differential equation analytically (‘algebraically’). The slope field (also known as vector field and directional field) will give us a qualitative analysis.

• The graph shows all the possible slopes in the form of a field.

• The arrows show the basic trend of how the slope changes. Using the initial condition, you can draw your solution.

• For the previous example, the slope field will be very simple to draw.

SLOPE FIELD FOR EXAMPLE 1

• Notice how slope field TRACES the tangent lines of points from the antiderivative from various constants.

• For the curve that is relevant with the correct, in our last problem C=2, connect those particular tangent lines and heavily bold it.

• I drew this by hand, so please forgive my sloppiness with the slopes. _/\_

SLOPE FIELDS

• The previous was so easy that a slope field was really not required.

• However, there are many differential equations that will not yield easily to form such a slope field.

HOW TO DRAW SLOPE FIELDS

• Consider dy/dx=-2xy. This is the formula for SLOPE

• To find the slope, you need both an (x,y) coordinate. For example, if you use (1,-1), then the slope = (-2)(1)(-1)=2.

DRAWING SLOPE FIELDS• Start from (1,-1) and make a

small line with the slope of 2. (Remember in high school, when you did lines, how did you do slope? Difference in y over difference in x).

• Thus, the solution of the differential equation with the initial condition y(1)=-1 will look similar to this line segment as long as we stay close to x=-1.

DRAWING SLOPE FIELDS

• However, simply drawing one line will not help us at all. You have to draw several lines. This what gets the Durvasa Muni out of the calculus students!

• Then connect the “lines” horizontally to fit a curve amongst the tangent lines. These lines are formed from various C values.

SLOPE FIELDS

• This topic of slope fields will be discussed highly in a college differential equations course. The AB Calculus exam, since 2002, has included slope fields in the curriculum, they have to know just as much about slope fields as BC Calculus.

• The college calculus teachers generally like to skip over such topics of differential equations, even the easy ones like the first example.

• Let’s consider the last example dy/dx=-2xy. Say we were the 2001 graduating class (that’s my graduating class ) and we didn’t learn slope fields. How would we such such an equation since there is a y there.

SEPERATION OF VARIABLES

• Such equations are known as separable differential equations. The way to go about solving such equations (raksasas lol ) is to round up your y terms with dy and round up your x terms with dx.

• When integrating dy/y, remember: the derivative of ln y is 1/y. Therefore the integral of 1/y is ln y.

yxydxdy

dxxydx

INTIAL VALUE PROBLEM

• Let’s say that y(1)=-1. We can find C that way.

• And finally, your exact answer.

AUTONOMOUS SEPARABLE DIFFERENTIAL EQUATIONS

(A.S.D.E.)• A differential equation that is autonomous means

that the derivative does not depend on the independent variable. For example, The equation below is an autonomous equation. Notice that there is no x involved.

SOLVING A.S.D.E

• You can still separate the y and bring dx to the right.

• The process is the same.

I have included the power point that now has the examples we were working in class included. Also, note that we assume that the subintervals are equal. This is the only way to factor out the h/2 or h/3 term. If the subintervals are not uniform then you have to compute each trapezoid separately because h is not the same. Also, note that for Simpson's rule, it assumes an even number of subintervals.

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.

af x dx

Trapezoidal RuleTo approximate , use

T = (y0 + 2y1 + 2y2 + …. 2yn-1 + yn)

af x dx

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

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

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

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

af x dx

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)

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.

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.

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.

Area limn

if P is a partition of the interval ,a b

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

The definite integral is then given by:

Leibnitz introduced a simpler notation for the definite integral:

limn b

f c x f x dx

Note that the very small change in x becomes dx.

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.

af x dx

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

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

If the velocity varies:

Distance:21

4s t t

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

After 4 seconds:1

1Area 1 3 4 8

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

Notice that the area is a trapezoid.

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.

24s t t

The area under the curve2

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

.....n

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

Turn UP! MAP

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

subinterval

partition

Subintervals do not all have to be the same size.

subinterval

partition

Let’s divide partition into 8 subintervals.

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

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.

Area limn

if P is a partition of the interval ,a b

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

The definite integral is then given by:

Leibnitz introduced a simpler notation for the definite integral:

limn b

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.

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

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]

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].

lim ( )n

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.

lim ( )n

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.”

af x dx

Using Definite integral notation

lim (3( ) 2 5)

(3 2 5)

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,

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

aA f x dx

Area of any integrable function

= (area above the x-axis) –

(area below x-axis)

af x dx

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:

2( 1)x dx

24 x dx

Evaluating Integrals using areas

Evaluate using areas:

4) (a<b)

(2 1)b

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

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

1sinx xdx

Evaluate Integrals on calculator

• Evaluate the following integrals numerically:

1) = approx. 3.14

2) = approx. .89

Rules for Definite Integrals

1) Order of Integration:

( ) ( )a b

b af x dx f x dx

2) Zero: ( ) 0a

af x dx

3) Constant Multiple:

( ) ( )

kf x dx k f x dx

f x dx f x dx

Any number k

4) Sum and Difference:

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

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

5) Additivity:

( ) ( ) ( )b c c

a b af x dx f x dx f x dx

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

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

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( ) 5f x dx

1( ) 2f x dx

1( ) 7h x dx

4( )f x dx

1( )f x dx

12 ( ) 3 ( )f x h x dx

0( )f x dx

2( )h x dx

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

The Fundamental Theorem of Calculus, Part I

af t dt

Antiderivative

Derivative

Applications of The Fundamental Theorem of Calculus, Part I

cos cosxd

tdt xdx

dx t x

1cos cos (2 ) 2 cos

xdtdt x x x x

Find dy/dx.

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

t t tx xdt dt dt

t tdt dt

1 1(2) (2 )

2 2x x

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

The Fundamental Theorem of Calculus, Part I

af t dt

Antiderivative

Derivative

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)

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

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

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

af x dx

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( ) ( )

af c f x dx

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 0x dx

Use fnInt

1( ) ( )

af c f x dx

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

01 cos xdx

Calculus Date: 3/12/2014 ID Check Obj: Objective: Properties of Definite Integrals Do Now: Pop Quiz...

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