1. HW: pg 331 #1-19 odds Calculus Date: 3/12/2014 ID Check
Obj: Objective: Properties of Definite Integrals
Do Now: Pop Quiz See Below
HW Requests: SM 165, 166, pg #57, 59
HW: pg 331 #1-19 odds
Announcements:
Saturday Tutoring 10-1 (Derivatives)
Mock AP Exam during ACT Testing
Maximize
Academic
Potential
Evaluate the integral analytically by using the Fundamental Theorem of Calculus.
Turn UP! MAP

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.

3. 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:

4. EXAMPLE 1 Find 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.

5. 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.

6. Solving for c.
Initial value

7. Differential Equations

8. Differential Equations
Example 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

9. 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.

10. 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. _/\_ 

11. 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.

12. 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.

13. 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.

14. 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.

15. 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.

16. 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.

17. INTIAL VALUE PROBLEM Let’s say that y(1)=-1. We can find C that way.
And finally, your exact answer.

18. 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.

19. SOLVING A.S.D.E
You can still separate the y and bring dx to the right.
The process is the same. 

20. 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.

22. Trapezoidal Rule To approximate , use
T = (y0 + 2y1 + 2y2 + …. 2yn-1 + yn)
where [a,b] is partitioned into n subintervals of equal length h = (b-a)/n.

25. Trapezoidal Rule To approximate , use
T = (y0 + 2y1 + 2y2 + …. 2yn-1 + yn)
where [a,b] is partitioned into n subintervals of equal length h = (b-a)/n.
Equivalently,
T = LRAMn + RRAMn 2
where LRAMn and RRAMn are the Riemann sums using
the left and right endpoints, respectively, for f for the
partition.

26. 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

27. EX 2: Trapezoidal Rule T = (y0 + 2y1 + 2y2 + …. 2yn-1 + yn)
where [a,b] is partitioned into n subintervals of equal length h = (b-a)/n.
Interval
[0,1]
[1,2]
[2,3]
[3,4] 4, X .5 1
1.5 2
Y = x2
.25
2.25 4
T = (y0 + 2y1 + 2y2 + 2y3 + y4)
T = ¼ (0 + 2(.25) + 2(1) + 2(2.25) + 4) = 11/4

28. 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

29. 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

30. EX 2: Simpson’s Rule Interval [0,1] [1,2] [2,3] [3,4] 4 X .5 1 1.5 2
Interval
[0,1]
[1,2]
[2,3]
[3,4] 4 X .5 1
1.5 2
Y = x2
.25
2.25

31. The Definite Integral

32. When we find the area under a curve by adding rectangles, the answer is called a Rieman sum.
The width of a rectangle is called a subinterval.
The entire interval is called the partition.
subinterval
partition
Subintervals do not all have to be the same size.

33. 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
As gets smaller, the approximation for the area gets better.
if P is a partition
of the interval

34. is called the definite integral of
over
If we use subintervals of equal length, then the length of a subinterval is:
The definite integral is then given by:

35. Leibnitz introduced a simpler notation for the definite integral:
Note that the very small change in x becomes dx.

36. variable of integration
upper limit of integration
Integration
Symbol
integrand
variable of integration
(dummy variable)
lower limit of integration
It is called a dummy variable because the answer does not depend on the variable chosen.

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

38. Since rate . time = distance:
In section 6.1, we considered an object moving at a constant rate of 3 ft/sec.
Since rate . time = distance:
If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.
time
velocity
After 4 seconds, the object has gone 12 feet.

39. If the velocity varies:
Distance:
(C=0 since s=0 at t=0)
After 4 seconds:
The distance is still equal to the area under the curve!
Notice that the area is a trapezoid.

40. 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.

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

42. Riemann Sums
Sigma notation enables us to express a large sum in compact form

43. Calculus Date: 2/18/2014 ID Check
Objective: SWBAT apply properties of the definite integral
Do Now: Set up two related rates problems from the HW Worksheet 6, 10
HW Requests: pg 276 #23, 25, 26,
Turn in #28 E.C
In class: Finish Sigma notation
Continue Definite Integrals
HW: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
Maximize
Academic
Potential

44. When we find the area under a curve by adding rectangles, the answer is called a Rieman sum.
The width of a rectangle is called a subinterval.
The entire interval is called the partition.
subinterval
partition
Subintervals do not all have to be the same size.

45. The width of a rectangle is called a subinterval.
The width of a rectangle is called a subinterval.
The entire interval is called the partition.
Let’s divide partition into 8 subintervals.
subinterval
partition
Pg 274 #9 Write this as a Riemann sum. 6 subintervals

46. 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
As gets smaller, the approximation for the area gets better.
if P is a partition
of the interval

47. is called the definite integral of
over
If we use subintervals of equal length, then the length of a subinterval is:
The definite integral is then given by:

48. Leibnitz introduced a simpler notation for the definite integral:
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.

49. variable of integration
upper limit of integration
Integration
Symbol
integrand
variable of integration
(dummy variable)
lower limit of integration
It is called a dummy variable because the answer does not depend on the variable chosen.

50. Calculus Date: 2/19/2014 ID Check
Objective: SWBAT apply properties of the definite integral
Do Now: Bell Ringer Quiz
HW Requests: pg 276 #25, 26, pg odds
In class: pg 276 #23, 28
Continue Definite Integrals
HW: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
Maximize
Academic
Potential

51. Bell Ringer Quiz (10 minutes)

52. 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]

53. 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].

54. 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.

55. 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.”

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

57. 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,
We can use integrals to calculate areas
and we can use areas to calculate integrals.

58. Nonpositive regions
If the graph is nonpositive from a to b then

59. Area of any integrable function
= (area above the x-axis) –
(area below x-axis)

60. Turn UP! MAP
Maximize
Academic
Potential

61. Integral of a Constant
If f(x) = c, where c is a constant, on the interval [a,b], then

62. 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)

63. Evaluating Integrals using areas
Evaluate using areas: 3)
4) (a<b)

64. 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

65. 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

66. Evaluate Integrals on calculator
Evaluate the following integrals numerically:
= approx. 3.14
= approx. .89

67. Rules for Definite Integrals
Order of Integration:

68. Rules for Definite Integrals
Zero:

69. Rules for Definite Integrals
Constant Multiple:
Any number k
k= -1

70. Rules for Definite Integrals
4) Sum and Difference:

71. Rules for Definite Integrals
5) Additivity:

72. Rules for Definite Integrals
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)

73. Rules for Definite Integrals
Domination: f(x) ≥ g(x) on [a,b]
f(x) ≥ 0 on [a,b] ≥ 0
(g =0)

74. Using the rules for integration
Suppose:
Find each of the following integrals, if possible:
b) c)
d) e) f)

75. Calculus Date: 2/27/2014 ID Check
Obj: SWBAT connect Differential and Integral Calculus
Do Now:
HW Requests: 145 #2-34 evens and 33
HW: Complete SM pg 156, pg 306 #1-19 odds
Announcements:
Mid Chapter Test Fri. Sect
Careful of units, meaning of area,
asymptotes, properties of integrals
Handout Inverses
Saturday 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
Maximize
Academic
Potential
Turn UP! MAP

76. The Fundamental Theorem of Calculus, Part I
Antiderivative
Derivative

77. Applications of The Fundamental Theorem of Calculus, Part I1. 2.

78. Applications of The Fundamental Theorem of Calculus, Part I

79. Applications of The Fundamental Theorem of Calculus, Part I

80. 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.

81. Applications of The Fundamental Theorem of Calculus, Part I=

82. Applications of The Fundamental Theorem of Calculus, Part I=

83. 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.

84. Applications of The Fundamental Theorem of Calculus, Part 2

85. End here

86. Calculus Date: 2/27/2014 ID Check
Obj: SWBAT connect Differential and Integral Calculus
Do Now:
HW Requests: 145 #2-34 evens and 33
HW: SM pg 156
Announcements:
Mid Chapter Test Fri. Sect
Careful of units, meaning of area,
asymptotes, properties of integrals
Handout Inverses
Saturday 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
Maximize
Academic
Potential
Turn UP! MAP

87. The Fundamental Theorem of Calculus, Part I
Antiderivative
Derivative

88. Applications of The Fundamental Theorem of Calculus, Part I

89. 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.

90. Antidifferentiation = F(x) + C If x = a, then 0 = F(a) + C C = -F(a)
A 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 = F(a) + C C = -F(a)
= F(x) – F(a)

91. Calculus Date: 3/3/2014 ID Check
Obj: SWBAT connect Differential and Integral Calculus
Do 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
Saturday 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
Maximize
Academic
Potential
Turn UP! MAP

92. The Fundamental Theorem
Applications of
The Fundamental Theorem
of Calculus, Part I

93. The Fundamental Theorem
Pg 307 #22 Construct a function of the form
Applications of
The Fundamental Theorem
of Calculus, Part I

94. Calculus Date: 3/4/2014 ID Check
Obj: SWBAT connect Differential and Integral Calculus
Do Now: Put up your designate problem from the final exam. Pg 306 #32, 34
HW Requests: SM pg 156; pg 306 #1-19 , odds if not completed odds
HW: pg 295 #11-17 odds, odds
Pg 307 #41-49 odds
Announcements:
Handout Inverses sent via
Saturday 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
Maximize
Academic
Potential
Turn UP! MAP

95. 2. Find the total area of the region
Net Area: Area below the x axis is counted as negative
Total Area:
Area below the x axis is counted
positive.
Pg 307 #42
Solve analytically and using fnint

96. Pg 307 #42

97. Average (Mean) Value
Pg 295 #12
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?

98. Mean Value Theorem for Definite Integrals
If f is continuous on [a,b], then at some point c in [a,b],

99. Applying the Mean Value
Av(f) = = 1/3 (3) = 1
f(x) = 4- x2
f(c) =1
4 – 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.
Use fnInt

101. 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