1. The Definite Integral

2. 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.

3. 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

4. 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:

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

6. 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.

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

8. 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.

9. 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.

10. 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.

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

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

13. 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:
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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
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14. 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.

15. 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

16. 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

17. 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:

18. 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.

19. 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.

20. 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
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21. Bell Ringer Quiz (10 minutes)

22. 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]

23. 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].

24. 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.

25. 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.”

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

27. 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.

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

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

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

32. 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)

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

34. 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

35. 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

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

37. Rules for Definite Integrals
Order of Integration:

38. Rules for Definite Integrals
Zero:

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

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

41. Rules for Definite Integrals
5) Additivity:

42. 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)

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

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

45. Calculus Date: 2/26/2014 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. 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

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

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

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

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

50. 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.

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

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

53. 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.

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

55. End here

56. 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

57. Average (Mean) Value

58. Applying the Mean Value
Av(f) =
= 1/3(3) = 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

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

60. 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)

61. 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.

62. 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

63. 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.

64. 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