# The Definite Integral.

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The Definite Integral

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.

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

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:

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

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.

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

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.

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.

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.

The area under the curve
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

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

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.

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

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

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:

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.

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.

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

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].

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.

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.”

Using Definite integral notation
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, 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

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

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

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)

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

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

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

Rules for Definite Integrals
Order of Integration:

Rules for Definite Integrals
Zero:

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

Rules for Definite Integrals
4) Sum and Difference:

Rules for Definite Integrals

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)

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

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

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

The Fundamental Theorem of Calculus, Part I
Antiderivative Derivative

Applications of The Fundamental Theorem of Calculus, Part I
1. 2.

Applications of The Fundamental Theorem of Calculus, Part I

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.

Applications of The Fundamental Theorem of Calculus, Part I
=

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.

Applications of The Fundamental Theorem of Calculus, Part 2

End here

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

Average (Mean) Value

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

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

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)

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.

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

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.

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