CHAPTER%20FOUR

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CHAPTER FOUR

• THE IDEA OF THE INTEGRAL

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SURVEY OF THIS CHAPTER

• Remember when I first introduced calculus to you?
  We said that there are two fundamental branches
  of calculus. One was differential calculus which
  included slopes of tangent lines, differentiation
  of functions, applications of derivatives,
  optimization, curve sketching, approximation and
  differentials.
• The other branch is called integral calculus
  which you might understand by the title of this
  presentation ? has to do with integrals. We are
  not with differential calculus yet. We have more
  to do with derivatives, however, in my humble
  opinion, this is a perfect place to take a break
  with differentiation.
• If you are not good at differentiating functions
  by now, look at those previous slides again and
  again and do those problems. You need to
  understand the concepts of differentiating. It
  would be really helpful if you knew how to
  differentiate very well.
• PRACTICE! PRACTICE! PRACTICE!

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PROBLEM

• After solving problem about tangent lines, Newton
  and Leibniz wanted to know how to find area under
  a curve.

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ANY IDEAS??

• A good way to find area is to think like a good
  math student. In other words, the way to find
  area under the curve is by FINDING a way to find
  a way to find the area under the curve.

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THE WAY TO GET AREA

• Try putting rectangles under the curve and find
  the area of all of those rectangles and thus, and
  add them all up. A l w OR
• This graph is y x2
• A Dx Dy
• A1 (1)(1) 1
• A2 (1)(4) 4
• A1 A2 5
• REAL ANSWER 8/3 or 2.6667
• There is a lot of error, since some of the
  rectangles werent exactly on the graph.
• Notice, the top right of each rectangle touches
  the graph. This method of finding area is called
  RRAM (Right Rectangular Approximation Method).

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LEFT HAND METHOD

• Lets try having the left top of the rectangle
  touch the graph. Perhaps, we will get a better
  answer.
• A (?x)(?y)
• A1 (0)(0) 0
• A2 (1)(1) 1
• A1 A2 1
• Still a lot of error.
• This method is called LRAM (Left Rectangular
  Approximation Method)

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MIDDLE?

• MRAM (Middle Rectangular Approximation Method)
  is having the graph touch the midpoint of the
  rectangle.
• Lets approximate
• A?x ?y
• A1 (1)(¼) ¼
• A2 (1)(2.25) 2.25
• A1 A2 2.50
• We are almost there. This method seems to be
  better, since the portions of the rectangle out
  of the graph take into account the missing
  portions of the graph, not taken into account
  for.
• In other words, the white part of the rectangle
  takes into account the blue part not included in
  any rectangle

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NOTICE

• For the area computations, we always held ?x as
  1. Lets try to make ?x smaller like ?x ½ using
  RRAM.
• A ?x?y
• A1 (½)(0)0
• A2 (½)(1)½
• A3 (½)(2.25)9/16
• A4 (½)(4) 2
• A1A2A3A4 3.0625
• Still off, but better than the RRAM with ?x 1

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MRAM and LRAM for ?x ½

• Here are the LRAM and MRAM respectively, and
  their areas.

A 1.75
A 2.59375
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ANALYSIS

• From the LRAM, RRAM, and MRAM and the
  computations of area, we have discovered some
  interesting patterns
• Of the three methods, MRAM was most accurate.
• Notice, when we made ?x smaller, we got a more
  better answer than before with less error.

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?x goes smaller

• If ?x gets smaller and smaller, there will be
  more and more rectangles that will be very very
  skinny. With the area of these rectangles added
  up, you will be able to compute a more accurate
  area.
• Georges Riemann thought of an idea using the
  concept of adding and skinny rectangles.

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Summation

• If you took New York State Course III, or some
  statistics class, you should be able to remember
  this summation notation.

- The sum of all values of f(x) from i (i 1) to
k. (i and k being integers)
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Example summation problem

• Solve this

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Reimann sum

• Think of the rectangle formula A ?x ?y
• Lets think of ?y y, because ?y y 0, since
  the initial point is on the x axis (y0). Lets
  even go further and say y f(x).
• Therefore A f(x) ?x

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Reimann Sums

• To express the sum of all those areas with a
  particular ?x value could be seen as?
• i and n are just for the number of rectangles.
• Now we want to make ?x really small

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How small should ?x go?

• Very small, very close very close to 0!
  Infinitesimally small!
• Time to use the limits again, since we ARE
  talking about the limit as ?x?0!
• If you want to talk in terms of how many
  rectangles, then you use the limit as n?8, rather.

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THE INTEGRAL

• These both yield to the same thing, the brand
  new, the interesting (sometimes annoying)
  INTEGRAL!

f(x) dx is the integrand
b is the second x boundary
Variable of integration
function
a is the first x boundary
f(x) must be continuous between a,b
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GRAPHICALLY

• The way to look at the integral graphically. This
  graph is y sin x. We want to find the area
  under the curve from a to b. You need to
  calculate

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THATS COOL, BUT HOW DO WE CALCULATE THE INTEGRAL?

• For me to know and for you to find out!
• We will discuss on evaluating such integrals
  later. Right now, lets focus on integral
  properties.
• Assuming f(x) is a continuous function and a,b,
  and c are various x limits and k is a constant,
  the following is proven true.

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PROPERTIES OF INTEGRALS
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GRAPHICAL PROPERTY

• Equal areas above the x axis and equal areas
  below the x axis cancels.
• Area below x axis lt 0 while area above x axis gt 0.

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OTHER WAYS OF FINDING AREAS

• You could use the trapezoidal rule which is just
  using trapezoids inside the region with the same
  ?x and use of the area formula A
  (½)(?x)(f(x0)f(x1)) where ?x x1 x0, and the
  trapezoid is made using x0 and x1. This also
  happens to be a better process. Even better than
  midpoint in some cases.

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FUNDAMENTAL THEOREM OF CALCULUS pt 1

• This theorem has two very crucial parts. One
  wont be used as much as the the other.
• From the book The Tour of the Calculus there is a
  very fancy proof to the following part of the FTC
  (fundamental theorem of calculus).
• However, you wont need to know this proof for
  this class.

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FTC Part 1

• Many books will tell you very long and boring
  ways of this part. Basically, differentiation and
  integration are inverses.
• Note, that x is a limit while the variable of
  integration is t. We could have the variable of
  integration as a, b, c, or any letter as long as
  f(variable of integration) is there. Due to this
  fact, the variable of integration is called the
  dummy variable.
• x as a limit vs. t as dummy variable is used
  to prevent confusion.

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FTC Part I

• If the top limit is not a plain simple x, then
  you must differentiate the top and thus multiply
  the f(x) on the right.
• If both limits are not x, then you have to split
  the integral and the same process as above.

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FTC Part 1

• Calculus teachers will test this on a class test.
  From looking and taking the AP Calculus exam,
  there are perhaps no or just one question on this
  theorem.
• KEY POINT Differentiation are integration are
  inverse functions.

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APPLICATION

• Recall from chapter 2 that velocity, v(t), was
  defined as the derivative of position, x(t).
  Acceleration, a(t), was the derivative of v(t),
  and the second derivative of x(t).
• So, the INTEGRAL of acceleration is velocity.
• Similarly the INTEGRAL of velocity is position.
• If you take the graph of v(t) vs. t and graphed
  it, you will see that the area between two times
  is the DISTANCE traveled.
• Similarly, the area under a(t) vs. t graph will
  give you the velocity.
• If you took physics, you were taught that taking
  the area under the curve will show you these
  quantities.

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FTC Part 2

• caps please
• THIS IS THE MOST IMPORTANT THEOREM IN CALCULUS.
  REMEMBER THIS THEOREM.
• YOU WILL, I REPEAT, YOU WILL BE ABLE TO DO ANY
  CALCULUS AFTER THIS POINT IF YOU DO NOT
  UNDERSTAND THIS THEOREM.

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FTC Part 2

• I guess I made my point clear ?!
• This theorem helps us evaluate integrals.

Generally, capital letters talk about
ANTIDERIVATIVES. Small letters talk about
functions.
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ANTIDIFFERENTIATION

• Antidifferention is basically doing the reverse
  of getting the derivative. No different from the
  concept of integral.
• Find the antiderivative of y 3x2

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Not too helpful eh? ?

• A better example Antidifferentiate (or get the
  integral of) y x2
• Power Rule
• Reverse Power Rule
• Answer

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ANTIDIFFERENTIATION

• Ask yourself, What function when I take its
  derivative will get me this?
• As you just saw, the reverse power rule in
  effect.
• You can easily come up with very basic rules on
  integrating (antidifferenting) functions.
• INTEGRAL ANTIDERIVATIVE. I will be using
  integral more to save my fingers from typing
  antiderivative over and over.

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GENERAL RULES

• With polynomials, use reversed power rule
• With trigonometric functions, remember your
  derivative rules
• y(x) sin x, then y(x) cos x
• y(x) cos x, then y(x) - sin x
• Basically it is the differentiation rules, just
  reversed.

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Evaluating Integral Example 1

• Evaluate the following
• GIVEN
• Sum of 2 integrals, pulled out constants
• Antidifferentiation
• FTC Part 2 used. Pulled out constants
• The area under the curve is

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EVALUATING NOTATION

• Notice how we evaluate antiderivatives in this
  format from the previous example

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AVERAGE VALUE

• Remember in Chapter 2 and 3, we discussed average
  rate. The average rate of distance with respect
  to time was average velocity. The limit as the
  change in the independent variable of such
  average rate goes to zero, then you get the
  instantaneous rate, or the derivative.
• Similarly, if you work backwards, you can get an
  average value, since you already have rate. For
  example, you have a velocity-time curve. If you
  want to find the average distance, you use the
  average value theorem.

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AVERAGE VALUE THEOREM

• The average value theorem is defined as the
  following.

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PROBLEM 1

• Devotees from New York wanted to go to New Jersey
  for the Ratha Yatra. A devotee who is part of the
  Gaudiya Mathematics group decided to plot speed
  and plot them on a graph. He saw that the
  velocity in m.p.h can be defined by v(t)2x-4.
  What is the net distance or displacement between
  their starting time and 3 hours later?
• What was their total distance?
• What is the average distance?

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DISTANCE VS. DISPLACEMENTSPEED VS. VELOCITY

• If you studied physics, you can skip this slide
  and go on to the solution. If you havent taken
  physics, you ought to read this.
• Distance is how much have you traveled over all,
  regardless of direction. It only deals with how
  much you traveled. Since it only has magnitude,
  it is called a scalar quantity.
• Displacement is a vector quantity. It has both
  magnitude as well as direction. The direction can
  be determined by a or sign, in this case,
  above or below the x axis. Velocity is also a
  vector, vs. speed is a scalar. Acceleration and
  jerk are also vector quantity.
• Since we are only discussing things in one
  dimension, you really have to know that vectors
  have direction by a /- sign.

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TO FIND DISPLACEMENT

• To calculate displacement is done by simply
  integrating v(t) with respect to t and using t0
  and t3 as limits.

Negative 3 miles???
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ZERO DISPLACEMENT

• Remember.. Integrals are areas! Equal areas with
  different signs ( and -) will cancel. For
  example Areas 8 and -8 will cancel. Therefore,
  equal areas cancelled out in the example.
• Negative displacements means that overall, they
  traveled in the negative direction. Remember,
  displacement is a vector!

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VECTOR AND SCALAR
DISTANCE SPEED
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DISTANCE and SPEED

• To find an absolute value of the velocity
  function, you find the zeros (i.e. the values of
  t when v(t)0) and find the absolute values of
  the areas separately. Then, add them up. Lets
  find how much they have traveled.

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TOTAL DISTANCE

• When we find the absolute value of the function,
  we are really flipping all the negative f(x)
  values and making them positive. In order to do
  that, we must find out where the f(x) values are
  0. Since v(t) is 0 when t2, then we must split
  it up into two integrals. One for 0,2 and the
  other for 2,3. One of them is going to be
  negative, therefore, you have to make each
  integral inside an absolute value bar.

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AVERAGE VALUE

• Finding average value is a piece of eggless cake.
  ?
• Just plug in the appropriate numbers and
  functions into the right spots, then integrate,
  evaluate, and calculate!

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CONCLUSION

• This chapter just merely dealt with the area
  problem. The integral is merely a sum of the
  areas infinite rectangles (with a really
  infinitesimally small width) under the curve. You
  can get good approximates with a small ?x, using
  RRAM, LRAM, and MRAM, which MRAM is most
  accurate.
• Differentiation and integration are inverses.
• The integral can be evaluated using the
  fundemental theorem of calculus.
• Velocity is the integral of acceleration.
  Position is the integral of velocity.
• Remember, net distance or displacement is merely
  the integral of v(t) with respect to t. The total
  distance is the absolute value of v(t) or you can
  break it down by solving for t when v(t) 0. and
  have that t be a limit. Remember to take the
  absolute values and add them together!
• Derivative is to integral as rate is to quantity.

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INTEGRAL

• Differential calculus, as mentioned earlier, was
  not finished, but we just took a good time to
  pause and reflect on ideas and concepts.
  Differential calculus is very short.
• Integral calculus, however, is very long and a
  tad difficult. Here is the mathematical reason
  why. You could integrate if f(x) g(x) is in the
  integrand. You could do the following

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YOU CANT DO THIS

• You cannot do that idea with products nor
  quotients.
• Product rule wasnt as easy as if you have
  h(x)f(x)g(x), then h(x)f(x)g(x) and same
  with the quotient rule too.
• Thus, the integration for such functions will
  discussed later on. If you are a BC Calculus
  (Calculus II) student, you will most likely learn
  most processes. If you are an AB Calculus
  (Calculus I) student, you will have to wait next
  year or semester for the fun ?
• This is why Calculus II was invented!

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THE ORIGINAL PARABOLA AREA PROBLEM

• Remember how I showed you the tilak (or parabola)
  graph earlier? You see how I got 2.66667 as my
  area..

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END OF CHAPTER FOUR