CHAPTER FOUR

1. CHAPTER FOUR THE IDEA OF THE INTEGRAL Date: January 31, 2002

2. 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!

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

4. 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.

5. 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 weren�t 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).

6. LEFT HAND METHOD Let�s 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)

7. MIDDLE? MRAM (Middle Rectangular Approximation Method) is having the graph touch the midpoint of the rectangle. Let�s approximate A=?x ?y A1= (1)(�) = � 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

8. NOTICE For the area computations, we always held ?x as 1. Let�s try to make ?x smaller like ?x= � using RRAM. A= ?x?y A1 = (�)(0)=0 A2 = (�)(1)=� A3 = (�)(2.25)=9/16 A4 = (�)(4)= 2 A1+A2+A3+A4 = 3.0625 Still off, but better than the RRAM with ?x = 1

9. MRAM and LRAM for ?x = � Here are the LRAM and MRAM respectively, and their areas.

10. 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.

11. ?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.

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

13. Example summation problem: Solve this:

14. Reimann sum Think of the rectangle formula� A = ?x ?y Let�s think of ?y = y, because ?y = y � 0, since the initial point is on the x axis (y=0). Let�s even go further and say y = f(x). Therefore A = f(x) ?x

15. 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�

16. 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.

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

18. 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:

19. THAT�S 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.

20. PROPERTIES OF INTEGRALS

21. GRAPHICAL PROPERTY Equal areas above the x axis and equal areas below the x axis cancels. Area below x axis < 0 while area above x axis > 0.

22. 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= (�)(?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.

23. FUNDAMENTAL THEOREM OF CALCULUS pt 1 This theorem has two very crucial parts. One won�t 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 won�t need to know this proof for this class.

24. 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.

25. 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.

26. 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.�

27. 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.

28. 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.

29. FTC Part 2 I guess I made my point clear ?! This theorem helps us evaluate integrals.

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

31. Not too helpful eh? ? A better example: Antidifferentiate (or get the integral of) y = x2 Power Rule Reverse Power Rule Answer

32. 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.

33. 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.

34. 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:

35. EVALUATING NOTATION Notice how we evaluate antiderivatives in this format from the previous example:

36. 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.

37. AVERAGE VALUE THEOREM The average value theorem is defined as the following.

38. 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?

39. DISTANCE VS. DISPLACEMENTSPEED VS. VELOCITY If you studied physics, you can skip this slide and go on to the solution. If you haven�t 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.

40. TO FIND DISPLACEMENT To calculate displacement is done by simply integrating v(t) with respect to t and using t=0 and t=3 as limits.

41. 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!

42. VECTOR AND SCALAR

43. 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. Let�s find how much they have traveled.

44. 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 t=2, 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.

45. 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!

46. 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.

47. 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:

48. YOU CAN�T DO THIS You cannot do that idea with products nor quotients. Product rule wasn�t 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!

49. 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..

50. LAST WORDS It is very important for you to understand the Reimann sum concept for it will be useful soon. If you are having difficulty antidifferentiating (getting the integral of) basic functions, you need to study the derivative rules. You have to practice these problems over and over again. I am available for help via e-mail at kksongs_1@hotmail.com. Please read help statement in introduction slideshow prior to sending e-mail. Thank you! Good luck! Hari Bol!