Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed.

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1 Preface Here are my online notes for my Calculus I course that I teach here at Lamar University. Despite the fact that these are my class notes, they shoul be accessible to anyone wanting to learn Calculus I or neeing a refresher in some of the early topics in calculus. I ve trie to make these notes as self containe as possible an so all the information neee to rea through them is either from an Algebra or Trig class or containe in other sections of the notes. Here are a couple of warnings to my stuents who may be here to get a copy of what happene on a ay that you misse. 1. Because I wante to make this a fairly complete set of notes for anyone wanting to learn calculus I have inclue some material that I o not usually have time to cover in class an because this changes from semester to semester it is not note here. You will nee to fin one of your fellow class mates to see if there is something in these notes that wasn t covere in class.. Because I want these notes to provie some more examples for you to rea through, I on t always work the same problems in class as those given in the notes. Likewise, even if I o work some of the problems in here I may work fewer problems in class than are presente here.. Sometimes questions in class will lea own paths that are not covere here. I try to anticipate as many of the questions as possible when writing these up, but the reality is that I can t anticipate all the questions. Sometimes a very goo question gets aske in class that leas to insights that I ve not inclue here. You shoul always talk to someone who was in class on the ay you misse an compare these notes to their notes an see what the ifferences are.. This is somewhat relate to the previous three items, but is important enough to merit its own item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!! Using these notes as a substitute for class is liable to get you in trouble. As alreay note not everything in these notes is covere in class an often material or insights not in these notes is covere in class.

2 Area Between Curves In this section we are going to look at fining the area between two curves. There are actually two cases that we are going to be looking at. In the first case we want to etermine the area between y = f ( x) an y g( x) [a,b]. We are also going to assume that f ( x) g( x). Take a look at the following sketch to get an iea of what we re initially going to look at. = on the interval In the Area an Volume Formulas section of the Extras chapter we erive the following formula for the area in this case. b ( ) ( ) A= f x -g x x (1) a The secon case is almost ientical to the first case. Here we are going to etermine the area between x f ( y) = an x= g( y) on the interval [c,] with f ( y) g( y).

3 In this case the formula is, c ( ) ( ) A= f y -g y y () Now (1) an () are perfectly serviceable formulas, however, it is sometimes easy to forget that these always require the first function to be the larger of the two functions. So, instea of these formulas we will instea use the following wor formulas to make sure that we remember that the area is always the larger function minus the smaller function. In the first case we will use, In the secon case we will use, b Û Ê upper ˆ Ê lower ˆ A= Ù Á - Á x, a x b ı Ëfunction Ëfunction a Û Ê right ˆ Ê left ˆ A= Ù Á - Á y, c y ı Ëfunction Ëfunction c () () Using these formulas will always force us to think about what is going on with each problem an to make sure that we ve got the correct orer of functions when we go to use the formula. Let s work an example. Example 1 Determine the area of the region enclose by y = x an y x =. First of all, just what o we mean by area enclose by. This means that the region we re intereste in must have one of the two curves on every bounary of the region. So, here is a graph of the two functions with the enclose region shae.

4 Note that we on t take any part of the region to the right of the intersection point of these two graphs. In this region there is no bounary on the right sie an so is not part of the enclose area. Remember that one of the given functions must be on the each bounary of the enclose region. Also from this graph it s clear that the upper function will be epenent on the range of x s that we use. Because of this you shoul always sketch of a graph of the region. Without a sketch it s often easy to mistake which of the two functions is the larger. In this case most woul probably say that y = x is the upper function an they woul be right for the vast majority of the x s. However, in this case it is the lower of the two functions. The limits of integration for this will be the intersection points of the two curves. In this case it s pretty easy to see that they will intersect at x = 0 an x = 1 so these are the limits of integration. So, the integral that we ll nee to compute to fin the area is, Û Ê upper ˆ Ê lower ˆ A= Ù Á -Á x ı Ëfunction Ëfunction b a = x - x x Ê 1 ˆ = Á x - x Ë = Before moving on to the next example, there are a couple of important things to note. First, in almost all of these problems a graph is pretty much require. Often the bouning region, which will give the limits of integration, is ifficult to etermine without a graph. Also, it can often be ifficult to etermine which of the functions is the upper function an which is the lower function without a graph. This is especially true in cases like the last example where the answer to that question actually epene upon the range of x s that we were using. Finally, unlike the area uner a curve that we looke at in the previous chapter the area between two curves will always be positive. If we get a negative number or zero we can be sure that we ve mae a mistake somewhere an will nee to go back an fin it. Note as well that sometimes instea of saying region enclose by we will say region boune by. They mean the same thing. Let s work some more examples. 1 0

5 Example Determine the area of the region boune by y = xe -x, y = x+ 1, x =, an the y-axis. In this case the last two pieces of information, x = an the y-axis, tell us the right an left bounaries of the region. Also, recall that the y-axis is given by the line x = 0. Here is the graph with the enclose region shae in. Here, unlike the first example, the two curves on t meet. Instea we rely on two vertical lines to boun the left an right sies of the region as we note above Here is the integral that will give the area. Û Ê upper ˆ Ê lower ˆ A= Ù Á - x function Á function ı Ë Ë 0 b a = x + x+ -x = x+ 1-xe x Ê1 1 Á e Ë -x - 7 e = + =.509 ˆ 0 Example Determine the area of the region boune by y = x + 10 an y x 16 = +. In this case the intersection points (which we ll nee eventually) are not going to be easily ientifie from the graph so let s go ahea an get them now. Note that for most of these problems you ll not be able to accurately ientify the intersection points from the graph an so you ll nee to be able to etermine them by han. In this case we can get the intersection points

6 by setting the two equations equal. x + 10= x+ 16 x -x- 6= 0 ( x )( x ) = 0 So it looks like the two curves will intersect at x=- 1 an x=. If we nee them we can get the y values corresponing to each of these by plugging the values back into either of the equations. We ll leave it to you to verify that the coorinates of the two intersection points on the graph are (,1) an (,8). Note as well that if you aren t goo at graphing knowing the intersection points can help in at least getting the graph starte. Here is a graph of the region. With the graph we can now ientify the upper an lower function an so we can now fin the enclose area. Û Ê upper ˆ Ê lower ˆ A= Ù Á -Á x ı Ëfunction Ëfunction b a 6 ( ) = x+ 16- x + 10 x = - x + x+ x Ê = Á Ë = 6 x x x Be careful with parenthesis in these problems. One of the more common mistakes stuents make with these problems is to neglect parenthesis on the secon term. ˆ

7 Example Determine the area of the region boune by y = x + 10, y = x+ 16, x =- an x = 5 So, the functions use in this problem are ientical to the functions from the first problem. The ifference is that we ve extene the boune region out from the intersection points. Since these are the same functions we use in the previous example we won t bother fining the intersection points again. Here is a graph of this region. Okay, we have a small problem here. Our formula requires that one function always be the upper function an the other function always be the lower function an we clearly o not have that here. However, this actually isn t the problem that it might at first appear to be. There are three regions in which one function is always the upper function an the other is always the lower function. So, all that we nee to o is fin the area of each of the three regions, which we can o, an then a them all up. Here is the area ( 16) 16 ( 10) 10 ( 16) A= x + - x+ x+ x+ - x + x+ x + - x+ x - 5 = x -x- 6x+ - x + x+ 6x+ x -x-6x - 5 x x 6x x x 6x x x 6x Ê ˆ Ê ˆ Ê ˆ = Á Á Á - - Ë Ë Ë = = -

8 Example 5 Determine the area of the region enclose by y = sin x, y = cos x, y-axis. p x =, an the First let s get a graph of the region. So, we have another situation where we will nee to o two integrals to get the area. The intersection point will be where sinx = cos x p in the interval. We ll leave it to you to verify that this will be x =. The area is then, p p 0 p A= cosx- sinxx+ sinx-cos xx ( sinx cosx ) ( cosx sin x) = ( ) = - 1+ = - = We will nee to be careful with this next example. p p 0 p 1 Example 6 Determine the area of the region enclose by x= y - an y = x- 1. Don t let the first equation get you upset. We will have to eal with these kins of equations occasionally so we ll nee to get use to ealing with them. As always, it will help if we have the intersection points for the two curves. In this case we ll get

9 the intersection points by solving the secon equation for x an then setting them equal. Here is that work, 1 y+ 1= y - y+ = y -6 = y - y- 0 8 ( y )( y ) 0= - + So, it looks like the two curves will intersect at y =- an y = or if we nee the full coorinates they will be : (,-) an (5,). Here is a sketch of the two curves. Now, we will have a serious problem at this point if we aren t careful. To this point we ve been using an upper function an a lower function. To o that here notice that there are actually two portions of the region that will have ifferent lower functions. In the range [-,] the parabola is actually both the upper an the lower function. To use the formula that we ve been using to this point we nee to solve the parabola for y. This gives, y =± x+ 6 where the + gives the upper portion of the parabola an the - gives the lower portion. Here is a sketch of the complete area with each region shae that we nee if we were going to use the first formula.

10 The integrals for the area woul then be, ( ) ( ) 5 A= x x+ 6 x+ x+ 6- x x - 5 = x+ 6x+ x+ 6- x+ 1x = x+ 6x+ x+ 6x+ - x+ 1x Ê ˆ 1 1 = u + u + Á- x + x Ë = 18 0 While these integrals aren t terribly ifficult they are more ifficult than they nee to be. Recall that there is another formula for etermining the area. It is, Û Ê right ˆ Ê left ˆ A= Ù Á - Á y, c y ı Ëfunction Ëfunction c an in our case we o have one function that is always on the left an the other is always on the right. So, in this case this is efinitely the way to go. Note that we will nee to rewrite the equation of the line since it will nee to be in the form x= f ( y) but that is easy enough to o. Here is the graph for using this formula.

11 The area is, Û Ê right ˆ Ê left ˆ A= Ù Á -Á y ı Ëfunction Ëfunction c Û Ê1 ˆ = Ù ( y+ 1) -Á y - y ı Ë - 1 = Û Ù ı - y y y Ê 1 1 = Á- + + Ë 6 = 18 y y y ˆ - This is the same that we got using the first formula an this was efinitely easier than the first metho. So, in this last example we ve seen a case where we coul use either formula to fin the area. However, the secon was efinitely easier. Stuents often come into a calculus class with the iea that the only easy way to work with functions is to use them in the form y = f ( x). However, as we ve seen in this previous example there are efinitely times when it will be easier to work with functions in the form x= f ( y). In fact, there are going to be occasions when this will be the only way in which a problem can be worke so make sure that you can eal with functions in this form.

12 Let s take a look at one more example to make sure we can eal with functions in this form. Example 7 Determine the area of the region boune by First, we will nee intersection points. y ( y ) = - - y + 10= y - y+ x 0= y -y-6 ( y )( y ) 0= =- y + 10 an x ( y ) The intersection points are y =- 1 an y =. Here is a sketch of the region. = -. This is efinitely a region where the secon area formula will be easier. If we use the first formula there woul be three ifferent regions that we have to look at. The area in this case is, Û Ê right ˆ Ê left ˆ A= Ù Á - y function Á function ı Ë Ë c ( ) = - y y- y = - y + y+ 6y Ê ˆ 6 = Á- y + y + 6y = Ë

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Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed. Preface Here are my online notes for my Algebra course that I teach here at Lamar University, although I have to admit that it s been years since I last taught this course. At this point in my career I

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