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MATH 141 (110)


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Mathematics & Statistics (Sci)
MATH 141
Antonio Lei

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 should be accessible to anyone wanting to learn Calculus I or needing a refresher in some of the early topics in calculus. I’ve tried to make these notes as self contained as possible and so all the information needed to read through them is either from an Algebra or Trig class or contained in other sections of the notes. 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. 1. Because I wanted to make this a fairly complete set of notes for anyone wanting to learn calculus I have included some material that I do not usually have time to cover in class and because this changes from semester to semester it is not noted here. You will need to find one of your fellow class mates to see if there is something in these notes that wasn’t covered in class. 2. Because I want these notes to provide some more examples for you to read through, I don’t always work the same problems in class as those given in the notes. Likewise, even if I do work some of the problems in here I may work fewer problems in class than are presented here. 3. Sometimes questions in class will lead down paths that are not covered 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 good question gets asked in class that leads to insights that I’ve not included here. You should always talk to someone who was in class on the day you missed and compare these notes to their notes and see what the differences are. 4. This is somewhat related 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 already noted not everything in these notes is covered in class and often material or insights not in these notes is covered in class. Area Between Curves In this section we are going to look at finding 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 determine the area bey = f ( ) and y = g ( ) on the interval [a,b]. We are also going to assume thf( )‡ g x( ). Take a look at the following sketch to get an idea of what we’re initially going to look at. In the Area and Volume Formulas section of the Extras chapter we derived the following formula for the area in this case. A= bf ( ) ( ) dx (1) òa The second case is almost identical to the first case. Here we are going to determine the area between x = f y and x = g y on the interval [c,d] wif y ‡ g y . ( ) ( ) ( ) ( ) In this case the formula is, A= d f y - g y dy (2) òc ( ) ( ) Now (1) and (2) 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, instead of these formulas we will instead use the following “word” 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, b ó æ upper ö æ lower ö A= ô çfunction÷ ç function ÷dx, a £ x £ b (3) ı a ł Ł ł In the second case we will use, ó æ right ö æ left ö A= ô ç ÷ ç ÷dy, c £ y £ d (4) ı c functionł Ł function ł Using these formulas will always force us to think about what is going on with each problem and to make sure that we’ve got the correct order of functions when we go to use the formula. Let’s work an example. 2 Example 1 Determine the area of the region enclosedy = x and y = x . Solution First of all, just what do we mean by “area enclosed by”. This means that the region we’re interested in must have one of the two curves on every boundary of the region. So, here is a graph of the two functions with the enclosed region shaded. Note that we don’t take any part of the region to the right of the intersection point of these two graphs. In this region there is no boundary on the right side and so is not part of the enclosed area. Remember that one of the given functions must be on the each boundary of the enclosed region. Also from this graph it’s clear that the upper function will be dependent on the range of x’s that we use. Because of this you should 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 would probably 2 say thaty = x is the upper function and they would 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 and x =1 so these are the limits of integration. So, the integral that we’ll need to compute to find the area is, b ó æ upper ö æ lower ö A= ô çfunction ÷ ç function÷ dx ı a ł Ł ł 1 2 = ò0 x - x dx æ 3 ö1 = ç2 x - x1 3÷ Ł3 3 ł 0 1 = 3 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 required. Often the bounding region, which will give the limits of integration, is difficult to determine without a graph. Also, it can often be difficult to determine which of the functions is the upper function and 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 depended upon the range of x’s that we were using. Finally, unlike the area under a curve that we looked 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 made a mistake somewhere and will need to go back and find it. Note as well that sometimes instead of saying region enclosed by we will say region bounded by. They mean the same thing. Let’s work some more examples. 2 Example 2 Determine the area of the region bounded by = xe-x , y = x+1 , x = 2, and the y-axis. Solution In this case the last two pieces of information, and the y-axis, tell us the right and left boundaries of the region. Also, recall that the y-axis is given by the line x = 0 . Here is the graph with the enclosed region shaded in. Here, unlike the first example, the two curves don’t meet. Instead we rely on two vertical lines to bound the left and right sides of the region as we noted above Here is the integral that will give the area. b ó æ upper ö æ lower ö A= ô ç ÷ ç ÷dx ı aŁfunction ł Ł function ł 2 2 = ò x+ - xe - x dx 0 2 = æ1 x + x+ e1 -x2 ö Ł2 2 ł 0 7 e-4 = + =3.5092 2 2 2 Example 3 Determine the area of the region bounded by = 2x +10 and y = 4x+16 . Solution In this case the intersection points (which we’ll need eventually) are not going to be easily identified from the graph so let’s go ahead and get them now. Note that for most of these problems you’ll not be able to accurately identify the intersection points from the graph and so you’ll need to be able to determine them by hand. In this case we can get the intersection points by setting the two equations equal. 2 2x +10 = 4x+16 2x2-4 x6 = 0 2( x+1 )(x-3 ) 0 So it looks like the two curves will intersect at x = -1 and x = 3. If we need them we can get the y values corresponding to each of these by plugging the values back into either of the equations. We’ll leave it to you to verify that the coordinates of the two intersection points on the graph are (-1,12) and (3,28). Note as well that if you aren’t good at graphing knowing the intersectio
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