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
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
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)
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,
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.
Example 1 Determine the area of the region enclosedy = x and y = x .
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
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
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,
ó æ upper ö æ lower ö
A= ô çfunction ÷ ç function÷ dx
ı a ł Ł ł
= ò0 x - x dx
æ 3 ö1
= ç2 x - x1 3÷
Ł3 3 ł
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
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.
ó æ upper ö æ lower ö
A= ô ç ÷ ç ÷dx
ı aŁfunction ł Ł function ł
= ò x+ - xe - x dx
= æ1 x + x+ e1 -x2 ö
Ł2 2 ł
= + =3.5092
Example 3 Determine the area of the region bounded by = 2x +10 and y = 4x+16 .
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.
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