Calculus I – Condensed Notes

Calculus I – Condensed Class Notes and Problems – Tristan Whalen Page 1
Calculus I – Condensed Notes
4.3 Local Extreme Values
A function has a local maximum at if it switches from increasing to decreasing there. The local max value is
( ).
A function has a local minimum at if it switches from decreasing to increasing there. The local min value is
( ).
Critical numbers are candidates for the locations of local extreme values. These are places where ( ) or
where ( ) is undefined.
There are two major steps to finding and classifying local extreme values: find the critical numbers, and then apply
either the first derivative test or the second derivative test to each critical number.
First Derivative Test:
1) Find where the function is undefined.
2) Find ( ) and see where it is undefined.
3) Set ( ) and solve for .
4) Mark the numbers from steps 1,2, and 3 on the number line. This breaks it up into intervals.
5) Check the sign of ( ) on each interval in step 4 by using a convenient “test point” in the interval. Mark
whether ( ) is positive or negative on each interval.
6) Look at each number from steps 2 and 3. If ( ) switches from positive to negative at , then ( ) is a
local maximum. If ( ) switches from negative to positive at , then ( ) is a local minimum. If ( ) is
undefined, then it cannot be a local max or min.
Second Derivative Test:
When to use this instead of the first derivative test:
If the function and its derivatives are defined everywhere and the second derivative is easy to find.
If you only need to classify one critical number for which ( ) and, perchance, don’t know the other
critical numbers.
1) This works only on the critical numbers where ( ) .
2) Find ( ) and plug in .
3) If ( ) , then ( ) is a local minimum. If ( ) , then ( ) is a local maximum. If ( ) then you
cannot draw any conclusion; use the first derivative test.
Problems:
Find and classify the local extreme values of the following functions:
1) ( )
2) ( )
3) ( ) | |
4) ( )
5) Suppose that is a critical number for and ( ) . Classify ( ) as a local
maximum, local minimum, or neither.

Calculus I – Condensed Class Notes and Problems – Tristan Whalen Page 2
4.4 Absolute Extreme Values
An absolute maximum value of a function is a value the function takes (so, a number in the range of the function)
that is bigger than or equal to any other value of the function.
An absolute minimum value of a function is a value the function takes (so, a number in the range of the function)
that is smaller than or equal to any other value of the function.
Examples:
( ) has no absolute maximum, and it has an absolute minimum of .
( )
( )
has no absolute maximum and no absolute minimum.
( ) has no absolute minimum, and it has an absolute maximum of
If we look at a function on a closed interval [ ], then ( ) and ( ) may be either an endpoint minimum or an
endpoint maximum. We figure out which using the first derivative. Use the steps of the first derivative test and
include and on the ends of the intervals for checking in step 4.
Examples:
( ) on [ ] has an endpoint minimum of ( ) and an endpoint maximum of ( )
( ) ( ) on [ ] has an endpoint maximum of ( ) and of ( )
A continuous function is guaranteed to have both an absolute maximum and an absolute minimum on a closed
interval [ ], so these are typical situations for problems involving absolute values.
STUDY TIP: Most of the steps below are the same as the steps for local extreme values.
Finding local, endpoint, and absolute extreme values of a continuous function on [ ]:
1) Find ( )
2) Set ( ) and solve for . Ignore any values that are not between and .
3) Make a number line with on the left end, on the right end, and every number from step 2 in between. This
breaks up the interval [ ] into some intervals to check (unless there are no numbers in step 2, and there is only
one interval to check).
4) Check the sign of ( ) on each interval in step 3 by using a convenient “test point” in the interval. Mark
whether ( ) is positive or negative on each interval.
[Steps 5 and 6 are wordy but easier to see on a picture]
5) Classify the left endpoint. If ( ) is positive immediately to the right of , then ( ) is an endpoint minimum.
If ( ) is negative immediately to the right of , then ( ) is an endpoint maximum.
6) Classify the right endpoint. If ( ) is positive immediately to the left of , then ( ) is an endpoint maximum.
If ( ) is negative immediately to the left of , then ( ) is an endpoint minimum.
7) Look at each number from steps 2. If ( ) switches from positive to negative at , then ( ) is a local
maximum. If ( ) switches from negative to positive at , then ( ) is a local minimum.
8) Finally, if you haven’t already, plug in the numbers to get the function values at each number from step 2 and
also get ( ) and ( ). The biggest number is the absolute maximum of on [ ]. The smallest number is the
absolute minimum of on [ ].
Problems:
Find the critical numbers and classify the extreme values of the function on the interval:
1) ( ) on [ ]:
2) ( ) on [ ]

Calculus I – Condensed Class Notes and Problems – Tristan Whalen Page 3
4.5 Max/Min Problems
These are word problems for finding minimum or maximum values.
1) What needs to be maximized or minimized? (the biggest volume, the smallest surface area, etc.)
2) What are the parameters or restrictions? (A certain perimeter, a certain amount of material, etc.)
3) Assign variables. Write the restriction in step 2 as an equation. Write the quantity in step 1 as an equation as
well (could involve area or volume formula, for example).
4) Using the restriction, get the quantity you’re trying to max/minimize in terms of one variable. Now it’s a
function that we can the derivative of and apply methods from before.
5) Take the derivative, find the critical number(s), among these find the location of the maximum or minimum.
6) Depending on the question, write the final answer. This could be the critical number itself where the max/min
occurs, or you may have to plug it in to the quantity in step 4.
Problems:
1) Find the minimum value of given that .
2) A rectangle is drawn with its base on the -axis and its top two corners on the graph of What’s the
largest possible area of such a rectangle?
3) (a) If you have feet of fencing to enclose a rectangular yard against a building, what’s the largest possible area
you could enclose?
3) (b) If you need to enclose a rectangular yard of size square feet against a building, what’s the least amount of
fencing you would need?
4) What point on the graph of is closest to the point (
5) If you need to make a rectangular box with square ends with a capacity of cubic feet, what’s the least amount
of material you’d need?
4.6 Concavity
The second derivative tells us about the concavity of a function. Wherever ( ) , is concave up. Wherever
( ) , is concave down.
Note: Since ( ) is the derivative of ( ), we can figure out where is concave up by finding where ( ) is
increasing. We can figure out where is concave down by finding where ( ) is decreasing.
Analyzing Concavity:
1) Make note of where is undefined.
2) Find ( ) and make note of where ( ) is undefined.
3) Set ( ) and solve for .
4) Mark the numbers from steps 1,2, and 3 on a number line. This breaks it up into intervals you have to check.
5) Pick a convenient “test point” from each interval in step 4 and plug it into ( ). All you need is whether ( )
is positive or negative on each interval.
6) If ( ) switches from positive to negative (or vice versa) at some point and ( ) is defined, then ( ( )) is
a point of inflection.
)?

Calculus I – Condensed Class Notes and Problems – Tristan Whalen Page 4
Problems:
Describe the concavity of the function and all points of inflection:
1) ( )
2) ( )
3) ( )
4) ( )
4.7 Asymptotes, Vertical Tangents, and Cusps
Vertical Asymptotes occur at points of infinite discontinuity. These are vertical lines with equation .
Examples:
1) ( )
2) ( )
has as a vertical asymptote
has as a vertical asymptote and no others.
Horizontal Asymptotes are horizontal lines at which the function’s graph “levels off”, that is, when one of the
limits exists:
( )
These limits can be determined easily for rational functions, that is, a polynomial divided by a polynomial:
( )
( )
( )
If (the degree of the top and bottom are the same) then
is the horizontal asymptote.
( )
If (the degree is smaller on top) then it is “bottom-heavy” and
is the horizontal asymptote.
If (the degree is bigger on top) then it is “top-heavy” and there is no horizontal asymptote.
Vertical Tangents occur wherever the function has a “vertical tangent line.” That is, places where:
( )
Examples:
( ) ( )
has a vertical tangent at .
( ) ( )
has a vertical tangent at .

Calculus I – Condensed Class Notes and Problems – Tristan Whalen Page 5
Vertical Cusps occur at sharp points in the graph of the function, wherever the derivative approaches from one
direction but from the other.
Examples:
has a vertical cusp at .
has a vertical cusp at .
( ) ( )
( ) ( )
To find vertical tangents or cusps, begin by taking the derivative and if necessary simplify it (into one big fraction,
for example). Usually you will look for where the derivative has division by 0, and then look at the power (the
numerator of the exponent). Even powers cancel negative signs, so in this situation there probably is a vertical
tangent. Odd powers do not cancel negative signs, so in this situation there is probably a vertical cusp. Note that
vertical tangents/cusps automatically only happen where the derivative does not exist (is undefined).
Problems:
1) Find the horizontal and vertical asymptotes (if any):
( )
( )
( )
( )
( )
2) Find the vertical tangents and vertical cusps (if any):
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )

Calculus I – Condensed Class Notes and Problems – Tristan Whalen Page 6
4.8 Curve Sketching
This combines the previous sections. Here is what is usually involved:
1) Domain of ( ) (note points where is not defined)
2) Vertical and horizontal asymptotes (section 4.7)
3) Use the first derivative to determine where is increasing or decreasing (section 4.2-4.3)
4) Use the second derivative to determine where is concave up or down (section 4.6)
6) Plot the asymptotes and the points of interest (places increasing/decreasing changes, places concavity changes,
etc). This includes getting the function value at the point so you know the -value to plot!
7) Sketch the graph by connecting the dots or asymptotes from step 6 with the appropriate shape of curve
(increasing and concave up; increasing and concave down; decreasing and concave up; decreasing and concave
down).
Notice that if you need local max or min values, you get these automatically with step 4 (the first derivative) and if
you need points of inflection, you get these automatically with step 5 (the second derivative)
You may also want to find the -intercepts (get ( )) or -intercepts (solve ( ) ). If applicable, determine
symmetry and periodicity (this will be clear, such as if you have a trig function). But focus on the steps above.
Problems:
1) Sketch a graph of the following functions:
( )
( )
( )
2) Here is a function and its first and second derivatives:
Sketch a graph of .
( )
( )
( )
( )
( )
( )
3) Which of the following is true about the graph of the function ( ) ?
A) has a point of inflection at the point ( )
B) is decreasing on the interval (
C) has a local minimum at the point ( )
D) has a local maximum at the point ( )
E) is increasing on the interval (
)
)

Calculus I – Condensed Class Notes and Problems – Tristan Whalen Page 7
5.2 The Area Problem and the Integral
When we want the area of a region that is not rectangular, we can approximate it using rectangles. If that region
is between a function’s graph and the -axis from to , these are the methods to approximate that area.
Chop up the interval [ ] into a “partition” . Each piece/slice of this partition is the base of a rectangle.
Put the top of the rectangle on each slice on the graph of the function using:
Left Endpoints: just put the top-left corner of each rectangle on the graph of the function
Right Endpoints: just put the top-right corner of each rectangle on the graph of the function
Midpoints: put the middle of the rectangle on the graph of the function
Upper Sum: each rectangle’s height is the largest function value over that slice
Lower Sum: each rectangle’s height is the smallest function value over that slice
Riemann Sum: each rectangle’s height is an arbitrary value of the function over that slice
All of this approximating is really in hopes of finding the exact, true area of the region, represented by the integral
symbol:
∫ ( )
This represents the number which is the area of the region between the graph of , the -axis, from to . Since
this is just a number, the variable in the integral is called a “dummy-variable.”
Problems:
1) Find the upper sum for ( ) with the partition {
2) Find the lower sum for ( ) with the partition {
}
}

Calculus I – Condensed Class Notes and Problems – Tristan Whalen Page 8
3) (a) Using the partition { }, draw the rectangles involved in the upper sum ( ) on
the graph of below. Then find the upper sum.
3) (b) Using the partition { }, draw the rectangles involved in the lower sum ( ) on the
graph of below. Then find the lower sum.
4) What is the actual area between the graph of and the -axis from to , where is graphed above for
problem 3? That is, find
∫ ( )

Calculus I – Condensed Class Notes and Problems – Tristan Whalen Page 9
5.3 The integral as a function
Given a function we can set
where the input is the upper limit .
We have that if is continuous and , then
If is continuous,
Another way to write that is:
( ) ∫ ( )
∫ ( ) ∫ ( ) ∫ ( )
( ) ( )
(∫ ( )
) ( )
If we compose with another function and want the derivative, then by the Chain Rule:
Putting all this together,
Examples:
(∫
Problems:
1) If you know the following
evaluate the following integrals:
( )
(∫ ( )
( )
( )
(∫ ( )
(∫ ( )
(∫
) ( ( )) ( )
) ( ( )) ( ) ( ( )) ( )
) ( ) ( ) ( )
)
) ( ( ) ( ) )( ) ( )( )
∫ ( )
∫ ( )
∫ ( )
∫ ( )
∫ ( )
∫ ( )
∫ ( )

Calculus I – Condensed Class Notes and Problems – Tristan Whalen Page 10
2) Find these derivatives:
(∫ √
(∫ ( )
3) Solve for the function . Then find ( )
4) If
)
)
( ) ∫
find (a) ( ), (b) ( ), (c) the critical numbers of ( ).
5.4 The Fundamental Theorem of Calculus
∫
( )
(∫ ( )
)
(∫ ( )
√
An antiderivative for is a function where ( ) ( ). The only difference between any two antiderivatives
for a given function is adding a constant.
If ( ) is continuous, then
∫ ( )
is an antiderivative of ( ) (where the lower limit is a constant), and if ( ) is any antiderivative of , then
∫ ( ) ( ) ( )
We can find many antiderivatives by “undoing” derivatives we already know.
Function Antiderivative
( ) ( )
( ) ( )
e ( ) ta ( )
e ( ) ta ( ) e ( )
( ) t( ) ( )
( ) t( )
)

Calculus I – Condensed Class Notes and Problems – Tristan Whalen Page 11
Evaluating a definite integral
1) Simplify the integrand as much as possible
2) Compare the integrand with the antiderivative chart
3) Using the chart and integration techniques (to be explored in section 5.7 and Calculus II) find the antiderivative
of the integrand.
4) Plug in the limits and subtract in the correct order: [antiderivative evaluated at upper limit] minus
[antiderivative evaluated at lower limit]
Examples:
∫
√
5.5 Area Problems
∫ √
[
( )
∫
|
( )
∫ e ( ) ta ( )|
[
( ) ] [
( )
]
[
( )
ta (
) ta (
]
( ) ] [
( )
) √
( ) ] [
( )
( ) ]
When the function is positive over the interval, then the integral gives
the area we want:
When the function is negative over the interval, then the integral gives us the area of the region but negative.
When we want the area of a region bounded by two functions, we will determine where each function is on top
or on bottom of the region, and subtract.

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To get the area of the region below ( ) (so is on top), above ( ) (so is on bottom), and between and ,
we use:
∫ ( ) ∫ ( )
Integrate the TOP function minus the BOTTOM function.
∫ ( ( ) ( ))
If we have two functions and they criss-cross, we just use more than one integral. If we have just one function and
the -axis, treat as the second function. Then this method of “top minus bottom” will always give us the
area.
Finding Area of Region bounded by two functions
1) Find the intersections of the two functions by setting them equal and solving for
(even if one of the functions is , the -axis).
2) If you have two points, you will need one integral. If you have three points, you will need two integrals. And so
on. Include in the points any boundaries given in the problem (a vertical line )
3) In between each pair of points from step 1, figure out which function is on top with a test value in between.
4) Set up each integral. The limits are, from left to right, each pair of points. The integrand is the top function
minus the bottom function for each pair of points.
5) Evaluate the integrals using the Fundamental Theorem of Calculus, then add them up.
Example:
Find the area of the region between the graph and the -axis.
1) Set equal to . Then we have , so ( ) , so we get .
2) We’ll need two integrals since we have 3 points.
3) Work from left to right. In between the first pair and , we check using
(
)
so is on top.
In between the second pair and , we check using
axis) is on top.
Next page…
that (
)
(
)
to see that (
)
so (the -

Calculus I – Condensed Class Notes and Problems – Tristan Whalen Page 13
4) For the first pair of points, we had to and on top, so:
∫ ( ) ( )
For the second pair of points, we had to and -axis on top, so:
5)
∫ ( ) ( )
And the area is
Problems:
∫ ( ) ( )
∫ ( ) [
∫ [
∫ ( ) ( )
]
[
]
)
] [(
[
] [
( )
1) Sketch and find the area of the region bounded by the graphs of and .
2) Sketch and find the area of the region bounded by the graphs of ( ) and ( ) .
3) Sketch and find the area of the region bounded by ( ) ( ) and the -axis between and .
4) Sketch and find the area of the region bounded by ( ) ( ) and ( ) ( ) between and
.
5) Sketch and find the area of the region bounded by ( ) ( ), ( ) ( ), AND the -axis between
and .
6) Sketch and find the area of the region bounded by ( ) and ( ) .
Still remaining are sections 5.6 through 5.9.
]
]
(
)

Calculus I – Condensed Class Notes and Problems – Tristan Whalen Page 14
Some practice problems:
Section 5.6
Section 5.7
∫( )
∫ ( )
∫ √
( )( √ )
∫
∫ ( )
∫ √
∫ ( ) ( )
∫ ( )