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Introduction to Mathematica – Calculus 2

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Introduction To Mathematica For Calculus II Students

Mathematica is a math programming language that allows us to do almost any

mathematical process you could imagine. In this Tutorial, you will learn how to do the

following.

1. How to Get Mathematica as an Olympic College Student. Pages 2 - 5

2. Initial Startup of the Mathematica Program. Pages 6 - 7

3. The Structure and Syntax of Mathematica Commands. Pages 8 - 9

4. Sigma Notation Pages 10 - 13

5. How to Estimate the Area under a curve Numerically. Pages 14 – 29

6. How to calculate the Exact Area under a curve. Pages 30 – 42

7. How to find the Area between two curves. Pages 43 – 45

8. How to find the Indefinite Integral of a function Pages 46 - 50

9. Volumes of Revolution Pages 51 – 54

10. Improper Integrals Pages 55 – 56

11. Arc Length Page 57

12. Surface Area Page 58

13. Application of Integrals Pages 59 – 64

14. Numerical Integration Pages 65 – 71

15. Introduction to Differential Equations Pages 72 – 76

16. Direction Fields Pages 77 – 79

17. Solving Differential Equations Pages 80 – 84

18. Applications of Differential Equations Pages 85 – 90


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1. How to Get Mathematica as an Olympic College Student. You can obtain Mathematica as a stand-alone program by following these instructions. First you must Create an Account with Wolfram

1. Create an account with Wolfram.

a. Go to https://user.wolfram.com/portal/login.html

b. Click "Create Wolfram ID"

c. Fill out the “Create a Wolfram ID” form.

d. In the “Your email address (this will be your Wolfram ID)” field, type your Olympic College @student.olympic.edu email address when creating your account.

e. After entering information in all the required fields, click “Create Wolfram ID”

f. You will receive a success message after creating your ID.

https://user.wolfram.com/portal/login.html


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2. Check your Olympic College email account.

a. Go to https://mail.student.olympic.edu/owa/ and log into your Olympic College email

account.

b. Open the “Validate your Wolfram ID” email from [email protected].

c. Click the link in the email to verify that your email address is valid.

d. After you have successfully verified your email address, you will receive a thank you

confirmation.

https://mail.student.olympic.edu/owa/

mailto:[email protected]


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3. Request a Mathematica Activation Key

1. Sign in to the Wolfram account that you created using your @student.olympic.edu email address.

a. Go to https://user.wolfram.com/portal/login.html

b. You will now be logged into the Wolfram User Portal

2. After signing in to the “Wolfram User Portal”, visit the “Wolfram Activation Key Request Form”

for Olympic College.

a. While signed into your “Wolfram User Portal”, copy and paste the following URL into a

new tab in your internet browser:

https://user.wolfram.com/portal/requestAK/b3d8fbb9b1c9930e57fadac88294b36eee36a3f1

b. Fill out the “Wolfram Activation Key Request Form”

c. In the “Email” field of the form, enter your Olympic College @student.olympic.edu email

address.

d. When the form has been completely filled out, click “Submit”

e. An Activation Key will be generated and will display on the screen. Make a note of this

activation key.


https://user.wolfram.com/portal/requestAK/b3d8fbb9b1c9930e57fadac88294b36eee36a3f1


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4. Download and Install Mathematica

1. Return to the “Wolfram User Portal” webpage: https://user.wolfram.com/portal/login.html

2. Click on the “My Products and Services” tab.

3. Under “My Products and Services”, click on “Mathematica for Students for Sites”

4. Under “Product Information”, click “Get Downloads”

5. Locate your operating system from the list and click "Download"

6. Verify that you have selected the correct version of the software and click “Start Download” when

ready.

7. Depending on the platform selected, you will be prompted to save a file to your local computer.

a. For example, if you selected to download Mathematica 10.1.0 for Windows, you will be

prompted to save a file called “Mathematica_10.1.0_WIN.zip”

8. When the file has finished downloading, unzip or extract the file if necessary.

a. For example, in Windows, right-click the .zip file and select “Extract All…”

b. Once the .zip file has been extracted, a “setup.exe” file will become available. This is the

Mathematica installer file for Windows.

9. Run the installer file on your computer. Follow the on-screen instructions and enter your

Activation Key when prompted.



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2. Initial Startup of the Mathematica Program. You will begin by loading up the Mathematica program its icon will look like The program will then load up and you will see a welcome screen similar to this one below. Click on the Tab to start a new notebook and you will have your first program ready to type.

The notebook screen will now look like this.


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A very useful addition is to the setup is to add the Basic Math Assistant Palette as this will allow you to

type many math characters and symbols in a more intuitive way.

To do this click on the Palettes Tab and Highlight Basic Math Assistant and click on it.

You will now see on the Right-Hand Corner the Basic Math Assistant.

As can be seen on the Righthand side.

The Basic Math Assistant contains an easy to use set of icons that will

allow you to type in the following items.

Fractions, Powers, Square roots.

Greek letters such as 𝜃 , ∅ , 𝜋.

Special characters such as ∞ , i = √−1 or ! (Factorial).

Common Functions such as Sin , Cos , ArcTan , ex , Log.

It also has Derivative and Integral Commands.

It also has an Advanced Tab that allows for even more choices.


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3. The Structure and Syntax of Mathematica Commands.

Mathematica is a specialized programming language and has a very particular structure and the rules of

how you type in command and their Syntax need to be followed carefully.

Rule1: Wolfram Language commands begin with capital letters

and are enclosed by square brackets [.....].

For Example, the Command

Factor[x2 – 5 x + 6] will factor the expression x2 – 5x + 6 and give you the output

(−3 + 𝑥)(−2 + 𝑥)

If however you type factor[x2 – 5 x + 6] or Factor(x2 – 5 x + 6) it will not work.

Also there are many predetermined Functions including Sin[x] , Cos[x], Exp[x], Log[x],

ArcSin[x], Pi, Infinity etc….

These Functions all start with a CAPITOL letter followed by square brackets [……]

So Sin[Pi/4] will give you the exact value of sin( 𝜋

4) =

√2

2

ArcTan[Infinity] will give you the exact value of tan-1(∞) = 𝜋

2

In Mathematica your program commands and output will look like this .

Note 1: If in the above two examples we did not use a Capital letter first such as sin[Pi/4] or

arctan[Infinity] or we did not use Square Brackets such as Sin(Pi/4) or ArcTan{Infinity}

then the command would not work. So, it is important that you get the rule that all

Commands start with a CAPITAL letter followed by square brackets. […..] for if you do

not follow this syntax exactly the programs often fail to work.

Note 2: When you have finished your Mathematica program then you must hit the Numeric

Keypad ENTER button Not the normal Keyboard ENTER button.

The Numeric Keyboard ENTER button tells Mathematica to run the program, while the

normal keyboard ENTER button just jumps to the next line in your program.


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Rule 2: All variable names must be in lower case. Variables in Mathematica can have any name you like as long as the variable name does not contain any

spaces and the first letter must always be in lower case.

The following are a list of possible variable names.

These are valid names for variables: x , y , z, max , profit, energy , velocity1, account1a

These are not valid names for variables: X , Max , xvalue 1, Sin

X is not valid as it is a Capitol Letter

Max is not valid as it is a Capitol Letter

xvalue 1 is not valid as it has a space in it.

Sin starts with a Capitol letter and is also a Mathematica command for the sine function.

There are some variable names that are allowed but they should not be used as they tend to cause

confusion when reading and interpreting a program.

For Example :

Don’t use variable names that have a similar name to a command such as using variable names like

factor, print or simplify. For although factor, print and simplify are acceptable variable names they can

easily be mistaken by someone reading your program as the commands Factor , Print and Simplify.

Don’t use variable names that are too long or have no obvious meaning as they make your program

difficult to read.

For Example, don’t use a variable name such as thelargestvalueofafunctionf or xyttppp2. Note: It is a good idea to name your variable, whenever possible, to reflect its meaning or content, so if

a variable is used to store the x coordinate of a point you should call it x or x-coord, if the

variable is the velocity of an object you could call it v or velocity.

So the rule of thumb here is that you should name your variables intelligently so that your

program is easy to read and to understand what your program is doing and the meaning behind

each variables value.

It is also good practice in your program to insert comments next to important parts of your

program such as telling the reader what each variables job is and what the program is doing at

specific points. This can be done by adding on the far right hand side of the same line of the

command the relevant comment, this can be done by using by using the comment feature in

Mathematica the comment feature has the syntax (* put comment here*)


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4. Sigma Notation

Sigma Notation allows us to express a sum of terms in a very concise way, in its general form it is

written as

n

i

if1

)( each part of the notation has a specific meaning.

For example we would read as “the sum from k = 1 to n of f(k)” .

∑ 𝑓(𝑘) = 𝑓(1) + 𝑓(2) + 𝑓(3) + ⋯ … 𝑓(𝑛)

𝑛

𝑘=1

In Mathematica, you can do Sigma Notation very intuitively by using the Sigma command.

First, you go to the Basic Math Template and look for 𝑑 ∫ ∑ then choose the Sigma ∑ ∎00=0 icon.

∑ 𝑓(𝑘)

𝑛

𝑘=1

The Sigma Icon


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You will now see the following

To complete the Sigma Command, Click on each part of the template above.

index will be replaced by the counting variable, normally k.

start will be replaced by first value of k that sigma starts with.

end will be replaced by the last value of k that sigma ends with.

expr will be replaced by the function f(k) with (……) round it.

So, for example if we wanted to evaluate )20(20

1

2

k

kk we would do the following

Get the sigma template

index will be replaced by k

start will be replaced with 1

end will be replaced by 20

expr will be replaced by (k2 + k + 2)

You then press the Number Pad ENTER Key.

The output will be 3120


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Here are some examples of sigma notation being used to simplify repeated addition situations.

(a)

5

1

2

k

k

= 12 +22 + 32 + 42 + 52 = 1 + 4 + 9 + 16 + 25 = 55

(b)

7

3

2

k

k = 32 + 42 + 52 + 62 + 72 = 9 + 16 + 25 + 36 + 49 = 135

(c)

5

0

2 5k

kk = (02 + 5*0) + (12 + 5*1) + (22 + 5*2) + ….. + (52 + 5*5) = 130

Example 1: Use Mathematica to evaluate following sums.

(a)

5

1

2

k

k

(b)

7

3

2

k

k (c)

5

0

2 5k

kk

Input 1(a):

5

1

2

k

k Output 55

Input 1(b):

7

3

2

k

k Output 135

Input 1(c):

5

0

2 )5(k

kk Output 130

Note: In example 1(c) To calculate

5

0

2 5k

kk we needed a set of brackets round the expression

(k2 + 5k) , if you did not do this you would get the correct result. So, for example if you used

5

0

2 5k

kk would get an output of 55 + 5𝑘 this is because Mathematica will interpret

5

0

2 5k

kk as

5

0

2 5)(k

kk .

The syntax rule is if there is more than one term in the expression when you are doing a sigma

sum you must put brackets round the expression.

It is possible to evaluate by hand these sums for extremely large number of terms, but it becomes

impractical in these situations so it is more effective to use Mathematica.

Example 2: Evaluate

100

1

2 153k

kk

Input: )153(100

1

2

k

kk Output 321 700


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We can generalize the sigma sums by replacing the end term by n

Example 3: Find the general formulas for the following sigma sums.

(a)

n

k 1

1 (b)

n

k

k1

(c)

n

k

k1

2 (d)

n

k

k1

3

Input 1(a):

n

k 1

1 Output 𝑛

Input 1(b):

n

k

k1

Output 1

2𝑛(1 + 𝑛)

Input 1(c):

n

k

k1

2 Output 1

6𝑛(1 + 𝑛)(1 + 2𝑛)

Input 1(d):

n

k

k1

3 0utput 1

4𝑛2(1 + 𝑛)2

It can be shown that the following properties involving the summation of terms using sigma notation are true but we will in this class just accept their validity.

Property 1: For any constant c nccn

k

1

Property 2: For any constant c

n

k

k

n

k

i acca11

Property 3: The addition property

n

k

k

n

k

kk

n

k

k baba111

Property 4: The subtraction property

n

k

k

n

k

kk

n

k

k baba111

Note 1: It is not true that

n

k

k

n

k

kk

n

k

k baba111

Note 2: It is not true that

n

k

k

n

k

kn

k k

k

a

a

b

a

1

1

1


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5. How to Estimate the Area under a curve Numerically.

The essential problem here is how to estimate the area under a curve using numerical methods..

Example 4: Estimate the area under the curve f(x) = 5 + 4𝑥 − 𝑥2 − 𝑥3 from x = 0 to 2 using 4 strips.

If you graph f(x) in the Domain [0,2] you get the following curve.

One approach is split the shape into 4 equally spaced strips in the interval [0,2] , We will call the size of

any one of these strips ∆𝑥 = 2−0

4 = 0.5 (∆𝑥 is called delta x). The 4 strips will generate five

x coordinates they will be called x0 = 0 , x1 = 0.5 , x2 = 1 , x3 = 1.5 and x4 = 2

We will then have a diagram that looks like the following.

0.0 0.5 1.0 1.5 2.0

2

4

6

8

0.0 0.5 1.0 1.5 2.0

2

4

6

8


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From these 4 strips, we can create 4 rectangles and the total area of these 4 rectangles will give us a

numerical approximation of the area under this curve.

We do however have a choice here because for each strip you could use either the left endpoint or the

right endpoint to define the rectangles heights.

So, the convention is to use the notation L4 to indicate we are using the Left endpoint and R4 to indicate

we are using the Right endpoint to define the heights of the rectangles.

These two choices result in different approximations to the area under the curve.

The value of L4 is an estimate for the area under the curve f(x) = 1 + 6x – x2 – x3 from x = 0 to x = 2

L4 = ∆𝑥𝑓(𝑥0) + ∆𝑥𝑓(𝑥1) + ∆𝑥𝑓(𝑥2) + ∆𝑥𝑓(𝑥3)

= 0.5𝑓(0) + 0.5𝑓(0.5) + 0.5𝑓(1.0) + 0.5𝑓(1.5)

= 0.5(5) + 0.5(6.625) + 0.5(7) + 0.5(5.375)

L4 = 12

Note 1: The width of each strip is found by calculating ∆𝑥 = 2−0

4 this can be generalized to n

strips. So, if you wish to find the width of a single strip ∆𝑥 when the interval from x = a

to x = b is split into n strips we use the formula

∆𝑥 = 𝑏−𝑎

𝑛

Note 2: We ca also generalize the values of the x coordinates x0 , x1 , etc… by using the formula

𝑥𝑘 = 𝑎 + 𝑘∆𝑥

0.0 0.5 1.0 1.5 2.0

2

4

6

8


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If we use the Right endpoints and calculate R4 as an estimate for the area under the curve we get the

situation below.

The value of R4 is an estimate for the area under the curve f(x) = 5 + 4𝑥 − 𝑥2 − 𝑥3 from x = 0 to 2.

R4 = ∆𝑥𝑓(𝑥1) + ∆𝑥𝑓(𝑥2) + ∆𝑥𝑓(𝑥3) + ∆𝑥𝑓(𝑥4)

= 0.5𝑓(0.5) + 0.5𝑓(1.0) + 0.5𝑓(1.5) + 0.5𝑓(2)

= 0.5(6.625) + 0.5(7) + 0.5(5.375) + 0.5(1)

R4 = 10

We could reduce the workload by letting Mathematica do most of the number crunching for us

We start by realizing that the general formula for calculating R4 is

R4 =

4

1

][k

kxfx where ∆𝑥 = 𝑏−𝑎

𝑛 and xk = a + ∆𝑥𝑘

In this example R4 =

4

1

]0[k

kxfx as ∆𝑥 = 2−0

4 and xk = 0 + ∆𝑥𝑘

The code for finding R4 is Input: 𝑓[x_]: = 5 + 4𝑥 − 𝑥2 − 𝑥3

deltax =2−0

4;

∑ deltax𝑓[0 + deltax 𝑘]4

𝑘=1

Output: 10

Note : We called the variable ∆x , deltax as all variables need to start with small alphanumeric

letters also we needed to add a space between the variables deltax and k to separate the two

variables.

0.0 0.5 1.0 1.5 2.0

2

4

6

8


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In a similar fashion the code for calculating L4 is.

Input: 𝑓[x_]: = 5 + 4𝑥 − 𝑥2 − 𝑥3

deltax =2−0

4;

3

0

]0[k

kdeltaxfdeltax

Output: 12

Note 1: There is a semicolon” ; “after the command deltax =2−0

4; this is just a suppressant

command, it tells Mathematica to do the command but not show the result as part of the

output. This is a useful trick as it can be a distraction to see all the outputs from various

calculations in a Mathematica program, when all you want to see if the final result.

Note 2: The estimates L4 = 12 and R4 = 10 are different, this is very common in using these

estimates and it is often not obvious that any one of the two is a better estimate than the

other.

Note 2: If you want to get a more accurate estimate then you will need to evaluate more

rectangles so instead of n = 4 strips you could use n = 10 strips or n = 400 strips.

Example 5: Estimate the area under the curve f(x) = 5 + 4𝑥 − 𝑥2 − 𝑥3 from x = 0 to 2 using right

and left endpoints for the following number of strips

(a) n = 10 strips. (b) n = 40 strips. (c) n = 1,000 strips

Input 5(a) 𝑓[x_]: = 5 + 4𝑥 − 𝑥2 − 𝑥3 𝑓[x_]: = 5 + 4𝑥 − 𝑥2 − 𝑥3

deltax =2−0

10; deltax =

2−0

10;

∑ deltax 𝑓[0 + deltax 𝑘]9

𝑘=0 ∑ deltax 𝑓[0 + deltax 𝑘]

10

𝑘=1

𝑁[∑ deltax 𝑓[0 + deltax 𝑘]9𝑘=0 ] 𝑁[∑ deltax 𝑓[0 + deltax 𝑘]9

𝑘=0 ]

Output: 292

25 Output:

272

25

Output: 11.68 Output: 10.88

Input 5(b) 𝑓[x_]: = 5 + 4𝑥 − 𝑥2 − 𝑥3 𝑓[x_]: = 5 + 4𝑥 − 𝑥2 − 𝑥3

deltax =2−0

40; deltax =

2−0

10;



40

𝑘=1

Output: 1143

100 Output:

1123

100

Output: 11.43 Output: 11.23


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Input 5(c) 𝑓[x_]: = 5 + 4𝑥 − 𝑥2 − 𝑥3 𝑓[x_]: = 5 + 4𝑥 − 𝑥2 − 𝑥3

deltax =2−0

1000; deltax =

2−0

1000;



1000

𝑘=1

Output: 708583

62500 Output:

1123

100

Output: 11.3373 Output: 11.3293

Example 6: Estimate the area under the curve f(x) = sin(𝑥) + 𝑥2 from x = 0 to 2𝜋 using 20 strips.

Input: 𝑓[x_]: = Sin[𝑥] + 𝑥2

deltax =2𝜋−0

20;


𝑘=0; (*the value of L20 *)

𝑁[∑ deltax 𝑓[0 + deltax 𝑘]9

𝑘=0]

Output: 77.5855

Input: 𝑓[x_]: = Sin[𝑥] + 𝑥2

deltax =2𝜋−0

20;


𝑘=1; (*the value of L20 *)

𝑁[∑ deltax 𝑓[0 + deltax 𝑘]20

𝑘=1]

Output: 88.988


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Example 7. Estimate the area under the graph f(x) = 9 – x2 from x = – 2 to x = 2 by calculating using 4 approximating rectangles L4 , R4

Solution: If we wish to do it by hand then we will do The following.

We start by calculating a table of values for the function f(x) = 9 – x2

starting with x = – 2 with a ∆x = 𝑏−𝑎

𝑛 =

2−(−2)

4 = 1

The next step is to estimate the area under the curve by calculation L4 , called the Left estimate, this is done by using the 4 left end points as the sample points.

The interval size will be ∆x = 𝑏−𝑎

𝑛 =

2−(−2)

4 = 1

The sample points will be x0 = – 2 x1 = – 1 x2 = 0 and x3 = 1

L4 = ∑ ∆𝑥𝑓(𝑥𝑘)3𝑘=0

= 1f(x0) + 1f(x1) +1f(x2) +1f(x3) = 1f(– 2) + 1f(– 1) + 1f(0) + 1f(1)

= 5 + 8 + 9 + 8 L4 = 30 If we use Mathematica the code will be as follows Input: 𝑓[x_]: = 9 − 𝑥2 𝑎 = −2; (* 𝑎 = −2 *) 𝑏 = 2; (* 𝑏 = 2 *) 𝑛 = 4; (* the number of strips 𝑛 = 4 *)

deltax =𝑏−𝑎

4;

∑ deltax 𝑓[𝑎 + deltax 𝑘]𝑛−1

𝑘=0; (* the value of L4 *)

𝑁[∑ deltax 𝑓[𝑎 + deltax 𝑘]𝑛−1

𝑘=0]

Output: 30 Note 1: We generalized the formula by creating the estimate L4 by using the variables called a , b , n and

delta x this allowed us to create a program that would be adaptable to many other situations. For example, if you wanted 10 strips – just change the value of n to n = 10.

If you wanted to change the start or finish values, just change the values of a and b in the code or you could even change the function by changing f[x_].

Note 2: Notice that for L4 the sigma notation is ∑ … … … . ]𝑛−1

𝑘=0

x0 x1 x2 x3 x4

x – 2 – 1 0 1 2

f(x) 5 8 9 8 5

x0 x1 x2 x3 x4

x – 2 – 1 0 1 2

f(x) 5 8 9 8 5


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Solution: We estimate the area under the curve by calculation R4 , called the Right estimate, this is done by using the 4 right end points as the sample points.


𝑛 =

2−(−2)

4 = 1

The sample points will be x1 = – 1 x2 = 0 x3 = 1 and x4 = 2

R4 = ∑ ∆𝑥𝑓(𝑥𝑘)4𝑘=1

= 1f(x1) +1f(x2) +1f(x3) + 1f(x4) = 1f(– 1) + 1f(0) + 1f(1) + 1f(2)

= 8 + 9 + 8 + 5 = 30 If we use Mathematica the code will be as follows Input: 𝑓[x_]: = 9 − 𝑥2 𝑎 = −2; (* 𝑎 = −2 *) 𝑏 = 2; (* 𝑏 = 2 *) 𝑛 = 4; (* the number of strips 𝑛 = 4 *)

deltax =𝑏−𝑎

4;

∑ deltax 𝑓[𝑎 + deltax 𝑘]𝑛

𝑘=1; (*the value of R4* )

𝑁[∑ deltax 𝑓[𝑎 + deltax 𝑘]𝑛

𝑘=1]

Output: 30 Note 1: We generalized the formula by creating the estimate R4 by using the variables called a , b , n

and delta x this allowed us to create a program that would be adaptable to many other situations. For example, if you wanted 10 strips – just change the value of n to n = 10.

If you wanted to change the start or finish values, just change the values of a and b in the code or you could even change the function by changing f[x_].

Note 2: Notice that for R4 the sigma notation is ∑ … … … . ]𝑛

𝑘=1

Note 3: Both these estimates of the area under the curve are identical , this is very unusual situation

and typically the two estimates will be different values.

x0 x1 x2 x3 x4

x – 2 – 1 0 1 2

f(x) 5 8 9 8 5


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Example 8. Estimate the area under the graph f(x) = √𝑥 from x = 0 to x = 8 by calculating the 4 approximating rectangles L4 , R4 and M4

We start by calculating a table of values

for the function f(x) = √𝑥 starting with x = 0

with a ∆x = 𝑏−𝑎

𝑛 =

8−0

4 = 2

Solution: We estimate the area under the curve by calculation L5 , called the Left estimate, this is done by

using the 4 left end points as the sample points. The interval size will be ∆x = 𝑏−𝑎

𝑛 =

8−0

2 = 4

The sample points will be x0 = 0 x1 = 2 x2 = 4 x3 = 6

L4 = ∑ ∆𝑥𝑓(𝑥𝑘−1)3𝑘=0

= 2.f(x0) + 2.f(x1) +2.f(x2) +2.f(x3)

= 2f(0) + 2f(2) + 2f(4) + 2f(6)

= 2[ 0 + √2 + 2 + √6 ] = 2[5.864] = 11.728

If we use Mathematica the code will be as follows

Input: 𝑓[x_]: = √𝑥 𝑎 = 0; (* 𝑎 = 0 *) 𝑏 = 8; (* 𝑏 = 8 *) 𝑛 = 4; (* the number of strips 𝑛 = 4 *)

deltax =𝑏−𝑎

4;


𝑘=1; (*the value of R4* )


𝑘=1]

Output: 11.7274

x0 x1 x2 x3 x4

x 0 2 4 6 8

f(x) 0 √2 2 √6 √8

x0 x1 x2 x3 x4

x 0 2 4 6 8

f(x) 0 √2 2 √6 √8


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𝑛 =

8−0

2 = 4

The sample points will be x1 = 2 x2 = 4 x3 = 6 x4 = 8

R4 = ∑ ∆𝑥𝑓(𝑥𝑘)4𝑘=1

= 2f(x1) +2f(x2) +2f(x3) + 2f(x4)

= 2[f(2) + f(4) + f(6) + f(8) ]

= 2[ √2 + 2 + √6 + √8 ]

= 2[8.692] = 17.384



deltax =𝑏−𝑎

4;


𝑘=1; (* the value of R4 * )


𝑘=1]

Output: 11.3843

x0 x1 x2 x3 x4

x 0 2 4 6 8

f(x) 0 √2 2 √6 √8


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Solution: We now estimate the area under the curve by calculation M4 , called the midpoint

estimate, this is done by using the 4 midpoints as the sample points.


𝑛 =

8−0

2 = 4

For example, the value of 𝑥1∗ is the midpoint between x0 and x1 and is found by

evaluating

𝑥1∗ =

𝑥0 + 𝑥1

2

In general, the kth midpoint 𝑥𝑘∗ will be between xk-1 and xk and its value is 𝑥𝑘

∗ = 𝑥𝑘−1+ 𝑥𝑘

2

Since xk-1 = 𝑎 + ∆𝑥 (𝑘 − 1) xk = 𝑎 + ∆𝑥 𝑘

𝑥𝑘∗ =

𝑎+∆𝑥 𝑘+𝑎+∆𝑥 𝑘(𝑘−1)

2

= 2𝑎+ ∆𝑥(2 𝑘−1)

2

𝑥𝑘∗ = 𝑎 +

∆𝑥(2 𝑘−1)

2

The sample midpoints will be 𝑥1

∗ = 1 𝑥2∗ = 3 𝑥3

∗ = 5 𝑥4∗ = 7

M4 = ∑ ∆𝑥𝑓(𝑥𝑘∗)4

𝑘=1

= 2f(𝑥1∗) +2f(𝑥2

∗) +2f(𝑥3∗) + 2f(𝑥4

∗)

= 2[f(1) + f(3) + f(5) + f(7) ]

= 2[1+ √3 + √5 + √7 ]

= 2[7.614]

= 15.288



deltax =𝑏−𝑎

4;

∑ deltax 𝑓[𝑎 +deltax (2𝑘−1)

2]

𝑛

𝑘=1; (* the value of R4 * )


𝑘=1]

Output: 15.2277

x0 𝑥1∗

x1 𝑥2∗ x2 𝑥3

∗ x3 𝑥4∗ x4

x 0 1 2 3 4 5 6 7 8

f(x) 0 1 √2 √6 2 √5 √6 √7 √8


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Example 9: An interesting question is to decide, which of any of these three estimates is the most accurate.

If you do a sketch of f(x) = √𝑥 you can see that it is an increasing function on the interval [0,8]

Since f(x) = √𝑥 is an increasing function on the interval [0,8], this means that R4 will be an overestimate of the true area and L4 will be an underestimate. So since L4 = 11.728 and R4 = 17.384 the exact area must be somewhere between 11.728 and 17.384 . As a reasonable compromise, we could choose the average of the two estimates.

Average = 11.728+17.384

2 = 14.556

It is known that the exact value of this area is 16

3√8 = 15.085….,

This means that the error in using the varying estimates are

L4 = 11.728 Error = 3.357

R4 = 17.384 Error = 2.299

Av = 14.556 Error = 0.525

M4 = 15.288 Error = 0.203

So, in this situation the value of M4 = 15.288 is the most accurate.

In general, the LEFT, RIGHT or MIDPOINT estimates vary in accuracy depending on

the shape of the curve whose area is being estimated and it is possible, in certain

circumstances, for any of the three to be the most accurate.

2 4 6 8

0.5

1.0

1.5

2.0

2.5


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Example 10: Estimate the area under the curve given by the table of values, using N= 6 (6 strips) and using

(a) L6 (b) R6 (c) M6

Solution: Since we are using n = 6 strips we will have ∆x = 𝑏−𝑎

𝑛 =

12−0

6 = 2 and the sample points in the

table will be. x0 = 0 x1 = 2 x2 = 4 x3 = 6 x4 = 8 x5 = 10 and x6 = 12

We estimate the area under the curve by calculation L6 , called the Left estimate, this is done by using the 6 left end points as the sample points.


𝑛 =

12−0

6 = 2

The sample points will be x0 = 0 x1 = 2 x2 = 4 x3 = 6 x4 = 8 and x5 = 10

L6 = ∑ ∆𝑥𝑓(𝑥𝑘)5𝑘=0

= 2f(x0) + 2f(x1) +2f(x2) +2f(x3) + 2f(x4) + 2f(x5 = 2f(0) + 2f(2) + 2f(4) + 2f(6) + 2f(8) + 2f(10)

= 2[f(0) +f(2) + f(4) + f(6) + f(8) + f(10) = 2[9 + 8.8 + 8.2 + 7.3 + 5.9 + 4.1] = 2[43.3] L6 = 86.6

The Mathematica code for this calculation is

Input: deltax =12−0

6;

left6 = deltax(9 + 8.8 + 8.2 + 7.3 + 5.9 + 4.1)

Output: 86.6

x0 x1 x2 x3 x4 x5 x6

x 0 1 2 3 4 5 6 7 8 9 10 11 12

f(x) 9 8.9 8.8 8.5 8.2 7.8 7.3 6.6 5.9 5.1 4.1 2.8 1

x0 x1 x2 x3 x4 x5 x6

x 0 1 2 3 4 5 6 7 8 9 10 11 12

f(x) 9 8.9 8.8 8.5 8.2 7.8 7.3 6.6 5.9 5.1 4.1 2.8 1

x 0 1 2 3 4 5 6 7 8 9 10 11 12

f(x) 9 8.9 8.8 8.5 8.2 7.8 7.3 6.6 5.9 5.1 4.1 2.8 1


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𝑛 =

12−0

6 = 2

The sample points will be x1 = 2 x2 = 4 x3 = 6 x4 = 8 x5 = 10 and x6 = 12

R6 = ∑ ∆𝑥𝑓(𝑥𝑘)6𝑘=1

= 2f(x1) +2f(x2) +2f(x3) + 2f(x4) + 2f(x5) + 2f(x6) = 2f(2) + 2f(4) + 2f(6) + 2f(8) + 2f(10) + 2f(12)

= 2[f(2) + f(4) + f(6) + f(8) + f(10) + f(12) = 2[8.8 + 8.2 + 7.3 + 5.9 + 4.1 + 1] = 2[35.3] = 70.6



6;

right6 = deltax(9 + 8.8 + 8.2 + 7.3 + 5.9 + 4.1)

Output: 70.6 Note: The Mathematica code for this calculation is not very efficient and the same calculation could

easily be done on a calculator

x0 x1 x2 x3 x4 x5 x6

x 0 1 2 3 4 5 6 7 8 9 10 11 12

f(x) 9 8.9 8.8 8.5 8.2 7.8 7.3 6.6 5.9 5.1 4.1 2.8 1


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Solution: We estimate the area under the curve by calculation M6 , called the Mid-point estimate, this is done by using the 6 mid-points as the sample points.


𝑛 =

12−0

6 = 2

The sample points will be 𝑥1

∗ = 1 𝑥2∗ = 3 𝑥3

∗ = 5 𝑥4∗ = 7 𝑥5

∗ = 9 and 𝑥6∗ = 11

M6 = ∑ ∆𝑥𝑓(𝑥𝑘∗)6

𝑘=1

= 2f(𝑥1∗) +2f(𝑥2

∗) +2f(𝑥3∗) + 2f(𝑥4

∗) + 2f(𝑥5∗) + 2f(𝑥6

∗) = 2f(1) + 2f(3) + 2f(5) + 2f(7) + 2f(9) + 2f(11)

= 2[f(1) + f(3) + f(5) + f(7) + f(9) + f(11)] = 2[8.9 + 8.5 + 7.8 + 6.6 + 5.1 + 2.8] = 2[39.7]

= 79.4 The Mathematica code for this calculation is


6;

mid6 = deltax(8.9 + 8.5 + 7.8 + 6.6 + 5.1 + 2.8)

Output: 79.4

Note 1: An interesting question is to decide, which of any of these three estimates is the most

accurate. By looking at the values of f|(x) we can assume that that the function f(x) is

probably decreasing.

This means that L6 is an overestimate since all the rectangles are larger than the area they

are estimating.

While R6 is an underestimate since all the rectangles are smaller than the area they are

estimating.

So, the exact area is probably somewhere between 70.6 and 79.4

Note 2: We could use 70.6+79.4

2 = 75 as another estimate of the true area under the curve.

x0 𝑥1∗

x1 𝑥2∗ x2 𝑥3

∗ x3 𝑥4∗ x4 𝑥5

∗ x5 𝑥6

∗ x6

x 0 1 2 3 4 5 6 7 8 9 10 11 12

f(x) 9 8.9 8.8 8.5 8.2 7.8 7.3 6.6 5.9 5.1 4.1 2.8 1


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Example 11: The velocity of a motorcycle over a 60 second period is given in the table below Find an estimate for the distance travelled by the motorcycle over 60 seconds Solution: The distance travelled by the motorcycle over 60 seconds is the same as the area under

the curve in the interval [0,60], as our first estimate we will use L5.

The interval size will be ∆t = 𝑏−𝑎

𝑛 =

60−0

5 =12

The sample points will be t0 = 0 t1 = 12 t2 = 24 t3 = 36 and t4 = 48

distance travelled = d ≈ L5

= ∑ ∆𝑡𝑉(𝑥𝑖−1)5𝑖=0

= 12V(t0) + 12V(t1) +12V(t2) +12V(t3) + 12V(t4)

= 12[V(0) + V(12) + V(24) + V(36) + V(48)] = 12[30 + 28 + 25 + 22 + 24] = 1548 feet



5;

left5 = deltax(30 + 28 + 25 + 22 + 241)

Output: 1548

t 0 12 24 36 48 60

V(t) 9 8.9 8.8 8.5 8.2 7.8

t0 t1 t2 t3 t4 t5

t 0 12 24 36 48 60

V(t) 9 8.9 8.8 8.5 8.2 7.8


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Solution: As our second estimate, we will use R5.


𝑛 =

60−0

5 = 12

The sample points will be x1 = 12 x2 = 24 x3 = 36 x4 = 48 and x5 = 60

distance travelled = d ≈ R5

= ∑ ∆𝑡𝑉(𝑡𝑖)5𝑖=1

= 12V(t1) +12V(t2) +12V(t3) + 12V(t4) + 12V(t5) = 12[V(12) + V(24) + V(36) + V(48) + V(60)] = 12[28 + 25 + 22 + 24 + 27] = 1512 feet



5;

right5 = deltax(28 + 25 + 22 + 24 + 27)

Output: 1512 So in conclusion we have two estimates for the total distance travelled is 1548 ft and 1512 ft since the motorcycle has a velocity that is decreasing we can assume that 1512 feet is an underestimate and that 1548 feet and so we can conclude that the exact value of the distance travelled is between 1512 and 1548 ft.

t0 t1 t2 t3 t4 t5

t 0 12 24 36 48 60

V(t) 9 8.9 8.8 8.5 8.2 7.8


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6. How to calculate the Exact Area under a curve.

We use the notation ∫ 𝑓(𝑥)𝑑𝑥𝑏

𝑎 to indicate the exact area under the curve y = f(x) from

x = a to x = b , this is called the Definite Integral of the function f(x)

The syntax for finding the area under the curve is very intuitive in Mathematica.

First you go to the Basic Math Template and look for 𝑑 ∫ ∑ then choose the Integral ∫ ∎𝑑 ⊡0

0 icon.

The Integral Icon


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To complete the Integral Command, Click on each part of the template above.

lower will be replaced by the value of a.

upper will be replaced by the value of b.

expr will be replaced by the function f(x) with (……) round it.

var Will be replaced by the variable (usually x).

Example 12: Find the exact value of ∫ (𝑥3 − 𝑥 + 1)𝑑𝑥5

1

Get the integral template,

lower will be replaced by 1.

upper will be replaced with 5.

expr will be replaced by (x3 – x +1)

var will be replaced by x

You then press the Number Pad ENTER Key. The output will be 148


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Example 13: Find the exact area under the curve for the following functions.

(a) ∫ (2𝑥3 − 𝑥2)𝑑𝑥3

1

(b) ∫ (sin(𝑥) + 2cos (𝑥))𝑑𝑥2𝜋

−𝜋

(c) ∫ (𝑥

𝑥+1) 𝑑𝑥

4

0

(d) ∫ (𝑥2 − 9)𝑑𝑥3

−3

Input (a): ∫ (2𝑥3 − 𝑥2) 𝑑𝑥3

1

Output: 94

3

Input (b): ∫ (Sin[𝑥] + 2Cos[𝑥]) 𝑑𝑥𝜋

−𝜋

Output: 0

Input (c): ∫ (𝑥

𝑥+1) 𝑑𝑥

4

0

Output: 4 − Log[5]

Input (d): ∫ (2𝑥3 − 𝑥2) 𝑑𝑥3

1:

Output: −36

Note: The area under a curve can be any value, positive negative or even zero.

This becomes clearer when you see the areas graphically.


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Example 14: Graph the area under the curve for the following functions.

(a) ∫ (2𝑥3 − 𝑥2)𝑑𝑥3

1 (b) ∫ (sin(𝑥) + 2cos (𝑥))𝑑𝑥

2𝜋

−𝜋

(c) ∫ (𝑥

𝑥+1) 𝑑𝑥

4

0 (d) ∫ (𝑥2 − 9)𝑑𝑥

3

−3

Input (a): Plot[2𝑥3 − 𝑥2, {𝑥, 1,3}, Filling → Axis]

Output:

Input (b): Plot[Sin[𝑥] + 2Cos[𝑥], {𝑥, −𝜋, 𝜋}, Filling → Axis]

Output:

Notice – that this integral

∫(2𝑥3 − 𝑥2)𝑑𝑥

3

1

Has a positive area as all the curve is above the x-axis


∫ (sin(𝑥) + 2cos (𝑥))𝑑𝑥

2𝜋

−𝜋

Has an area of zero the positive area above the x-axis matches the negative area below the x-axis


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Input (c): Plot[𝑥

𝑥+1, {𝑥, 0,4}, Filling → Axis]

Output:

Input (b): Plot[𝑥2 − 9, {𝑥, −3,3}, Filling → Axis]

Output:


∫ (𝑥

𝑥+1) 𝑑𝑥

4

0

Has a positive area as all the curve is above the x-axis


∫(𝑥2 − 9)𝑑𝑥

3

−3

Has a negative area as all of the curve is below the x-axis


of 90

A. How to calculate the Errors in Estimating the Area under a curve Numerically.

There are two measurements of error that are typically used they are called the Absolute Error and the

Relative Error they are defined as follows.

Absolute Error = |𝐸𝑥𝑎𝑐𝑡 𝑉𝑎𝑙𝑢𝑒 − 𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝑉𝑎𝑙𝑢𝑒|

Relative Error = |𝐸𝑥𝑎𝑐𝑡 𝑉𝑎𝑙𝑢𝑒−𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝑉𝑎𝑙𝑢𝑒|

𝐸𝑥𝑎𝑐𝑡 𝑉𝑎𝑙𝑢𝑒𝑋 100

Example 15: What are the Absolute and Relative Errors in using L10 and L1000 to estimate the area

∫ (5 + 4𝑥 − 𝑥2 − 𝑥3)2

0dx.

Input: 𝑓[x_]: = 5 + 4𝑥 − 𝑥2 − 𝑥3

exact = ∫ 𝑓[𝑥] 𝑑𝑥2

0;

deltax =2−0

10;

estimate = 𝑁[∑ deltax𝑓[0 + deltax𝑘]9

𝑘=0];

abserror = Abs[exact − estimate]

relerror = Abs[(exact−estimate

exact)100]

Output: 0.346667

Output: 3.05882

Input: 𝑓[x_]: = 5 + 4𝑥 − 𝑥2 − 𝑥3

exact = ∫ 𝑓[𝑥] 𝑑𝑥2

0;

deltax =2−0

1000;

estimate = 𝑁[∑ deltax𝑓[0 + deltax𝑘]999

𝑘=0];

abserror = Abs[exact − estimate]

relerror = Abs[(exact−estimate

exact)100]

Output: 0.0039947

Output: 0.0352471

Note 1: As you increase the number of strips the accuracy of the estimate becomes greater so

L1000 = 11.3373 has only an Absolute error of approximately 0.004 which is a more

accurate estimate than L10 = 11.68 which has an Absolute error of 0.35

Note 2: In general, if you want a very accurate estimate of the area under a curve choose a value

of n as large as is practical.


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B. How to Find the Exact Area under a curve using Sigma Notation.

To find the exact under a curve we generalise the concept of splitting the shape into a large number of

rectangular strips and then adding up all these strips together.

For example, suppose we want to know the area under the curve y = f(x) from x = a to x = b.

as shown in the diagram below.

The method used is to approximate this area by splitting the area into n equal strips of width ∆𝑥 .

The width of each rectangle ∆𝑥 =𝑏−𝑎

𝑛 we then construct rectangles using these strips.

We next start with the first point with x coordinate x0 = a

the next point will have x coordinate x1 = a + ∆𝑥

x2 = a + 2∆𝑥

The coordinate of the general x-coordinate xk is xk = a + 𝑘∆𝑥

and so on until we reach the last x coordinate xn = b.

These n + 1 points x0 , x1 , x2 , ……. , xn are called the ordinates.

The distance between any two consecutive ordinates say xk to xk+1 = ∆𝑥

We must now find the area of each strip and then add them all together to get estimated area under the

curve. Each strips area is approximated by using a rectangle and the method we use has three common

variations.

y = f(x)

y

x X=a X=b


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The three common choices for which rectangle to use to estimate the area of a strip are, we can either

use the left end point of each subinterval or the right end point or we could use a value in the middle of

the subinterval to calculate the height of the rectangle.

Each choice we make will give a slightly different form to the expression for calculating the area.

Choice 1: If we choose the left end point of each sub-interval we get the following situation.

Notice that the rectangles use the left ordinate for the height of the rectangle. If we then add all

these rectangles together we can obtain an estimate for the area under the curve.

Left End point Area using n strips = ∆𝑥 f(x0) + ∆𝑥 f(x1) + ∆𝑥 f(x2) + ...+∆𝑥 f(xn-1)

Ln = ∆𝑥 [f(x0) + f(x1) + f(x2) + f(x3) + …. + f(xn-1)]

Ln = xxfn

k

k

1

0

)(

The Mathematica code for this last line would be

1

0

)(n

k

kdeltaxafdeltax

y = f(x)

y

x X0=a Xn=b X1 X2 X3 ………………………….

Xn-1 Xn-2 Xn-3

y = f(x)

y

x X0=a Xn=b X1 X2 X3 ………………………….

Xn-1 Xn-2 Xn-3

∆𝑥 ∆𝑥 ∆𝑥 ∆𝑥 ∆𝑥 ∆𝑥 ∆𝑥 ∆𝑥


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Choice 2: If we choose the right end point of each sub-interval we get the following situation.

Notice that the rectangles use the right ordinate for the height of the rectangle.

If we then add all these rectangles together we can obtain an estimate for the area under

the curve.

Right End Point Area using n strips = ∆𝑥 f(x1) + ∆𝑥 f(x2) + ∆𝑥 f(x3) + ...+∆𝑥 f(xn)

Rn = ∆𝑥 [f(x1) + f(x2) + f(x3) + f(x4) + ….. + f(xn)]

Rn = xxfn

k

k 1

)(

The Mathematica code for this last line would be

n

k

kdeltaxafdeltax1

)(

y = f(x)

y

x X0=a Xn=b X1 X2 X3 ………………………….

Xn-1 Xn-2 Xn-3

∆𝑥 ∆𝑥 ∆𝑥 ∆𝑥 ∆𝑥 ∆𝑥 ∆𝑥 ∆𝑥


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Choice 3: If we choose the midpoint of each sub-interval we get the following situation.

The midpoint of any subinterval xk-1 to xk is called 𝑥𝑘∗ =

𝑥𝑘−1+𝑥𝑘

2

xk-1 = 𝑎 + ∆𝑥 (𝑘 − 1)

xk = 𝑎 + ∆𝑥 𝑘

𝑥𝑘∗ =

𝑎+∆𝑥 𝑘+𝑎+∆𝑥 𝑘(𝑘−1)

2

= 2𝑎+ ∆𝑥(2 𝑘−1)

2

𝑥𝑘∗ = 𝑎 +

∆𝑥(2 𝑘−1)

2

Notice that the rectangles use the midpoint of each subinterval for the height of the rectangle.

If we then add all these rectangles together we can obtain an estimate for the area under

the curve.

Midpoint Area using n strips = ∆𝑥𝑓(𝑥1∗) + ∆𝑥𝑓(𝑥1

∗) + ∆𝑥𝑓(𝑥1∗) + ⋯ + ∆𝑥𝑓(𝑥𝑛

∗ )

Mn = ∆𝑥[𝑓(𝑥1∗) + 𝑓(𝑥1

∗) + 𝑓(𝑥1∗) + ⋯ + 𝑓(𝑥𝑛

∗ )]

Mn = xxfn

k

k 1

* )(

The Mathematica code for this last line would be ∑ deltax𝑓[𝑎 +deltax(2𝑘−1)

2]

𝑛

𝑘=1

In general, the greater the number of strips that are used the more accurate the estimate of the area under

the curve will be. So, for example, if we use n = 200 strips it will generate a more accurate estimate than

if we only use n = 10 strips.

∆𝑥 ∆𝑥 ∆𝑥 ∆𝑥 ∆𝑥 ∆𝑥 ∆𝑥 ∆𝑥

y = f(x)

y

x X0=a Xn=b X1 X2 X3 ………………………….

Xn-1 Xn-2 Xn-3


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We can now generalise this problem to come up with the following definitions for finding the area under

the curve A by making the number of strips tend to infinity.

Definition: The area exact area A that lies under the curve y = f(x) from x = a to x = b can be found

by taking an infinite number of strips and using either left, right or midpoints.

So A = lim𝑛→∞

𝐿𝑛 = lim𝑛→∞

∑ 𝑓(𝑥𝑘)∆𝑥𝑛−1𝑖=0

A = lim𝑛→∞

𝑅𝑛 = lim𝑛→∞

∑ 𝑓(𝑥𝑘)∆𝑥𝑛𝑖=1

A = lim𝑛→∞

𝑀𝑛 = lim𝑛→∞

∑ 𝑓(𝑥𝑘∗ )∆𝑥𝑛

𝑖=1

Where ∆𝑥 = 𝑏−𝑎

𝑛

xk = a + ∆𝑥 k

𝑥𝑘∗ = 𝑎 +

∆𝑥(2𝑘−1)

2

Example 16: Find the exact area of the integral ∫ (2𝑥3 − 𝑥2)𝑑𝑥2

0 using Ln , Rn and Mn

Input: 𝑓[x_]: = 2𝑥3 − 𝑥2

deltax =2−0

𝑛;

Limit[∑ deltax𝑓[0 + deltax𝑘]𝑛−1

𝑘=0, 𝑛 → ∞] (* using limit of Ln as 𝑛 → ∞*)

Output: 16

3

Input: 𝑓[x_]: = 2𝑥3 − 𝑥2

deltax =2−0

𝑛;

Limit[∑ deltax𝑓[0 + deltax𝑘]𝑛


Output: 16

3

Input: 𝑓[x_]: = 2𝑥3 − 𝑥2

deltax =2−0

𝑛;

Limit[∑ deltax𝑓[0 +deltax(2𝑘−1)

2]

𝑛


Output: 16

3

Note: All three methods generate the same result so we typically use Rn in these type of questions as it

has the simplest form


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In some situations, it is possible to generalise the process by using the variables a and b.

Example 17: Find the exact area of the integral ∫ (2𝑥3 − 𝑥2)𝑑𝑥𝑏

𝑎 using Rn.

Input: 𝑓[x_]: = 2𝑥3 − 𝑥2

deltax =𝑏−𝑎

𝑛;

Limit[∑ deltax𝑓[0 + deltax𝑘]𝑛


Output: 1

6(2𝑎3 − 3𝑎4 + 𝑏3(−2 + 3𝑏))

We can alter the look of the output by using this code

𝑓[x_]: = 2𝑥3 − 𝑥2

deltax =𝑏−𝑎

𝑛;

Expand[

Limit [∑ deltax𝑓[0 + deltax𝑘]

𝑛

𝑘=1

, 𝑛 → ∞]

]

Output: 𝑎3

3−

𝑎4

2−

𝑏3

3+

𝑏4

2


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C. Using Your Calculator to calculate areas under a curve We can use our calculator to check our estimates for the area under the curve y = f(x) by using the option

∫ 𝑓(𝑥)𝑑𝑥 on the TI 84 calculator.

Example Use your calculator to find an estimate for ∫ 𝑐𝑜𝑠𝑥. 𝑑𝑥𝜋

20

Solution: We can use our TI 84 calculator to get the area below the curve y = cos x from x = 0 to x = 𝜋

2

In order to do this we follow these steps. 1) Press y = button

2) Type in the desired function in this case Y1 = cos(x) 3) Press the WINDOWS button and choose

Xmin = – 1 Xmax = 2 Xscl = 1 Ymin = – 1 Ymax = 1 Yscl = 1 4) Press 2nd Trace to get the CALC menu 5) choose option 7:∫ 𝑓(𝑥)𝑑𝑥 (this is the numerical integration menu) 6) We tell the computer the Lower Limit by typing 0 and pressing ENTER.

We tell the computer the Lower Limit by typing x =𝜋

2 = 1.57 (or as close as you wish to the true

value 𝜋

2 of ) and pressing ENTER.

The screen will now have the desired area in this case A = 1.0165799 (Approximately)

Example Use your calculator to find an estimate for ∫ ln (𝑥). 𝑑𝑥4

1

Solution: We can use our TI 84 calculator to get the below the curve y = ln(x) from x = 1 to x = 4 In order to do this we follow these steps. 1) Press y = button 2) Type in the desired function in this case Y1 = ln(x) 3) Press the WINDOWS button and choose

Xmin = – 1 Xmax = 5 Xscl = 1 Ymin = – 1 Ymax = 3 Yscl = 1 4) Press 2nd Trace to get the CALC menu 5) choose option 7:∫ 𝑓(𝑥)𝑑𝑥 (this is the numerical integration menu) 6) We tell the computer the Lower Limit by typing 1 and pressing ENTER.

We tell the computer the Lower Limit by typing 4 and pressing ENTER.

The screen will now have the desired area in this case A = 2.6041663 (Approximately)


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7. How to find the Area between two curves

To find the area between two curves y = f(x) and y = g(x) from x = a to x = b you

evaluate the integral.

∫(𝑓(𝑥) − 𝑔(𝑥))𝑑𝑥

𝑏

𝑎

Where f(x) is larger than g(x) in the domain [a,b]

Example 18: Find the area between the two curves f(x) = x2 and g(x) = √𝑥

from x = 0 to x = 1

Solution: We can draw a graph of f(x) and g(x) to see which is greater.

Input: 𝑓[x_]: = 𝑥2;

𝑔[x_]: = √𝑥; Plot[{𝑓[𝑥], 𝑔[𝑥]}, {𝑥, 0,1}, PlotLegends → "Expressions"]

Output:

Note: From the graph above we can see that g(x) = √𝑥 is the larger of the two

functions and so we will use (g(x) – f(x)) in evaluating the area

Input: ∫ (𝑔[𝑥] − 𝑓[𝑥]) 𝑑𝑥1

0

Output: 1

3

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0


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It is possible to find the area between two curves that intersect but these situations require

more work.

Example 19: Find the area between the two curves f(x) = sin(x) and g(x) = 𝑐𝑜𝑠(𝑥)

from x = 0 to x = 𝜋

Solution: We start by graphing the functions to see the interaction of the two

functions.

Input: 𝑓[x_]: = Sin[𝑥]; 𝑔[x_]: = Cos[𝑥]; Plot[{𝑓[𝑥], 𝑔[𝑥]}, {𝑥, 0, 𝜋}, PlotLegends → "Expressions"]

Output:

Note: From the graph above we can see that g(x) = cos (𝑥) is larger than

f(x) = sin(x) up to the point of intersection and from that point onwards until x

= 𝜋 that f(x) = sin(x) is larger than g(x) = cos (𝑥).

This means that we need to find the x-coordinate of the point of intersection.

Input: 𝑓[x_]: = Sin[𝑥]; 𝑔[x_]: = Cos[𝑥]; Solve[{Sin[𝑥] == Cos[𝑥]&&0 ≤ 𝑥 ≤ 𝜋}, {𝑥}, Reals]

Output: {{𝑥 → −2ArcTan[1 − √2]}}

The point of intersection is at x = −2ArcTan[1 − √2] = 𝜋

4


of 90

So, the area between these two curves will be.

Area = ∫ (𝑔(𝑥) − 𝑓(𝑥)) 𝑑𝑥𝜋

40

+ ∫ (𝑔(𝑥) − 𝑓(𝑥)) 𝑑𝑥𝜋

𝜋

4

= ∫ (𝑐𝑜𝑠(𝑥) − 𝑠𝑖𝑛(𝑥)) 𝑑𝑥𝜋

40

+ ∫ (𝑠𝑖𝑛(𝑥) − 𝑐𝑜𝑠(𝑥)) 𝑑𝑥𝜋

𝜋

4

The Mathematica code is

Input: 𝑓[x_]: = Sin[𝑥]; 𝑔[x_]: = Cos[𝑥];

∫ (𝑔[𝑥] − 𝑓[𝑥]) 𝑑𝑥𝜋

40

+ ∫ (𝑓[𝑥] − 𝑔[𝑥]) 𝑑𝑥𝜋

𝜋

4

Output: 2√2

We can see the area between the curves by using the following code.

Plot[ {𝑓[𝑥], 𝑔[𝑥]}, {𝑥, 0, 𝜋},

Ticks → {{0,𝜋

4,𝜋

2,3𝜋

4, 𝜋} , {−1,0,1}},

Filling → {1 → {2}},

PlotLabels → Expressions

]


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8. How to find the Indefinite Integral of a function.

We have already seen how to find the definite integral of a function (see pages 30 – 32)

We use the notation ∫ 𝑓(𝑥)𝑑𝑥𝑏

𝑎 to indicate the exact area under the curve y = f(x) from

x = a to x = b , this is called the Definite Integral of the function f(x)

The syntax for finding the area under the curve is very intuitive in Mathematica.

First you go to the Basic Math Template and look for 𝑑 ∫ ∑ then choose the Integral icon.

The Integral Icon


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To complete the Sigma Command, Click on each part of the template above.

expr will be replaced by the function f(x) with (……) round it.

var Will be replaced by the variable (usually x).

Example 18: Find the Indefinite Integral ∫ (𝑥3 − 2𝑥 + 7) 𝑑𝑥

Get the integral template,

expr will be replaced by (𝑥3 − 2𝑥 + 7)

var will be replaced by x

You then press the Number Pad ENTER Key. The output will be 7𝑥 − 𝑥2 +𝑥4

4

Note 1: You need to put brackets round the function (𝑥3 − 2𝑥 + 7)

Note 2: The output is 7𝑥 − 𝑥2 +𝑥4

4 but technically the proper form as 7𝑥 − 𝑥2 +

𝑥4

4+ 𝐶

Where C is called the Arbitrary constant.


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Example 20: Find the following simple Indefinite Integrals.

(a) ∫(𝑥3 − 5𝑥 + 7)𝑑𝑥 (A Polynomial Function)

(b) ∫ (𝑥4−16

2𝑥2+8𝑥) 𝑑𝑥 (A Rational Function)

(c) ∫(√𝑥2 − 6𝑥 + 1)𝑑𝑥 (A Radical Function)

(d) ∫(100𝑒3𝑥+7)𝑑𝑥 (An Exponential Function)

Input (a): ∫ (𝑥3 − 5𝑥 + 7) 𝑑𝑥

Output: 7𝑥 −5𝑥2

2+

𝑥4

4

Input (b): ∫ (𝑥4−16

2𝑥2+8𝑥) 𝑑𝑥

Output: 8𝑥 − 𝑥2 +𝑥3

6− 2Log[𝑥] − 30Log[4 + 𝑥]

Input (c): ∫ (√𝑥2 − 6𝑥 + 1) 𝑑𝑥

Output: 1

2(−3 + 𝑥)√1 − 6𝑥 + 𝑥2 − 4Log[3 − 𝑥 − √1 − 6𝑥 + 𝑥2]

Input (a): ∫ (100𝑒3𝑥+7) 𝑑𝑥

Output: 100

3𝑒7+3𝑥

Note: All of the integrals are missing the arbitrary constant C , so you will need to

remember to add that yourself.

For example, ∫ (𝑥3 − 5𝑥 + 7) 𝑑𝑥 = 7𝑥 −5𝑥2

2+

𝑥4

4+ 𝐶


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We can find complex integrals using this command as well as can be seen in the example

below.

Example 21: Find the following complex Indefinite Integrals.

(a) ∫(𝑥3𝑒2𝑥4)𝑑𝑥 (U-Substitution method)

(b) ∫(𝑥2𝐿𝑛(𝑥))𝑑𝑥 (Integration by parts)

(c) ∫(𝑠𝑖𝑛2𝑥 𝑐𝑜𝑠7𝑥)𝑑𝑥 (Trig Integration)

(d) ∫ (1

√𝑥2−25) 𝑑𝑥 (X-Substitution method)

(e) ∫ (24

(𝑥−3)(𝑥+6)) 𝑑𝑥 (Integration by parts)

Input (a): ∫ (𝑥3𝑒2𝑥4) 𝑑𝑥

Output: 𝑒2𝑥4

8

Input (b): ∫ (𝑥2Log[𝑥]) 𝑑𝑥

Output: −𝑥3

9+

1

3𝑥3Log[𝑥]

Input (c): ∫ (Sin[𝑥]^2Cos[𝑥]^7) 𝑑𝑥

Output: 7Sin[𝑥]

128−

1

160Sin[5𝑥] −

5Sin[7𝑥]

1792−

Sin[9𝑥]

2304

Input (d): ∫ (1

√𝑥2−25) 𝑑𝑥

Output: Log[𝑥 + √−25 + 𝑥2]

Input (e): ∫ (24

(𝑥−3)(𝑥+6)) 𝑑𝑥

Output: 24(1

9Log[−3 + 𝑥] −

1

9Log[6 + 𝑥])

Note : The integral ∫(𝑠𝑖𝑛2𝑥 𝑐𝑜𝑠7𝑥)𝑑𝑥 had to be written using the syntax ∫ (Sin[𝑥]^2Cos[𝑥]^7) 𝑑𝑥


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Since there are an infinite number of possible functions to Integrate, it will happen, from

time to time that the result of an Indefinite Integral may not be Analytically possible to

find or that the integral may be given in terms of a function that you do not know.

Example 22: Find the following complex Indefinite Integrals.

(a) ∫(𝑒2𝑥4)𝑑𝑥

(b) ∫(𝑒2𝑥4)𝑑𝑥

(c) ∫(𝑒sin (𝑥))𝑑𝑥

Input (a): ∫ (𝑒2𝑥4) 𝑑𝑥

Output: −𝑥Gamma[

1

4,−2𝑥4]

421 4⁄ (−𝑥4)1 4⁄

Note: In this example, the solution is given in terms of the Gamma Function.

Input (b): ∫ (𝑒𝑥2) 𝑑𝑥

Output: 1

2√𝜋Erfi[𝑥]

Note: In this example, the solution is given in terms of the Erfi Function - the imaginary error function

Input (c): ∫ (𝑒Sin[𝑥]) 𝑑𝑥

Output: ∫ 𝑒Sin[𝑥] 𝑑𝑥

Note: In this example, the solution is the original Integral , this means that there is no algebraic

expression that is the integral of esin(x).


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9. Volumes of Revolution

A. Volume of Integration (Disk Method)

Example 23: Find the volume of the solid generated when you rotate the curve f(x) = 3x2 + 1 about the

x-axis, from x = 1 to x = 3

Solution: For this situation, we will use the formula. V = dxxf

b

a

2

)(

V = dxx

3

1

22 13

V = dxxx

3

1

24 169

V = dxxx

3

1

24 69

V = 3

1

35 35

9

x

xxxx

V =

)1()1(3)1(

5

9)3()3(3)3(

5

9 3535

V = 5

2448

The Mathematica Code for this question is.

Input: 𝑓[x_]: = 3𝑥2 + 1

𝑎 = 1; 𝑏 = 3;

∫ 𝜋(𝑓[𝑥])2 𝑑𝑥𝑏

𝑎

Output: 2448𝜋

5


of 90

Example 24: (a) Find the volume of the solid generated when you rotate the area between the

curves f(x) = 10 – x2 and g(x) = sin(x) about the x-axis, from x = 0 to x = 𝜋

4.

(b) Show this information graphically.

Solution (a): For this situation, we will use the formula. V = dxxgxf

b

a

])()([22

Input: 𝑓[x_]: = 10 − 𝑥2

𝑔[x_]: = Sin[𝑥] 𝑎 = 0;

𝑏 =𝜋

4;

∫ 𝜋((𝑓[𝑥])2 − (𝑔[𝑥])2) 𝑑𝑥𝑏

𝑎

Output: 𝜋

4+

7𝜋2

8−

𝜋4

48+

𝜋6

5120

So, the volume of revolution = 𝜋

4+

7𝜋2

8−

𝜋4

48+

𝜋6

5120

Solution (b): The code to show this information graphically.

Input: 𝑓[x_]: = 10 − 𝑥2

𝑔[x_]: = Sin[𝑥] 𝑎 = 0;

𝑏 =𝜋

4;

Plot[{𝑓[𝑥], 𝑔[𝑥]}, {𝑥, 𝑎, 𝑏}, PlotRange → {−2,2}, PlotLegends → {"f(x) = 2 − "x2"", "Sin[x]"}]

Output:


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B. Volumes by Cylindrical Shells.

Example 25: Find the volume created by rotating the curve f(x) = x2 about the y-axis

from x= 0 to x = 1.

Solution: Since we are rotating about the y – axis we use the formula in terms of x.

See diagram below.

V = b

a

dxheightradius ))((2

V = b

a

dxxxf2

= 1

0

2.2 dxxx

= 1

0

3.2 dxx

= 1

0

4

2

1

x

xx

= 𝜋

2


Input: 𝑓[x_]: = 𝑥2

𝑎 = 0; 𝑏 = 1;

∫ (2𝜋𝑥𝑓[𝑥]) 𝑑𝑥𝑏

𝑎

Output: 𝜋

2

Note: By setting up the code in this general format it is easy to adapt the code it to other similar

situations. So, if the function was f(x) = x3 + x2 – 1 we only need to change one line of

code , namely we change 𝑓[x_]: = 𝑥2 and replace it with 𝑓[x_] : = 𝑥3 + 𝑥2 − 1

y axis

x = 1 0

y = x2

Radius = x

Height = y = f(x)


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Example 26: Find the volume created by rotating the region between the two curves f(x) = 3 + 2x – x2

and g(x) = 3 – x about the y-axis

Solution: Since we are rotating about the y – axis we use the formula in terms of x.

So, radius of a typical shell is x and the

RaDius = x

Height = f(x) – g(x)

= (3 + 2x – x2 ) – (3 – x)

= 3 + 2x – x2 – 3 + x

= 3x – x2

The values of a and b are found by finding where f(x) = 3 + 2x – x2 and g(x) = 3 – x

intersect. In this case, a = 0 and b = 3, See diagram below.

V = b

a

dxheightradius ))((2

V =

b

a

dxxgxfx ))()((2

=

3

0

23.2 dxxxx

=

3

0

32 )2.6( dxxx

= 1

0

43

2

12

x

xxx

V = 27𝜋

2


Input: 𝑓[x_]: = 3 + 2𝑥 − 𝑥2

𝑔[x_]: = 3 − 𝑥

Solve[𝑓[𝑥] == 𝑔[𝑥], {𝑥}, Reals]

Output: {{𝑥 → 0}, {𝑥 → 3}}

Input: 𝑓[x_]: = 3 + 2𝑥 − 𝑥2

𝑔[x_]: = 3 − 𝑥

𝑎 = 0; 𝑏 = 3;

∫ (2𝜋𝑥(𝑓[𝑥] − 𝑔[𝑥])) 𝑑𝑥𝑏

𝑎

Output: 27𝜋

2

Radius = x

y axis

3 0 g(x) = 3 – x

f(x) = 3 + 2x – x2

Height = f(x) – g(x)


of 90

10. Improper Integrals

There are two types of Improper Integrals – Improper Integrals of Type 1 have the property that at least

one of the two end-points of integration is infinity.

For example, the following three integrals are all Improper Integrals of Type 1.

∫10

𝑥3 𝑑𝑥∞

1 ∫

𝑥

𝑥2+1𝑑𝑥

5

−∞ ∫

4

𝑥2+1𝑑𝑥

∞

−∞

The method used is to use limits to accommodate the infinities.

∫10

𝑥3 𝑑𝑥∞

1 = lim

𝑏→∞(∫

10

𝑥3 𝑑𝑥𝑏

1)

∫𝑥

𝑥2+1𝑑𝑥

5

−∞ = lim

𝑏→−∞(∫

𝑥

𝑥2+1𝑑𝑥

5

𝑏)

∫4

𝑥2+1𝑑𝑥

∞

−∞ = lim

𝑏→∞(∫

𝑥

𝑥2+1𝑑𝑥

𝑏

0) + lim

𝑏→−∞(∫

𝑥

𝑥2+1𝑑𝑥

0

𝑏)

Mathematica copes with Improper Integrals in a very intuitive way as the following examples show

Example 27: Find the value of the following three Type 1 improper Integrals.

(𝑎) ∫10

𝑥3 𝑑𝑥∞

1 (b) ∫

4

𝑥2+1𝑑𝑥

∞

−∞ (𝑐) ∫

𝑥

𝑥2+1𝑑𝑥

5

−∞

Input (a) 𝑓[x_]: =10

𝑥3

∫ 𝑓[𝑥] 𝑑𝑥∞

1 Output: 5

Input (b) 𝑓[x_]: = 𝑓[x_]: =4

𝑥2+1


1 Output: 4𝜋

Input (c) 𝑓[x_]: =10

𝑥3


1

Output: Integrate: Integral of 𝑥

𝑥2+1 does not converge on {1,∞}.

Note: The Output statement Integrate: Integral of 𝑥

𝑥2+1 does not converge on {1,∞} in this

situation is essentially saying that the area under the curve is infinitely large.


of 90

Improper Integrals of Type 2 have the property that there is a vertical asymptote at one of the endpoints

or between the endpoints.

For example, the following three integrals are all Improper Integrals of Type 2.

∫𝑥

(𝑥−3)𝑑𝑥

3

1 ∫

𝑥

(𝑥+4)3 𝑑𝑥6

−4 ∫

1

𝑥𝑑𝑥

𝜋

−𝜋

The method used is to use limits to accommodate the vertical asymptotes.

∫𝑥

(𝑥−3)𝑑𝑥

3

1 = lim

𝑏→3−(∫

𝑥

(𝑥−3)𝑑𝑥

𝑏

1)

∫𝑥

(𝑥+4)3 𝑑𝑥6

−4 = lim

𝑏→4+(∫

𝑥

(𝑥+4)3 𝑑𝑥6

𝑏)

∫1

𝑥𝑑𝑥

𝜋

−𝜋 = lim

𝑏→0−(∫

1

𝑥𝑑𝑥

𝑏

−𝜋) + lim

𝑏→0+(∫

1

𝑥𝑑𝑥

𝜋

𝑏)

Mathematica copes with Improper Integrals in a very intuitive way as the following examples show

Example 28: Find the value of the following three Type 2 improper Integrals.

(𝑎) ∫𝑥

(𝑥−3)𝑑𝑥

3

1 (b) ∫

𝑥

√6−𝑥𝑑𝑥

6

−4 (𝑐) ∫

1

𝑥2 𝑑𝑥𝜋

−𝜋

Input (a) 𝑓[x_]: =𝑥

𝑥−3

∫ 𝑓[𝑥] 𝑑𝑥3

1

Output: Integrate: Integral of 𝑥

𝑥−3 does not converge on {1,3}.

Input (b) 𝑓[x_]: =𝑥

√6−𝑥

∫ 𝑓[𝑥] 𝑑𝑥6

−4 Output:

16√10

3

Input (c) 𝑓[x_]: =10

𝑥3

∫ 𝑓[𝑥] 𝑑𝑥𝜋

−𝜋

Output: Integrate: Integral of 10

𝑥3 does not converge on {1,∞}.

Note: The Output statement Integrate Integrate: Integral of 𝑥

𝑥−3 does not converge on {1,3}.in

this situation is essentially saying that the area under the curve is infinitely large.


of 90

11. Arc Length

To find the length of the arc along the curve y = f(x) from x = a to x = b is

L = ∫ √1 + (𝑓′(𝑥))2𝑑𝑥𝑏

𝑎

Example 29: Find the exact value of the arc length for the curve f(x) = 4x + 7 from x = 1 to x = 3

Input: 𝑓[x_]: = 4𝑥 + 7

𝑎 = 1; 𝑏 = 3;

∫ √1 + (𝑓′[𝑥])2 𝑑𝑥𝑏

𝑎

Output: 2√17

Example 30: Find the exact value of arc length for the curve f(x) = 𝑒𝑥 +1

4𝑒−2𝑥 from x = 0 to x = ln(4)

Input: 𝑓[x_]: = 2𝑒𝑥 +1

8𝑒−𝑥

𝑎 = 0; 𝑏 = 𝐿𝑜𝑔[4];

∫ √1 + (𝑓′[𝑥])2 𝑑𝑥𝑏

𝑎

Output: 195

32

Example 31: Find the numerical estimate for the arc length for the curve f(x) = 𝑥 + sin (𝑥)

from x = 0 to x = 1

Input: 𝑓[x ]: = 𝑥 + 𝑆𝑖𝑛[𝑥]

𝑎 = 0; 𝑏 = 1;

𝑁[∫ √1 + (𝑓′[𝑥])2 𝑑𝑥, 4𝑏

𝑎]

Output: 2.097

Note : There are situations where you will not be able to find the exact arc length and in these

situations, you can only find a numerical estimate.

Example 31 is such a situation, in fact it takes Mathematica about 1 minute to perform

this numerical estimate.


of 90

12. Surface Area

To find the area of the surface generated by rotating the graph of the function y = f(x) about the x-axis

from x = a to x = b is

S = ∫ 2𝜋𝑓(𝑥)√1 + (𝑓′(𝑥))2𝑑𝑥𝑏

𝑎

Example 32: Find the exact value of the surface area when the graph of f(x) = 6 – 2x is rotated about

the x-axis from x = 0 to x = 3

Input: 𝑓[x_]: = 6 − 2𝑥

𝑎 = 0; 𝑏 = 3;

∫ 2𝜋𝑓[𝑥]√1 + (𝑓′[𝑥])2 𝑑𝑥𝑏

𝑎

Output: 18√5𝜋

Example 33: Find the exact value of arc length for the curve f(x) = 𝑒𝑥 +1

4𝑒−2𝑥 from x = 0 to x = ln(2)

Input: 𝑓[x_]: = 2𝑒𝑥 +1

8𝑒−𝑥

𝑎 = 0; 𝑏 = 𝐿𝑜𝑔[4];

∫ √1 + (𝑓′[𝑥])2 𝑑𝑥𝑏

𝑎

Output: 𝜋(61455

1024+ Log[4])

Example 34: Find the numerical estimate for the arc length for the curve f(x) = 𝑥 + sin (𝑥)

from x = 0 to x =

Input: 𝑓[x_]: = Sin[𝑥] + Cos[𝑥] 𝑎 = 0;

𝑏 =𝜋

2;

𝑁[∫ 2𝜋𝑓[𝑥]√1 + (𝑓′[𝑥])2 𝑑𝑥𝑏

𝑎, 3]

Output: 14.4

Note : There are situations where you will not be able to find the exact value of the Surface

Area. Example 34 is such a situation, in fact it takes Mathematica about 1 minute to

perform this numerical estimate.


of 90

13. Application of Integrals

There are many applications of Integrals, here are some practical examples.

A. Finding velocity and position functions

B. Average Value of a Function

C. Work Done Problems

D. Hydrostatic Problems

C. Finding velocity and position functions

Example 35: The acceleration of a moving object is given by the formula a(t) = t2 + 6t , find the

velocity function v(t) and the position function s(t) when v(1) = 5 and s(1) = 0

Solution: We use the fact that v(t) = ∫ 𝑎(𝑡)𝑑𝑡 and that s(t) = ∫ 𝑣(𝑡)𝑑𝑡

v(t) = ∫ 𝑎(𝑡)𝑑𝑡 = ∫(𝑡2 + 6t )𝑑𝑡

Input: ∫ (𝑡2 + 6𝑡) 𝑑𝑡 Output: 3𝑡2 +𝑡3

3

This means that v(t) = 3𝑡2 +𝑡3

3+ 𝐶 so to find the value of C we solve v(1) = 5.

Input: 𝑣[t_]: = 3𝑡2 +𝑡3

3+ 𝐶

Solve[𝑣[1] == 5] Output: {{𝐶 →5

3}}

This means that v(t) = 3𝑡2 +𝑡3

3+

5

3 so to find the value s(t)

We now find s(t) = ∫ 𝑣(𝑡)𝑑𝑡 = ∫(3𝑡2 +𝑡3

3+

5

3)𝑑𝑡

Input: ∫ (3𝑡2 +𝑡3

3+

5

3) 𝑑𝑡 Output:

5𝑡

3+ 𝑡3 +

𝑡4

12

This means that S(t) = 5𝑡

3+ 𝑡3 +

𝑡4

12+ 𝐶 so to find the value of C we solve s(1) = 0.

Input: 𝑠[t_]: =5𝑡

3+ 𝑡3 +

𝑡4

12+ 𝐶

Solve[𝑠[1] == 0] Output: {{𝐶 → −11

4}}

This means that s(t) = 5𝑡

3+ 𝑡3 +

𝑡4

12−

11

4


of 90

D. Average Value of a function

Example 36: (a) Find the average value of the function f(x) = 2

√𝑥+1 from x = 0 to x = 3

(b) Find the value(s) of x where the function is equal to the average value.

(c) Show this information graphically.

Solution (a): The formula for the average value of a function f(x) from x = a to x = b is:-

fave = 1

𝑏−𝑎∫ 𝑓(𝑥)𝑑𝑥

𝑏

𝑎

In this question fave = 1

3−0∫ (

2

√𝑥+1) 𝑑𝑥

3

0

Input: 𝑓[x_]: =2

√𝑥+1

𝑎 = 0; 𝑏 = 3;

fave =1

𝑏−𝑎∫ (

2

√𝑥+1) 𝑑𝑥

𝑏

𝑎

Output: 4

3 So, the average value of the function is fave =

4

3

Solution (b): To find the value(s) of x where the function is equal to the average value we need to

solve for what values of x is f(x) = fave

In this question, we need to solve f(x) = 4

3

Input: 𝑓[x_]: =2

√𝑥+1

𝑎 = 0; 𝑏 = 3;

fave =1

𝑏−𝑎∫ (

2

√𝑥+1) 𝑑𝑥

𝑏

𝑎

Solve[𝑓[𝑥] == fave&&𝑎 ≤ 𝑥 ≤ 𝑏, {𝑥}, Reals]

Output: {{𝑥 →5

4}}

So, x = 5

4 is the point where the function f(x) has the same value as the

average value fave.


of 90

Solution (c): To show this information graphically we will graph y = f(x) and y = 4

3

Input: 𝑓[x_]: =2

√𝑥+1 (* this is f(x) *)

𝑔[x ]: =4

3 (* this is g(x) *)

Plot[ (* start of Plot Command *)

{𝑓[𝑥], 𝑔[𝑥]}, (* two functions to graph *) {𝑥, 0,3}, (* the range of x- values *)

PlotRange → {0,2}, (* the range of x- values *)

PlotLegends → {"f(x) =2

√x+1", "g(x)=

4

3"} (* legend at side of graph *)

Ticks → {{0,1

4,

1

2,

3

4, 1,

5

4,

3

2,

7

4, 2,

9

4,

5

2,

11

4, 3}, {0,

1

3,

2

3, 1,

4

3,

5

3, 2}} (* scales on x and y-axes *)

GridLines → {{0,1

4,

1

2,

3

4, 1,

5

4,

3

2,

7

4, 2,

9

4,

5

2,

11

4, 3}, {0,

1

3,

2

3, 1,

4

3,

5

3, 2}} (* gridlines on x and y-axes *)

] (* end of Plot Command *)

Output:

Note 1: The point of intersection is at (5

4,

4

3)

Note 2: The code for the graphing of the function contains Plot Options

such as PlotRange , PlotLegends,Ticks and GridLines. These are optional but they do add

some detail to the graphs to help with the overall appearance.


of 90

E. Work Done Problems

Example 37: A heavy rope 50ft long, weighs 0.5 pounds per foot and hangs over the edge of a building 120 ft high. How much work is done in pulling the rope to the top of the building?

The total height of the building (200ft ) is not important - only The work done in pulling the rope up is important.

We break the problem into a smaller simpler situation first. Suppose we have the rope at a height of xi above the ground and we raise it up by a small amount called ∆𝑥 then the work done by pulling the rope up this extra ∆𝑥 is given by the formula. W = FD = 0.5xi∆𝑥.

If we sum this up over the entire 50 ft we get

W = lim𝑛→∞

∑ 0.5xi∆𝑥𝑛𝑖=1

which becomes the integral W = ∫ 0.5𝑥𝑑𝑥50

0

W = [0.25𝑥2]𝑥=0𝑥=50

W = 625 ft lb

The Mathematica Code for this question is. Input: 𝑤[x_]: = 0.5𝑥 𝑎 = 0; 𝑏 = 50;

∫ (𝑤[𝑥]) 𝑑𝑥𝑏

𝑎

Output: 625.

Example 38: A cable weighs 2lb per foot is used to lift 800 pounds of coal up a mine shaft 500 ft deep.

Find the work done. Input: 𝑤[x ]: = 800 + 2𝑥 𝑎 = 0; 𝑏 = 500;

∫ (𝑤[𝑥]) 𝑑𝑥𝑏

𝑎

Output: 650000

Note: w(x) is the work function it is made up of a fixed 800 pounds plus x ft of rope that weigh 2 lb per foot so the total weight at point x is w(x) = 800 + 2x

50 ft

xi xi +∆𝑥


of 90

F. Hydrostatic Problems

Example 39: Find the hydrostatic pressure on the rectangular plate shown below.

Water has a weight of 62.5 pounds per cubic feet.

Solution: The pressure function p(x) is the pressure at depth xk.

P(xk) = Pressure at depth xk

= 62.5xk (62.5x when feet)

A(xk) = Area of a typical strip at depth xk

= 6∆𝑥

Force Function at xk F(xk) = P(xk)A(xk)

= (62.5xk )( 6∆𝑥)

= 375𝑥𝑘∆𝑥

Total Force F = lim𝑛→∞

∑ 0.5x𝑘 ∆𝑥𝑛𝑘=1

= ∫ 375𝑥𝑑𝑥6

2

= (375

2𝑥2)

𝑥 = 6𝑥 = 2

= 6750 – 750

Total Force F = 6000 lb


Input: 𝑎 = 2; 𝑏 = 6; 𝑓[x_]: = 375𝑥

∫ 𝑓[𝑥] 𝑑𝑥𝑏

𝑎

Output: 6000

∆𝑥

xk


of 90

Example 40: Find the hydrostatic pressure on the rectangular plate shown below.

Water has a weight of 62.5 pounds per cubic feet.

P(x) = Pressure at depth xk = 62.5x (125

2 x when feet , 9810x when in m)

A(x) = Area of a typical strip at position d = w∆𝑥

We use similar triangles to find w in terms of d the distance from the top of the triangle

3

4 =

3−𝑑

𝑤

3w = 4(3 – d)

3w = 12 – 4d

w = 12 – 4d

3

w = 4 −4

3𝑑

Since d = x – 1 we can calculate the

width of a typical strip in terms of x.

w = 4 −4

3(𝑥𝑘 − 1)

w = 4 −4

3𝑥𝑘 +

4

3

w = 16

3−

4

3𝑥𝑘

Area of a typical strip at depth xk A(xk ) = 𝑤∆𝑥 = (16

3−

4

3𝑥𝑘)∆𝑥

Force Function at xk

F(xk) = P(xk )A(xk ) = (125

2 xk)(

16

3−

4

3𝑥𝑘)∆𝑥 = (

1000

3xk –

250

3𝑥𝑘

2)∆𝑥

Total Force F = lim𝑛→∞

∑1000

3 xk –

250

3 𝑥𝑘

2)∆𝑥𝑛𝑘=1 = ∫ (

1000

3x –

250

3 𝑥2)𝑑𝑥

4

1

= (500

3𝑥2 −

250

9𝑥3)

𝑥 = 4𝑥 = 1

= 750 lb


Input: 𝑎 = 1; 𝑏 = 4;

𝑓[x_]: =1000

3𝑥 −

250

3𝑥2

∫ 𝑓[𝑥] 𝑑𝑥𝑏

𝑎

Output: 750

d w ∆𝑥

x


of 90

14. Numerical Integration

We have three methods for approximating the Area under a curve the y are Ln , Rn and Mn

These three methods use the technique of splitting the area into rectangles and can be very accurate

if you take enough strips. Two other techniques can also be used to estimate the area under a curve.

These numerical methods are called the Trapezoid Rule and Simpsons Rule.

A. Trapezoid Rule.

We now find a method for getting the approximate area under a curve.

We begin by splitting the area into n strips as shown above, each of width ∆𝑥 = 𝑏−𝑎

𝑛 and xk = a + k∆𝑥 .

To find the area of one these strips we find the area of a trapezium similar to the ones drawn above.

The area of one trapezium AK from xk-1 to xk = 1

2∆𝑥[f(xk-1) + f(xk)]

The estimate of the total Area Tn is

Tn = A1 + A2 + A3 + ….. + An-1 + An

= 1

2∆𝑥[f(x0) + f(x1)]+

1

2∆𝑥[f(x1) + f(x2)] + …….+

1

2∆𝑥[f(xn-1) + f(xn)]

= 1

2∆𝑥[f(x0) + f(x1)+ f(x1) + f(x2) + f(x2) + f(x3) + … + f(xn-1) + f(xn-1) + f(xn)]

= 1

2∆𝑥[f(x0) +2f(x1)+ 2f(x2) + 2f(x3) + ……. + 2f(xn-1) + f(xn)]

Note: This formula is sometimes called the Trapezoidal Rule , and the accuracy of your estimate

depends on the number of strips taken. There are two main situations where we would use this

technique the first is when the information we are given about the function f(x) is not algebraic

but in the form of a table or when f(x) is too complex or is impossible to integrate as in

y

y = f(x)

x xn1 xn2 xn x2 x1 x0 ∆𝑥

𝑓(𝑥0) 𝑓(𝑥1) 𝑓(𝑥2) 𝑓(𝑥𝑛−2)

… … … … … … … ….

𝑓(𝑥𝑛−1) 𝑓(𝑥𝑛)


of 90

Example 41: Find an estimate for ∫2

𝑥2+1𝑑𝑥

4

0 using the Trapezoid rule with n = 8 strips.

Solution: The first step is to create a table of values for our function.

Using a = 0 b = 4 n = 8 ∆𝑥 = 𝑏−𝑎

𝑛 =

4−0

8 =

1

2

The Mathematica Code that does this is

Input: 𝑓[x_]: =2

𝑥2+1

𝑎 = 0; 𝑏 = 4; 𝑛 = 8;

deltax =𝑏−𝑎

𝑛;

Table[𝑓[𝑥], {𝑥, 𝑎, 𝑏, deltax}]

Output: {2,8

5, 1,

8

13,

2

5,

8

29,

1

5,

8

53,

2

17}

We then calculate T8 using the following formula.

T8 = 1

2∆𝑥[f(x0) +2f(x1)+ 2f(x2) + 2f(x3)+ 2f(x4) + 2f(x5) + 2f(x6) + 2f(x7) + f(x8)]

T8 = 1

2(

1

2)[2 + 2(

8

5) + 2(1) + 2(

8

13) + 2(

2

5) + 2(

8

29) + 2(

1

5) + 2(

8

53) + 2(

2

17)]

T8 = 1

4[2 +

16

5 + 2 +

16

13 +

4

5 +

16

29 +

2

5 +

16

53 +

4

17]

T8 = 1.32525


Input: 𝑓[x_]: =1

𝑥2+1

𝑎 = 0; 𝑏 = 4; 𝑛 = 8;

deltax =𝑏−𝑎

𝑛;

area = 𝑁[1

2deltax𝑓[𝑎] + ( ∑

𝑘=1

𝑛−1

deltax𝑓[𝑎 + deltax𝑘]) +1

2deltax𝑓[𝑏]]

Output: 1.32525

x0 x1 x2 x3 x4 x5 x6 x7 x8

Xk 0 0.5 1 1.5 2 2.5 3 3.5 4

f(xk) 2 8

5 1

8

13

2

5

8

29

1

5

8

53

2

17


of 90

Example 42: Find the relative error in using the Trapezoid rule to estimate ∫1

𝑥2+1𝑑𝑥

4

0 with

n = 4,8,16,50 and 100 strips.


Input: 𝑓[x_]: =2

𝑥2+1

𝑎 = 0; 𝑏 = 4; 𝑛 = 4;

deltax =𝑏−𝑎

𝑛;

area = 𝑁[1


𝑘=1

𝑛−1


2deltax𝑓[𝑏]]

exactarea = ∫ 𝑓[𝑥] 𝑑𝑥𝑏

𝑎;

rerror = 𝑁[Abs[(exactarea−area

exactarea)100]]

Output: 0.2710856

Input: 𝑓[x_]: =2

𝑥2+1

𝑎 = 0; 𝑏 = 4; 𝑛 = 8;

deltax =𝑏−𝑎

𝑛;

area = 𝑁[1


𝑘=1

𝑛−1


2deltax𝑓[𝑏]]


𝑎;


exactarea)100]]

Output: 0.0425595

In a similar fashion we get

Change the number of strips to n = 16 the new Output is 0.0108674



Note 1: The relative error decreases when the number of strips gets larger

Note 2: The relative error in these examples are given as percentages, so for example when n = 8

strips the relative error is 0.0425595%


of 90

Example 43: (a) Find the relative error in using the Trapezoid rule to estimate ∫ sin (𝑥)𝑑𝑥𝜋

0 with

n = 3,4,5,6,7 and 8

(b) Show the results graphically.

Input (a): 𝑓[x_]: = 𝑆𝑖𝑛[𝑥] 𝑎 = 0; 𝑏 = 𝜋; 𝑛 = 2;

deltax =𝑏−𝑎

𝑛;

area = 𝑁[1


𝑘=1

𝑛−1


2deltax𝑓[𝑏]]

exactarea = ∫ 𝑓[𝑥] 𝑑𝑥𝜋

0;


exactarea)100]]

The Results are

Input (b) ListPlot[

{{3,9.31003}, {4,5.19406}, {5,3.31172}, {6,2.29514}, {7,1.68417}, {8,1.28842}},

AxesLabel → {number of strips n,relative error (%)}

]

n 3 4 5 6 7 8

Relative error

9.31003 5.19406 3.31172 2.29514 1.68417 1.28842

3 4 5 6 7 8number of strips n0

2

4

6

8

relative error


of 90

A. Simpsons Rule.

The final numerical technique is called Simpsons Rule; we start by splitting the curve into n equal width

strips.

∆𝑥 = 𝑏−𝑎

𝑛 and xk = a + k∆𝑥.

The major difference in Simpsons rule compared to the others is the shape of the strip.

For Left, Right and Midpoints Ln , Rn and Mn the strips will be rectangles.

For the Trapezoid Rule, Tn the strips will be trapezoids.

For Simpsons Rule, Sn the strips will have at the top a quadratic curve joining (xk-1 ,f(xk-1) ) to (xk ,f(xk))

The formula for Simpsons Rule is:-

Sn = 1

3∆𝑥[f(x0) +2f(x1)+ 4f(x2) + 2f(x3)+ 4f(x4) +……… 4f(xn-2) + 2f(xn-1) + f(xn)]

So one strip would look like the diagram shown here.

The blue is the function; The orange is the quadratic curve.

These “quadratic shaped” strips are done for each of the n strips

and their areas are then added together to give an estimate for the

area under the curve.

y

y = f(x)

x xn1 xn2 xn x2 x1 x0 ∆𝑥

𝑓(𝑥0) 𝑓(𝑥1) 𝑓(𝑥2) 𝑓(𝑥𝑛−2)

… … … … … … … ….

𝑓(𝑥𝑛−1) 𝑓(𝑥𝑛)

Xk-1 Xk

∆𝑥

f(Xk)

f(Xk-1)


of 90

Example 44: Find an estimate for ∫2

𝑥2+1𝑑𝑥

4

0 using the Simpsons Rule with n = 8 strips.

Solution: The first step is to create a table of values for our function.

Using a = 0 b = 4 n = 8 ∆𝑥 = 𝑏−𝑎

𝑛 =

4−0

8 =

1

2


Input: 𝑓[x_]: =2

𝑥2+1

𝑎 = 0; 𝑏 = 4; 𝑛 = 8;

deltax =𝑏−𝑎

𝑛;

Table[𝑓[𝑥], {𝑥, 𝑎, 𝑏, deltax}]

Output: {2,8

5, 1,

8

13,

2

5,

8

29,

1

5,

8

53,

2

17}

We then calculate S8 using the following formula.

T8 = 1

3∆𝑥[f(x0) +4f(x1)+ 2f(x2) + 4f(x3)+ 2f(x4) + 4f(x5) + 2f(x6) + 4f(x7) + f(x8)]

T8 = 1

3(

1

2)[2 + 4(

8

5) + 2(1) + 4(

8

13) + 2(

2

5) + 4(

4

29) + 2(

1

5) + 4(

4

53) + 2(

1

17)]

T8 = 1

6[1 +

32

5 + 2 +

32

13 +

4

5 +

16

29 +

2

5 +

16

53 +

2

17]

T8 = 1.59053


Input: 𝑓[x_]: =1

𝑥2+1

𝑎 = 0; 𝑏 = 4; 𝑛 = 8;

deltax =𝑏−𝑎

𝑛;

area =1

3deltax(𝑓[0] + 2𝑓[0.5] + 4𝑓[1] + 2𝑓[1.5] + 4𝑓[2] + 2𝑓[2.5] + 4𝑓[3] + 2𝑓[3.5] + 𝑓[4])

Output: 1.59053

x0 x1 x2 x3 x4 x5 x6 x7 x8

Xk 0 0.5 1 1.5 2 2.5 3 3.5 4

f(xk) 2 8

5 1

8

13

2

5

8

29

1

5

8

53

2

17


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Example 45: Find the relative error in using the Trapezoid rule to estimate ∫1

𝑥2+1𝑑𝑥

4

0 with

n = 4,8 and 16 strips.


Input: 𝑓[x_]: =1

𝑥2+1

𝑎 = 0; 𝑏 = 4; 𝑛 = 4;

deltax =𝑏−𝑎

𝑛;

area =1

3deltax(𝑓[0] + 4𝑓[1] + 2𝑓[2] + 4𝑓[3] + 𝑓[4]);


𝑎;


exactarea)100]]

Output: 2.98255

In a similar fashion, we get:-



If we compare the estimates using the Trapezium Rule and Simpsons Rule we get the following.

Note 1: When you compare the two methods, the Trapezium rule is more accurate than Simpsons

Rule for a small number of strips ( n = 4 and n = 8) but when you do a large number of

strips (n = 16) Simpsons Rule is more accurate.

Note 2: In general, as long as there are a reasonably large number of strips, Simpsons Rule will

always give a more accurate result than the Trapezium Rule. For example, if we had used

n = 40 strips the relative error for the Trapezium rule would be 0.00173973% and if we

used Simpsons Rule the relative error is 0.000072225% a factor of 24 times more

accurate!!!

Number of strips 4 8 16

Trapezium Rule

(Relative Error) 0.2710856% 0.0425595% 0.0108674%

Simpsons Rule

(Relative Error) 2.98255% 0.1471078% 0.000303374%


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15. Introduction to Differential Equations

A Differential is an equation that contains derivatives.

If the highest derivative is a first derivative the we call this a First Order Differential Equation or for

short a 1st order D.E. and if the highest derivative in the equation is a second derivative then we call it a

2nd order D.E.

So for example, these are all 1st Order D.E. 1. 𝑥𝑑𝑦

𝑑𝑥+ 5𝑦 = 𝑥2

2. 𝑦′(𝑡) + 4𝑦(𝑡) = 0

3. 𝑓′(𝑥) − 7𝑓(𝑥) = sin (𝑥)

So for example, these are all 2nd Order D.E. 4. 𝑑2𝑦

𝑑𝑥2 + 5𝑑𝑦

𝑑𝑥+ 6𝑦 = 2

5. 𝑦′′(𝑡) + 4𝑦′(𝑡) + 3𝑦(𝑡) = 𝑒𝑡

6. 𝑓′′(𝑥) − 7𝑓(𝑥) = 0

When you solve differential equation, you are trying to find any function that satisfies the equation.

It is very often the case that the solution to a D.E. is not unique, many solutions contain arbitrary

constants.

For example, the first order D.E. 𝑥𝑑𝑦

𝑑𝑥− 2𝑥2 − 2𝑥 = 0 has a solution y = x2 + 2x + C where C is any

arbitrary constant.

The solution to the differential equation that contains an arbitrary constant is called the General Solution

to the differential equation

Sometimes you are given a D.E. with some extra information called the initial condition (also called the

boundary condition) and when this happens, you are able to find value(s) of the arbitrary constant C.

The solution to the D.E. will no longer contain any constants and this solution is called the Particular

Solution to the Differential Equation.

For example if we are given the D.E. 𝑥𝑑𝑦

𝑑𝑥− 2𝑥2 − 2𝑥 = 0 and we are also told the initial condition

that y(1) = 5 then we will get the Particular Solution to the D.E. of y = x2 + 2x + 2

In order to solve any differential equation we use the DSolve[…..] Command, the syntax for Solving

D.E. is as follows


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DSolve[𝑑𝑖𝑓𝑓𝑒𝑟𝑛𝑡𝑖𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛, 𝑦[𝑥], 𝑥]

Where differential equation = The differential equation with == instead of just one =

Example 46: Find the General Solutions to the following 1st Order differential equations.

(a) 𝑥𝑑𝑦

𝑑𝑥+ 5𝑦 = 𝑥2

(b) 𝑦′(𝑡) + 4𝑦(𝑡) = 0

(c) 𝑓′(𝑥) − 7𝑓(𝑥) = sin (𝑥)

Solution (a): Input: ClearAll[𝑦, 𝑥] DSolve[𝑥𝑦′[𝑥] + 5𝑦[𝑥] == 𝑥2, 𝑦[𝑥], 𝑥]

Output: {{𝑦[𝑥] →𝑥2

7+

𝐶[1]

𝑥5 }}

Note: The General Solution to this D.E. is y = 𝑥2

7+

𝐶

𝑥5

Solution (b): Input: ClearAll[𝑦, 𝑡] DSolve[𝑦′[𝑡] + 4𝑦[𝑡] == 0, 𝑦[𝑡], 𝑡]

Output: {{𝑦[𝑡] → 𝑒−4𝑡𝐶[1]}}

Note: The General Solution to this D.E. is y = 𝐶𝑒−4𝑡

Solution (c): Input: ClearAll[𝑦, 𝑥] DSolve[𝑓′[𝑥] − 7𝑓[𝑥] == Sin[𝑥], 𝑓[𝑥], 𝑥]

Output: {{𝑓[𝑥] → 𝑒7𝑥𝐶[1] +1

50(−Cos[𝑥] − 7Sin[𝑥])}}

Note: The General Solution to this D.E. is y = 𝐶𝑒7𝑥 +1

50(−Cos 𝑥 − 7Sin 𝑥)


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Example 47: Find the General Solutions to the following 2nd Order differential equations.

(a) 𝑑2𝑦

𝑑𝑥2 + 5𝑑𝑦

𝑑𝑥+ 6𝑦 = 2

(b) 𝑦′′(𝑡) + 4𝑦′(𝑡) + 3𝑦(𝑡) = 𝑒𝑡

(c) 𝑓′′(𝑥) − 7𝑓(𝑥) = 0

Solution (a): Input: ClearAll[𝑦, 𝑥] DSolve[𝑦''[𝑥] + 5𝑦′[𝑥] + 6𝑦[𝑥] == 𝑥2, 𝑦[𝑥], 𝑥]

Output: {{𝑦[𝑥] →1

108(19 − 30𝑥 + 18𝑥2) + 𝑒−3𝑥𝐶[1] + 𝑒−2𝑥𝐶[2]}}

Note: The General Solution to this D.E. is y = 1

108(19 − 30𝑥 + 18𝑥2) + 𝐶𝑒−3𝑥 + 𝐷𝑒−2𝑥

Where C and D are arbitrary Constants.

Solution (b): Input: ClearAll[𝑦, 𝑡] DSolve[𝑦′[𝑡] + 4𝑦[𝑡] + 3𝑦[𝑡] == 𝑒𝑡, 𝑦[𝑡], 𝑡]

Output: {{𝑦[𝑡] →𝑒𝑡

8+ 𝑒−3𝑡𝐶[1] + 𝑒−𝑡𝐶[2]}}


8𝑒𝑡 + 𝐶𝑒−3𝑡 + 𝐷𝑒−𝑡


Solution (c): Input: ClearAll[𝑦, 𝑥] DSolve[𝑓''[𝑥] − 7𝑓[𝑥] == 0, 𝑓[𝑥], 𝑥]

Output: {{𝑓[𝑥] → 𝑒√7𝑥𝐶[1] + 𝑒−√7𝑥𝐶[2]}}


8𝑒𝑡 + 𝐶𝑒−3𝑡 + 𝐷𝑒−𝑡


All of these differential equations have General solutions, in general a 1st order D.E. will have one

arbitrary constant typically called C and 2nd order D.E. will have two arbitrary constants typically called

C and D.

When you are given a D.E. with boundary conditions, we can solve the D.E. BY finding suitable

value(s) for the arbitrary constants and so find Particular Solutions to the given D.E.


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In order to solve any differential equation with boundary conditions we use the DSolve[…..] Command,

the syntax for Solving D.E. is as follows

DSolve[{𝑑𝑖𝑓𝑓𝑒𝑟𝑛𝑡𝑖𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛, 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛(𝑠)}, 𝑦[𝑥], 𝑥]

Where differential equation = The differential equation with == instead of just one =

𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛(𝑠) = The given boundary Conditions

Example 48: Find the Particular solutions to the following 1st Order differential equations.

(a) 𝑑𝑦

𝑑𝑥+ 5𝑦 = 2𝑥2 with y(6) = 8

(b) 𝑦′(𝑡) − 𝑦(𝑡) = sin (𝑡) with y(0)=0

(c) 𝑓′(𝑥) − 𝑓(𝑥) = x with y(𝜋) = 1

Solution (a): Input: ClearAll[𝑦, 𝑥] DSolve[{𝑦′[𝑥] + 5𝑦[𝑥] == 2𝑥2, 𝑦[6] == 8}, 𝑦[𝑥], 𝑥]


125𝑒−5𝑥(−342𝑒30 + 2𝑒5𝑥 − 10𝑒5𝑥𝑥 + 25𝑒5𝑥𝑥2)}}

Note: The Particular Solution to this D.E. is

y = 2

125𝑒−5𝑥(−342𝑒30 + 2𝑒5𝑥 − 10𝑒5𝑥𝑥 + 25𝑒5𝑥𝑥2)}}

Solution (b): Input: ClearAll[𝑦, 𝑡] DSolve[{𝑦′[𝑡] − 𝑦[𝑡] == Sin[𝑡], 𝑦[0] == 0}, 𝑦[𝑡], 𝑡]

Output: {{𝑦[𝑡] →1

2(𝑒𝑡 − Cos[𝑡] − Sin[𝑡])}}

Note: The Particular Solution to this D.E. is y = 1

2(𝑒𝑡 − Cos 𝑡 − Sin 𝑡)

Solution (c): Input: ClearAll[𝑦, 𝑥] DSolve[{𝑓′[𝑥] − 𝑓[𝑥] == 𝑥, 𝑓[𝜋] == 1}, 𝑓[𝑥], 𝑥]

Output: {{𝑓[𝑥] → −𝑒−𝜋(𝑒𝜋 − 2𝑒𝑥 − 𝑒𝑥𝜋 + 𝑒𝜋𝑥)}}

Note: The Particular Solution to this D.E. is y =−𝑒−𝜋(𝑒𝜋 − 2𝑒𝑥 − 𝑒𝑥𝜋 + 𝑒𝜋𝑥)


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Example 49: Find the Particular Solutions to the following 2nd Order differential equations.

(a) 𝑑2𝑦

𝑑𝑥2 − 2𝑦 = 𝑥 𝑦′(0) = 2 and y(0) = 1

(b) 𝑦′′(𝑡) + 4𝑦′(𝑡) + 3𝑦(𝑡) = 0 𝑦′(1) = 1 and y(0) = 0

(c) 𝑢′′(𝑥) − 𝑢′(𝑥) = sin 𝑥 𝑢′(𝜋) = 2 and u(𝜋) = 1

Solution (a): Input: ClearAll[𝑦, 𝑥] DSolve[{𝑦''[𝑥] − 2𝑦′[𝑥] == 𝑥, 𝑦′[0] == 2, 𝑦[0] == 1}, 𝑦[𝑥], 𝑥]


8(−1 + 9𝑒2𝑥 − 2𝑥 − 2𝑥2)}}

Note: The Particular Solution to this D.E. is y = 1

8(−1 + 9𝑒2𝑥 − 2𝑥 − 2𝑥2)

Solution (b): Input: ClearAll[𝑦, 𝑡] DSolve[{𝑦''[𝑡] + 4𝑦′[𝑡] + 3𝑦[𝑡] == 0, 𝑦′[1] == 1, 𝑦[0] == 0}, 𝑦[𝑡], 𝑡]

Output: {{𝑦[𝑡] → −𝑒3−3𝑡(−1+𝑒2𝑡)

−3+𝑒2 }}

Note: The Particular Solution to this D.E. is y = −𝑒3−3𝑡(−1+𝑒2𝑡)

−3+𝑒2

Solution (c): Input: ClearAll[𝑢, 𝑥] DSolve[{𝑢''[𝑥] − 𝑢′[𝑥] == Sin[𝑥], 𝑢′[𝜋] == 2, 𝑢[𝜋] == 1}, 𝑢[𝑥], 𝑥]

Output: {{𝑢[𝑥] →1

2𝑒−𝜋(3𝑒𝑥 + 𝑒𝜋Cos[𝑥] − 𝑒𝜋Sin[𝑥])}}

Note: The Particular Solution to this D.E. is u(x) =1

2𝑒−𝜋(3𝑒𝑥 + 𝑒𝜋cos 𝑥 − 𝑒𝜋sin 𝑥)


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16. Direction Fields

A direction field is a way to visualize a first order differential equation by showing visually the value of

the derivative on the Cartesian Grid. The command that draws Direction Fields is called VectorPlot [….]

and has the Syntax

VectorPlot [{1,𝑑𝑦

𝑑𝑥}, {𝑥, 𝑥𝑚𝑖𝑛, 𝑥𝑚𝑎𝑥}, {𝑦, 𝑦𝑚𝑜𝑛, 𝑦𝑚𝑎𝑥}, 𝑂𝑝𝑡𝑖𝑜𝑛𝑠]

Where:

𝑑𝑦

𝑑𝑥 Is the first order D.E. written in the form

𝑑𝑦

𝑑𝑥= 𝑓(𝑥, 𝑦)

{x,xmin,xmax} are the interval range for the x values

{y,ymin,ymax} are the interval range for the y values

Options Are various options that can be added to the VectorPlot such as.

VectorScale → Automatic which controls the size of the arrow head on each

vector it can be a number such as VectorScale → 0.2 or a general value such as

VectorScale → Automatic or VectorScale → Large

Axes → True which draws the x and y axis onto the Direction Field

Example 50: Draw a Direction Field for the first order D.E. 𝑑𝑦

𝑑𝑥=

𝑥+𝑦

𝑥2+𝑦2+1 using values of x in the

interval [-2,2] and values of y in the interval [-2,2]

Input: VectorPlot[

{1,𝑥 + 𝑦

𝑥2 + 𝑦2 + 1},

{𝑥, −2,2}, {𝑦, −2,2}, VectorScale → {0.04, Automatic, None}, VectorStyle → {{Red, Arrowheads[0]}},

Axes → True

]

Note 1: The Option

VectorScale→ {0.04, Automatic, None}

Makes all the lines the same length, 0.04

Note 2: The Option

VectorStyle → {{Red, Arrowheads[0]}}

makes the lines red in color and it removes

the arrowhead at the end of each line.

2 1 0 1 2

2

1

0

1

2


of 90

Example 51: (a) Draw a Direction Field for the first order D.E. 𝑑𝑦

𝑑𝑥=

𝑥+𝑥2

𝑦+1 using values of x in

the interval [−3, 3] and values of y in the interval [-3,3]

(b) Find the Particular Solution to the first order D.E. 𝑑𝑦

𝑑𝑥=

𝑥+𝑥2

𝑦+1 , with y(0)=1

(c) Graph the Direction Field and the Particular Solution on the same diagram.

Input (a): VectorPlot[

{1,𝑥 + 𝑥2

𝑦 + 1},

{𝑥, −3,3}, {𝑦, −3,3}, VectorScale → {0.04, Automatic, None},

VectorStyle → {{Red, Arrowheads[0]}},

Axes → True

]

Note 1: The Option

VectorScale → {0.04, Automatic, None}

Makes all the lines the same length, 0.04

Note 2: The Option

VectorStyle → {{Red, Arrowheads[0]}}

makes the lines red in color and it removes

the arrowhead at the end of each line.

Input (b): ClearAll[𝑦, 𝑥]

DSolve[{𝑦′[𝑥] ==𝑥+𝑥2

𝑦[𝑥]+1, 𝑦[𝜋] == 1}, 𝑦[𝑥], 𝑥]


3(−3 + √3√12 + 3𝑥2 + 2𝑥3)}}

Note 1: The Particular Solution to this D.E is y = 1

3(−3 + √3√12 + 3𝑥2 + 2𝑥3)

Note 2: The output DSolve: For some branches of the general solution, the given boundary

conditions lead to an empty solution. This means that there are some versions of the

General Solution to this D.E. will not exist with the value y(0)=1.

Note 3: There is an interesting feature at the point (−1, −1) the derivative will be

infinite at this point, its like an infinite singularity in the derivative function.

3 2 1 0 1 2 3

3

2

1

0

1

2

3


of 90

Example 51(c) Graph the direction field and the Particular Solution on the same diagram.

Input (c): ClearAll[𝑦, 𝑥]

DSolve[{𝑦′[𝑥] ==𝑥+𝑥2

𝑦[𝑥]+1, 𝑦[0] == 1}, 𝑦[𝑥], 𝑥];

Show[ VectorPlot[

{1,𝑥 + 𝑥2

𝑦 + 1},

{𝑥, −3,2}, {𝑦, −3,2}, VectorScale → {0.04, Automatic, None},

VectorStyle → {{Red, Arrowheads[0]}},

Axes → True], Plot[

1 3⁄ (−3 + √3√(12 + 3𝑥^2 + 2𝑥^3)), {𝑥, −2,2}

], PlotRange → ({−3,2}, {−3,3}}

]

Output:

Note: The Particular Solution y = 1

3(−3 + √3√12 + 3𝑥2 + 2𝑥3) only exists in part of

the diagram this is because the domain of the solution is [– 2.477,∞) and so

the curve should stop at x = – 2.477

3 2 1 0 1 2

3

2

1

0

1

2

3


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17. Solving Differential Equations There are three main types of Differential Equations that we will solve in Calculus II they are

A. Simple Differential Equations

B. Variables Separable Differential Equations

C. Linear Differential Equations

A. Simple Differential Equations

A Simple Differential Equation where you can find the solution by repeated Integration.

The following are all examples of Simple D.E.

1. 𝑑𝑦

𝑑𝑥 = 𝑥3 + 5𝑥 + 7

2. 𝑓′′(𝑥) = 𝑠𝑖𝑛(𝑥) − cos2 (𝑥)

3. 𝑢′′′(𝑡) = 𝑒2𝑡 + ln (t)

To solve a Simple D.E. you integrate until you get to the solution.

In Mathematica we use the DSolve[….] command as seen earlier.

Example 52: Find the General Solutions to the following Simple D.E.

(a) 𝑑𝑦

𝑑𝑥 = 𝑥3 + 5𝑥 + 7

(b) 𝑓′′(𝑥) = 𝑠𝑖𝑛(𝑥) − cos2 (𝑥)

(c) 𝑢′′′(𝑡) = 𝑒2𝑡 + ln (t)

Input (a) DSolve[𝑦′[𝑥] == 𝑥3 + 5𝑥 + 7, 𝑦[𝑥], 𝑥]

Output {{𝑦[𝑥] → 7𝑥 +5𝑥2

2+

𝑥4

4+ 𝐶[1]}}

Solution y = 7𝑥 +5𝑥2

2+

𝑥4

4+ 𝐶

Input (b) DSolve[𝑦''[𝑥] == Sin[𝑥] − Cos[𝑥]^2, 𝑦[𝑥], 𝑥]

Output {{𝑦[𝑥] → −𝑥2

4+ 𝐶[1] + 𝑥𝐶[2] +

Cos[𝑥]2

4− Sin[𝑥]}}

Solution y = −𝑥2

4+ 𝐶 + 𝐷𝑥 +

Cos[𝑥]2

4− Sin 𝑥

Input (c) DSolve[𝑢′[𝑡] == 𝑒2𝑡 + Log[𝑡], 𝑢[𝑡], 𝑡]

Output {{𝑢[𝑡] → 𝐶[1] + 𝑡𝐶[2] + 𝑡2𝐶[3] +1

4(

𝑒2𝑡

2−

11𝑡3

9+

2

3𝑡3Log[𝑡])}}

Solution y = 𝐶 + 𝐷𝑡 + 𝐸𝑡2 +1

4(

𝑒2𝑡

2−

11𝑡3

9+

2

3𝑡3ln 𝑡


of 90

Example 53: Find the Particular Solutions to the following Simple D.E.

(a) 𝑑𝑦

𝑑𝑥 = 𝑥3sin (𝑥) y(0) = 1

(b) 𝑓′′(𝑥) = 2𝑥

𝑥2+1 𝑓′(

𝜋

4) = 1 and 𝑓(0) = 0

(c) 𝑢′′′(𝑡) = 𝑒𝑡𝑠𝑖𝑛𝑡 𝑢′′(0) = 2, 𝑢′(0) = 1 𝑎𝑛𝑑 𝑢(0) = −1

Input (a) DSolve[{𝑦′[𝑥] == 𝑥3Sin[𝑥], 𝑦[0] == 1}, 𝑦[𝑥], 𝑥]

Output {{𝑦[𝑥] → 1 + 6𝑥Cos[𝑥] − 𝑥3Cos[𝑥] − 6Sin[𝑥] + 3𝑥2Sin[𝑥]}}

Solution y = 1 + 6𝑥 cos 𝑥 − 𝑥3 cos 𝑥 − 6 sin 𝑥 + 3𝑥2 sin 𝑥

Input (b) DSolve[{𝑦''[𝑥] ==2𝑥

𝑥2+1, 𝑦′[

𝜋

4] == 1, 𝑦[0] == 0}, 𝑦[𝑥], 𝑥]

Output {{𝑦[𝑥] → −𝑥 + 2ArcTan[𝑥] − 𝑥Log[1 +𝜋2

16] + 𝑥Log[1 + 𝑥2]}}

Solution y = −𝑥 + 2 tan−1𝑥 − 𝑥ln(1 +𝜋2

16} + 𝑥ln(1 + 𝑥2)

Input (c) DSolve[{𝑢'''[𝑡] == 𝑒𝑡Sin[𝑡], 𝑢''[0] == 2, 𝑢′[0] == 1, 𝑢[0] == −1}, 𝑢[𝑡], 𝑡]

Output {{𝑢[𝑡] →1

4(−3 + 6𝑡 + 5𝑡2 − 𝑒𝑡Cos[𝑡] − 𝑒𝑡Sin[𝑡])}}

Solution y = 1

4(−3 + 6𝑡 + 5𝑡2 − 𝑒𝑡cos 𝑡 − 𝑒𝑡 sin 𝑡


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B. Variables Separable Differential Equations

A Variables Separable Differential Equation is one that can be written in the form 𝑑𝑦

𝑑𝑥 =

𝑓(𝑥)

𝑔(𝑦)

The following are all examples of Variables Separable D.E.

1. 𝑑𝑦

𝑑𝑥 =

4𝑥−7

2𝑦+1

2. 𝑑𝑦

𝑑𝑥 =

𝑥 𝑠𝑖𝑛𝑥

2𝑒𝑦

3. 𝑑𝑢

𝑑𝑡 =

tan 𝑡

√𝑢+1



Example 54: Find the General Solution to the following Variables Separable D.E.

(a) 𝑑𝑦

𝑑𝑥 =

4𝑥−7

2𝑦+1 (b)

𝑑𝑦

𝑑𝑥 =


2𝑒𝑦 (c) 𝑑𝑢

𝑑𝑡 =

𝑡2

√𝑢+1

Input (a) DSolve[𝑦′[𝑥] ==4𝑥−7

2𝑦[𝑥]+1, 𝑦[𝑥], 𝑥]

Output {{𝑦[𝑥] →1

2(−1 − √1 − 28𝑥 + 8𝑥2 + 4𝐶[1])}, {𝑦[𝑥] →

1

2(−1 + √1 − 28𝑥 + 8𝑥2 + 4𝐶[1])}}

Solution y = 1

2(−1 ± √1 − 28𝑥 + 8𝑥2 + 4𝐶)

Input (b) DSolve[𝑦''[𝑥] == Sin[𝑥] − Cos[𝑥]^2, 𝑦[𝑥], 𝑥]

Output {{𝑦[𝑥] → Log[𝐶[1] +1

2(−𝑥Cos[𝑥] + Sin[𝑥])]}}

Solution y = ln (𝐶 +1

2(−𝑥 cos 𝑥 + sin 𝑥))

Input (c) DSolve[𝑢′[𝑡] ==𝑡2

√𝑢[𝑡]+1, 𝑢[𝑡], 𝑡]

Output {{𝑢[𝑡] → −1 +(𝑡6+6𝑡3𝐶[1]+9𝐶[1]2)1 3⁄

22 3⁄ }, {𝑢[𝑡] → −1 −(1−𝑖√3)(𝑡6+6𝑡3𝐶[1]+9𝐶[1]2)1 3⁄

222 3⁄ }, {𝑢[𝑡] → −1 −(1+𝑖√3)(𝑡6+6𝑡3𝐶[1]+9𝐶[1]2)1 3⁄

222 3⁄ }}

Solution u = −1 +(𝑡6+6𝑡3𝐶[1]+9𝐶2)1 3⁄

22 3⁄

Note: There is only one Real Function Solution exists for 𝑑𝑢

𝑑𝑡 =

𝑡2

√𝑢+1

u = −1 +(𝑡6+6𝑡3𝐶[1]+9𝐶2)1 3⁄

22 3⁄ the other two are Complex functions as they contain i = √−1


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Example 55: Find the Particular Solution to the following Variables Separable D.E.

(a) 𝑑𝑦

𝑑𝑥 =

4𝑥−7

2𝑦+1 with y(0) = 10

(b) 𝑑𝑦

𝑑𝑥 =


2𝑒𝑦 with y(𝜋) = 0

(b) 𝑑𝑢

𝑑𝑡 =

𝑡2

√𝑢+1 with u(0) = 0

Input (a) DSolve[{𝑦′[𝑥] ==4𝑥−7

2𝑦[𝑥]+1, 𝑦[0] == 0}, 𝑦[𝑥], 𝑥]

Output {{𝑦[𝑥] →1

2(−1 + √1 − 28𝑥 + 8𝑥2)}}

Solution y = 1

2(−1 + √1 − 28𝑥 + 8𝑥2)

Input (b) DSolve[{𝑦′[𝑥] ==𝑥Sin[𝑥]

2𝑒𝑦[𝑥] , 𝑦[𝜋] == 0}, 𝑦[𝑥], 𝑥]

Output {{𝑦[𝑥] → Log[2−𝜋

2+

1

2(−𝑥Cos[𝑥] + Sin[𝑥])]}}

Solution y = ln[2−𝜋

2+

1

2(−𝑥 cos 𝑥 + sin 𝑥)

Input (c) DSolve[{𝑢′[𝑡] ==𝑡2

√𝑢[𝑡]+1, 𝑢[0] == 0}, 𝑢[𝑡], 𝑡]

Output {{𝑢[𝑡] →1

2(−2 + 21 3⁄ ((−2 + 𝑡3)2)1 3⁄ )}, {𝑢[𝑡] →

1

2(−2 + 21 3⁄ ((2 + 𝑡3)2)1 3⁄ )}}

Solution u = 1

2(−2 + 21 3⁄ ((−2 + 𝑡3)2)1 3⁄ )

u = (−2 + 21 3⁄ ((2 + 𝑡3)2)1 3⁄ )

Note: There are two possible Particular solutions to the D.E 𝑑𝑢

𝑑𝑡 =

𝑡2

√𝑢+1 with u(0) = 0


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C. Linear Differential Equations

A linear Differential Equation has the form 𝑑𝑦

𝑑𝑥= 𝑘𝑦 + 𝑐

The following are all examples of Variables Separable D.E.

1. 𝑑𝑦

𝑑𝑥 = 5y + 4

2. 𝑦′(𝑡) = 0.03y + 200

3. 𝑝′(𝑡) = – 0.2p + 50



Example 56: Find the General Solution to the following Linear D.E.

(a) 𝑑𝑦

𝑑𝑥 = 5y + 4

(b) 𝑦′(𝑡) = 0.03y + 200

(c) 𝑝′(𝑡) = – 0.2p + 50

Input (a) DSolve[𝑦′[𝑥] == 5𝑦[𝑥] + 4, 𝑦[𝑥], 𝑥]

Output {{𝑦[𝑥] → −4

5+ 𝑒5𝑥𝐶[1]}}

Solution y = −4

5+ 𝐶𝑒5𝑥

Input (b) DSolve[𝑦′[𝑡] == 0.03𝑦[𝑡] + 4, 𝑦[𝑡], 𝑡]

Output {{𝑦[𝑡] → −133.33333333333334 + 𝑒0.03𝑡𝐶[1]}}

Solution y = −133.333 + 𝐶𝑒0.03𝑡

Input (c) DSolve[𝑝′[𝑡] == −0.2𝑝[𝑡] + 50, 𝑝[𝑡], 𝑡]

Output {{𝑝[𝑡] → 250. + 𝑒−0.2𝑡𝐶[1]}}

Solution u = 250 + 𝐶𝑒−0.2𝑡


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18. Applications of Differential Equations There are many practical applications in using Differential Equations , it is by far, the most adaptable

method for modelling situations in Engineering , Sciences , Business etc…. We will use population

models to show the versatility in using Differential equations.

Example 57: Exponential Growth Population Models

(a) A town has population grows at the rate of 2.1% per year , the initial

population of the town is 5000 , find a formula for the population of the

town in t years.

(b) What will the population of the town be in 12 years?

(c) How long before the population reaches 20,000 people.

Solution (a): The D.E. that models this situation is 𝑝′(𝑡) = 0.021𝑝 with p(0) = 500

The Mathematica code that solves this equation is as follows

Input: DSolve[{𝑝′[𝑡] == 0.021𝑝[𝑡], 𝑝[0] == 500}, 𝑝[𝑡], 𝑡]

Output: {{𝑝[𝑡] → 500. 𝑒0.021𝑡}}

The population of this town in t years will be p(t) = 500𝑒0.021𝑡

Solution (b): We need to evaluate p(12)


Input: 𝑝[t_]: = 500𝑒0.021𝑡

Output: 643.298

The population of this town in 12 years will be p(12) = 643 (approx)

Solution (c): We need to solve the equation p(t) = 20,000


Input: 𝑝[t_]: = 500𝑒0.021𝑡

Output: {{𝑡 → 175.66092638637792}}

The population will reach 20,000 in 175.7 years (approx.)


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Example 58: Exponential Decay Models

(a) A radioactive element decays at the rate of 0.005% per year , if you start

with 40 grams how much will be left in t years.

(b) What is the half-life of this radioactive material?

(c) Graph this information.

Solution (a): The D.E. that models this situation is 𝑎′(𝑡) = −0.00005𝑎 with a(0) = 40

(0.005% = 0.00005 as a decimal , it is negative since the rate is decreasing)

Input: DSolve[{𝑎′[𝑡] == −0.00005𝑎[𝑡], 𝑎[0] == 40}, 𝑎[𝑡], 𝑡]

Output: {{𝑎[𝑡] → 40. 𝑒−0.00005𝑡}}

The amount of radioactive material left after t years will be p(t) = 40. 𝑒−0.00005𝑡

Solution (b): We now need to solve when is a(t) = 20

Since the half-life is the time it takes to end up with half of the radioactive

material you started with, in this case half of 40 g is 20 g.

Input: DSolve[{𝑎′[𝑡] == −0.00005𝑎[𝑡], 𝑎[0] == 40}, 𝑎[𝑡], 𝑡]

Output: {{𝑡 → 13862.943611198907}}

The half-life of this radioactive material is 13,863 years (approx).

Solution (c): Graph this information.

𝑎[t_]: = 40𝑒−0.00005𝑡

ℎ[t_]: = 20

Plot[{𝑎[𝑡], ℎ[𝑡]}, {𝑡, 0,20000}, AxesLabel → {Time(𝑡years), Amount(𝑎)}]

Output:

5000 10000 15000 20000t Time years

20

25

30

35

40

a Amount


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Example 59: Extended Population Models

A country has a population of 300 million and a birth rate of 3% and a death rate

of 2.5%. Every year there are another 2 million people who come to the country

as immigrants.

(a) What will the population of this country be in t years.

(b) What will the population of this country be in 30 years.

(c) How long before this country’s population reaches 350 million?

Solution (a): The D.E. that models this situation is

𝑝′(𝑡) = (𝑏𝑖𝑟𝑡ℎ 𝑟𝑎𝑡𝑒 − 𝑑𝑒𝑎𝑡ℎ 𝑟𝑎𝑡𝑒)𝑝 + 𝑖𝑚𝑚𝑖𝑔𝑎𝑟𝑡𝑖𝑜𝑛

𝑝′(𝑡) = (0.03 − 0.025)𝑝 + 2

𝑝′(𝑡) = 0.005𝑝 + 2

Input: DSolve[{𝑝′[𝑡] == 0.005𝑝[𝑡] + 2, 𝑝[0] == 300}, 𝑝[𝑡], 𝑡]

Output: {{𝑝[𝑡] → 700. (−0.5714285714285714 + 1. 𝑒0.005𝑡)}}

The population of the country is

p(t) = 700. (−0.5714285714285714 + 1. 𝑒0.005𝑡)

p(t) = 700𝑒0.005𝑡 − 400 (approx.)

Solution (b): The population of the country in 30 years will be p(30)

Input: 𝑝[t_]: = 700𝑒0.005𝑡 − 400

𝑝[30]

Output: 413.284

The population of the country in 30 years will be 413 million (approx.)

Solution (c): We need to solve when is p(t) = 350

Input: 𝑝[t_]: = 700𝑒0.005𝑡 − 400

Solve[𝑝[𝑡] == 350, {𝑡}, Reals]

Output: {{𝑡 → 13.79857429739029}}

The population of the country will reach 350 million in 13.8 years (approx.)


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Example 60: Logistic Growth Models

(a) An island has enough food to maintain a population of 2000 deer (this number is called the carrying capacity and is usually denoted by K = 2000).

The deer have a growth rate of 12%

(this is usually denoted by r = 0.12).

At present the island has a population of 250 deer

(this is usually called p0 = 250)

What is the population model p(t) for the number of deer in t years.

(b) How many deer will be on the island in 22 years?

Solution (a): This situation is called a logistic model and its D.E. takes the form

𝑝′(𝑡) = 𝑟𝑝(1 −𝑝

𝑘)

The solution to the logistic model is

p(t) = 𝑝0𝑘

𝑝0+(𝑘−𝑝0)𝑒−𝑟𝑡

The population of deer in t years will be p(t) = 𝑝0𝑘


p(t) = 250(2000)

250+(2000−250)𝑒−0.12𝑡

p(t) = 500 000

250+1750𝑒−0.12𝑡

Solution (b):

Input: 𝑝[t_]: =500000

250+1750𝑒−0.12𝑡

p[22]

Output: 1333.75

In 22 years there will be 1334 deer on island (approx.)


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Example 61: Graphing Logistic Growth Models

An island has enough food to maintain a population of 450 rabbits. (this number is called the carrying capacity and is usually denoted by K = 450).

The rabbits have a growth rate of 20%

(this is usually denoted by r = 0.2).

At present the island has a population of p0

Graph the Direction field for this logistic model along with the following initial

Populations of rabbits.

(a) 20 rabbits , (b) 200 rabbits (c) 700 rabbits

Solution: The D.E. that models this situation is 𝑝′(𝑡) = 𝑟𝑝(1 −𝑝

𝑘)

𝑝′(𝑡) = 0.2𝑝(1 −𝑝

450)

The solution is p(t) = 𝑝0𝑘


p(t) = 𝑝0(450)

𝑝0+(450−𝑝0)𝑒−0.12𝑡

(a) When p0 = 20 rabbits

p(t) = 𝑝0(450)

𝑝0+(450−𝑝0)𝑒−0.12𝑡

= (20)(450)

20+(450−20)𝑒−0.12𝑡

= 9000

20+430𝑒−0.12𝑡

(b) When p0 = 200 rabbits

p(t) = 𝑝0(450)

𝑝0+(450−𝑝0)𝑒−0.12𝑡

= (200)(450)

20+(450−20)𝑒−0.12𝑡

= 90 000

200+250𝑒−0.12𝑡

(c) When p0 = 700 rabbits

p(t) = 𝑝0(450)

𝑝0+(450−𝑝0)𝑒−0.12𝑡

= (700)(450)

700+(450−700)𝑒−0.12𝑡

= 315000

700−250𝑒−0.12𝑡


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Here is the diagram showing the Direction Field and three particular Solutions to the logistic Model.

VectorPlot[

{1,0.2𝑦 (1 −𝑦

450)},

{𝑥, 0,50}, {𝑦, 0,700}, VectorScale → {0.04, Automatic, None}, VectorStyle → {{Red, Arrowheads[0]}}, Axes → True],

Plot [{9000

20 + 430𝑒−0.12𝑥} , {𝑥, 0,50}],

Plot [{90000

200 + 250𝑒−0.12𝑥} , {𝑥, 0,50}],

Plot [{315000

700 − 250𝑒−0.12𝑥} , {𝑥, 0,50}],

PlotRange → {{0,50}, {0,700}}

]

0 10 20 30 40 50

0

100

200

300

400

500

600

700

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