**Applied Analysis is a one-level course that covers the fundamentals of various topics, such as functions, derivatives, and integrals.**

It is also known as "*baby bill*' and deals with various topics that are also**Part of a math course.**. In this topic, we will discuss applied calculus, its similarities and differences with calculus, and related examples.

This topic should not be seen as a book on applied calculus, as we will only discuss it.**specific topics together with some examples of applied analysis**S. In addition, we will study the basic concepts of functions, derivatives, and integrals in the context of applied analysis.

## What is applied analysis?

Applied calculus, also known as "children's calculus or business calculus," is a**Introductory course covering the basics of various topics.**as functions, derivatives and integrals.

It does not include trigonometry or advanced algebra, which are covered in Calculus I and II. High School Algebra may be considered a prerequisite for Applied Calculus.

## Analysis applied vs. analysis

The main difference between Calculation Applied and Calculation is Calculation Applied**covers the basics of functions, derivatives, and integrals, but skips advanced topics**related to derivatives and integration that fall under Calculus. Applied calculus is simple and does not involve the high-level calculus that scientists and engineers study.

Students who choose to study calculus are mostly**Engineering or science students.**, and they study calculus in two parts; Calculus - I and Calculus - II Both courses are completed in two semesters or one year. On the other hand, applied analysis is studied mainly by economics and business administration students, since its subject does not involve complex analysis.

The general content of the course Applied Analysis, Precalculus, Analysis - I and Analysis - II is shown below.

### invoice applied

ES**does not contain trigonometry topics**. It has the fewest theorems compared to other calculus topics and does not contain a discussion of complex algebraic functions.

*Key topics in applied analysis include:*

- functions
- griffins
- Derivatives Applications
- easy integration
- Simple multivariate calculation

### Vorcálculo

As the name suggests, Pre-Calculus is just that.**Prerequisite for Applied Analysis, Analysis -I and Analysis -II**. Precalculus deals only with functions, and topics related to precalculus will be reviewed before the applied calculus course begins. Both precalculus and applied calculus therefore involve a discussion of procedures.

*The main topics of the preliminary calculation are:*

- linear functions
- inverse functions
- operations on functions
- Complex numbers and roots.
- polynomial functions

### Calculus - I

The focus of Calculus is on**Frontiers, continuous functions, differentiation and applications**in connection with differentiations such as mean value theorems, von Rolle's theorem, extreme value theorem, etc.

*The main themes of Analysis-I are:*

- griffins
- Limits and applications of derivatives
- partial differentiation
- Integration
- integration applications

### Calculus – II

Calculus-II is an advanced form of Calculus-I and contains topics specific to**Curriculum for engineering and science students.**. Calculus-II is used to study changes or continuous motion represented in terms of functions.

*The main topics of Analysis-II include:*

- Differential equations and their applications.
- complex functions
- serie binomial
- Sequences, series and geometric functions
- Analytic geometry

Fundamental subject-specific differences in course content contained in Applied Analysis and Analysis are shown in the table below. The table can be used as**a direct comparison of the course outline**between applied calculation and calculation.

**subjects****invoice applied****calculation****Advanced or Analytical Geometry**Not includedIncluded**Trigonometry**Not includedIncluded**functions**Linear, quadratic, and polynomial functions are included. Sometimes basic level logarithmic and exponential functions are also included. Polynomial, linear, logarithmic, exponential, and integral functions are included.**griffins**Simple algebraic derivations, chain rule and applied optimization included**extended differential equations**Not includedIncluded**Integration**Basic integration, antiderivatives and calculation of area and volume by integrationAlgebraic integration, advanced integration by substitution method**Limits and continuous functions**Basic and numerical graphicsAdvanced graphics, numerical and algebraic functions.

### History of infinitesimal calculus

The modern infinitesimal calculus was developed by none other than**Sir Isaac Newton and Gottfried Leibniz**. These scientists studied the continuous movements of the planets and moons, hence the name "*Calculation of the infinitesimal*" was coined. Calculus means studying continuous changes using mathematics.

Since the development of calculus in the 17th century, many other scientists have contributed to calculus and it has evolved. Many new methods, theorems, and hypotheses have been presented, and now it is calculus.**applied in physics, biology, economics and engineering**.

The beauty of calculus is that it is easy to understand and presents some simple basic ideas that we can apply to many everyday scenarios. If we use calculus for**simple real life problems, it becomes applied calculus**.

## Who should study applied analysis?

We have discussed the similarities and differences between applied analysis and calculus, so now a question arises:*Who should study applied analysis?*Applied analytics has its uses, and even if"*baby bill*," Hay**The importance of studying this course cannot be denied.**.

He**List of schools/colleges**where applied calculation is preferred over calculation is given below:

- medical preschoolers
- pharmaceutical schools
- business and management schools
- Non-Research Graduate Programs
- Applications of applied analysis

The next question that comes to the mind of the students is: “*Is the applied calculus difficult?*' The answer to that question is this**it is simpler and easier compared to calculus -I and II**. The applications of applied calculus differ significantly from those of calculus. Engineers and scientists use calculus to solve advanced geometric problems, find volumes and distances of complex functions, derive theorems, and solve advanced multivariate infinitesimal problems.

On the contrary, the calculus applied is primary.**used by economic and business personnel**to determine maximum or minimum profits, find or calculate elasticity of demand, and calculate income streams and breakeven points in cash flows using basic calculus.

## Topics of applied analysis

We have discussed applied calculus in detail and how it differs from calculus; let's study now**some of the course content**applied analysis and its numerical examples.

### function

The function is defined in calculus as**the relationship between two variables**where one variable is dependent and the other is independent. The value of the dependent variable varies according to the value of the independent variable.*For example*, the functional equation is represented as follows when “x” is the independent variable and “y” is the dependent variable:

$y = f(x)$

In general we can say that**The output of the function depends on the input.**. For example, we want to make a hamburger. If we just add lettuce, tomatoes, pickles, and olives, we'll get a veggie burger, but if we want to make a zinger burger, we'll have to add chicken. As you can see, the entered ingredients define the type of burger.

Thus, the type of hamburger is a dependent variable while the ingredients are the independent variables. He**Mapping from inputs to outputs**is called function.

### lineal funtion

A linear function is widely used in the field of economics. It is popular in business because it is easy to use and the graphics are easy to understand. Variables in linear functions will be without the exponents; This means that**All variables have the power "1".**

*The following equations are examples of a linear function:*

- $y = 3x$
- $y = 3x +2$
- $y = 6x -2$

### nonlinear functions

A nonlinear function is also a**Relationship between dependent and independent variables**, but unlike a linear function, it does not form a straight line. Quadratic functions, cubic functions, exponential functions, and logarithmic functions are examples of nonlinear functions. The following equations are examples of a nonlinear function.

- $y = 3x^{2}$
- $y = e^{2x}$
- $y = \dfrac{1}{x^{3}}$
- $y = ln(3x)$

### scope of a function

The domain of a function is defined as**the set of all possible inputs of the function**. It can also be defined as all possible values of the independent variable.

We'll see*An example*— for the function $y = \dfrac{1}{x}$, the value of "$y$" is infinite or undefined if $x = 0$. Other than that, it will have some value. Thus, the domain of the function includes all values of "$x$", that is, h all real numbers except $x = 0$.

### scope of a function

The range of a function is defined as t**The set of all possible outputs of a function.**. It can also be defined as all possible values of the dependent variable. If we take the same numerical example $y = \dfrac{1}{x}$, then the range of values of the function is also any non-zero value. The following graph shows the values of "$x$" and "$y$", and from the curve it can be seen that "$y$" can have any value except "$0$".

### Open interval of a function

The open interval can be defined as an interval that**contains all points within the specified boundary except the two endpoints**, and is denoted by ( ). For example, if the function $y = 3x +2$ is defined for the interval $(2, 4)$, the value of $x$ includes all points greater than $2$ and less than $4$.

### Closed interval of a function

The closed interval can be defined as an inclusive interval**all points within the given limit, and is represented by [ ]**. For example, if the function y = 3x +2 is defined for the interval $[2, 4]$, the value of "x" contains all values greater than or equal to $2$ and less than or equal to $4$ .

### Example 1:

From the data given below, determine the value of $f(3)$ for the function $y = f(x)$

X$1$$2$$3$$4$$5$Y$2$$4$$6$$8$$10$

### Solution:

From the table we can clearly see that $f(3) = 6$.

### Example 2:

Express the equation $6x – 3y = 12$ as a function $y = f(x)$.

### Solution:

$6x – 3y = $12

3 $ (2x-y) = 12 $

$ 2x – y = \dfrac{12}{3}$

$ 2x – y = 4 $

$y = f(x) = 2x – 4$

### Example 3:

Solve the function $f(x) = 6x +12$, in $x = 3$

### Solution:

$f(x) = 6x +12$

$f(3) = 6 (3) +12$

$f(3) = 18 + 12 = 30$

### Example 4:

Solve the function $f(x) = 6x^{2} +14$, in $x = 2$

### Solution:

$f(x) = 6x^{2} + 14$

$f(2) = 6 (2)^{2} + 14$

$f(2) = 6 (4) + 14$

$f(2) = 24 + 14 = 38$

### Example 5:

Find out the domain and scope of the following features.

- $f(x) = 2x + 4$
- $f(x) = \sqrt{x+4}$
- $f(x) = \dfrac{6}{4x – 8}$

### Solution:

1) For the function $f(x) = 2x + 4$,**no restrictions**. The variable "$x$" can take any value and the result will always be a real number, so the domain of the function is $(-\infty, \infty)$.

The scope of the function is also unrestricted, since for any value of "$x$" the function can take any real value, i.e.**the range of functions is too**$(-\infty, \infty)$.

2) It is an irrational function, and**We cannot take or solve for the square root of a negative number**. Therefore, the value of "x" must be greater than or equal to $-4$, so the domain of the function is given as $[-4, \infty)$. We start the domain with a closed parenthesis and end it with an open parenthesis, so $x$ can be any value greater than $-4$ and less than infinity.

We need to look at the minimum and maximum possible output of the function to determine the range. The function can reach values from "$0$" to infinity for the specified domain. With that,**the range of functions is**$[0, \infty)$.

3) The function will consist of real values, except $x = 2$, which will be indeterminate. Therefore, the domain of the function $( – \infty , 2) U (2, \infty)$. So, for this domain, the output of the function will never be zero.**The scope of the function will be**$(-\infty, 0) U (0, \infty)$.

### inverse function

Heinverse of a functionIt's fundamental**the reciprocal of the original function**. If the original function is $y = f(x)$ then its inverse is $x = f(y)$. The inverse function is denoted as $f^{-1}$.

We have studied most of the basic concepts about functions along with numerical examples. Now let's look at a real world example related to functions.

### Example 6:

Steve has a library of $400 books at home. He buys books for $10 a month and adds them to his collection. He must write the formula for the total number of books (in the form of function $y = f(x)$). Is the function for the number of books linear or nonlinear? He must also determine the total number of books at the end of $2$ years.

### Solution:

In this example, we have a constant value of $400 books already in the library. Steve adds $10 books monthly, so those $10 books represent the exchange rate and "$ x $" is the number of months.

*So we can write the equation as:*

$y = 400 + 10 (x)$

We can see that from the above equation**is a linear function**. We need to determine the total number of books at the end of $2$ years.

$x = 2$ years $= 24$ months.

$y = 400 + 10 (24) = 400 + 240 = $640 books

### Example 7:

Let's modify the previous example. Let's say Steve is pretty picky when it comes to buying books and he has the money to buy books for $0 to $10 a month. His library already contains $400 books. He writes the number of books "$ and $" at the end of the year in equation form and determines the domain and range of the function.

### Solution:

*We can write the function as:*

$y = 400 +12x$

Here $12$ is the number of months in a year.

The value of "$x$" can range from $0$ to $10$, so the domain of the function is $[0,10]$.**The scope of the function will be**$[400, 520]$.

### derived

In mathematics, especially differential calculus, the derivative is defined as**the rate of change of a function for a given variable**. The derivative of a function $f(x)$ is denoted by $f'(x)$.

We can easily explain the idea of a derivative using a slope as an example. If we draw a straight line in the $x-y$ plane, changing the value of "$y$" as the value of "x" changes gives the slope.

La pendiente del punto A al B se da como m $= \dfrac{y_2\hspace{1mm}-\hspace{1mm}y_1}{x_2\hspace{1mm}-\hspace{1mm}x_1}$

So, if we take into account the definition of slope,*then we can define the derivative as:*

1. The derivative is the slope of the tangent of the function $y = f(x)$ at a given point $(x,y)$ or $(x,f(x))$.

2. The derivative can also be defined as the slope of the curve of the function $y = f(x)$ at the point $(x,y)$ or $(x, f(x))$.

### limits and continuity

The limit of a function is used when the variable is used in the function**has no specific value**; instead, it is close to a certain value. Suppose that the function $f(x)$ is defined for an open interval near the number "$c$". So, when "x" approaches "$c$", the value of the function is, say, "$L$". Then the symbolic representation of this function is given as:

$\lim_{x\a\c} f(x)=L$

The above equation tells us that as "$x$" gets closer to "$c$", $f(x)$ gets closer and closer to $L$.

**right edge**:

For the right limit applies**we will write**$\lim_{x\a\c^{+}} f(x) = M$. This means that the value of the function $f(x)$ tends to "$M$" when "x" tends to "$c$" to the right, that is to say h the value of "$x$" will always be very close to "$c$", but will always be greater than "$c$".

**left edge**:

The left limit exists when the value of the function is**determined by approaching the variable from the left**. It is written as $\lim_{x \to \c^{-}} f(x) = L$, so the value of $f(x)$ is close to $L$ when "$x$" is " $c$" from the left, ie "$x$" is close to but smaller than "$c$".

#### continuity of a function:

A function is said to be continuous at $x = c$ if*meets the following three conditions:*

1. The value $f (c)$ is defined.

2. $\lim_{x \to \c} f(x)$ also exist, also $\lim_{x \to \c^{-}}f(x) = \lim_{x \to \c^ { +}}f(x)$

3. $\lim_{x\to\c} f(x) = f(c)$

### Example 8:

Determine if $\lim_{x \to \ 3} f(x)$ exists for a given function:

$f(x) = \begin{casos}& 3x+2 \quad 0<x<3 \\ & 14-x \quad 3<x<6 \end{casos}$

### Solution:

The left limit of the function is written as:

$\lim_{x \a\3^{-}} f(x) = \lim_{x \a\3^{-}} (3x+2)$

$\lim_{x\to\3^{-}} (3x+2) = {3(3) + 2} = 11$

$\lim_{x\a\3^{+}} f(x) = \lim_{x\a\3^{-}} (14-x)$

$\lim_{x\to\3^{-}} (14-x) = 14 – 3 = 11$

Also gives $\lim_{x \a\3^{-}}f(x) = \lim_{x \a\3^{+}} f(x)$

Given $\lim_{x \a\3} f(x)$**it exists and it's the same**$11$

### Example 8:

Discuss whether the function $f(x) = 4x^{2} + 6x -7$ at $x = 2$ is continuous or not.

### Solution:

$\lim_{x\a\2} f(x) = \lim_{x\a\2} (4x^{2} + 6x -7)$

$\lim_{x\a\2} (4x^{2} + 6x -7) = 4(2)^{2}+ 6(2) -7) = 16 +12 -7 = 21$

$f(2) = (4x^{2} + 6x -7) = 4(2)^{2}+ 6(2) -7) = 21$

$\lim_{x\to\2} f(x) = f(2)$

With that,**the function is continuous at**$ x = 2 $.

### Example 9:

Discuss whether the given function $f(x)$ is continuous at $x = 2$ or not.

$f(x) = \begin{casos}& 3x-4 \quad x<2 \\ & 10-x \quad 2 \leq x \end{casos}$

### Solution:

The left limit of the function is written as:

$\lim_{x\a\2^{-}} f(x) = \lim_{x\a\2^{-}} (3x-4)$

$\lim_{x\to\2^{-}} (3x-4) = {3(2) – 4} = 2$

$\lim_{x\a\2^{+}} f(x) = \lim_{x\a\2^{+}} (10-x)$

$\lim_{x\a\2^{+}} (10-x) = 10 – 2 = 8$

Since $\lim_{x\to\2^{-}}f(x)\neq\lim_{x\to\2^{+}}f(x)$, condition II is not satisfied and the function f (X)**is not continuous in**$ x = 2 $.

**differentiation of a function**

In calculus, the derivative of a continuous real-valued function is defined as**the change in the function with respect to the change in the independent variable**. As you noticed, we use the word continuous in the definition because the function can only be differentiated if it is continuous. The derivative of a function is denoted as $f'(x)$ and*its formula is:*

$\dfrac{d}{dx}f(x) = \dfrac{df}{dx}= \dfrac{dy}{dx}$

The algebraic representation of the derivative of a function with respect to the limit*can be specified as:*

$f'(x) = \lim_{c \a\0} \dfrac{f(x+c)-f(x)}{c}$

**prove**:

Consider a**continuously**(real – appreciated)**function**„$f$“**in an interval**$(x, x_1)$. The average rate of change for this function for the given points*It can be written as:*

Rate of change $= \dfrac{f(x_1)-f(x)}{x_1 – x}$

If the variable "$x_1$" is close to "$x$", we can say that "$x_1$" is getting closer to "$x$".

*So we can write:*

$\lim_{x \a\x_1} \dfrac{f(x_1)-f(x)}{x_1 – x}$

We have assumed that the function is continuous, so this limit will exist since it is one of the conditions for a function to be continuous. If the limit exists**we can write this function as**$f'(x)$

If $x_1- x = c$, since "$x_1$" is close to "$x$", the value of "$c$" should be zero and*we can write:*

$\lim_{c \a\0} \dfrac{f(x+c)-f(x)}{c}$

So if this limit exists, let's say your instantaneous rate of change from "$x$" to "$x$" and is**denoted by**$f’(x)$.

**Steps to find the derivative**:

Given a continuous real-valued function "$f$", then $f' (x)$ can be determined by*according to the given steps:*

1. Find $f(x+h)$.

2. Solve for $f(x+h) – f(x)$.

3. Divide the equation from step 2 by "h".

4. Solve for $\lim_{h \to \0} \dfrac{f(x+h)-f(x)}{h}$.

### Example 10:

Find the derivative of the function $y = x^{3}- 3x + 6$ at $x = 3$ using the limit method.

### Solution:

$= (x+h)^{3}-3(x+h) +6$

$= {(x+h)^{3}-3(x+h) +6} – (x^{3}- 3x + 6)$

$= [(x+h)^{3}- x^{3} ] – [3 {(x+h) – x} ] + [6 – 6]$

$= [(x+h) – x ] [(x+h)^{2}+ x^{2} + (x+h) x] -3h$

Divide both sides by "h" and set the border to h*Approaching zero:*

$f'(x) = \lim_{h \to \0} \dfrac{[(x+h) – x ] [(x+h)^{2}+ x^{2} + (x+h) x] -3h {h}$

$f'(x) = \lim_{h \to \0}\dfrac{h [(x + h)^{2}+ (x + h) x + x^{2}] -3h }{h} ps

$f'(x) = \lim_{h \to \ 0}\dfrac{h ([(x + h)^{2}+ (x + h) x + x^{2}] – 3) }{ h}$

$f'(x) = \lim_{h \to \0}{ ([(x + h)^{2}+ (x + h) x + x^{2}] – 3) }$

$f'(x) = (x)^{2}+ (x). (x) + x^{2} – 3$

$f'(x) = 3x^{2} – 3$

$f'(3) = 3 (3) ^{2} – 3 = 27 – 3 = 24$

**Differential function rules**

There are different types of functions, and we can find the derivative of each function by**with different differential rules**. With the limit method we can do that.*Define the following rules for the differential of a function:*

1. Differentiation of a constant function

2. Derivation of a power function, also called a power rule

3. Derivation of a product function (product rule)

4. Differentiation of the exponential function

5. Differentiation of addition and subtraction functions

6. Derivation of a quotient function (quotient rule)

Let's see some examples.

### Example 11:

Calculate the derivative of the constant function $f(c) = 6$.

### Solution:

The derivative of a constant function is always zero.

$f'(c) = \dfrac{dy}{dx} 6 = 0$

### Example 12:

Find the derivative of the function $f(x) = 4x ^{\dfrac{3}{4}}$.

### Solution:

$f(x) = 4x ^{\dfrac{3}{4}}$.

Derivation with respect to the variable "$x$"

$f'(x) = 4 \times (\dfrac{3}{4}) x ^{(\dfrac{3}{4})-1}$ ( Potenzregel)

$f'(x) = 3 x ^{\dfrac{3}{4}-1}$

$f'(x) = \dfrac{3}{x}$

### Example 13:

Let's take the same function from example 10 again and check the answer with different derivation rules.

### Solution:

$f(x) = x^{3}- 3x + 6$

We will use**the combination of addition, subtraction and power rule**of derivatives to solve this function.

Derivation on both sides with respect to "$x$":

$f'(x) = 3x^{2} – 3 + 0$

We need to calculate the value of $f'(x)$ at $x = 3$.

$f'(3) = 3(3)^{2} – 3$

$f'(3) = 27 – 3 = 4$

The limits and continuity of the function are used to define derivatives, and then we set up some rules to quickly solve problems related to the derivative of functions. now consider**some real examples of derivatives**.

### Example 15:

The function or formula for the height of an object is $d(t) = -8t^{2}+ 36 t +30$, where t is the time in seconds and d is the distance in meters. Suppose the object is thrown 30 meters above the ground with a velocity of $50 \dfrac{m}{sec}$. What is the maximum height of the object?

### Solution:

Velocity is defined as the rate of change of an object's position over time. So if an entity travels a distance from one point to another in terms of time, and if we take the derivative of this function,**it will give us speed**.

So if we take the derivative of $d(t) = -8t^{2}+ 36 t +30$, we get the velocity.

$v = d'(t) = -16t + 36$

The velocity of an object at its highest point is**is equal to zero**.

$v = d'(t) = -16t + 36 = 0$

$-16t +36 = 0$

$t = \dfrac{9}{4} = 2,25$ sek

So the highest point or distance traveled above the ground*through the object will be:*

$d(2,25) = -8(2,25)^{2}+ 36 (2,25) +30 = -40,5 + 81 + 30 = 70,5$ Metro

### Example 16:

Suppose a company $XYZ$ makes soap. The demand for your product can be given as a function $f(x) = 400 - 5x - 5 x^{2}$, where "$x$" is the price of the product. What will be the marginal revenue of the product if it has a price of $5?

### Solution:

The marginal revenue of the product is calculated by**take the derivative of the rotation function**.

Product turnover is equal to the product of price and quantity. If $f(r)$ is the revenue function,*then it is written as:*

$f(r) = f(x) . x$

$f(r) = [400 – 5x – 5x^{2}]. x$

$f(r) = 400x -5x^{2} – 5x^{3}$

$f'(r) = 400 – 10x – 5x^{2}$

$f'(r) = 400 – 10 (5) – 5 (5)^{2}$

$f'(r) = 400 – 50 – 125 = 225 $

That is, if the price of the product is set at $5,**then income increases**$225$.

### Example 17:

Allan is studying mathematics and recently accepted a position with the national health system. Allan is tasked with estimating the growth of the coronavirus in one of the largest cities in the country. The virus growth rate function is $g(x) = 0.1e^{\dfrac{x}{2}}+ x^{2}$, where $x$ is in days. Allan needs to calculate the growth rate from the first week to the end of the second week.

### Solution:

Allan needs to calculate the growth rate at the end of the first week and then at the end of the second week. After that,**taking the ratio of both growth rates**, Allan will be able to tell how fast the virus is growing.

$g (x) = 0,1e^{\dfrac{x}{2}}+ x^{2}$

$g'(x) = \dfrac{0.1}{2} e^{\dfrac{x}{2}} + 2x$

$g'(7) = 0,05 y ^{\dfrac{7}{2}} + 2 (7) = 15,66$

$g'(14) = 0,05 y ^{\dfrac{14}{2}} + 2 (14) = 82,83$

$\dfrac{ g'(14)}{ g'(7)} = 5$ aprox.

So the growth rate of the coronavirus will be $5**times greater at the end**$14$**carry**(second week) vs end of $7$ days (first week).

### integral calculus

Integral calculus takes some getting used to**Study integrals and related properties.**. The integral calculus combines smaller parts of a function and then combines them as a whole.

*How do we find the area under the curve? Given the derivative of a function, can we determine the original function? How can we add infinitely small functions?*The integral calculus provides the answers to all these questions, so we can say that the integral calculus is one**is used to find the antiderivative of**$f’(x)$.

We find the area under the curve for each function.

**Integration**

Integration is defined as**the antiderivative of a function**. If the derivative was used to break a complicated function into smaller parts, then integration is the inverse of the derivative, in that it combines the smaller elements into a whole. Its main use is to find the area under the curve.

*There are two types of integration:*

1. Definite integrals

2. Indefinite integrals

**definite integrals**

The definite integral is the type of integration that**follows a certain limit or limits during the integration calculation**. Certain integrals establish the upper and lower limits of the independent variable of the function.

$\int_{a}^{b}f(x).dx = F(b) – F(a)$

**indefinite integrals**

The indefinite integral is defined as the type of integration that**does not use upper and lower bounds**. This integration results in a constant value that is added to the antiderivative, and*is presented as follows:*

$\int f(x).dx = F(x) + c$

**important integral formulas**

This section deals with important integral formulas.**for definite and indefinite integrals**used in applied analysis. Since the applied analysis does not involve trigonometry, we will not include trigonometry formulas.

1. $\int x^{n}.dx = \dfrac{x^{n+1}}{n+1} + c$

2. $\int (ax+b)^{n}.dx = \dfrac{(ax+b)^{n+1}}{a(n+1)} + c$

3. $\int 1. dx = x + c$

4. $\int e^{x}. dx = e^{x} + c$

5. $\int b^{x}.dx = (\dfrac{b^{x}}{log b})$

6. $\int_{a}^{b}f'(x).dx = f(b) – f(a)$

7. $\int_{a}^{b}f(x).dx = – \int_{a}^{b}f(x).dx $

8. $\int_{-a}^{a}f(x).dx = 2 \int_{0}^{a}f(x).dx$, provided that the function is even

9. $\int_{-a}^{a}f(x).dx = 0$, provided the function is odd

### Example 18:

Evaluate the following integral functions:

- $\int (x^{2} – 3x + 6) dx$
- $\int (\dfrac{x}{x+4}) dx$ , $(x >4)$
- $\int (6x^{5} – 14\sqrt{x} + 18) dx$

### Solution:

1.

$\int (x^{2} – 3x + 6) dx$ = $\int x^{2}.dx – \int 3x.dx + \int 6.dx$

$= \dfrac{x^{3}}{3} – 3 \dfrac{x^{2}}{2} + 6x + c $

2.$\int (\dfrac{x}{x+4}) dx$ = $\int (\dfrac{x+ 4 – 4}{x+4}) dx$

= $\int 1 – \dfrac{4}{x+4} dx$

= $\int 1.dx – 4 \int (x+4)^{-1}.dx$

= $x – 4 ln (x+4) + c$

3.

$\int (6x^{5} – 14\sqrt{x} + 18) dx$

$= \int 6x^{5}.dx -\int 14 \sqrt{x}.dx + \int 18.dx$

$= \int 6x^{5}.dx -\int 14 x^{\dfrac{1}{2}}.dx + \int 18.dx$

$= 6 \dfrac{x^{6}}{6} – 14 x^{\dfrac{3}{2}} + 18x + c$

### Example 19:

Evaluate the following integral functions:

- $\int_{1}^{4}(3+x). dx$
- $\int_{-1}^{4}x^{4} +3x^{2}. dx$

### Solution:

1.

$\int_{1}^{4}(3+x). dx$

= $\int_{1}^{4}3.dx + \int_{1}^{4}x.dx$

= $[3x] _ {1}{4} + [ \dfrac{x^{2}}{2}] _ {1}{4}]$

= $[ 3(4) – 3(1) ] + [ \dfrac{4^{2}}{2} -\dfrac{1^{2}}{2} ]$

= $(12 – 3) + [(\dfrac{16}{2}) – \dfrac{1}{2}]$

= $9 + (8 – \dfrac {1}{2} )$

= $9 – \dfrac{15}{2} = \dfrac{3}{2}$

2.

$\int_{-1}^{4}x^{4} +3x^{2}. dx$

= $\int_{-1}^{4}x^{4}.dx + \int_{-1}^{4} 3x^{2}.dx$

= $[\dfrac{x^{5}}{5}] _ {-1}{4} + 3 [ \dfrac{x^{3}}{3}] _ {-1}{4}]$

= $[ \dfrac{4^{5}}{5}- \dfrac{(-1)^{5}}{5}] + 3 [ \dfrac{4^{3}}{3} -\dfrac {(-1)^{3}}{3} ]$

= $[\dfrac{1024}{5} + \dfrac{1}{5}] + 3 [ \dfrac{64}{3} + \dfrac{1}{3} ]$

= 205 $ + 65 = 270 $

### Example 20:

Find the value of the highlighted area below the graph for the function $y = x +1$.

### Solution:

The blue area below the chart has the lower limit of "$1$" and the upper limit of "$4$". The integral function of the graph.*It can be written as:*

$\int_{1}^{4} (x+1).dx$

Bereich $= \int_{1}^{4} x. dx + \int_{1}^{4} 1.dx$

= $[\dfrac{x^{2}}{2}] _{1}^{4} + [x] _ {1}^{4}$

= $[ \dfrac{16}{2}- \dfrac{1}{2}] + (4-1)$

= $(8- \dfrac{1}{2}) + 3$

= $\dfrac{15}{2} + 3$

= $\dfrac{21}{2}$ square units

### Example 21:

Mason studies the rate of decline of bacterial infection in patients. The infection is decreasing at a rate of $-\dfrac{12}{(t + 3)^{2}}$ per day. On the third day of her treatment, the percentage of infection in the patients was 3 (ie 300%). What is the percentage of infection on day 15?

### Solution:

Let "y" be the percentage of infection and the variable "t" the number of days.

The infection rate of change is given as $\dfrac{dy}{dt} = -\dfrac{6}{(t + 3)^{2}}$.

$\int dy = -12 \int (t+3)^{-2} dt$

$y = 12 (t+3)^{-1}+c$

$y = \dfrac{12}{t+3} + c$

On the third day we know $t = $3 and $y = $3

$3 = \dfrac{12}{3+3} + c$

3 $ = 2 + c$

$c = 1 $

so now we can**Calculate the percentage of infection on day 1**.

$y = \dfrac{12}{15 + 3} + 1$

$y = \dfrac{12}{18} + 1$

$y = \dfrac{2}{3} + 1 = 0,6 + 1$ = $1,6$ o $160\%$

He**Infection rate reduced by**$140 \%$ .

*practice questions:*

1. Suppose Simon throws a ball upward with an initial velocity of $40 \dfrac{m}{s}$ while standing on the ground. Considering gravity, find the data given below:

- The time it takes for the ball to hit the ground
- The maximum height of the ball.

2. The number of coronavirus patients in the city $XYZ$ for the year $2019$ was 3000$$; the number of patients is expected to double in 4 and a half years. Write the function y for the number of patients in $t$ years. After building the function, you also need to find:

- The total number of patients in $4$ years (after function formation)
- The time it would take to reach $60,000 in patients

*answer key*

1.

- $8$ themselves. California.
- $81,6$ Metro

2.

The function can be written as $y = 3,000. 2^{\dfrac{t}{4}}$

- $6,000 patients
- $17.14$ years approx.