Indefinite integral

An integral can be defined for all values (that is, for all $X$ ). An example of this type of function is the polynomial - which we will study in the coming years.

However, there are integrals that are not defined for all values (all $X$ ), since if we place certain $X$ or a certain range of values of $X$ we will receive an expression considered "invalid" in mathematics. The values of $X$ for which integration is undefined cause the discontinuity of a function.

Detailed explanation

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Test yourself on domain of a function!

$\frac{6}{x+5}=1$

What is the field of application of the equation?

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An example of this is a function with a fraction with values $X$ in the denominator.
For example $1\over x$
According to mathematical rules, the denominator of a fraction cannot be zero since it is not possible to divide by zero. Therefore, when there is a possibility that the denominator equals zero, the integral cannot be defined for the values of $X$ that could cause the denominator to be zero.

Another example is a square root function. For example
According to the algebraic rules, the expression under the square root cannot be negative, that is, it must be positive or zero, but in no way negative. Therefore, The integral will be undefined for a range of values of $X$ that cause the expression under the square root to be negative. $f(x)=\sqrt{x^2-x-5}$

Example of a negative square root function

If you are interested in this article, you might also be interested in the following articles:

Graphical representation of a function

Algebraic representation of a function

Function notation

Domain of a function

Numeric value assignment in a function

Variation of a function

Increasing function

Decreasing function

Constant function

Functions for seventh grade

Intervals of increase and decrease of a function

On the Tutorela website, you will find a variety of articles with interesting explanations about mathematics

Examples and exercises with solutions of indefinite integral

Exercise #1

$\frac{6}{x+5}=1$

What is the field of application of the equation?

Video Solution

Step-by-Step Solution

To solve this problem, we will determine the domain, or field of application, of the equation $\frac{6}{x+5} = 1$ .

Step-by-step solution:

Step 1: Identify the denominator. In the given equation, the denominator is $x+5$ .
Step 2: Determine when the denominator is zero. Solve for $x$ by setting $x+5 = 0$ .
Step 3: Solve the equation: $x+5 = 0$ gives $x = -5$ .
Step 4: Exclude this value from the domain. The domain is all real numbers except $x = -5$ .

Therefore, the field of application of the equation is all real numbers except where $x = -5$ .

Thus, the domain is $x \neq -5$ .

Answer

$x\operatorname{\ne}-5$

Exercise #2

$\frac{x+y:3}{2x+6}=4$

What is the field of application of the equation?

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps to find the domain:

Step 1: Recognize that the expression $\frac{x+y:3}{2x+6}=4$ involves a fraction. The denominator $2x + 6$ must not be zero, as division by zero is undefined.
Step 2: Set the denominator equal to zero and solve for $x$ to find the values that must be excluded: $2x + 6 = 0$ .
Step 3: Solve $2x + 6 = 0$ :

$2x + 6 = 0$
$2x = -6$
$x = -3$

Step 4: Conclude that the domain of the function excludes $x = -3$ , meaning $x \neq -3$ .

Thus, the domain of the given expression is all real numbers except $x = -3$ . This translates to:

$x\operatorname{\ne}-3$

Answer

$x\operatorname{\ne}-3$

Exercise #3

$\frac{3x:4}{y+6}=6$

What is the field of application of the equation?

Video Solution

Step-by-Step Solution

To determine the field of application of the equation $\frac{3x:4}{y+6}=6$ , we must identify values of $y$ for which the equation is defined.

The denominator of the given expression is $y + 6$ . In order for the expression to be defined, the denominator cannot be zero.
This leads us to solve the equation $y + 6 = 0$ .
Solving $y + 6 = 0$ gives us $y = -6$ .
This means $y = -6$ would make the denominator zero, thus the expression would be undefined for this value.

Therefore, the field of application, or the domain of the equation, is all real numbers except $y = -6$ .

We must conclude that $y \neq -6$ .

Comparing with the provided choices, the correct answer is choice 3: $y \neq -6$ .

Answer

$y\operatorname{\ne}-6$

Exercise #4

$22(\frac{2}{x}-1)=30$

What is the domain of the equation above?

Video Solution

Step-by-Step Solution

To find the domain of the given function $22\left(\frac{2}{x} - 1\right) = 30$ , follow these steps:

Identify critical terms: The term $\frac{2}{x}$ is undefined when $x = 0$ because division by zero is undefined.
We need to exclude $x = 0$ from the domain to ensure the function remains defined.
The correct domain for the equation is all real numbers except $x = 0$ .

Thus, the domain of the equation is $x \neq 0$ .

Therefore, the solution to the problem is $x \neq 0$ .

Answer

x≠0

Exercise #5

$2x+\frac{6}{x}=18$

What is the domain of the above equation?

Video Solution

Step-by-Step Solution

To solve this problem and find the domain for the expression $2x + \frac{6}{x}$ , we apply the following steps:

Step 1: Identify when the fraction $\frac{6}{x}$ is undefined. This occurs when the denominator $x$ equals zero.
Step 2: To find the restriction, set the denominator equal to zero: $x = 0$ .
Step 3: Solve for $x$ to find the values excluded from the domain. Here, $x \neq 0$ .

Since $\frac{6}{x}$ is undefined for $x = 0$ , the value $x = 0$ must be excluded from the domain.
Hence, the domain of the equation is all real numbers except zero.