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!

Consider the following function:

$\frac{3x+4}{2x-1}$

What is the domain of the function?

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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

Does the given function have a domain? If so, what is it?

$\frac{9x}{4}$

Video Solution

Step-by-Step Solution

Since the function's denominator equals 4, the domain of the function is all real numbers. This means that any one of the x values could be compatible.

Answer

No, the entire domain

Exercise #2

Given the following function:

$\frac{5}{x}$

Does the function have a domain? If so, what is it?

Video Solution

Step-by-Step Solution

Since the unknown is in the denominator, we should remember that the denominator cannot be equal to 0.

In other words, $x\ne0$

The domain of the function is all those values that, when substituted into the function, will make the function legal and defined.

The domain in this case will be all real numbers that are not equal to 0.

Answer

Yes, $x\ne0$

Exercise #3

Given the following function:

$\frac{65}{(2x-2)^2}$

Does the function have a domain? If so, what is it?

Video Solution

Step-by-Step Solution

The denominator of the function cannot be equal to 0.

Therefore, we will set the denominator equal to 0 and solve for the domain:

$(2x-2)^2\ne0$

$2x\ne2$

$x\ne1$

In other words, the domain of the function is all numbers except 1.

Answer

Yes, $x\ne1$

Exercise #4

Look at the following function:

$\frac{5+4x}{2+x^2}$

Does the function have a domain? If so, what is it?

Video Solution

Step-by-Step Solution

Since the denominator is positive for all $x$ , the domain of the function is the entire domain.

That is, all values of $x$ . Therefore, there is no domain limits.

Answer

No, the entire domain

Exercise #5

Find the area of domain (no need to solve)

$9(4x-\frac{5}{x})=20(3x-\frac{6}{x+1})$

Video Solution

Step-by-Step Solution

The domain of the equation is the set of domain values (of the variable in the equation) for which all algebraic expressions in the equation are well defined,

From this, of course - we exclude numbers for which arithmetic operations are not defined,

In the expression on the left side of the given equation:

$9(4x-\frac{5}{x})=20(3x-\frac{6}{x+1})$

there is multiplication between fractions whose denominators contain algebraic expressions that include the variable of the equation (which we are looking for when solving the equation),

Of course, these fractions are defined as long as the expressions in their denominators are not equal to zero (since division by zero is not possible),

Therefore, the domain of definition of the variable in the equation will be obtained from the requirement that these expressions (in the denominators of the fractions) do not equal zero, meaning:

For the fraction in parentheses in the expression on the left side we get:

$\boxed{ x\neq0}$

For the fraction in parentheses in the expression on the right side we get:

$x+1\neq0 \\$ Let's solve the second inequality above (in the same way as solving an equation):

$x+1\neq0 \\ \boxed{x\neq-1}$

Therefore, the correct answer is answer A.

Note:

It should be noted that the above inequality is a point inequality and not a trend inequality (meaning it negates equality: $(\neq)$ and does not require a trend: $(<,>,\leq,\geq)$ ) which is solved exactly like solving an equation, this is unlike solving a trend inequality where different solution rules apply depending on the type of expressions in the inequality, for example: solving a first-degree inequality with one variable (which has only first-degree and lower algebraic expressions), is solved almost identically to solving an equation, however any division or multiplication of both sides by a negative number requires reversing the trend.