These properties are used to carry out transformations of the integral in order to reduce it to one of the elementary integrals and further calculation.
1. The derivative of the indefinite integral is equal to the integrand:
2. The differential of the indefinite integral is equal to the integrand:
3. The indefinite integral of the differential of a certain function is equal to the sum of this function and an arbitrary constant:
4. The constant factor can be taken out of the integral sign:
Moreover, a ≠ 0
5. The integral of the sum (difference) is equal to the sum (difference) of the integrals:
6. Property is a combination of properties 4 and 5:
Moreover, a ≠ 0 ˄ b ≠ 0
7. Invariance property of the indefinite integral:
If , then
8. Property:
If , then
In fact, this property is special case integration using the variable change method, which is discussed in more detail in the next section.
Let's look at an example:
First we applied property 5, then property 4, then we used the table of antiderivatives and got the result.
The algorithm of our online integral calculator supports all the properties listed above and can easily find detailed solution for your integral.
These properties are used to carry out transformations of the integral in order to reduce it to one of the elementary integrals and further calculation.
1. The derivative of the indefinite integral is equal to the integrand:
2. The differential of the indefinite integral is equal to the integrand:
3. The indefinite integral of the differential of a certain function is equal to the sum of this function and an arbitrary constant:
4. The constant factor can be taken out of the integral sign:
Moreover, a ≠ 0
5. The integral of the sum (difference) is equal to the sum (difference) of the integrals:
6. Property is a combination of properties 4 and 5:
Moreover, a ≠ 0 ˄ b ≠ 0
7. Invariance property of the indefinite integral:
If , then
8. Property:
If , then
In fact, this property is a special case of integration using the variable change method, which is discussed in more detail in the next section.
Let's look at an example:
First we applied property 5, then property 4, then we used the table of antiderivatives and got the result.
The algorithm of our online integral calculator supports all the properties listed above and will easily find a detailed solution for your integral.
This article talks in detail about the main properties of the definite integral. They are proved using the concept of the Riemann and Darboux integral. The calculation of a definite integral takes place thanks to 5 properties. The remaining ones are used to evaluate various expressions.
Before moving on to the main properties of the definite integral, it is necessary to make sure that a does not exceed b.
Basic properties of the definite integral
Definition 1The function y = f (x) defined at x = a is similar to the fair equality ∫ a a f (x) d x = 0.
Evidence 1
From this we see that the value of the integral with coinciding limits is equal to zero. This is a consequence of the Riemann integral, because every integral sum σ for any partition on the interval [ a ; a ] and any choice of points ζ i equals zero, because x i - x i - 1 = 0 , i = 1 , 2 , . . . , n , which means we find that the limit of integral functions is zero.
Definition 2
For a function that is integrable on the interval [a; b ] , the condition ∫ a b f (x) d x = - ∫ b a f (x) d x is satisfied.
Evidence 2
In other words, if you swap the upper and lower limits of integration, the value of the integral will change to the opposite value. This property is taken from the Riemann integral. However, the numbering of the partition of the segment starts from the point x = b.
Definition 3
∫ a b f x ± g (x) d x = ∫ a b f (x) d x ± ∫ a b g (x) d x applies to integrable functions of type y = f (x) and y = g (x) defined on the interval [ a ; b ] .
Evidence 3
Write down the integral sum of the function y = f (x) ± g (x) for partitioning into segments with a given choice of points ζ i: σ = ∑ i = 1 n f ζ i ± g ζ i · x i - x i - 1 = = ∑ i = 1 n f (ζ i) · x i - x i - 1 ± ∑ i = 1 n g ζ i · x i - x i - 1 = σ f ± σ g
where σ f and σ g are the integral sums of the functions y = f (x) and y = g (x) for partitioning the segment. After passing to the limit at λ = m a x i = 1, 2, . . . , n (x i - x i - 1) → 0 we obtain that lim λ → 0 σ = lim λ → 0 σ f ± σ g = lim λ → 0 σ g ± lim λ → 0 σ g .
From Riemann's definition, this expression is equivalent.
Definition 4
Extending the constant factor beyond the sign of the definite integral. Integrated function from the interval [a; b ] with an arbitrary value k has a fair inequality of the form ∫ a b k · f (x) d x = k · ∫ a b f (x) d x .
Proof 4
The proof of the definite integral property is similar to the previous one:
σ = ∑ i = 1 n k · f ζ i · (x i - x i - 1) = = k · ∑ i = 1 n f ζ i · (x i - x i - 1) = k · σ f ⇒ lim λ → 0 σ = lim λ → 0 (k · σ f) = k · lim λ → 0 σ f ⇒ ∫ a b k · f (x) d x = k · ∫ a b f (x) d x
Definition 5
If a function of the form y = f (x) is integrable on an interval x with a ∈ x, b ∈ x, we obtain that ∫ a b f (x) d x = ∫ a c f (x) d x + ∫ c b f (x) d x.
Evidence 5
The property is considered valid for c ∈ a; b, for c ≤ a and c ≥ b. The proof is similar to the previous properties.
Definition 6
When a function can be integrable from the segment [a; b ], then this is feasible for any internal segment c; d ∈ a ; b.
Proof 6
The proof is based on the Darboux property: if points are added to an existing partition of a segment, then the lower Darboux sum will not decrease, and the upper one will not increase.
Definition 7
When a function is integrable on [a; b ] from f (x) ≥ 0 f (x) ≤ 0 for any value x ∈ a ; b , then we get that ∫ a b f (x) d x ≥ 0 ∫ a b f (x) ≤ 0 .
The property can be proven using the definition of the Riemann integral: any integral sum for any choice of points of partition of the segment and points ζ i with the condition that f (x) ≥ 0 f (x) ≤ 0 is non-negative.
Evidence 7
If the functions y = f (x) and y = g (x) are integrable on the interval [ a ; b ], then the following inequalities are considered valid:
∫ a b f (x) d x ≤ ∫ a b g (x) d x , f (x) ≤ g (x) ∀ x ∈ a ; b ∫ a b f (x) d x ≥ ∫ a b g (x) d x , f (x) ≥ g (x) ∀ x ∈ a ; b
Thanks to the statement, we know that integration is permissible. This corollary will be used in the proof of other properties.
Definition 8
For an integrable function y = f (x) from the interval [ a ; b ] we have a fair inequality of the form ∫ a b f (x) d x ≤ ∫ a b f (x) d x .
Proof 8
We have that - f (x) ≤ f (x) ≤ f (x) . From the previous property we found that the inequality can be integrated term by term and it corresponds to an inequality of the form - ∫ a b f (x) d x ≤ ∫ a b f (x) d x ≤ ∫ a b f (x) d x . This double inequality can be written in another form: ∫ a b f (x) d x ≤ ∫ a b f (x) d x .
Definition 9
When the functions y = f (x) and y = g (x) are integrated from the interval [ a ; b ] for g (x) ≥ 0 for any x ∈ a ; b , we obtain an inequality of the form m · ∫ a b g (x) d x ≤ ∫ a b f (x) · g (x) d x ≤ M · ∫ a b g (x) d x , where m = m i n x ∈ a ; b f (x) and M = m a x x ∈ a ; b f (x) .
Evidence 9
The proof is carried out in a similar way. M and m are considered to be the largest and lowest value function y = f (x) defined from the segment [ a ; b ] , then m ≤ f (x) ≤ M . It is necessary to multiply the double inequality by the function y = g (x), which will give the value of the double inequality of the form m g (x) ≤ f (x) g (x) ≤ M g (x). It is necessary to integrate it on the interval [a; b ] , then we get the statement to be proved.
Consequence: For g (x) = 1, the inequality takes the form m · b - a ≤ ∫ a b f (x) d x ≤ M · (b - a) .
First average formula
Definition 10For y = f (x) integrable on the interval [ a ; b ] with m = m i n x ∈ a ; b f (x) and M = m a x x ∈ a ; b f (x) there is a number μ ∈ m; M , which fits ∫ a b f (x) d x = μ · b - a .
Consequence: When the function y = f (x) is continuous from the interval [ a ; b ], then there is a number c ∈ a; b, which satisfies the equality ∫ a b f (x) d x = f (c) b - a.
The first average formula in generalized form
Definition 11When the functions y = f (x) and y = g (x) are integrable from the interval [ a ; b ] with m = m i n x ∈ a ; b f (x) and M = m a x x ∈ a ; b f (x) , and g (x) > 0 for any value x ∈ a ; b. From here we have that there is a number μ ∈ m; M , which satisfies the equality ∫ a b f (x) · g (x) d x = μ · ∫ a b g (x) d x .
Second average formula
Definition 12When the function y = f (x) is integrable from the interval [ a ; b ], and y = g (x) is monotonic, then there is a number that c ∈ a; b , where we obtain a fair equality of the form ∫ a b f (x) · g (x) d x = g (a) · ∫ a c f (x) d x + g (b) · ∫ c b f (x) d x
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Solving integrals is an easy task, but only for a select few. This article is for those who want to learn to understand integrals, but know nothing or almost nothing about them. Integral... Why is it needed? How to calculate it? What are definite and indefinite integrals?
If the only use you know of for an integral is to use a crochet hook shaped like an integral icon to get something useful out of hard-to-reach places, then welcome! Find out how to solve the simplest and other integrals and why you can’t do without it in mathematics.
We study the concept « integral »
Integration was known back in Ancient Egypt. Of course not in modern form, but still. Since then, mathematicians have written many books on this topic. Especially distinguished themselves Newton And Leibniz , but the essence of things has not changed.
How to understand integrals from scratch? No way! To understand this topic you will still need a basic understanding of the basics. mathematical analysis. We already have information about , necessary for understanding integrals, on our blog.
Indefinite integral
Let us have some function f(x) .
Indefinite integral function f(x) this function is called F(x) , whose derivative is equal to the function f(x) .
In other words, an integral is a derivative in reverse or an antiderivative. By the way, read about how in our article.
An antiderivative exists for all continuous functions. Also, a constant sign is often added to the antiderivative, since the derivatives of functions that differ by a constant coincide. The process of finding the integral is called integration.
Simple example:
In order not to constantly calculate antiderivatives elementary functions, it is convenient to summarize them in a table and use ready-made values.
Complete table of integrals for students
Definite integral
When dealing with the concept of an integral, we are dealing with infinitesimal quantities. The integral will help to calculate the area of a figure, the mass of a non-uniform body, the distance traveled during uneven movement, and much more. It should be remembered that an integral is an infinite sum large quantity infinitesimal terms.
As an example, imagine a graph of some function.
How to find the area of a figure bounded by the graph of a function? Using an integral! Let us divide the curvilinear trapezoid, limited by the coordinate axes and the graph of the function, into infinitesimal segments. This way the figure will be divided into thin columns. The sum of the areas of the columns will be the area of the trapezoid. But remember that such a calculation will give an approximate result. However, the smaller and narrower the segments, the more accurate the calculation will be. If we reduce them to such an extent that the length tends to zero, then the sum of the areas of the segments will tend to the area of the figure. This is a definite integral, which is written like this:
Points a and b are called limits of integration.
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Rules for calculating integrals for dummies
Properties of the indefinite integral
How to solve an indefinite integral? Here we will look at the properties of the indefinite integral, which will be useful when solving examples.
- The derivative of the integral is equal to the integrand:
- The constant can be taken out from under the integral sign:
- The integral of the sum is equal to the sum of the integrals. This is also true for the difference:
Properties of a definite integral
- Linearity:
- The sign of the integral changes if the limits of integration are swapped:
- At any points a, b And With:
We have already found out that a definite integral is the limit of a sum. But how to get a specific value when solving an example? For this there is the Newton-Leibniz formula:
Examples of solving integrals
Below we will consider the indefinite integral and examples with solutions. We suggest you figure out the intricacies of the solution yourself, and if something is unclear, ask questions in the comments.
To reinforce the material, watch a video about how integrals are solved in practice. Don't despair if the integral is not given right away. Contact a professional student service, and any triple or curvilinear integral on a closed surface you will be able to do it.
Let the function y = f(x) is defined on the interval [ a, b ], a < b. Let's perform the following operations:
1) let's split [ a, b] dots a = x 0 < x 1 < ... < x i- 1 < x i < ... < x n = b on n partial segments [ x 0 , x 1 ], [x 1 , x 2 ], ..., [x i- 1 , x i ], ..., [x n- 1 , x n ];
2) in each of the partial segments [ x i- 1 , x i ], i = 1, 2, ... n, choose an arbitrary point and calculate the value of the function at this point: f(z i ) ;
3) find the works f(z i ) · Δ x i , where is the length of the partial segment [ x i- 1 , x i ], i = 1, 2, ... n;
4) let's make up integral sum functions y = f(x) on the segment [ a, b ]:
WITH geometric point From a visual perspective, this sum σ is the sum of the areas of rectangles whose bases are partial segments [ x 0 , x 1 ], [x 1 , x 2 ], ..., [x i- 1 , x i ], ..., [x n- 1 , x n ], and the heights are equal f(z 1 ) , f(z 2 ), ..., f(z n) accordingly (Fig. 1). Let us denote by λ length of the longest partial segment:
5) find the limit of the integral sum when λ → 0.
Definition. If there is a finite limit of the integral sum (1) and it does not depend on the method of partitioning the segment [ a, b] to partial segments, nor from the selection of points z i in them, then this limit is called definite integral from function y = f(x) on the segment [ a, b] and is denoted
Thus,
In this case the function f(x) is called integrable on [ a, b]. Numbers a And b are called the lower and upper limits of integration, respectively, f(x) – integrand function, f(x ) dx– integrand expression, x– integration variable; line segment [ a, b] is called the integration interval.
Theorem 1. If the function y = f(x) is continuous on the interval [ a, b], then it is integrable on this interval.
The definite integral with the same limits of integration is equal to zero:
If a > b, then, by definition, we assume
2. Geometric meaning of the definite integral
Let on the segment [ a, b] a continuous non-negative function is specified y = f(x ) . Curvilinear trapezoid is a figure bounded above by the graph of a function y = f(x), from below - along the Ox axis, to the left and right - straight lines x = a And x = b(Fig. 2).
Definite integral of non-negative function y = f(x) from a geometric point of view is equal to the area of a curvilinear trapezoid bounded above by the graph of the function y = f(x) , left and right – line segments x = a And x = b, from below - a segment of the Ox axis.
3. Basic properties of the definite integral
1. The value of the definite integral does not depend on the designation of the integration variable:
2. The constant factor can be taken out of the sign of the definite integral:
3. The definite integral of the algebraic sum of two functions is equal to the algebraic sum of the definite integrals of these functions:
4.If function y = f(x) is integrable on [ a, b] And a < b < c, That
5. (mean value theorem). If the function y = f(x) is continuous on the interval [ a, b], then on this segment there is a point such that
4. Newton–Leibniz formula
Theorem 2. If the function y = f(x) is continuous on the interval [ a, b] And F(x) is any of its antiderivatives on this segment, then the following formula is valid:
which is called Newton–Leibniz formula. Difference F(b) - F(a) is usually written as follows:
where the symbol is called a double wildcard.
Thus, formula (2) can be written as:
Example 1. Calculate integral
Solution. For the integrand f(x ) = x 2 an arbitrary antiderivative has the form
Since any antiderivative can be used in the Newton-Leibniz formula, to calculate the integral we take the antiderivative that has the simplest form:
5. Change of variable in a definite integral
Theorem 3. Let the function y = f(x) is continuous on the interval [ a, b]. If:
1) function x = φ ( t) and its derivative φ "( t) are continuous for ;
2) a set of function values x = φ ( t) for is the segment [ a, b ];
3) φ ( a) = a, φ ( b) = b, then the formula is valid
which is called formula for changing a variable in a definite integral .
Unlike the indefinite integral, in this case not necessary to return to the original integration variable - it is enough just to find new limits of integration α and β (for this you need to solve for the variable t equations φ ( t) = a and φ ( t) = b).
Instead of substitution x = φ ( t) you can use substitution t = g(x) . In this case, finding new limits of integration over a variable t simplifies: α = g(a) , β = g(b) .
Example 2. Calculate integral
Solution. Let's introduce a new variable using the formula. By squaring both sides of the equality, we get 1 + x = t 2 , where x = t 2 - 1, dx = (t 2 - 1)"dt= 2tdt. We find new limits of integration. To do this, let’s substitute the old limits into the formula x = 3 and x = 8. We get: , from where t= 2 and α = 2; , where t= 3 and β = 3. So,
Example 3. Calculate
Solution. Let u= log x, Then , v = x. According to formula (4)