 ИНТЕГРАЛЫ НЕКОТОРЫХ СПЕЦИАЛЬНЫХ ФУНКЦИЙ - Студенческий научный форум # VI Международная студенческая научная конференция Студенческий научный форум - 2014

## ИНТЕГРАЛЫ НЕКОТОРЫХ СПЕЦИАЛЬНЫХ ФУНКЦИЙ

Дайынова А.О. 1, Куттыкожаева Ж.К. 1, Куспан З.А. 1, Куспан З.А. 1
1Актюбинский региональный государственный университет им. К. Жубанова
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1.1. Integral of absolute value function.

For integrating absolute value functions firstly we will find the positive and negative parts of given function. For the intervals where f(x) change its sign we will use property of definite integral and by using this method we will change negative parts to positive.

Example 1. Let us consider this integral:

Solution. To find the value of integral, we will use the definition of absolute value of function. Then:

Then we divide the interval [-3, 2] to positive and negative parts. So:

Therefore, the solution of given integral is equal to .

Example 2. Calculate the value of following integral:

Solution.At first we define zeroes of function under integrals: x – 2 = 0; x = 2 is the root of х – 2.

So

Now the interval of [0, 3] consider as the sum of two intervals: [0, 2] and [2, 3]. So we will calculate the given integral as the sum of two integrals. Then

.

Example 3. Calculate the integral: .

Solution: To calculate this integral, first we define the value of function under integral at intervals [0, ]. So

If we use the value specified at the top, during the solution of the integral, then we get the integral as the sum of two integrals. So, the result will be in the following form:

1.2. Integral of integer value function.

[f(x)] is the greatest integer value which is not greater than f(x). If f(x)  Z, [f(x)] is not continuous for the points of discontinuity (i.e. for critical points) there is not integral. For this reason we will divide the interval into subintervals by using discontinuity points.

Example 1. See the following integral:

=?

Solution. In order to integrate this function, we divide the interval [-1, 3] into subintervals:

if - 1 ≤ x < 0 then ([x - 1]) = - 2;

if 0 ≤ x < 1 then ([x - 1]) = - 1;

if 1 ≤ x < 2 then ([x - 1]) = 0;

if 2 ≤ x < 3 then ([x - 1]) = 1.

Now, using the obtained values we calculate this integral. Then

Example 2. Let us consider this integral:

.

Solution.In this case, is only one interval. If, 1 ≤ x < 2, then . Now put the value 3 in the integral and integrate. Then

Example 3. Let us consider this integral:

Solution. To calculate this integral, from the beginning divide the interval [-1, 3] on the subintervals:

if then

if then

if then

if then .

Using the values of functions, obtained in subintervals, we compute the integral, then

.

1.3. Integral of signum function.

F(x) = sgn g(x) =

In these type of questions we will divide the interval into two parts as negative and positive intervals. Because f(x) is not continuous at the point where g(x) = 0.

Example 1.

Solution. x + 1 = 0 => x = - 1. At first, we should find critical points. Now, we define value of interval of function under integral:

sgn(x + 1) =

Values which are given above are we use to integrate the function. Then

Example2. Given this integral: .

Solution. As in the example, at first we will find critical points: Then, the value of sgn(- 4) we will find as in the below:

Keeping in mind given values we compute the integral. From this

Example3. We will define the value of integral.

Solution. We compute this integral dividing into steps. The first step is to find the critical points x – 3 = 0 => x = 3. The second step is to use getting values in integrating given function:

Then

The list of the literatures:

1. V. V. Konev “The elements of mathematics”, Text Book, Изд. Томского политехнического университета, 2009.

2. Ahmet Çakir “Integral”, Zambak publishing, 2005.

3. J. C. Burkill “A first course in mathematical analysis”, Cambridge University Press, 1962.

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