Integral Calculus

The definite integrals have a pre-existing value of limits, thus making the final value of an integral, definite. if f(x) is a function of the curve, then \(\int\limits_a^b f(x) dx = f(b) – f(a)\)

Müəyyən integral

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Chapter 5 : Integrals

In this chapter we will be looking at integrals. Integrals are the third and final major topic that will be covered in this class. As with derivatives this chapter will be devoted almost exclusively to finding and computing integrals. Applications will be given in the following chapter. There are really two types of integrals that we’ll be looking at in this chapter : Indefinite Integrals and Definite Integrals. The first half of this chapter is devoted to indefinite integrals and the last half is devoted to definite integrals. As we will see in the last half of the chapter if we don’t know indefinite integrals we will not be able to do definite integrals.

Here is a quick listing of the material that is in this chapter.

Indefinite Integrals – In this section we will start off the chapter with the definition and properties of indefinite integrals. We will not be computing many indefinite integrals in this section. This section is devoted to simply defining what an indefinite integral is and to give many of the properties of the indefinite integral. Actually computing indefinite integrals will start in the next section.

Computing Indefinite Integrals – In this section we will compute some indefinite integrals. The integrals in this section will tend to be those that do not require a lot of manipulation of the function we are integrating in order to actually compute the integral. As we will see starting in the next section many integrals do require some manipulation of the function before we can actually do the integral. We will also take a quick look at an application of indefinite integrals.

Substitution Rule for Indefinite Integrals – In this section we will start using one of the more common and useful integration techniques – The Substitution Rule. With the substitution rule we will be able integrate a wider variety of functions. The integrals in this section will all require some manipulation of the function prior to integrating unlike most of the integrals from the previous section where all we really needed were the basic integration formulas.

More Substitution Rule – In this section we will continue to look at the substitution rule. The problems in this section will tend to be a little more involved than those in the previous section.

Area Problem – In this section we start off with the motivation for definite integrals and give one of the interpretations of definite integrals. We will be approximating the amount of area that lies between a function and the \(x\)-axis. As we will see in the next section this problem will lead us to the definition of the definite integral and will be one of the main interpretations of the definite integral that we’ll be looking at in this material.

Definition of the Definite Integral – In this section we will formally define the definite integral, give many of its properties and discuss a couple of interpretations of the definite integral. We will also look at the first part of the Fundamental Theorem of Calculus which shows the very close relationship between derivatives and integrals

Computing Definite Integrals – In this section we will take a look at the second part of the Fundamental Theorem of Calculus. This will show us how we compute definite integrals without using (the often very unpleasant) definition. The examples in this section can all be done with a basic knowledge of indefinite integrals and will not require the use of the substitution rule. Included in the examples in this section are computing definite integrals of piecewise and absolute value functions.

Substitution Rule for Definite Integrals – In this section we will revisit the substitution rule as it applies to definite integrals. The only real requirements to being able to do the examples in this section are being able to do the substitution rule for indefinite integrals and understanding how to compute definite integrals in general.

Integral Calculus

Integral calculus helps in finding the anti-derivatives of a function. These anti-derivatives are also called the integrals of the function. The process of finding the anti-derivative of a function is called integration. The inverse process of finding derivatives is finding the integrals. The integral of a function represents a family of curves. Finding both derivatives and integrals form the fundamental calculus. In this topic, we will cover the basics of integrals and evaluating integrals.

1.	What is Integral Calculus?
2.	Fundamental Theorems of Integrals
3.	Types of Integrals
4.	Properties of Integrals
5.	Integrals Formulas
6.	Methods of Integrals
7.	Applications of Integrals
8.	FAQs on Integrals

What is Integral Calculus?

Integrals are the values of the function found by the process of integration. The process of getting f(x) from f'(x) is called integration. Integrals assign numbers to functions in a way that describe displacement and motion problems, area and volume problems, and so on that arise by combining all the small data. Given the derivative f’ of the function f, we can determine the function f. Here, the function f is called antiderivative or integral of f’.

Example: Given: f(x) = x 2 .

Derivative of f(x) = f'(x) = 2x = g(x)

if g(x) = 2x, then anti-derivative of g(x) = ∫ g(x) = x 2

Definition of Integral

F(x) is called an antiderivative or Newton-Leibnitz integral or primitive of a function f(x) on an interval I. F'(x) = f(x), for every value of x in I.

Integral is the representation of the area of a region under a curve. We approximate the actual value of an integral by drawing rectangles. A definite integral of a function can be represented as the area of the region bounded by its graph of the given function between two points in the line. The area of a region is found by breaking it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summed up. We specify an integral of a function over an interval on which the integral is defined.

Fundamental Theorems of Integral Calculus

We define integrals as the function of the area bounded by the curve y = f(x), a ≤ x ≤ b, the x-axis, and the ordinates x = a and x =b, where b>a. Let x be a given point in [a,b]. Then \(\int\limits_a^b f(x) dx\) represents the area function. This concept of area function leads to the fundamental theorems of integral calculus.

• First Fundamental Theorem of Integral Calculus
• Second Fundamental Theorem of Integral Calculus

First Fundamental Theorem of Integrals

A(x) = \(\int\limits_a^b f(x) dx\) for all x ≥ a, where the function is continuous on [a,b]. Then A'(x) = f(x) for all x ϵ [a,b]

Second Fundamental Theorem of Integrals

If f is continuous function of x defined on the closed interval [a,b] and F be another function such that d/dx F(x) = f(x) for all x in the domain of f, then \(\int\limits_a^b f(x) dx\) = f(b) -f(a). This is known as the definite integral of f over the range [a,b], a being the lower limit and b the upper limit.

Types of Integrals

Integral calculus is used for solving the problems of the following types.

a) the problem of finding a function if its derivative is given.

b) the problem of finding the area bounded by the graph of a function under given conditions. Thus the Integral calculus is divided into two types.

• Definite Integrals (the value of the integrals are definite)
• Indefinite Integrals (the value of the integral is indefinite with an arbitrary constant, C)

Indefinite Integrals

These are the integrals that do not have a pre-existing value of limits; thus making the final value of integral indefinite. ∫g'(x)dx = g(x) + c. Indefinite integrals belong to the family of parallel curves.

Definite Integrals

The definite integrals have a pre-existing value of limits, thus making the final value of an integral, definite. if f(x) is a function of the curve, then \(\int\limits_a^b f(x) dx = f(b) – f(a)\)

Properties of Integral Calculus

Let us study the properties of indefinite integrals to work on them.

• The derivative of an integral is the integrand itself. ∫ f(x) dx = f(x) +C
• Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent. ∫ [ f(x) dx -g(x) dx] =0
• The integral of the sum or difference of a finite number of functions is equal to the sum or difference of the integrals of the individual functions. ∫ [ f(x) dx+g(x) dx] = ∫ f(x) dx + ∫ g(x) dx
• The constant is taken outside the integral sign. ∫ k f(x) dx = k ∫ f(x) dx, where k ∈ R.
• The previous two properties are combined to get the form: ∫ [k\(_1\)f\(_1\)(x) + k\(_2\)f\(_2\)(x) +. k\(_n\)f\(_n\)(x)] dx = k\(_1\)∫ f\(_1\)(x)dx + k\(_2\)∫ f\(_2\)(x)dx+ . k\(_n\) ∫ f\(_n\)(x)dx

Integrals Formulas

We can remember the formulas of derivatives of some important functions. Here are the corresponding integrals of these functions that are remembered as standard formulas of integrals.

• ∫ x n dx=x n+1 /n+1+C, where n ≠ -1
• ∫ dx =x+C
• ∫ cosxdx = sinx+C
• ∫ sinx dx = -cosx+C
• ∫ sec 2 x dx = tanx+C
• ∫ cosec 2 x dx = -cotx+C
• ∫ sec 2 x dx = tanx+C
• ∫ secx tanxdx = secx+C
• ∫ cscx cotx dx = -cscx+C
• ∫1/(√(1-x 2 ))= sin -1 x + C
• ∫-1/(√(1-x 2 ))= cos -1 x + C
• ∫1/(1+x 2 )= tan -1 x + C
• ∫-1/(1+x 2 )= cot -1 x + C
• ∫1/(x√(x 2 -1))= sec -1 x + C
• ∫-1/(x√(x 2 -1))= cosec -1 x + C
• ∫ e x dx=e x + C
• ∫dx/x=ln|x| + C
• ∫ a x dx=a x /ln a + C

Methods to Find Integrals

There are several methods adopted for finding the indefinite integrals. The prominent methods are:

• Finding integrals by integration by substitution method
• Finding integrals by integration by parts
• Finding integrals by integration by partial fractions.

Finding Integrals by Substitution Method

A few integrals are found by the substitution method. If u is a function of x, then u’ = du/dx.

∫ f(u)u’ dx = ∫ f(u)du, where u = g(x).

Finding Integrals by Integration by Parts

If two functions are of the product form, integrals are found by the method of integration by parts.

∫f(x)g(x) dx = f(x)∫ g(x) dx – ∫ (f'(x) ∫g(x) dx) dx.

Finding Integrals by Integration by Partial Fractions

Integration of rational algebraic functions whose numerator and denominator contain positive integral powers of x with constant coefficients is done by resolving them into partial fractions.

To find ∫ f(x)/g(x) dx, decompose this improper rational function to a proper rational function and then integrate.

∫f(x)/g(x) dx = ∫ p(x)/q(x) + ∫ r(x)/s(x), where g(x) = a(x) . s(x)

Applications of Integral Calculus

Using integration, we can find the distance given the velocity. Definite integrals form the powerful tool to find the area under simple curves, the area bounded by a curve and a line, the area between two curves, the volume of the solids. The displacement and motion problems also find their applications of integrals. The area of the region enclosed between two curves y = f(x) and y = g(x) and the lines x =a, x =b is given by

Area = \(\int\limits_a^b (f(x) -g(x))dx\)

Let us find the area bounded by the curve y = x and y = x 2 that intersect at (0,0)and (1,1).

The given curves are that of a line and a parabola. The area bounded by the curves = \(\int\limits_0^1 (y_2 -y_1)dx\)

Area = \(\int\limits_0^1 (x-x^2)dx\)

Important Notes

• The primitive value of the function found by the process of integration is called an integral.
• An integral is a mathematical object that can be interpreted as an area or a generalization of area.
• When a polynomial function is integrated the degree of the integral increases by 1.

☛ Also Check:

• Integration of uv formula
• Definite integral formula

Integral Calculus Examples

Example 1. Find the integral of e 3x Solution: ∫ d/dx(f(x)) = ∫ d/dx( e 3x ) We know this is of the form of integral, ∫ d/dx( e ax ) = 1/a e ax + C ∫ d/dx( e 3x ) = 1/3 e 3x + C Answer: The integral of e 3x = 1/3 e 3x + C

Example 2. Find the integral of cos 3x. Solution: ∫ d/dx(f(x)) =∫ cos 3x Let 3x = t thus x = t/3 dx = dt/3 The given integral becomes ∫1/3(cos t) dt = 1/3(sin t) + C = 1/3 sin (3x) + C Answer: The integral of cos 3x = 1/3 sin (3x) + C

Example 3. Evaluate the integral i = \(\int\limits_2^3\) (x+1) dx Solution: By the 2nd theorem of fundamentals of integrals we know that \(\int\limits_a^b F(x) dx = f(b) – f(a)\) \(\int\limits_2^3\) (x+1) dx = f(3) -f(2) f(x) = x 2 /2 + x + C f(3) = 3 2 /2 +3 = 9/2 + 3 = 15/2 f(2)= 2 2 /2 + 2 = 4/2 + 2 = 4 f(3) -f(2) = 15/2 – 4 = 7/2 Answer: The value of the given integral I = 7/2

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Онлайн калькулятор. Решение интегралов.

Онлайн-калькулятор позволяет решать математические выражения любой сложности с выводом подробного результата решения по шагам.

Решить интеграл онлайн по фото ▶
Интегралы онлайн ▶
Онлайн интеграл ▶
Калькулятор невизначених інтегралів ▶
Решить интегральное уравнение онлайн ▶

Также универсальный калькулятор умеет вычислять определенные и неопределенные интегралы.

Онлайн калькулятор интегралов

\int _1^\left(\frac>\right)dx
Выражение скопировано в буфер обмена.
Ввод степени. Для завершения ввода выражения в степени используйте кнопку → .

Ввод дроби. Для перемещения указателя между числителем и знаменателем дроби используйте кнопки ← и → .

$$\textbf <Вычисление определенного интеграла:>\newline \int f(x) dx = 2$$

Пояснения к калькулятору

1. Для решения математического выражения необходимо набрать его в поле ввода с помощью предложенной виртуальной клавиатуры и нажать кнопку ↵ .
2. Управлять курсором можно кликами в нужное местоположение в поле ввода или с помощью клавиш со стрелками ← и → .
3. ⌫ – удалить в поле ввода символ слева от курсора.
4. C – очистить поле ввода.
5. При использовании скобок ( ) в выражении в целях упрощения может производится автоматическое закрытие, ранее открытых скобок.
6. Для того чтобы ввести смешанное число или дробь необходимо нажать кнопку ½ , ввести сначала значение числителя, затем нажать кнопку со стрелкой вправо → и внести значение знаменателя дроби. Для ввода целой части смешанного числа необходимо установить курсор перед дробью с помощью клавиши ← и ввести число.
7. Ввод числа в n-ой степени и квадратного корня прозводится кнопками a b и √ соответственно. Завершить ввод значения в степени или в корне можно клавишей → .

Решение интегралов

Онлайн калькулятор предоставляет инструменты для интегрирования функций. Вычисления производятся как с неопределенными, так и с определенными интегралами. Ввод интегралов в поле калькулятора осуществляется вызовом групповой кнопки f(x) и далее:
∫ f(x) – для неопределенного интеграла;
b _a∫ f(x) – для определенного интеграла.

В определенном интеграле кроме самой функции необходимо задать нижний и верхний пределы.

Примеры вычислений интегралов: