Understanding Indefinite Integrals in Calculus

There are two important processes in calculus: differentiation and integration. Differentiation refers to the process of finding the derivative of a function, whereas integration is the opposite process of differentiation. Indefinite integral meaning is that when a function f is given, you find a function F in a way that F’ = f. Finding indefinite integrals is an important process when it comes to calculus. It is used as a method for obtaining the area under a curve and for obtaining many physical and electrical equations which scientists and engineers use in their day to day lives. In this article, you would learn about the indefinite integral definition, indefinite integral formulas, and indefinite integral problems.

Indefinite Integral Definition

Indefinite integral refers to an integral that does not have any upper and lower limit.

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Here, f(x) is integrated and is represented by:

\[\int\] f(x) dx = F(x) + C

This is the indefinite integral notation.

Here, with respect to x, the integral of f(x) is given on the R.H.S.

F(x) is termed as antiderivative or primitive

f(x) is called as the integrand

dx is the integrating agent

C is the arbitrary constant of integration

x is called the variable of integration.

Indefinite Integral Formula

For a function f(x), the set of all the antiderivatives are called as the indefinite integrals of the function f(x) that is denoted as:

\[\int\] f(X) dx = F(x) + C

Indefinite Integrals Properties

Given below are some of the properties of indefinite integrals.

Property 1

Integration and differentiation are inverse processes to one another because

\[\frac{d}{dx}\] \[\int\] f(x) dx = f(x)

And

\[\int\] f'(x) dx = F(x) + C

Here, C is any given arbitrary constant.

Proof:

Consider a given function f such that its anti-derivative is noted by F, that is,

\[\frac{d}{dx}\] F(x) = f(x)

Hence, 

\[\int\] f(x) dx = F(x) + C

When you differentiate it on both the sides with respect to x, you get

\[\frac{d}{dx}\] \[\int\] f(x) dx = \[\frac{d}{dx}\] (F(x) + C)

Since the derivative of any given constant function is equal to zero, you get

\[\frac{d}{dx}\] \[\int\] f(x) dx = \[\frac{d}{dx}\] (F(x) + C) = f(x)

The derivative of the given function f is denoted by f’(x), and hence,

f'(x)= \[\frac{d}{dx}\] f(x)

Therefore,

\[\int\] f'(x) dx = f(x) + C

where C is the arbitrary constant known as the constant of integration.

Property 2

Two indefinite integrals that have the same derivative have the same family of integrals or curves and hence they are said to be equivalent.

Proof:

Consider two given functions f and g in x in a way that

\[\frac{d}{dx}\] \[\int\] f(x) dx =  \[\frac{d}{dx}\] \[\int\] f(x) dx

You can also write this as

 \[\frac{d}{dx}\] \[\int\] f(x) dx - \[\frac{d}{dx}\] \[\int\] f(x) dx= 0

When you integrate on both the sides, you get

\[\int\] f(X) dx - \[\int\] g(x) dx = c

where C is any given real number

From the above equation, you can say that the family of the curves of [ ∫ f(x)dx + c3, c3∈ R] and [ ∫ g(x)dx + c2, c2∈ R] are same. Hence, you can say that, ∫ f(x)dx = ∫ g(x)dx.

Property 3

The integral of the sum of two functions equals the sum of the integrals of the given functions, that is,

\[\int\][f(x) dx + g(x)] dx = \[\int\] f(x)dx + \[\int\] g(x) dx

Proof:

Using the first property of the integrals, you have

\[\frac{d}{dx}\] [f(x) dx  + g(x) dx] =f(x) + g(x)...(1)

You also have

 \[\frac{d}{dx}\] \[\int\] [f(x) dx + g(x)dx] =  \[\frac{d}{dx}\] \[\int\] f(x) dx+ \[\frac{d}{dx}\] \[\int\] dx = f(x) + g(x)...(2)

Hence, from the equations 1 and 2, you have

\[\int\] [f(X) + g(x)] dx = \[\int\] f(x) dx + \[\int\] g(x) dx

Property 4

For any given real value of p,

\[\int\] pf(x) dx = p\[\int\] f(x) dx

Proof:

From the first property of integrals you can say that 

\[\frac{d}{dx}\] \[\int\] pf(x) dx = pf(x)

You also have

\[\frac{d}{dx}\] [ p \[\int\] f(x) dx] = P\[\frac{d}{dx}\]f(x) dx = pf(x)

From the second property of integrals you can say that

\[\int\] pf(x) dx = p \[\int\] f(x) dx

Property 5

For a given finite number of functions f1, f2…. fn and the given real numbers p1, p2…pn,

∫[p1f1(x) + p2f2(x)…. + pnfn(x) ]dx

= p1∫f1(x)dx + p2∫f2(x)dx +…..+  pn∫fn(x)dx

Indefinite Integral Examples

Let us now look at solving indefinite integrals

Example 1

Evaluate the following the indefinite integral.

\[\int\] (3x² - 6x + 2cosx) dx

Solution:

From the integral properties 1 and 2, you have

I = \[\int\] (3x² - 6x + 2Cosx) dx = \[\int\]  (3x² dx - \[\int\] 6xdx + \[\int\] 2cosxdx

= 3 \[\int\] x² dx - 6 \[\int\] xdx + 2 \[\int\] cosxdx

You can evaluate all the three integrals when you use the integration table. This would give you

I = 3\[\frac{x^{3}}{3}\] - 6 \[\frac{x^{2}}{2}\]+ 2 sin x + C 

= x² - 3x² + 2sin x + c

Example 2

Find the indefinite integral of the following

 \[\int\]  \[\frac{x+1}{\sqrt{x}}\] dx

Solution:

For the given function, you need to write the integrals in the form of the sum of two different integrals and then calculate each of them separately.

 \[\int\] \[\frac{x+1}{\sqrt{x}}\] dx = \[\int\] (\[\frac{x}{\sqrt{x}}\] +  \[\frac{x}{\sqrt{x}}\]) dx = \[\int\] ( \[\sqrt{x}\] + \[\frac{x+1}{\sqrt{x}}\]) dx = \[\int\] \[\sqrt{x}\]dx +  \[\int\] \[\frac{dx}{\sqrt{x}}\]

= \[\frac{x^{\frac{2}{2}}}{\frac{3}{2}}\] + 2\[\sqrt{x}\] + C = \[\frac{2\sqrt{x^{3}}}{3}\] + 2\[\sqrt{x}\] + C