A **antiderivative** *f(x)* of a function *F*(x) is also called primitive or simply the indefinite integral of said function, if in a given interval *Yo*It is true that *F´(x) = f(x)*

For example, let’s take the following function:

f(x) = 4×3

An antiderivative of this function is F(x) = x4, since when differentiating F(x) using the derivation rule for powers:

One obtains precisely f(x) = 4×3.

However, this is only one of the many antiderivatives of f(x), since this other function: G(x) = x4 + 2 is also one, because differentiating G(x) with respect to x, the same is obtained from turn f(x).

Let’s check it:

Remember that the derivative of a constant is 0. Therefore, any constant can be added to the term x4 and its derivative will continue to be 4×3.

We conclude that any function of the general form F(x) = x4 + C, where C is a real constant, serves as an antiderivative of f(x).

The above illustrative example can be expressed as follows:

dF(x) = 4x3dx

The antiderivative or indefinite integral is expressed with the symbol ∫, therefore:

F(x) = ∫4×3 dx = x4 + C

Where the function f(x) = 4×3 is called *integrating*and C is the *constant of integration*.

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**Examples of antiderivatives**

Finding an antiderivative of a function is easy in some cases where the derivatives are well known. For example, let the function be f(x) = sin x, an antiderivative for it is another function F(x), such that by differentiating it, f(x) is obtained.

That function can be:

F(x) = – cos x

Let’s check that it is true:

F´(x) = (- cos x)´= – (-sin x) = sin x

Therefore we can write:

∫sin x dx = -cos x + C

In addition to knowing the derivatives, there are some basic and simple integration rules to find the antiderivative or indefinite integral.

Let k be a real constant, then:

1.- ∫*kdx = k ∫dx =kx + C*

2.- *∫kf(x)dx = k ∫f(x)dx *

If a function h(x) can be expressed as the addition or subtraction of two functions, then its indefinite integral is:

3.- *∫h(x)dx = ∫[f(x) ± g(x)]dx = ∫f(x)dx ± ∫g(x)dx*

This is the property of linearity.

The *power rule* For integrals it can be established in this way:

This rule has an obvious restriction: since the denominator *n+1* cannot be made 0, therefore n ≠ -1.

For the case of n = -1 the following rule is used:

5.- ∫*x -1 dx = ln x +C*

It is easy to show that the derivative of *ln x* it is precisely *x-1*.

**Differential equations**

A differential equation is one in which the unknown is found as a derivative.

Now, from the previous analysis, it is easy to realize that the inverse operation of the derivative is the antiderivative or indefinite integral.

Let f(x) = y´(x), that is, the derivative of a certain function. We can use the following notation to indicate such a derivative:

It immediately follows that:

*dy = f(x)dx*

The unknown of the differential equation is the function y(x), the one whose derivative is f(x). To solve it, the previous expression is integrated on both sides, which is equivalent to applying the antiderivative:

*∫dy = ∫f(x)dx*

The left integral is solved by the integration rule 1, with k = 1 and thus the unknown is solved:

*y(x)* = *∫f(x) dx = F(x) + C*

And since C is a real constant, to know which is appropriate in each case, the statement must contain enough additional information to calculate the value of C. This is called *initial condition*.

We will see application examples of all this in the next section.

**antiderivative exercises**

**– Exercise 1**

Apply the rules of integration to obtain the following antiderivatives or indefinite integrals of the given functions, simplifying the results as much as possible. It is convenient to verify the result by derivation.

**Solution to**

We first apply rule 3, since the integrand is the sum of two terms:

∫ (x+7) dx = ∫ xdx + ∫7dx

For the first integral the power rule is applied:

∫ xdx = (x2 /2)+C1

In the second integral, rule 1 is applied, where k = 7:

∫7dx = 7∫dx = 7x + C2

And now the results are added. The two constants are grouped into one, generically called C:

∫ (x+7) dx = (x2 /2) + 7x + C

**solution b**

Due to linearity, this integral is decomposed into three simpler integrals, to which the power rule will be applied:

∫(x3/2 + x2 + 6) dx = ∫x3/2 dx + ∫x2 dx +∫6 dx =

= (2/5)x5/2 + (1/3)x3 + 6x + C

Note that for each integral there is a constant of integration, but they are gathered in a single call C.

**solution c**

In this case it is convenient to apply the distributive property of multiplication to develop the integrand. Then the power rule is used to find each integral separately, as in the previous exercise.

∫(x+1)(3x-2)dx = ∫(3×2-2x+3x-2)dx = ∫(3×2 + x – 2)dx

The attentive reader will observe that the two central terms are similar, therefore they are reduced before integrating:

∫(x+1)(3x-2) dx = ∫3×2 dx + ∫ x dx + ∫- 2 dx = x3 + (1/2)x2 – 2x + C

**solution and**

One way to solve the integral would be to expand the power, as was done in example d. However, since the exponent is higher, it would be convenient to make a change of variable, so as not to have to do such a long development.

The change of variable is as follows:

u = x + 7

Deriving on both sides this expression:

du = dx

The integral is transformed to a simpler one with the new variable, which is solved with the power rule:

∫ (x+7)5 dx = ∫ u5 du = (1/6)u6 + C

Finally, the change is returned to return to the original variable:

∫ (x+7)5 dx = (1/6)(x+7)6 + C

**– Exercise 2**

A particle is initially at rest and moves along the x axis. Its acceleration for t > 0 is given by the function a

**Solution**

Since acceleration is the first derivative of velocity with respect to time, we have the following differential equation:

a

It follows that:

v

On the other hand, we know that the velocity is in turn the derivative of the position, therefore we integrate again:

x

The constants of integration are determined from the information given in the statement. First, it says that the particle was initially at rest, therefore v(0)=0:

v(0) = sin 0 + C1 = 0

C1 = 0

Then we have x(0) = 3:

x(0) = – cos 0 + C1 0 + C2 = – 1 + C2 = 3 → C2 = 3+1=4

The velocity and position functions definitely look like this:

v

x

**References**

Engler, A. 2019. Integral Calculus. National University of the Coast. Larson, R. 2010. Calculus of a variable. 9na. Edition. McGraw Hill. Mathematics Free Texts. Antiderivatives. Retrieved from: math.liibretexts.org. Wikipedia. Antiderivative. Retrieved from: en.wikipedia.org. Wikipedia. Indefinite integration. Recovered from: es.wikipedia.org.