Antiderivative formulas and equations, examples, exercises

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 I, It is true that F '(x) = f (x)

For example let's take the following function:

f (x) = 4x³

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

We obtain precisely f (x) = 4x³.

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

Let's check it out:

Remember that the derivative of a constant is 0. Therefore the term x⁴ you can add any constant and its derivative will remain 4x³.

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

The illustrative example above can be expressed like this:

dF (x) = 4x³ dx

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

F (x) = ∫4x³ dx = x⁴ + C

Where the function f (x) = 4x³  it is called integrating, and C is the constant of integration.

Article index

• 1 Examples of antiderivatives
  • 1.1 Differential equations
• 2 Antiderivative exercises
  • 2.1 - Exercise 1
  • 2.2 - Exercise 2
• 3 References

Examples of antiderivatives

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

That function can be:

F (x) = - cos x

Let's check that it is true:

F '(x) = (- cos x)' = - (-sen x) = sin x

Therefore we can write:

∫sen 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

two.- ∫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 rule of powers for integrals it can be set this way:

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 to 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 this 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, thus solving the desired unknown:

y (x) = ∫f (x) dx = F (x) + C

And since C is a real constant, to know which one is appropriate in each case, the statement must contain sufficient 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 integration rules 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 apply rule 3 first, since the integrand is the sum of two terms:

∫ (x + 7) dx = ∫ xdx + ∫7dx

For the first integral the rule of powers applies:

∫ xdx = (x^two / 2) + C₁

Rule 1 applies to the second integral, where k = 7:

∫7dx = 7∫dx = 7x + C_two

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

∫ (x + 7) dx = (x^two / 2) + 7x + C

Solution b

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

∫ (x^3/2 + x^two + 6) dx = ∫x^3/2 dx + ∫x^two dx + ∫6 dx =

Note that a constant of integration appears for each integral, but they meet 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 = ∫ (3x^two-2x + 3x-2) dx = ∫ (3x^two + x - 2) dx

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

∫ (x + 1) (3x-2) dx = ∫3x^two dx + ∫ x dx + ∫- 2 dx = x³ + (1/2) x^two - 2x + C

Solution e

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

The change of variable is as follows:

u = x + 7

Deriving this expression to both sides:

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 (t) = cos t. It is known that at t = 0, the position is x = 3, all in units of the International System. It is asked to find the velocity v (t) and the position x (t) of the particle.

Solution

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

a (t) = v '(t) = cos t

It follows that:

v (t) = ∫ cos t dt = sin t + C₁

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

x (t) = ∫ v (t) dt = ∫ (sin t + C₁) dt = ∫sen t dt + ∫C₁ dt = - cos t + C₁ t + C_two

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 that x (0) = 3:

x (0) = - cos 0 + C₁ 0 + C_two = - 1 + C_two = 3 → C_two = 3 + 1 = 4

The velocity and position functions are definitely like this:

v (t) = sin t

x (t) = - cos t + 4

References

1. Engler, A. 2019. Integral Calculus. National University of the Litoral.
2. Larson, R. 2010. Calculation of a variable. 9th. Edition. Mcgraw hill.
3. Mathematics Free Texts. Antiderivatives. Recovered from: math.liibretexts.org.
4. Wikipedia. Antiderivative. Recovered from: en.wikipedia.org.
5. Wikipedia. Indefinite integration. Recovered from: es.wikipedia.org.