CONSTANT OF INTEGRATION: MEANING, CALCULATION AND EXAMPLES - MATHS - 2024

The constant of integration is an added value to the calculation of antiderivatives or integrals, it serves to represent the solutions that make up the primitive of a function. It expresses an inherent ambiguity where any function has an infinite number of primitives.

For example, if we take the function: f (x) = 2x + 1 and we get its antiderivative:

∫ (2x + 1) dx = x ² + x + C; Where C is the constant of integration and represents graphically the vertical translation between the infinite possibilities of the primitive. It is correct to say that (x ² + x) is one of the primitives of f (x).

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Similarly we can define (x ² + x + C) as the primitive of f (x).

Reverse property

It can be noted that when deriving the expression (x ² + x) the function f (x) = 2x + 1 is obtained. This is due to the inverse property existing between the derivation and integration of functions. This property allows to obtain integration formulas starting from the differentiation. Which allows the verification of integrals through the same derivatives.

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However (x ² + x) is not the only function whose derivative is equal to (2x + 1).

1. d (x ² + x) / dx = 2x + 1
2. d (x ² + x + 1) / dx = 2x + 1
3. d (x ² + x + 2) / dx = 2x + 1
4. d (x ² + x + 3) / dx = 2x + 1
5. d (x ² + x + C) / dx = 2x + 1

Where 1, 2, 3 and 4 represent particular primitives of f (x) = 2x + 1. While 5 represents the indefinite or primitive integral of f (x) = 2x + 1.

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The primitives of a function are achieved through the antiderivation or integral process. Where F will be a primitive of f if the following is true

• y = ∫ f (x) dx = F (x) + C; C = constant of integration
• F '(x) = f (x)

It can be seen that a function has a single derivative, unlike its infinite primitives resulting from integration.

The indefinite integral

∫ f (x) dx = F (x) + C

It corresponds to a family of curves with the same pattern, which experience incongruity in the value of the images of each point (x, y). Each function that fulfills this pattern will be an individual primitive and the set of all functions is known as an indefinite integral.

The value of the constant of integration will be the one that differentiates each function in practice.

The constant of integration suggests a vertical shift in all graphs representing the primitives of a function. Where the parallelism between them is observed, and the fact that C is the value of the displacement.

According to common practices, the constant of integration is denoted by the letter "C" after an addend, although in practice it is indifferent whether the constant is added or subtracted. Its real value can be found in various ways under different initial conditions.

Other meanings of the constant of integration

It has already been discussed how the constant of integration is applied in the branch of integral calculus; Representing a family of curves that define the indefinite integral. But many other sciences and branches have assigned very interesting and practical values ​​of the constant of integration, which have facilitated the development of multiple studies.

In physics the constant of integration can take multiple values ​​depending on the nature of the data. A very common example is knowing the function V (t) that represents the velocity of a particle versus time t. It is known that when calculating a primitive of V (t) the function R (t) is obtained that represents the position of the particle versus time.

The constant of integration will represent the value of the initial position, that is, at time t = 0.

In the same way, if the function A (t) that represents the acceleration of the particle versus time is known. The primitive of A (t) will result in the function V (t), where the constant of integration will be the value of the initial velocity V ₀.

In economics, by obtaining by integration the primitive of a cost function. The constant of integration will represent the fixed costs. And so many other applications that merit differential and integral calculus.

How is the constant of integration calculated?

To calculate the constant of integration, it will always be necessary to know the initial conditions. Which are in charge of defining which of the possible primitives is the corresponding one.

In many applications it is treated as an independent variable at time (t), where the constant C takes the values ​​that define the initial conditions of the particular case.

If we take the initial example: ∫ (2x + 1) dx = x ² + x + C

A valid initial condition can be to condition that the graph passes through a specific coordinate. For example, we know that the primitive (x ² + x + C) passes through the point (1, 2)

F (x) = x ² + x + C; this is the general solution

F (1) = 2

We substitute the general solution in this equality

F (1) = (1) ² + (1) + C = 2

From where it easily follows that C = 0

In this way the corresponding primitive for this case is F (x) = x ² + x

There are several types of numerical exercises that work with constants of integration. In fact, the differential and integral calculus does not stop being applied in current investigations. At different academic levels they can be found; from initial calculation, through physics, chemistry, biology, economics, among others.

It is also appreciated in the study of differential equations, where the integration constant can take different values ​​and solutions, this due to the multiple derivations and integrations that are carried out in this matter.

Examples

Example 1

1. A cannon located 30 meters high fires a projectile vertically upwards. The initial velocity of the projectile is known to be 25 m / s. Decide:

• The function that defines the position of the projectile with respect to time.
• The time of flight or instant of time when the particle hits the ground.

It is known that in a rectilinear motion uniformly varied the acceleration is a constant value. This is the case of the projectile launch, where the acceleration will be gravity

g = - 10 m / s ²

It is also known that the acceleration is the second derivative of the position, which indicates a double integration in the resolution of the exercise, thus obtaining two integration constants.

A (t) = -10

V (t) = ∫A (t) dt = ∫ (-10t) dt = -10t + C ₁

The initial conditions of the exercise indicate that the initial velocity is V ₀ = 25 m / s. This is the velocity at the instant of time t = 0. In this way it is satisfied that:

V (0) = 25 = -10 (0) + C ₁ and C ₁ = 25

With the velocity function defined

V (t) = -10t + 25; The similarity can be observed with the MRUV formula (V _f = V ₀ + axt)

In a homologous way, we proceed to integrate the velocity function to obtain the expression that defines the position:

R (t) = ∫V (t) dt = ∫ (-10t + 25) dt = -5t ² + 25t + C ₂

R (t) = -5t ² + 25t + C ₂ (position primitive)

The initial position R (0) = 30 m is known. Then the particular primitive of the projectile is calculated.

R (0) = 30m = -5 (0) ² + 25 (0) + C ₂. Where C ₂ = 30

Example 2

1. Find the primitive f (x) that satisfies the initial conditions:

• f '' (x) = 4; f '(2) = 2; f (0) = 7

With the information of the second derivative f '' (x) = 4 the antiderivation process begins

f '(x) = ∫f' '(x) dx

∫4 dx = 4x + C ₁

Then, knowing the condition f '(2) = 2, we proceed:

4 (2) + C ₁ = 2

C ₁ = -6 and f '(x) = 4x - 8

We proceed in the same way for the second constant of integration

f (x) = ∫f '(x) dx

∫ (4x - 8) dx = 2x ² - 8x + C ₂

The initial condition f (0) = 7 is known and we proceed:

2 (0) ² - 8 (0) + C ₂ = 7

C ₂ = 7 and f (x) = 2x ² - 8x + 7

• f '' (x) = x ²; f '(0) = 6; f (0) = 3

In a similar way to the previous problem, we define the first derivatives and the original function from the initial conditions.

f '(x) = ∫f' '(x) dx

∫ (x ²) dx = (x ³ /3) + C ₁

With the condition f '(0) = 6 we proceed:

(0 ^3/3) + C ₁ = 6; Where C ₁ = 6 and f '(x) = (x ³ /3) + 6

Then the second constant of integration

f (x) = ∫f '(x) dx

∫ dx = (x ⁴ /12) + 6x + C ₂

The initial condition f (0) = 3 is known and we proceed:

+ 6 (0) + C ₂ = 3; Where C ₂ = 3

Thus we obtain the primitive particular

f (x) = (x ⁴ /12) + 6x + 3

Example 3

1. Define the primitive functions given the derivatives and a point on the graph:

• dy / dx = 2x - 2 which passes through the point (3, 2)

It is important to remember that derivatives refer to the slope of the line tangent to the curve at a given point. Where it is not correct to assume that the graph of the derivative touches the indicated point, since this belongs to the graph of the primitive function.

In this way we express the differential equation as follows:

∫dy = ∫ (2x - 2) dx

Applying the initial condition:

2 = (3) ² - 2 (3) + C

C = -1

It is obtained: f (x) = x ² - 2x - 1

• dy / dx = 3x ² - 1 that passes through the point (0, 2)

We express the differential equation as follows:

Applying the initial condition:

2 = (0) ² - 2 (0) + C

C = 2

We obtain: f (x) = x ³ - x + 2

Proposed exercises

Exercise 1

1. Find the primitive f (x) that satisfies the initial conditions:

• f '' (x) = x; f '(3) = 1; f (2) = 5
• f '' (x) = x + 1; f '(2) = 2; f (0) = 1
• f '' (x) = 1; f '(2) = 3; f (1) = 10
• f '' (x) = -x; f '(5) = 1; f (1) = -8

Exercise 2

1. A balloon ascending at a velocity of 16 ft / s drops a bag of sand from a height of 64 ft above ground level.

• Define the flight time
• What will the vector V _{f be} when it hits the ground?

Exercise 3

1. The figure shows the acceleration-time graph of a car moving in the positive direction of the x-axis. The car was traveling at a constant speed of 54 km / h when the driver applied the brakes to stop in 10 seconds. Determine:

• The initial acceleration of the car
• The speed of the car at t = 5s
• The displacement of the car during braking

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Exercise 4

1. Define the primitive functions given the derivatives and a point on the graph:

• dy / dx = x that passes through the point (-1, 4)
• dy / dx = -x ² + 1 which passes through the point (0, 0)
• dy / dx = -x + 1 which passes through the point (-2, 2)

References

1. Integral calculus. The indefinite integral and integration methods. Wilson, Velásquez Bastidas. Magdalena University 2014
2. Stewart, J. (2001). Calculation of a variable. Early transcendentals. Mexico: Thomson Learning.
3. Jiménez, R. (2011). Mathematics VI. Integral calculus. Mexico: Pearson Education.
4. Physics I. Mc Graw hill