# How to find the constant in this indefinite integration?

• Istiak
In summary, According to the problem, the book says that the velocity is constant, but the author provides a different equation that assumes the velocity is not constant.
Istiak
Homework Statement
Find constant of a indefinite integration for classical mechanics
Relevant Equations
$\int$
$$x(t)=\int \dot{x}(t)\mathrm dt=vt+c$$

That's what I did. But, book says

$$x(t)=\int \dot{x}(t)\mathrm dt=x_0+v_0 t+ \frac{F_0}{2m}t^2$$

Seems like, $$x_0 + \dfrac{a_0}{2}t^2$$ is constant. How to find constant is equal to what?

To find the value of the constant, you will need a initial condition, often ##x(0)=0## or in general ##x(t_0)=x_0##. Then insert it in your equation and solve for ##c##.

The reason is the following:
A differential equation has - depending on c - many solutions. Each solution corresponds to a flow through a vector field which is defined by the differential equation. The initial condition tells us where to start the flow, such that it will be the unique one that solves the equation plus its initial condition.

ref.: https://mathematica.stackexchange.com/questions/22190/why-are-these-flow-lines-cut-short

What is ##\dot{x}(t)## as a function of t? Don't assume it's a constant.

fresh_42 said:
To find the value of the constant, you will need a initial condition, often x(0)=0 or in general x(t0)=c0. Then insert it in your equation and solve for c.
If I try to do this way then, I get c=-vt

0=vt+c
c=-vt

Istiakshovon said:
is velocity... It's obviously constant [as far as I know]
Why do you assume that? Your book didn't!

Doc Al said:
Why do you assume that? Your book didn't!
What my book didn't do?

Istiakshovon said:
What my book didn't do?
It looks to me as if the problem assumes a constant acceleration, not a constant velocity.

Istiakshovon said:
If I try to do this way then, I get c=-vt

0=vt+c
c=-vt
If you set ##t=0## then you get ##c=x(0)## in the first equation, and ##x(0)=x_0## in the second.

Initial conditions could as well be at any time and of any value, e.g. ##x(5)=7## but it is usually ##x(0)## we are interested in because it makes the equations easier to solve for ##c##. It is also where the name comes from: initial as in starting point which is usually ##t=0.##

Istiak
fresh_42 said:
If you set ##t=0## then you get ##c=x(0)## in the first equation, and ##x(0)=x_0## in the second.

Initial conditions could as well be at any time and of any value, e.g. ##x(5)=7## but it is usually ##x(0)## we are interested in because it makes the equations easier to solve for ##c##. It is also where the name comes from: initial as in starting point which is usually ##t=0.##
So, I can find $$x(t)=vt+x(0)$$ but, my book assumes that $$x(t)=x(0)+vt+\frac{at^2}{2}$$

Istiakshovon said:
So, I can find $$x(t)=vt+x(0)$$ but, my book assumes that $$x(t)=x(0)+vt+\frac{at^2}{2}$$
You assume that the velocity is constant, but your book does not.

Delta2
Doc Al said:
You assume that the velocity is constant, but your book does not.
How? If I integrate velocity than, I get position not acceleration...

Istiakshovon said:
How? If I integrate velocity than, I get position not acceleration...
You're trying to find the position.

Do this: Start with a constant acceleration. Then integrate to find the velocity, and then integrate again to find the position.

Istiak
Istiakshovon said:
So, I can find $$x(t)=vt+x(0)$$ but, my book assumes that $$x(t)=x(0)+vt+\frac{at^2}{2}$$
So? Your book says ##x(0)=v\cdot 0 +x(0)## in the first and ##x(0)=x(0) + v\cdot 0 +\frac{a\cdot 0^2}{2}=x(0)## in the second equation. But what does it say in the problem statement before ##c## occurs?

Without initial condition ##x(t_0)=vt_0+x_0## you get infinitely many solution from ##\dot x =v## and with initial condition only one solution: ##x(t)=v\cdot t +x_0##.

And in order to solve ##\ddot x = a## you need even two initial conditions because you integrate twice and each integral has a ##c##.

Istiak
Istiakshovon said:
I got confused... $$\dot{x}=\int \ddot{x} \mathrm dt=\dot{x} t +c$$
That's a good start. Now solve for that constant. Looks like they assume the velocity is v0 when t = 0.

Doc Al said:
You're trying to find the position.

Do this: Start with a constant acceleration. Then integrate to find the velocity, and then integrate again to find the position.
Ohh! Thanks... got it...

$$\dot{x} (t)=\int \ddot{x} t dt=\ddot{x}t+c$$
$$\dot{x}(t)=\ddot{x}t+\dot{x}(0)$$
$$x(t)=\int \dot{x} (t) dt$$
$$=\int \ddot{x}t+\dot{x_0} dt$$
$$=\dot{x_0}t+\frac{\ddot{x}}{2}t^2+c$$

Istiakshovon said:
Ohh! Thanks... got it...

$$\dot{x} (t)=\int \ddot{x} t dt=\ddot{x}t+c$$
$$\dot{x}(t)=\ddot{x}t+\dot{x}(0)$$
$$x(t)=\int \dot{x} (t) dt$$
$$=\int \ddot{x}t+\dot{x_0} dt$$
$$=\dot{x_0}t+\frac{\ddot{x}}{2}t^2+c$$
OK, much better!

And it looks like your book assumed a constant acceleration (because of a constant force) equal to F0/m.

Istiak
Doc Al said:
OK, much better!

And it looks like your book assumed a constant acceleration (because of a constant force) equal to F0/m.
Hum..!

I don't understand the confusion. We start with a constant force along the x-axis, ##F_0##. Newton's law says,
$$F=ma=m\frac{d^2x}{dt^2}=F_0$$
$$m\frac{d^2x}{dt^2}=m\frac{d\dot x }{dt}$$
$$\frac{d\dot x }{dt}=\frac{F_0}{m}$$
Assuming ##t_0=0## we integrate
$$\int_{\dot {x}_0}^{\dot x}d\dot {x}'=\int_0^t\frac{F_0}{m}dt'$$
with the primes indicating dummy variables of integration.
$$\dot x - \dot {x}_0=\frac{F_0}{m}t$$
$$\dot {x}_0=v_0$$
$$\dot x=\frac{dx}{dt}$$
$$\frac{dx}{dt}= v_0+\frac{F_0}{m}t$$
Integrate again
$$\int_{x_0}^x dx'=\int_{0}^t (v_0+\frac{F_0}{m}t')dt'$$
$$x=x_0 + v_0t+ \frac{F_0t^2}{2m}$$
Does this help or am I missing something?

Fred Wright said:
We start with a constant force along the x-axis
How do you know that? I don't see any links in this thread to the original question.

## 1. What is the constant in indefinite integration?

The constant in indefinite integration, also known as the constant of integration, is a term that is added to the antiderivative of a function to account for all possible solutions. It is represented by the letter "C" and is necessary because the derivative of a constant is always zero.

## 2. How do I find the constant in indefinite integration?

To find the constant in indefinite integration, you can use the initial condition of the problem or the boundary conditions to determine its value. You can also solve for the constant by setting the given function equal to its antiderivative and then solving for the constant.

## 3. Do all indefinite integrals have a constant?

Yes, all indefinite integrals have a constant. This is because indefinite integration is the reverse process of differentiation, and the derivative of a constant is always zero. Therefore, a constant is always added when finding the antiderivative of a function.

## 4. Can the constant in indefinite integration be any value?

Yes, the constant in indefinite integration can be any real number. This is because the constant is added to the antiderivative to account for all possible solutions. However, when solving a specific problem, the value of the constant may be determined by the initial or boundary conditions.

## 5. Is the constant in indefinite integration always necessary?

Yes, the constant in indefinite integration is always necessary. This is because the derivative of a constant is always zero, and without the constant, the antiderivative may not be a complete solution. Therefore, the constant is essential in representing all possible solutions in indefinite integration.

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