Definite integral and constants

[Chemist@]

Homework Statement

If acceleration=dV/dt, V=def.integal from t₁ to t₂ of a*dt,=at+C. At t=0, C=V₀. Why do we have a constant C, as a definite integral was calculated. Won't C₂ and C₁ cancel out?

Homework Equations

Definite integral from x₁ to x₂ of dx is x₂-x₁ without any constant, right?

The Attempt at a Solution

I am really confused.

 

[ShayanJ]

The answer to a definite integral is a definite number and, yes, the constant of integration cancels out. Only the answer to an indefinite integral contains a constant of integration.
The point is, in deriving the kinematical equations, there are two equivalent ways. Taking definite integrals or taking indefinite integrals and giving the constants appropriate values. There is no difference, they're the same thing.

 

[Ray Vickson]

If $a(t)$ is the acceleration at time $t$, the velocity at time $T$ is
V(T) = V_0 + \int_{t=0}^T a(t) \, dt,
Note that $\int_0^0 a(t) \, dt = 0$, so $V_0$ is the initial "starting" velocity; it may or may not be zero. For example, if you throw a ball upwards, $V_0$ is the ball's speed just when it leaves your hand.

 

[Chemist@]

How did you arrive at that equation from the one I posted?

Shyan: "Taking definite integrals or taking indefinite integrals and giving the constants appropriate values."
How does taking a definite integral give constant an appropriate value?

 

[PcumP_Ravenclaw]

Hello Chemist,

Lets say the velocity function is given as in graph.png (replace f(x) with v and x represents time, ignore negative x).
$v = 5x^3 + 3$
The acceleration function is differentiation of the velocity function as shown in graph(1).png
$a = 15x^2$
Now let's go in reverse. Say you are given the acceleration function $a = 15x^2$ how will you find the velocity function?
using integration of course.

$v = \int a \, dx = \int 15x^2 \, dx = 5x^3$

do you notice that the above velocity function is missing something (i.e. +3) we want our velocity function to be $v = 5x^3 + 3$ but if we derive the velocity function from acceleration function we cannot know what the constant will be so we just add an unknown c. so it should be $v = 5x^3 + c$

The constant c can be calculated from knowing initial conditions of the velocity function. e.g. if the question says, at time = 0 (i.e. x = 0) v was 3 then by substituting these values into the above equation ($v = 5x^3 + c$) you will calculate c as 3 ($c = 3 - 5(0)^3$).

what we just performed is indefinite integral (without any boundary to integrate like x₁ to x₂ ).

Now let's say, given the velocity function can we find the distance traveled in-between any point in time? e.g. from x = 0.2 to 0.5
$v = 5x^3 + 3$ first we integrate and we will get the distance function, graph(2).png $d = \frac{5}{4}x^4 + 3x$ then find the difference in distance (i.e. d(0.5) - d(0.2)) for x = 0.5 and x = 0.2. notice I did not add +c in the above $d = \frac{5}{4}x^4 + 3x$ equation because when I subtract d(0.5) - d(0.2) the c will cancel out as it remains unaltered by x. graph(2).png has c = 0 because of the initial condition d = 0 for x = 0.

when you integrate between two points you get the area under the curve between the two points. When the velocity function was integrated it gave the distance traveled by an object between two points although the velocity was changing instantaneously.

Graphs plotted with
http://rechneronline.de/function-graphs/

 

[ShayanJ]

How did you arrive at that equation from the one I posted?

Shyan: "Taking definite integrals or taking indefinite integrals and giving the constants appropriate values."
How does taking a definite integral give constant an appropriate value?

Oh...Sorry. That was wrong! I got confused myself!
What we actually do, is taking an indefinite integral and so we will have some integration constants. The only point is that, we give those integration constants, proper interpretations in terms of initial values of different quantities.

 

[Ray Vickson]

How did you arrive at that equation from the one I posted?

Shyan: "Taking definite integrals or taking indefinite integrals and giving the constants appropriate values."
How does taking a definite integral give constant an appropriate value?

Which message are you responding to? Please use the "quote" button, which helps to keep things straight.

 

[Chemist@]

Hello Chemist,

Lets say the velocity function is given as in graph.png (replace f(x) with v and x represents time, ignore negative x).
$v = 5x^3 + 3$
The acceleration function is differentiation of the velocity function as shown in graph(1).png
$a = 15x^2$
Now let's go in reverse. Say you are given the acceleration function $a = 15x^2$ how will you find the velocity function?
using integration of course.

$v = \int a \, dx = \int 15x^2 \, dx = 5x^3$

do you notice that the above velocity function is missing something (i.e. +3) we want our velocity function to be $v = 5x^3 + 3$ but if we derive the velocity function from acceleration function we cannot know what the constant will be so we just add an unknown c. so it should be $v = 5x^3 + c$

The constant c can be calculated from knowing initial conditions of the velocity function. e.g. if the question says, at time = 0 (i.e. x = 0) v was 3 then by substituting these values into the above equation ($v = 5x^3 + c$) you will calculate c as 3 ($c = 3 - 5(0)^3$).

what we just performed is indefinite integral (without any boundary to integrate like x₁ to x₂ ).

Now let's say, given the velocity function can we find the distance traveled in-between any point in time? e.g. from x = 0.2 to 0.5
$v = 5x^3 + 3$ first we integrate and we will get the distance function, graph(2).png $d = \frac{5}{4}x^4 + 3x$ then find the difference in distance (i.e. d(0.5) - d(0.2)) for x = 0.5 and x = 0.2. notice I did not add +c in the above $d = \frac{5}{4}x^4 + 3x$ equation because when I subtract d(0.5) - d(0.2) the c will cancel out as it remains unaltered by x. graph(2).png has c = 0 because of the initial condition d = 0 for x = 0.

when you integrate between two points you get the area under the curve between the two points. When the velocity function was integrated it gave the distance traveled by an object between two points although the velocity was changing instantaneously.

Graphs plotted with
http://rechneronline.de/function-graphs/

Okay, so an indefinite integral should be used in the case I wrote.

 

[PcumP_Ravenclaw]

Okay, so an indefinite integral should be used in the case I wrote.

Yes use indefinite integral if you are deriving the velocity function (add +c to it) from acceleration function. if you want to calculate the difference in velocity between two points in time use definite integral which cancels out the constant c.