Definite vs Indefinite Integrals in Physics

[coarse vortex]

how do i know when to use either one?

Example question:

A body of mass 5kg moving horizontally is subject to a resistance of v+2v^2 N, where v is the velocity. If the initial velocity is 2ms^-1, find the distance travelled when v=1ms^-1
my solution used a definite integral but my teacher used an indefinite one then solved for c

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[coarse vortex]

$can i use latex here$

[ebon anchor]

Kahoneki

[coarse vortex]

cool

my equation was this

$\int_0^s{ds}=-5\int_{v_0}^v{\frac{v}{2v^2+1}dv}$

[ebon anchor]

Kahoneki

[coarse vortex]

my teacher's was this

$\int{ds}=5\int{\frac{v}{2v^2+1}dv}$

[ebon anchor]

Kahoneki

[coarse vortex]

does it make a difference?

we got different answers and im not sure if this is the reason why or if it's something else

[coarse vortex]

oh this should have a minus sign in front of the 5 btw, typo sdfjkhskd

[subtle valve]

I mean you do definite integration when initial and final boundary limits are known(idk I only know like basic shit) so you don't have to account for the +C, but if you do indefinite integration you need to add the +C, which then must be solved using some additional information as maybe provided in the question.

[modest solar]

Can you elaborate on how you set up the integral?

[coarse vortex]

is it not kinda the same though? using starting and final values for the bounds or using them to solve for c?

[spark delta]

how did your teacher solve for c

[coarse vortex]

his working's a bit all over the place sorry 😭

he used the fact that initial velocity is given to be 2ms^-1

[spark delta]

i see

what abt ur method? you got steps?

[coarse vortex]

like steps after this @spark delta?

[spark delta]

yeah

[spark delta]

uhh didnt u get the same answer?

[coarse vortex]

i dont think so? where r u seeing it?

[spark delta]

multiply the part inside the bracket with that minus one. then simplify the natural log...
u get (5/2) ln (5/2v+1)

which is 5/2 ln(5/3) same as what he got

i might be wrong im not very good

[coarse vortex]

huh

yea

why doesn't the way i did it work?

[spark delta]

it does, you did it correctly but simplified it differently

both are numerically same answers just expressed in different forms

like, ln(5/3) = - ln(3/5)
so u just have to simplify ur last step by removing that negative sign

[coarse vortex]

oh my god im so stupid

when i was checking it i set v=2

[spark delta]

ah i see

[coarse vortex]

thanks @spark delta 🙃

-# (sorry)

[signal bluff]

@coarse vortex has given 1 rep to @spark delta

[modest solar]

Ah, I think I see. Let me try that approach.
We have Newton's second law:
F = ma
In our case F is proportional to 2v^2 + v, and a = dv/dt, as usual. So:
-(k2 v^2 + k1 v) = mdv/dt
We can then use the chain rule:
dv/dt = (dv/dx)(dx/dt) = v(dv/dx)
So:
-(k2 v^2 + k1 v) = mvdv/dx
dx = mdv/(k2 v + k1)
So, I think your mistake was that you accidentally wrote v/(2v^2 + 1) instead of v/(2v^2 + v).

[coarse vortex]

where did i write 2v^2+1?

[coarse vortex]

i think it was just this that was wrong

[modest solar]

Here.

[spark delta]

seems like an error while doing latex

the steps in the picture they sent dont have that error

[modest solar]

Oh, ok.

Anyway, let me try continuing it.

Corrected the above a bit to account for dimensionality of constants in the force.

[modest solar]

So, we have:
dx = -mdv/(k2 v + k1)
Suppose the initial and final velocity are v0 and v1. Then we can integrate:
x = -m∫(dv/(k2 v + k1), v0, v1) = -(m/k2)ln((k2 v1 + k1)/(k2 v0 + k1))
In our case m = 5 kg, k1 = 1 kg/s, k2 = 2 kg/m, v0 = 2 m/s. So:
x = -(5 kg/(2 kg/m))ln((2 kg/m*v1 + 1 kg/s)/(2 kg/m*2 m/s + 1 kg/s)) = -2.5 m*ln(1/5 + (2/5) s/m*v1)
Thus, you just substitute the needed values of v1.

So yeah, I agree with your final answer.

[coarse vortex]

[subtle valve]

I mean yeah

[coarse vortex]

actually?

i guess it makes sense if it's like, the indefinite integral gives you a function with an unknown that can be solved for if you know some state, giving you the complete function

whereas you don't need that for definite integrals because it's not returning a function, you're just giving a function and asking what the area under it is between two values

so yea i think it makes sense...

[subtle valve]

Definite integration gives you the net area below the below the function while indefinite integration represents a family of curves corresponding to the antiderivatives of the function.

[coarse vortex]

right yea, and those curves are different by the constant C?

[subtle valve]

Yeah

And that indefinite integration can be applied to any function whose anti derivative does exist not necessarily requiring continuity, while definite integration required the function to be integrable on the interval, typically necessitating continuity.

[coarse vortex]

hm yea i think i get that, not really done much on continuity yet but it makes sense ! (i think)

thanks @subtle valve :]

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@coarse vortex has given 1 rep to @subtle valve

[coarse vortex]

+close