Numerical Integration twice (acceleration to displacement)

[Pcmath]

Hello everyone

I have the following question regarding numerical integration twice from acceleration to displacement.

Suppose that a particle has acceleration function of a = t^t (which has non-elementary integral), to find the velocity it is easy as I can use Simpson's rule for numerical integration. But now I would like to find the displacement of the particle after some time, let say 5 seconds, how do I calculate it?

Note that I need very accurate answer, so I usually integrate numerically using Excel with thousand interval.

I prefer to use Newton-Cotes formula if possible.

EDIT
I tried using integration by parts, but I always reach a point where I can't continue.
a = t^t
Let v = ∫ t^t
s = ∫ v
= t*v - ∫ t*t^t
= t ∫ t^t - ∫ t*t^t

For example, to find the displacement after t = 5 seconds
∫ t*t^t from 0 to 5 is easy to integrate numerically
But the problem is t ∫ t^t I don't know how to do it because there is variable outside the integral.

For those who don't know, I am using integration by parts formula for definite integral, it includes find the value of
( t ∫ t^t ) from 0 to 5

 

[BvU]

Hello pcmath, : welcome:

You have $\ a = t^t\$ and with $\ a = \displaystyle {dv\over dt} \$ you get $\ v = \int a\;dt\ = \int t^t\; dt$.

Then , with $\ v= \displaystyle {ds\over dt} \$ you can write $\ s = \int \left (\ \int t^t\; dt\ \right ) dt'$. Written with bounds:$$
s(t) = \int_{t'= 0}^t \left (\ \int_{t''=0}^{t' } t''^{(t'')} \; dt''\ \right ) dt' $$
That is your double integral.

s = ∫ v = t*v - ∫ t*t^t = t ∫ t^t - ∫ t*t^t

is something I cannot agree with.

You could do a double numerical integration using forward Euler
$$ v(t+\Delta t) = v(t) + a(t) * \Delta t \\ s(t+\Delta t) = s(t) + v(t) * \Delta t \ $$ and you would indeed have to take very small steps because a(t) is so steep.

 

[mfb]

If you just need this once do it with tiny integration steps (or check WolframAlpha). If you need it many times, it is worth implementing a better integration scheme - Verlet integration or one of the other methods discussed there.

 

[Pcmath]

Thank you guys for your help, I will try it.

is something I cannot agree with.

This is actually correct as I have tried it. Assuming that a = x^2 then you can find s using that way, it gives the same value as integrate twice.

 

[BvU]

This is actually correct as I have tried it

Note that one successful example does not a proof constitute. It looks like integration by parts and then the question is: what is the time derivative of $t^t$

 

[nathal]

Hello everyone

I have the following question regarding numerical integration twice from acceleration to displacement.

Suppose that a particle has acceleration function of a = t^t (which has non-elementary integral), to find the velocity it is easy as I can use Simpson's rule for numerical integration. But now I would like to find the displacement of the particle after some time, let say 5 seconds, how do I calculate it?

Note that I need very accurate answer, so I usually integrate numerically using Excel with thousand interval.

I prefer to use Newton-Cotes formula if possible.

EDIT
I tried using integration by parts, but I always reach a point where I can't continue.
a = t^t
Let v = ∫ t^t
s = ∫ v
= t*v - ∫ t*t^t
= t ∫ t^t - ∫ t*t^t

For example, to find the displacement after t = 5 seconds
∫ t*t^t from 0 to 5 is easy to integrate numerically
But the problem is t ∫ t^t I don't know how to do it because there is variable outside the integral.

For those who don't know, I am using integration by parts formula for definite integral, it includes find the value of
( t ∫ t^t ) from 0 to 5

Hallo

You can use precise cubatures of Gauss type according: "Computation of definite integral over repeated integral" Tatra mountains, publ. 75 (2018). The Newton-Cotes is in preparation now. Shortly, for order 5 you can use standard quadrature Newton-Cotes: v5= v0+5h/288(19a0+75a1+50a2+50a3+75a4+19a5)
and cubature for displacement: s5=s0+v0*(5*h)+25h*h/2016(122a0+475a1+100a2+250a3+50a4+11a5)
The polynomial precision is 5.