How Is the Number of Particles Calculated in a 1D Diffusion Slab?

[poiuy]

The solution to the diffusion equation in 1D may be written as follows:
n'(x,t) = N/sqrt(4piDt) * exp(-x^2/4DT)

where n'(x,t) is the concentration of the particles at position x at time t, N is the total number of particles and D is the diffusion coefficient.

Write down an expression for the number of particles in a slab of thickness dx located at position x.

I assumed it would be the integral of the function between x and x+dx with respect to x.

However exp(-x^2/4Dt) can't be integrated between these values. I have a standard integral for exp(-ax^2) which is 0.5 sqrt (pi/a) but this only applies to integrating between zero and infinity.If anybody could point me in the right direction it would be greatly appreciated, I think I am missing something obvious here and this is a really simple question.

Thanks

 

[nrqed]

The solution to the diffusion equation in 1D may be written as follows:



n'(x,t) = N/sqrt(4piDt) * exp(-x^2/4DT)

where n'(x,t) is the concentration of the particles at position x at time t, N is the total number of particles and D is the diffusion coefficient.

Write down an expression for the number of particles in a slab of thickness dx located at position x.


I assumed it would be the integral of the function between x and x+dx with respect to x.

However exp(-x^2/4Dt) can't be integrated between these values. I have a standard integral for exp(-ax^2) which is 0.5 sqrt (pi/a) but this only applies to integrating between zero and infinity.


If anybody could point me in the right direction it would be greatly appreciated, I think I am missing something obvious here and this is a really simple question.

Thanks

Welcome to the forums!

Note that the integrale of any function f(x) between x and x +dx is simply f(x) dx!

\int_x^{x+dx} f(x') dx' \approx f(x) dx

 

[poiuy]

Wow thanks, incredible that I could have had 14 years of education and never been taught that, thanks very much.

 

[poiuy]

Actually thinking about it, it's incredible that I couldn't work that out for myself.