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Mic :) said:Hi!
Could someone please integrate the expression (with intention of finding the normalisation constant / value of A).
Thanks a lot!
vela said:You need to tell us the complete problem first. You haven't said what variable you're integrating with respect to. You haven't specified the limits of integration, or perhaps it's an indefinite integral you're looking for. You also need to review basic properties of exponentials. In particular, you should know that ##e^a e^b = e^{a+b}##. Your rearrangement isn't equal to the original.
Once you get the details down, what's stopping you from evaluating it? Have you even tried to do it on your own yet?
The formula for integrating A exp( (− 3 R^2)/(2Na^2)) is: ∫A exp( (− 3 R^2)/(2Na^2)) dR = A (πNa^2/3)^(1/2) erf((√3R)/(Na)) + C
The variable A represents the amplitude or magnitude of the function. R represents the variable of integration and is typically a distance or position. N represents the number of particles or molecules in the system. And a represents the characteristic length scale of the system.
This formula is commonly used in physics and chemistry to calculate the probability of finding particles or molecules at a certain distance in a system. It is also used in statistics and probability to model the distribution of random variables.
The negative exponent in the formula indicates an exponential decay, meaning that the function decreases rapidly as the distance or position increases. This is often seen in physical systems where particles tend to move away from each other.
Yes, this formula can be applied to any system where the variables A, R, N, and a are relevant and the function being integrated has an exponential decay with distance or position.