Integral

[Logarythmic]

If the total luminosity is given by

L_T = \int_0^{\infty} e^{-(r/a)^{1/4}} r^2 dr

estimate the radius [tex]r(a)[/itex] corresponding to half of the total luminosity.

This would be to integrate from zero to r and get a function r(a,L) but this is impossible so anyone got an idea on how to estimate this?

[quasar987]

Well the primitive of the integrand exists, you knew that right?

The difficult part is solving for r_{1/2} in \int_0^{r_{1/2}} e^{-(r/a)^{1/4}} r^2 dr=L_T/2. But computers are good with this.