I am working on some problems from a book. I want to know how equation
$ s[Y] = 2 \int ^{Z}_{0} z (\dfrac {1} {4z^2} (Y')^2 - w^2Y^2 ) dz $
Where $ Z = X^{1/2} $
simplifies to this one ?
$ s[Y]= \dfrac{1}{2} \int ^{Z}_{0} (\dfrac{1}{z}(Y')^2 - 4zw^2Y^2) dz $
how 2 outside of $\int$ sign became 1/2 and how second term inside $\int$ has 4 now ?
Thank you
Just multiply everything by $1=\frac{4}{4}$ where you multiply by 4 inside the integral and by $\frac{1}{4}$ outside of it.