## anonymous 5 years ago let g(x)=f(2x). show that the average value of f on the interval [0,2b] is the same as the average value of g on the interval [0,b].

1. ash2326

average value of a function say q(x) in the interval [l,m] is given by the relation $Average=(1/(m-l))*(\int\limits_{x=l}^{x=m} q(x) dx)$

2. ash2326

now let's find the average value of g(x) in [0,b], it's given as $(\int\limits_{0}^{b} g(x) dx)/(b-0)$ now average value of f(2x) in interval [0,2b] $(\int\limits_{0}^{2b} f(2x) dx)/(2b-0)$ substitute x/2=t dx/2=dt or dx=2dt so the limits will be from 0 to b $(\int\limits_{0}^{b} f(t) dt)/(b)$ replace t---> x $(\int\limits_{0}^{b} f(x) dx)/(b)$ and we know that f(x) is same as g(x) so this is equal to $(\int\limits_{0}^{b} g(x) dx)/(b$ hence the average value is same

3. Zarkon

$g_{ave}=\frac{1}{b}\int\limits_{0}^{b}g(x)dx=\frac{1}{b}\int\limits_{0}^{b}f(2x)dx$ let $$u=2x$$ then $$du=2dx$$ and $$\frac{du}{2}=dx$$ $=\frac{1}{b}\int\limits_{u(0)}^{u(b)}f(u)\frac{du}{2}$ $=\frac{1}{2b}\int\limits_{0}^{2b}f(u)du=f_{ave}$