Question

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].

Answer 1

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

Answer 2

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

Answer 3

\[g_{ave}=\frac{1}{b}\int\limits_{0}^{b}g(x)dx=\frac{1}{b}\int\limits_{0}^{b}f(2x)dx\]

let \(u=2x\) then \(du=2dx\) and \(\frac{du}{2}=dx\)

\[=\frac{1}{b}\int\limits_{u(0)}^{u(b)}f(u)\frac{du}{2}\]

\[=\frac{1}{2b}\int\limits_{0}^{2b}f(u)du=f_{ave}\]