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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].
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].

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ash2326
 5 years ago
Best ResponseYou've already chosen the best response.1average value of a function say q(x) in the interval [l,m] is given by the relation \[Average=(1/(ml))*(\int\limits_{x=l}^{x=m} q(x) dx)\]

ash2326
 5 years ago
Best ResponseYou've already chosen the best response.1now let's find the average value of g(x) in [0,b], it's given as \[(\int\limits_{0}^{b} g(x) dx)/(b0)\] now average value of f(2x) in interval [0,2b] \[(\int\limits_{0}^{2b} f(2x) dx)/(2b0)\] 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

Zarkon
 5 years ago
Best ResponseYou've already chosen the best response.1\[g_{ave}=\frac{1}{b}\int\limits_{0}^{b}g(x)dx=\frac{1}{b}\int\limits_{0}^{b}f(2x)dx\] let \(u=2x\) then \(du=2dx\) and \(\frac{du}{2}=dx\) \[=\frac{1}{b}\int\limits_{u(0)}^{u(b)}f(u)\frac{du}{2}\] \[=\frac{1}{2b}\int\limits_{0}^{2b}f(u)du=f_{ave}\]
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