A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 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
 4 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].

This Question is Closed

ash2326
 4 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
 4 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
 4 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 \(\frac{du}{2}=dx\) \[=\frac{1}{b}\int\limits_{u(0)}^{u(b)}f(u)\frac{du}{2}\] \[=\frac{1}{2b}\int\limits_{0}^{2b}f(u)du=f_{ave}\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.