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Bladerunner1122
Group Title
Find the average value of the function f on the given interval.
f(x)=√x on the interval [1,9]
Please explain the process.
 one year ago
 one year ago
Bladerunner1122 Group Title
Find the average value of the function f on the given interval. f(x)=√x on the interval [1,9] Please explain the process.
 one year ago
 one year ago

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abb0t Group TitleBest ResponseYou've already chosen the best response.1
You are dealing with the fundamental theorem of calculus pt. II. According to FToC II Suppose f(x) is a contunuous function on an interval [a, b] which in your case is [1,9]. and supposed that F(x) is any antiderviateive for f(x). Then: \[\int\limits_{1}^{9}\sqrt{x}dx = F(x)\] where F(x) is evalauted on the inteval [1,9] meaning: \[F(b)  F (a) = F(9)F(1)\]
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.1
@amoodarya that's very helpful! Good job, but I think you should of let him try it out on his own first as this is a Calculus course which requires more critical thinking and application.
 one year ago

amoodarya Group TitleBest ResponseYou've already chosen the best response.1
sorry for hurry
 one year ago

Bladerunner1122 Group TitleBest ResponseYou've already chosen the best response.0
I'm trying to understand how to get the answer. Thanks both of you. :)
 one year ago

Bladerunner1122 Group TitleBest ResponseYou've already chosen the best response.0
(emphasis on "how to") ;)
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.1
If there's any confusion, please ask. It's best to ask all the questions now and know for certain than be uncertain during an exam.
 one year ago

Bladerunner1122 Group TitleBest ResponseYou've already chosen the best response.0
Thanks, I think I'm good for now, but I may come back to ask more questions.
 one year ago

Bladerunner1122 Group TitleBest ResponseYou've already chosen the best response.0
It just occurred to me that the answer is wrong because I wanted it on the interval [1,9] not [0,9].
 one year ago
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