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(a) Find the average value of $ f $ on the given …

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Problem 9 Easy Difficulty

(a) Find the average value of $ f $ on the given interval.
(b) Find $ c $ such that $ f_{ave} = f(c) $.
(c) Sketch the graph of $ f $ and a rectangle whose area is the same as the area under the graph of $ f $.

$ f(x) = (x - 3)^2 $ , $ [2, 5] $

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 6

Applications of Integration

Section 5

Average Value of a Function

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Video Transcript

and this problem. We're talking about the applications of integration, one being where we can find the average value of our function. And so just remember, let's review how we find that on the clothes interval A B The average value of of our function is equivalent to one over B minus. A times are integral from A to B A. Pardon me of ethics in D. X and remember, we were given our interval and our function. So we'll find that the average value of our function will be equivalent toe 1/5 minus two times the interval from 2 to 5, and we'll take the integral of X months. Three squared in DX. So now we can use a U substitution will say that u equals X minus three because that's what we have in the quantity squared. So that means do you equals D X? Now we can change our limits of integration by simply subtracting three in this case, because that is what are you is. So we'll subtract three from our lower and upper limit of integration, and we'll find that our new limits of integration will be from negative 1 to 2 now we can solve our integral well, multiply our entire integral by 1/3 will take the integral of negative 1 to 2 of you squared in do you? And if we take the anti derivative, we'll get 1/3 times You cubed over three. And remember, we're evaluating this from negative 1 to 2. And then we'll get 1/3 times 8/3 plus 1/3, and we'll get one. So the average value of our function is simply one. Now, the next part of this question is asking us to find C. So simply all we do is take C and plug it in for X. So we know that C minus three squared is equivalent toe one because we just found that the average value was one. So we'll get C squared minus six c plus nine is equal to one will move over that one value will get C squared minus 60 plus eight is equal to zero. Now we just simply factor that quadratic and we'll get C minus four as one binomial and C minus two is our second. So that means C is equal to four or C is equal to two and now finally were asked to sketch the graph of what we really just analyzed. So this is our function, Remember? We're evaluating it from 2 to 5, and we found the average value was one, so that would be the height of our rectangle. And this portion of our graph is what we just analyze, remember from 2 to 5. So I hope that this problem helped you understand how we can find the average value of a function using integration, then finding our see value and then connecting it back to the graph of our function.

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Calculus: Early Transcendentals

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