Finding the Average Value of a Function

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https://imgur.com/a/Maq1N

1. Homework Statement

Attached above is the problem statement. Essentially, they've given two different volumes involving some unknown function f(x) and ask that you find its average value over the given closed interval.

Homework Equations


Relevant equations include the fundamental theorem of calculus as well as the mean value theorem for integrals.

The Attempt at a Solution


As you may or may not see, depending on how hard my work is to follow, I've attempted to use the given information in such a way that I might find the value for "F(7) - F(1)". I'm not sure if what I did was valid or not. I took the sqrt(f(7)2 - f(1)2 = 2) to get f(7) - f(1) = sqrt(2). It may be that I need to actually solve for F(7) and F(1)? Any help is greatly appreciated as I have no way to check my answer.
 

Answers and Replies

  • #2
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https://imgur.com/a/Maq1N

1. Homework Statement

Attached above is the problem statement. Essentially, they've given two different volumes involving some unknown function f(x) and ask that you find its average value over the given closed interval.

Homework Equations


Relevant equations include the fundamental theorem of calculus as well as the mean value theorem for integrals.
Neither of these is relevant for this problem. What is relevant, but missing, is the formula for the average value of a function.
domabo said:

The Attempt at a Solution


As you may or may not see, depending on how hard my work is to follow, I've attempted to use the given information in such a way that I might find the value for "F(7) - F(1)". I'm not sure if what I did was valid or not. I took the sqrt(f(7)2 - f(1)2 = 2) to get f(7) - f(1) = sqrt(2). It may be that I need to actually solve for F(7) and F(1)? Any help is greatly appreciated as I have no way to check my answer.
That's not how you find the average value of a function. Your textbook surely has the formula for this.
For this problem, you will need to use the given information about the two volumes of revolution.
 
  • #3
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Neither of these is relevant for this problem. What is relevant, but missing, is the formula for the average value of a function.

That's not how you find the average value of a function. Your textbook surely has the formula for this.
For this problem, you will need to use the given information about the two volumes of revolution.
Okay. I wrote the formula for the average value function but said mean value theorem. I'm sure you saw that, though. How is the FTC not relevant? Isn't it always relevant when taking definite integrals...

And, as I've said, I have attempted to use the information given with the volumes in order to find a value for F(7) and F(1) since f(x) is not explicitly given. You have not told me whether I am way off or on the right track?
 
  • #4
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Okay. I wrote the formula for the average value function but said mean value theorem. I'm sure you saw that, though.
The mean value theorem for integrals is different terminology (which I forgot) for average value of a function.
How is the FTC not relevant? Isn't it always relevant when taking definite integrals...
I suppose so, but it's not used very much in this problem. See below.
And, as I've said, I have attempted to use the information given with the volumes in order to find a value for F(7) and F(1) since f(x) is not explicitly given. You have not told me whether I am way off or on the right track?
Since you aren't explicitly given the function f here, there's no way to get F, the antiderivative of f. For that reason, the FTC plays almost no role here.So there's no way to evaluate F(1) and F(7).

In your work, you have integrals for the two volumes of revolution, which are given as ##6\pi## and ##10\pi##, respectively. If you subtract the two integrals you will end up with an equation involving ##\int_1^7 f(t)dt##, which is close to what you want for the average value of f on the interval [1, 7].
 
  • #5
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The mean value theorem for integrals is different terminology (which I forgot) for average value of a function.
I suppose so, but it's not used very much in this problem. See below.

Since you aren't explicitly given the function f here, there's no way to get F, the antiderivative of f. For that reason, the FTC plays almost no role here.So there's no way to evaluate F(1) and F(7).

In your work, you have integrals for the two volumes of revolution, which are given as ##6\pi## and ##10\pi##, respectively. If you subtract the two integrals you will end up with an equation involving ##\int_1^7 f(t)dt##, which is close to what you want for the average value of f on the interval [1, 7].
Okay, but you fail to share your reasoning behind why this works graphically or conceptually. Why is it that when I subtract a volume of some disk from the volume of some washer that I somehow end up with the average value of the function?
 
  • #6
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Okay, but you fail to share your reasoning behind why this works graphically or conceptually. Why is it that when I subtract a volume of some disk from the volume of some washer that I somehow end up with the average value of the function?
You don't "end up" with the average value of the function. The difference between the two volumes is ##4\pi##, not the average value of ##f##. You could just as well say "Why do we end up with the area under the curve ##y=f(x)##?". Of course, what happened is that the integrals of ##f^2## terms cancel out, leaving an expression in terms of ##\int_1^7 f(x)~dx##, which allows you to calculate that without knowing ##f##. I don't think there is any deep geometrical principle going on here.
 
  • #7
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You don't "end up" with the average value of the function. The difference between the two volumes is ##4\pi##, not the average value of ##f##. You could just as well say "Why do we end up with the area under the curve ##y=f(x)##?". Of course, what happened is that the integrals of ##f^2## terms cancel out, leaving an expression in terms of ##\int_1^7 f(x)~dx##, which allows you to calculate that without knowing ##f##. I don't think there is any deep geometrical principle going on here.
Ok. Fair enough. So, you can't just get the anti-derivative because I could be describing infinitely many different functions? And, in the interest of seeing whether or not I could properly do what you both have described... does this now look better? https://i.imgur.com/bbxsgy8.jpg
 
  • #8
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Ok. Fair enough. So, you can't just get the anti-derivative because I could be describing infinitely many different functions?
Right.
omabo said:
And, in the interest of seeing whether or not I could properly do what you both have described... does this now look better? https://i.imgur.com/bbxsgy8.jpg
Looks OK to me.
 
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  • #10
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Thanks for the heads up about formatting; I'll make sure to take a look at that. What I did was find the outer radius to be R(x) = f(x) - (-2) and the inner radius to be 0 - (-2), and then multiplied out and simplified. For washers, it's pi * integral of ( R(x)^2 - r(x)^2). How is my integral wrong then?
 
  • #11
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Thanks for the heads up about formatting; I'll make sure to take a look at that. What I did was find the outer radius to be R(x) = f(x) - (-2) and the inner radius to be 0 - (-2), and then multiplied out and simplified. For washers, it's pi * integral of ( R(x)^2 - r(x)^2). How is my integral wrong then?
Your work is fine -- I set up my integral incorrectly. I've edited my earlier post.
 
  • #12
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Your work is fine -- I set up my integral incorrectly. I've edited my earlier post.
Okay. Thank you. Those pi's on the outside of the integral would cancel one another out through subtraction right? Or should I first, before subtraction, divide, for both integrals, both sides by pi? Then, it seems my answer would be the integral of f(x) = 1 and so the average value would be 1/6.
 
  • #13
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Okay. Thank you. Those pi's on the outside of the integral would cancel one another out through subtraction right? Or should I first, before subtraction, divide, for both integrals, both sides by pi? Then, it seems my answer would be the integral of f(x) = 1 and so the average value would be 1/6.
Yes, that's what I get, too.
Using LaTeX,
##\int_1^7 f(x) dx = 1 \Rightarrow \frac 1 {7 - 1} \int_1^7 f(x) dx = \frac 1 6##

Here's the unrendered script for the above: \int_1^7 f(x) dx = 1 \Rightarrow \frac 1 {7 - 1} \int_1^7 f(x) dx = \frac 1 6

Notice that I used some shortcuts. For fractions and limits of integration and some other things, if the expression is two or more characters, you have to surround them with a pair of braces -- { }. If it's only one character, you don't need the braces.
 
  • #14
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Yes, that's what I get, too.
Using LaTeX,
##\int_1^7 f(x) dx = 1 \Rightarrow \frac 1 {7 - 1} \int_1^7 f(x) dx = \frac 1 6##

Here's the unrendered script for the above: \int_1^7 f(x) dx = 1 \Rightarrow \frac 1 {7 - 1} \int_1^7 f(x) dx = \frac 1 6

Notice that I used some shortcuts. For fractions and limits of integration and some other things, if the expression is two or more characters, you have to surround them with a pair of braces -- { }. If it's only one character, you don't need the braces.
Thank you
 

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