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Let f be the function defined by f(w) 522 x What is the average value of f on the interval [~1, 1] written in simplest form?...

Question

Let f be the function defined by f(w) 522 x What is the average value of f on the interval [~1, 1] written in simplest form?

Let f be the function defined by f(w) 522 x What is the average value of f on the interval [~1, 1] written in simplest form?



Answers

Find the average value of the function on the given interval. $$f(x)=5 x^{2} ; \quad[1,4]$$

Okay, so we're gonna be using this formula I have written down um which is the average value of our function F of X is equal to one divided by b minus a times the interval or the integral. Um Of ffx from A to B and A. Is the lower bound of the interval that we're looking at and be as the upper bound. So in our case this average value is going to be one divided by um one minus negative one, which would be one half And then multiplied by the integral from negative 1-1 of one minus X squared D. X. And so now we just need to find this integral and it's going to be the integral of one which is going to be equal to X. And then minus the integral of X squared which is equal to x cubed divided by three. And we're looking from negative 1 to 1. So when X is equal to one we get one minus one third, We're going to get 2/3 and we still have this one half out here and then when X is equal to negative one we get negative one -23 or negative one plus one third. And so negative one plus one third is negative two thirds, serving it one half times two thirds minus negative two thirds, which is equal to one half times two thirds plus two thirds. It's four thirds, Which is then just equal to 2/3.

In a girl That would be the length of the interval times three in a row of the function on that interval. So we end up with one, huh? Uh, let's see. You can see the riveters fundamental there. Let's keep Bill between minus X evaluated at one and bring you end up with 1/2. Uh, but seen three squared is nine land and sea contract one minus one when it was minus 2/3 here, plus defense. Six. So six must. This is a you know, withering. So 20/3, divided by 2 10/3 Okay, that's the average, really?

They want us to find The average value of the function at the X is equal to X squared minus one on the interval under three, well, average value. That's divided by the length of the interval, and we're going to multiply by the value of the function. When you think that after the interval on that, interval it so we'll have X squared minutes. One the x. So we'll have 1/2 and then by the fundamental Terry calculus. Exe cute over three minus X evaluated at 31 So let's see what we got. IM have apply three to it, one of with three squared, which was nine minus street. And then we're going to subtract Let's see 1/3 minus one. So the sign flips because we're subtracting that on the both 1/2 of let's see six plus one minus one day. So that's seven minus wondered car. In other words, 21 minus wondered 21/3 minus wondered. So end up with 20/3, but 1/2 of 20/3 is just 10/3

Before this problem or to find the average value of the function F. Of X, which is equal to 1/2 X over the interval 1 to 4. And we want to find the values of extra which the function equals its average value. Now the average value of the function F of X over an interval A B is equal to one over B -8 times the integral from A to B of F. Of X. Dx. And so using this we have f average that's equal to One over that's 4 -1 times the integral from 24 of the function 1/2 x. And then the X. Now this is just the same as 1/3 times you can take out the one half And then we multiply this by the integral from 1 to 4 of one over X. Dx. Now the Anti derivative of one over X is just L. And absolute value of X. And so we get 1/6 times Ln absolute value of X. This evaluated from 1 to 4. And so evaluating the get 1/6 times Ln absolute value of when X is four. That's Elena four minus when X is one. We have Ln of one now Ellen of one is just zero. And so we get Ellen of four Over six as the average value of the function. Now we need to find the value of X. For which the function 1/2 x equals this value. And so we get 1/2 X. This is just L. N four over X. Or that's the same as Six over Elena four, that's equal to two x. And if we divide both sides by two we get X is equal to three over LNL. four. And so this is the value of X that we want. And this is valid because this is approximately Um 2.1640. Which is just in between the interval 1- four.


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