Average distance on a parabola What is the av...

Question

Average distance on a parabola What is the average distance between the parabola y = 30𝓍 (20 ― 𝓍 ) and the 𝓍-axis on the interval [0, 20] ?

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to find the average distance between the parabola Y is equal to 18 X, multiplied by the quantity of 12 minus X in quantity, and the X axis on the interval from 0 to 12. And we're given 4 possible choices as our answers. For choice A, we have 584. For choice B, we have 432. For choice C, we have 864, and for choice D, we have 518. So this problem wants us to find the average distance between the parabola and the x-axis. So what we really need to do is to find the average value of the function for our parabola. And so, recall your formula for finding the average value of a function on an interval. That's going to be our average value, which is F average, is going to be equal to 1 divided by the quantity of B minus A in quantity, multiplied by the interval from A to B, of F of X. DX. Now, since we're looking at the interval from 0 to 12, That means that A is going to be equal to 0, B is going to be equal to 12. And our function F of X is just going to be our function for Y. And so that means that F of X is going to be equal to 18X multiplied by the quantity of 12 minus X in quantity. So substituting these values into our formula for the average value, we see that our average value F average is going to be equal to 1 divided by the quantity of 12 minus 0 in quantity, multiplied by the integral from 0 to 12 of 18 X, multiplied by the quantity of 12 minus X in quantity DX. Now, to integrate this function, we first want to distribute the 18 X into the quantity of 12 minus X. In doing so, we see that this is going to be equal to 1 divided by 12, multiplied by the integral from 0 to 12. Of the quantity of 216 multiplied by X. Minus 18 X 2 in quantity DX. So now we just need to integrate each of these terms individually. So, in doing so, we see that this is going to be equal to 1 divided by 12, multiplied by the quantity of 216, multiplied by, and when we integrate X with respect to X, we get X squared, divided by 2. And then we'll have minus. 18, multiplied by, and here we're integrating X2d with respect to X, and so we get X cubed, divided by 3. In quantity, and this needs to be evaluated from 0 to 12. So substituting in our limits of our integration, we see that this function is going to be equal to 1 divided by 12, multiplied by the quantity, and here we'll have 216 divided by 2, which is 108. And then that gets multiplied by, for the upper limit, X12 X squad, and so X is going to be equal to 12, so we'll have 12 squad. Then we'll have minus, we'll have 18 divided by 3, which is 6, and then that gets multiplied by X cubed, and so when X is equal to 12, we will have 12 cubed. Then we'll have minus the quantity, and here for our lower bound, we will have 108 multiplied by 0 squared. Then we'll have minus 6. Multiplied by 0 cubed in quantity. And then they'll find this expression. This is going to be equal to 1 divided by 12. Multiplied by the quantity of 15,552. Minus 10,368. Minus 0 + 0 in quantity. And finally, Performing this calculation, we see that this is going to be equal to 432. So that is the answer that we're looking for for this problem. And we compare our answer here to our answer choices, we see that the correct answer choice corresponds to answer choice B. So I hope that this explanation has been useful, and I'll see you in the next video.

Average distance on a parabola What is the average distance between the parabola y = 30𝓍 (20 ― 𝓍 ) and the 𝓍-axis on the interval [0, 20] ?

Key Concepts

Average Value of a Function

Definite Integral

Parabola Properties

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