Mean Value Theorem for Integrals Find or appr...

Question

Mean Value Theorem for Integrals Find or approximate all points at which the given function equals its average value on the given interval.

ƒ(𝓍) = 1 ― |𝓍| on [―1, 1]

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says let F of X be equal to 3 minus 2 multiplied by the absolute value of X on the closed interval from -1 to 2. At which points on the closed interval from -1 to 2, does F of X equal its average value on this interval? And we're given 4 possible choices as our answers. For choice A, we have X contained in a set of minus 5 divided by 6. For choice B, we have X contained in the set of 5 divided by 6. For choice C, we have X contained in the set of minus 5 divided by 6 and 5 divided by 6. And for choice D, we have X contained in the set of minus 6 divided by 5 and 6 divided by 5. So, the first thing we need to do to solve this problem is to figure out the average value of F of X on this interval. I recall to find the average value, F average, that that is going to be equal to 1, divided by the quantity of B minus A. In quantity multiplied by the integral from A to B of F of X D X. Well, where in our case, A is going to be equal to -1 and B is going to be equal to 2. So substituting in those values, we see that the average F is going to be equal to 1 divided by the quantity of 2 minus minus 1 in quantity, multiplied by the integral from -1 to 2 of FX, which we were given in the problem, as the quantity of 3 minus 2 multiplied by the absolute value of X in quantity DX. Now, in order to evaluate this interval, we actually want to write the absolute value of X as a piece-wise function. So, on the interval from -1 to 0, We can write the absolute value of X as being equal to minus X. So this means that F of X. It's going to be equal to. 3 minus 2 multiplied by minus X, which is equal to 3 + 2 X. Now, on the interval from 0 to 2, We can write the absolute value of X as being equal to X. So that means that F of X. It's going to be equal to. 3 minus 2 X. So, what we're going to do is write our single integral here into two different integrals. With different limits So this means that our average F. is going to be equal to 1 divided by 3, multiplied by the quantity of the integral from -1 to 0. Of the quantity of 3 + 2 X. In Quanti DX. Plus the integral from 0 to 2 of the quantity of 3 minus 2 X in quantity DX in quantity. So next we'll just perform the integration, so this is going to be equal to 1 divided by 3. Multiplied by the quantity, and here for the first interval, when I integrate 3 with respect to X, I get 3 X. They'll have + 2. Multiplied by, and here I'm going to be integrating X with respect to X, and that will give me X2 divided by 2. And then that whole quantity is going to be evaluated over the interval from -1 to 0. Then we'll have plus for the next interval. We're again going to be integrating 3, so that will give us 3 X. Then we'll have -2, and again, we're integrating X, that'll give us X2d divided by 2. And then that quantity is going to be evaluated from 0 to 2. And so now, if I uh plugging in the upper and lower limits, this is going to be equal to 1 divided by 3, multiplied by the quantity. Of 3, multiplied by 0. Plus, and here we can simplify the expression a little bit since we have 2 divided by 2, that'll just give us 1, so we really just have X squad for the second term, and so we'll have 0 squared. Then we'll have minus the quantity, and for a lower limit, we'll have 3 multiplied by -1. Plus minus 1 squared. In quantity Then plus or the second interval, or the upper limit of 2, we'll have 3 multiplied by 2. Minus, and again, here we can simplify the second expression to just X2. So we'll have minus. 2 squared. Minus the quantity of 3, multiplied by 0 for the lower limit, minus 0 squared in quantity. And when we evaluate this expression, this is going to be equal to 4 divided by 3. So now we just need to find the X values, where F of X is equal to 4 divided by 3, and we're gonna have to do this on the two intervals that we broke our function up into. So, for the first interval, we're going to look at From -1 to 0. And on this interval, we will have 4 divided by 3. Being equal to 3 + 2 X. And so we just need to solve this expression for X. And so the first step is to subtract 3 from both sides of the equation. In doing so, we'll have minus 5 divided by 3, and that's going to be equal to 2 X. And it's all for X, we'll just divide both sides of the equation by 2, and that means that X is going to be equal to minus 5 divided by 6. So that is one possible location. But we need to check our other interval, so we need to check from 0 to 2. And so here we will have 4 divided by 3, and this is going to be equal to 3 minus 2 X. And so again, we need to solve for X. So subtracting 3 from both sides, we'll get minus 5 divided by 3 is equal to -2 X, and divided both sides by minus 2, we get that X is equal to 5 divided by 6. That is another possible location, and these are actually the only two possible locations, since we now covered the entire interval. And so that means that x. is going to be contained in a set of minus 5 divided by 6. And 5 divided by 6. This here is the answer that we're looking for, for this problem. And if we compare this to our answer choices, We see that the correct answer choice corresponds to answer choice C. So I hope that this explanation has been useful, and I'll see you in the next video.

Mean Value Theorem for Integrals Find or approximate all points at which the given function equals its average value on the given interval.

ƒ(𝓍) = 1 ― |𝓍| on [―1, 1]

Key Concepts

Mean Value Theorem for Integrals

Average Value of a Function

Continuous Functions

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