WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, and welcome back to AP Calculus.
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Today, we are going to be talking about the average value of a function.
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Let us jump right on in, let us work in blue here.
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Let f(x) be defined by the following table of values.
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We have x and then we have different values for f(x).
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Let us take 1, 2, 3, 4, 5, and then, 6, 7, 8, 9, and 10.
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We have 10 discreet points and our values are going to be 2, 2.7, 3.6, let us say 2.5, 6.8.
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I have 9.7, 10.2, 10, and 9 is 13.4, we have 11.7.
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What is the average value of this function?
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In other words, what is the average value of f(x)?
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It is very simple, like any average value, you add them all up and you divide by the number of points that there are.
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What is the average value of f(x)?
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It is exactly what you think.
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The average = 1/10 because there are 10 points.
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The sum as x goes from 1 to 10 of f(x).
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f(1)+ f(2) + f(3) + f(4), all divided by 10, like any other average.
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Now the question is what if the domain of f(x) is not finite?
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It is a finite domain, I have 10 numbers in it.
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I take an interval, like from 0 to 5.
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The length of that interval is 5 but the number of points in that interval is infinite,
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because there is an infinite number of points on the real number line.
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How do we find the average value of the function then?
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Such as, if I have the function f(x) is equal to x².
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If I have x is in the interval from 2 to 5.
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What is going to be the average value of the function?
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Between 2 and 5, f(x) is going to take on an infinite number of values.
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How do I find the average?
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It is the same thing, we add and we divide by the length, so to speak.
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Same thing here, except adding when we are dealing with an infinite domain becomes integration.
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Because we are adding smaller and smaller things, we passed the integration, when we passed the limit.
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We integrate the function and then we divide by the length which in this case is going to be 5/2.
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Average value, here the average value = 1/ the length, which is going to be b – a.
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This is the b, this is the a.
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The integral from a to b of f(x) dx, that is it, that is the definition of the average value of a function.
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It is going to be the integral of the function from a to b, the lower limit to the upper limit divided by the length of the interval.
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It is essentially what we did here.
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We are adding up all the values of the function and we are dividing by the number there are.
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Here we are adding up all values of the function, the integral.
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And we are dividing by the number there are which is essentially going to be the length of the interval.
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Let us calculate the average value for f(x) = x², on the interval from 2 to 5.
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Let me go back to blue here.
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The average value is going to be 1/ 5 - 2 × the integral from 2 to 5 of x² dx.
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It is just going to equal 1/3 x³/ 3 from 2 to 5
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which is going to equal 1/3, 125/3 - 8/3 which is going to equal 13.
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What does this mean?
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The average value of f(x)² from 2 to 5 is 13, what is that mean?
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What does this mean? It means this.
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I know my x² function look something like that.
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If I have 2 and if I have 5, from 2 the value of f(x) is going to be 4.
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At 5, the f(x) value is going to be 25.
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What this means is that, when x goes from 2 to 5, f(x) takes on values from 4 to 25.
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From 4 to 25, it takes on all the values from 4 to 25.
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The average of those values is 13, that is it.
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The average of those values given the weight of the function is not linear.
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It is not y = x, the average of those values is 13.
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Now you have an average, you have a way of finding an average of a finite number of terms.
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Now you have a way of finding the average of an infinite number of terms.
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That is all we are doing, we are using integration.
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That is all, very simple.
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The hardest issue is going to be integration itself.
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We can find the average of an infinite sum, if you will, an infinite sum of numbers.
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We just passed the integration.
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Again, for f(x) such that x is in the interval from a to b.
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The average value of f/ that interval is equal to 1/ b - a × the integral from a to b of f(x) dx.
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Now we know what the integral from a to b of f(x) dx is, it is the area under the curve from a to b.
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It is the area under the curve from a to b.
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It turns out, we have something called the mean value theorem for integrals.
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We have the mean value theorem for derivatives, way back when.
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The mean value theorem for integrals says if f(x) is continuous on a closed interval ab,
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then there exist a number c which is contained in ab.
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There is a number c between a and b, such that f of that c is actually equal to the average value
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or another way of saying this, I can move this a/b to the other side.
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There is a c such that f of that c × the length of the interval actually = the area under the curve.
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What that means is the following.
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Something like that.
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This is a, this is b.
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Let us go ahead and use our example that we did.
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This is our x² function, y = x².
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This is 4, this is 25, this was our 13.
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What the mean value theorem for integrals is saying is that there is some number c here.
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There is some number c such that, let us draw this out actually.
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Let me draw that again.
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I have got 2, this is not a and b, this was 2 and this was 5.
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Let us make it specific, 5 and 13.
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The mean value theorem says that the f(c), there is some c.
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In this particular case, it is going to be √13 because f (c) is 13.
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The average value of the function, f(c) which was 13 × 5 - 2 = the integral from 2 to 5 of the x² dx.
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In other words, f(c) × b - a = the integral from a to b of f(x) dx.
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What this means geometrically is that the area under the curve, this area right here,
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there is some c such that f(c) × f(c) which is in this case 13 × the length.
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This is the 13, that is our height.
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This is the 5 – 2, that is 3, where the area underneath that rectangle is the area underneath the curve.
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That is it, average value, that is all that means.
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Here we have the area under the curve.
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Here is the area of the interval length × the average value.
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That is it, mean value theorem for integrals.
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Let us do some examples.
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Find the average value of the given function over the given interval y = t³ e⁻²⁴ from 0 to 4, very simple.
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Let us go ahead and work in blue here.
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We know that our f average is equal to 1/ b – a, the integral from a to b of f(x) dx.
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Here that is equal to 1/ 4 – 0, the integral from 0 to 4 of t³ e ⁻t⁴ dt.
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That is all, that is all we are doing, this is just an integration problem.
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This is going to equal 1/4 × the integral from 0 to 4 of,
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Let me do it over here first.
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I’m going to do a u substitution.
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I’m going to call u - t⁴, du is going to be -3t³ dt.
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Therefore, t³ dt is going to equal -1/3 du.
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The du = -1/4 t³, there we go dt.
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Take a breath, slow down, there we go.
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-4t³ dt, there we go.
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Now we have t³ dt is equal to -du/4.
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Now I'm going to go ahead and put these in.
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It is going to be ¼, that is that one, the integral, t³ dt.
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t³ dt is going to be - du/4 × e ⁺u = -1/16 0⁴ e ⁺u du = -1/16 e ⁺u
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= -1/16 e ^- t⁴, from 0 to 4 = -1/16 × 1/ e ⁺256 – e⁰ which is 1.
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If you want, I can go ahead and just leave it that way.
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The graph is going to look as follows.
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This is the function, this is y = t³ e ⁻t⁴.
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The average value is a particular something.
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We went from 0 to 4, past here.
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Clearly, even though the majority of it is you are getting up to like 0.38, 0.39,
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but we are actually taking all of these average values from about 1.6, 1.7, all the way to 4 is virtually 0,
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which is why the average value of this function from here to 4 is actually going to be very small.
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Find the average value of the given function over the given interval,
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then find the number c in the given interval such that f(c) = the f average.
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Our function is this and our interval is 3,8.
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Let us see, our f average is equal to 1/ b - a which is 8,3, 3 to 8 of 5 × cos(x) – sin(5x) dx.
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This is going to equal 1/5, the integral from 3 to 8 of 5 cos x - 1/5 the integral from 3 to 8 of sin 5x and my dx.
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5 comes out, we are left with just the integral of cos x which is going to be sin x from 3 to 8.
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This is going to be -1/5 × 1/5.
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The integral of sin is –cos, this is going to be × a - cos 5x from 3 to 8.
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Let us see what I have got.
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When I actually do this, I should have sin x, from 3 to 8 this is going to be, we have +1/25.
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That is fine, 1/25.
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You know what, I do not need simplify it, just go ahead and put these in.
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I have got sin 8 - sin 3 -, when I multiply all these out, putting 8 and 3, keeping this negative sign here,
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I left the negative sign in here, I just pulled off the 1/5 to turn it into 1/25.
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I left it like that.
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What you are going to end up getting once you expand is going to be cos of 40/25 - cos 15/25.
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I ended up with the value of 0.852, that is the average value of this function.
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Here is what it looks like.
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We are going from 3 to 8, this is the actual graph itself.
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This is y = 5 cos x – sin 5x.
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We are going from 3 to 8.
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Some negative values, some positive values, some negative values.
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On average, we are going to end up with 0.852.
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We want to find the c such that f(c) = the average value.
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f(c) is just 5 × cos(c) - sin 5c = the average value which is 0.852.
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When I solve this equation for c, I move this over to the left to set it equal to 0.
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Use my calculator or whatever graphical utility I have, I get c is equal to 4.685 and c is equal to 7.611.
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The interval was from 3 to 8.
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Both of those numbers are in the interval from 3 to 8.
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Therefore, both of these, there are two numbers.
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It does not have to be just one, you can have more.
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There are two numbers c, such that f of these c’s is equal to 1/ b - a × the integral from a to b of f(x) dx.
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Example 3, find the number a such that the average value of the function = 2/ the interval from -1 to a.
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The average value of the function 1/ b – a, a - -1 × the integral from -1 to a of -4x² + 8x + 4 dx.
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We want this average value to equal 2.
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Solve this equation for a.
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I get 1/ a + 1, the integral here is -4x³/ 3 + 8x²/ 2.
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This is going to be + 4x² + 4x from -1 to a is equal to 2 – 4a³/ 3
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+ 4a² + 4a - 4/3 + 4 - 4 = 2 × a + 1,
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multiplied through, solve for that.
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I end up with -4a/ 3 + 4a² + 4a - 4/3 + 2a + 2.
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Combine terms, a’s, numbers, I get – 4a³/ 3.
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I simplify this equation, I bring things together, multiply by 3.
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My final equation I get is, once I have gotten rid of the fractions, -4a³ + 12a² + 6a - 10 = 0.
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When I solve this, I get a = -1, I get a = 0.775, and I get a = 3.225.
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-1 to -1, we can ignore this one.
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We have -1 to 0.775, two possible values, and -1 to 3.225.
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There you go, that function is the graph of the function.
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We are going from -1, this is one value of a and this is our second value of a.
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That makes this average value equal 2.
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That is it, we are just using the definition of average value.
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The linear density of a rod 10 m long is a function of the distance from one end.
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The density function is this, 17/ √3x + 5 kg/m, find the average density of the rod.
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The average density is 1/, what does this rod looks like?
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We have a rod which is going to be 10 m long.
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This is 0, this is 10.
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They are telling me as I move along the x axis, as x increases from 0 to 10, the density changes.
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It is a variable density, as we move along the rod.
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The average value is just , the density average = 1/ 10 - 0
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the integral from 0 to 10 of the functions 17/ √3x + 5 dx.
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How are we going to solve that integral?
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Let us go ahead and do this.
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Let us do u is equal to 3x + 5¹/2, that is going to make du = ½ 3x + 5 ^½ × 3 dx,
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which gives me 2 du divided by 3 is equal to dx/ √3x + 5.
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Perfect, that means our average density is equal to 1/10 from 0 to 10,
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2 du/ 3 because now dx/ 3x + 5 is that right there.
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That is just 2/3 du, then × 17.
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We are going to get 2 × 17 is 34/30.
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We are going to get 34/30 × the integral du from 0 to 10 which = 34/30 u from 0 to 10 = 34 ⁺30.
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u is equal to √3x + 5, from 0 to 10.
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When we solve this, we end up with 4.17 kg/m.
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That is it, the average value of the depth, they have given you a function.
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All this means is that when we move from 0 to 10, this is the density.
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As you move along the rod, it is more dense here, towards the left end of the rod.
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As you move to the right, the density starts to decrease.
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That is it, that is all that is going on here.
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The average density is going to be the average value of the function.
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That is all hope, I hope that makes sense.
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Once again, the average value of the function over the interval from a to b is equal to 1/ b – a, the integral from a to b of f(x) dx.
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Thank you so much for joining us here at www.educator.com.
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We will see you next time, bye.