 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# Average Value of a Function - Problem 2

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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The formula for average value of f(x) from x=a to x=b is the integral of f(x) from a to b, divided by (b - a). b - a is the length of the interval over which you are finding an average value.

In order to calculate the average value, simply plug in values of a, b, and f(x) into the formula. Integrate f(x) using the usual rules or methods. Evaluate at the limits of integration a and b, and take the difference of the value of the integral at b and at a.

Let’s find the average value of another function. Let’s find the average value of f(x) equals 12x over x² plus 16 on the interval from 2 to 8. Here is a graph of that function, just the part that we’re interested in, between 2 to 8. This shaded region represents the area under that graph which will be given by the integral of this function from 2 to 8.

Remember that’s how you get the average value. You find the area underneath the function of the interval you’re interested in and then divide by the length of the interval which is 6. That’s going to give us this formula; f bar is 1 over (8 minus 2), the length of the interval, times the integral from 2 to 8 of f(x), in this case; 12x over x² plus 16dx. This is what I’m going to evaluate.

Let’s make the observation that this 1/6, I can move it inside the integral and multiply it by the 12 and get 2x. So this whole thing simplifies to the integral from 2 to 8 of 2x over x² plus 16dx. Let me evaluate that up here. The integral from 2 to 8 of 2x over x² plus 16. This looks like a substitution integral, so I’m going to substitute for the x² plus 16 in the denominator. Another clue that it’s a substitution is that, this 2x is the derivative of my x² plus 16. I’ll need that in a second. I have to calculate dw and I get 2xdx. It’s a perfect substitution. Dw will take the place of the 2x and the dx.

I also want to substitute for the limits. This lower limit x equals 2 becomes 2², 4 plus 16, 20 so w equals 20. And the upper limit of 8, x equals 8 becomes 8², which is 64 plus 16, 80. So my new integral is going to be the integral form 20 to 80 2xdx that’s dw over x² plus 16 and that’s w. Really easy integral. The integral of 1/w is the natural log of the absolute value of w. So this is going to be ln of the absolute value of w from 20 to 80. So the natural log of the absolute value of 80, which is just natural log 80 minus the natural log of the absolute value of 20 which is ln20.

You can use the properties of logs to combine this into a single logarithm. Remember the difference of logarithms is the log of a quotient; 80/20. And that’s just 4, ln 4. That’s the average value of my function. Now if I wanted to get an idea of exactly what this was in decimal, it’s approximately 1.386. Let’s take a look back at our graph. This function, which happens to start and end at about 0.2, and has a maximum value of 1.5 on this interval, is going to have an average value of about 1.386 which is nearly 1.4, about here.

Let’s just see what that looks like. So right about there. And that means that if this were a tank of water and this were the surface of the water, after the water settled, this is the level it would settle to. Remember, the way we define average value; it’s the height that the rectangle needs to be to have the same area as the area under the curve.