Average Value of a Function - Problem 3
Let’s find the average value of another function. Here we want to find the average of x to the 2/3 times the quantity 8 minus x on the interval from 0 to 8. I’ve got a graph of that function here. This is the curve x to the 2/3 times 8 minus x. It crosses the x axis at 0 and at 8. Actually it just touches down here and bounces up but I didn’t show that part. It goes up to a maximum value of about 10.4, just to give you some bearings. Remember that the average value of this function is going to be this area; the area underneath the curve, divided by the length of this interval. That’s what the formula give us.
F bar, 1 over the length of the interval, 8 minus 0 times the area under the curve. And that’s the integral from 0 to 8 of this function. X to the 2/3 times the quantity (8 minus x)dx. Let’s take this calculation up here. We have 1/8, the integral from 0 to 8, x to the 2/3 times 8 minus x.
The first thing I would do is distribute this x to the 2/3 over these two terms, so I could just get powers of x in my integrand. I get 8 times x to the 2/3 minus, and x to the 2/3 times x. You add the exponents, you get 2/3 plus 1 is 5/3, so it's x to the 5/3. Now I just have powers of x that I can use the power rule for anti derivatives to anti differentiate these.
I have 1/8 and the anti derivative of this guy is going to be 8 times x to the 5/3. I have to add one to the exponent, 2/3 plus 1 is 5/3 and I have to divide by 5/3, which is the same as multiplying by 3/5. So 3/5x to the 5/3 minus. And here I also have to add one to the exponent, 5/3 plus 1 is 8/3. So it's x to the 8/3 divided by 8/3 which is the same as multiplying by 3/8. I have to evaluate this anti derivative from 0 to 8.
I’m going to actually pull this one 1/8 inside. It will cancel with this 8. We’re going to have 3/5 x to the 5/3 minus 3/64 x to the 8/3, and I have to evaluate this from 0 to 8. When I plug 8 in, I get 3/5, 8 to the 5/3 which I’ll evaluate in a second. Minus 3/64, 8 to the 8/3, that will be fun. When I plug 0 in, both of these terms are going to be 0, so I have minus 0.
What do we have here and here? Well, 8 to the 5/3, since 8 is a perfect cube, I’m going to take the cube root first and I’ll get 2. Cube root of 8 is 2. 2 to the 5th. So this first term is 3/5 times 32, that’s 2 to the 5th power. Minus 3/64 times, and the cube root of 8 is 2. 2 to the 8th is the same as 4 to the 4th, 256. 256/64 is 4. Let me just write 2856 for now.
Next step, this is 3 times 32, 96 over 5 minus; and here I have 256/64 which is 4, times 3, 12 in 5ths is 60/5. So this is 96/5 minus 60/5. This all becomes 36/5, which is the same as 72/10, 7.2. That’s the average value of my function.
Let’s take a look at the graph and see if that makes sense. Since I did graph this, I can plot 7.2, it will be somewhere around here. And that does kind of make sense. If you imagine that this was the surface of a tank of water, it was disturbed and forced into this bubble shape. After it’s settled, 7.2 seems like a reasonable value for the final depth. That’s the average value of this function.