Average Value of a Function - Problem 1
Let’s find the average value of a function. We have this formula for the average value. 1 over the length of the interval over which you’re averaging, times the area underneath the curve that you’re averaging. Let’s apply this to find the average value of f(x) equals root x on the interval from 0 to 9.
The integral on the right hand side of the formula represents this area here. And I need to divide that area by the length of this interval and that would give me the average value. Let’s calculate this up here.
Average value is 1 over 9 minus 0, that’s the width of the interval, times the integral from 0 to 9 of f(x). Let me just write in square root of xdx. So that’s 1/9 integral from 0 to 9. And let me change the square root of x to x to the ½. I want to represent it as a power because that’s the way I antidifferentiate it. So 1/9, and the antiderivative of x to the ½ is I raise this number by 1, I get 3/2 and I divide by 3/2 which is the same as multiplying by 2/3. It’s 2/3x to the 3/2, and I evaluate that from 0 to 9. I can actually pull this 1/9 in, it doesn’t really do much one way or the other what’s 9 to the 3/2, because but 2/27x to the 3/2 from 0 to 9.
Let me evaluate this at 9 first. 2/27 times 9 to the 3/2 minus 2/27 times 0 to the 3/2, that’s just going to be 0. What’s 9 to the 3/2? Because 9 is a perfect square I’m going to take the square root first I’ll get 3 and then I’ll cube the result and get 27. So this is 2/27 times 27 which is 2. 2 is the average value of the square root function over the interval form 0 to 9.
Let’s take a look at the graph and see what that looks like. The maximum value it reaches is 3 at 9, 2 is right about here, this is the average value. Think of it this way, if this were a tank of water, and this curve represented the surface of the water after the water settled, this is the level it would reach, the level 2.