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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Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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. 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.
Unit
The Definite Integral