Video Transcript
Find the average value of 𝑓 of 𝑥 equals 𝑥 squared over 𝑥 cubed minus five all squared on the closed interval negative one to one.
Remember, the formula for the average value of a function over a closed interval 𝑎 to 𝑏 is given as one over 𝑏 minus 𝑎 times the integral of 𝑓 of 𝑥 with respect to 𝑥 evaluated between 𝑎 and 𝑏. In this case, we can see that our function is equal to 𝑥 squared over 𝑥 cubed minus five all squared. And we see our closed interval is from negative one to one. So we let 𝑎 be equal to negative one and 𝑏 be equal to one. And this means the average value of our function is given as one over one minus negative one multiplied by the integral of 𝑥 squared over 𝑥 cubed minus five all squared with respect to 𝑥 evaluated between negative one and one.
One minus negative one is two. But how do we evaluate this definite integral? Well, we need to notice that the numerator is a scalar multiple of the derivative of part of the denominator. The derivative of 𝑥 cubed minus five is three 𝑥 squared. And this tells us we can use integration by substitution. We let 𝑢 be equal to 𝑥 cubed minus five, and d𝑢 by d𝑥 is therefore equal to three 𝑥 squared. We know that d𝑢 by d𝑥 is not a fraction. But for the purposes of integration by substitution, we do treat it a little like where we say that this is equivalent to a third d𝑢 equals 𝑥 squared d𝑥.
We’re now in a position to replace the various parts of our integral. We replace 𝑥 squared d𝑥 with a third d𝑢. And we replace 𝑥 cubed minus five with 𝑢. And we see that the average value of our function is equal to a half times the integral of a third times one over 𝑢 squared with respect to 𝑢. But what do we do about these limits?
Well, we use our definition of 𝑢. We said that 𝑢 was equal to 𝑥 cubed minus five. So for our upper limit, when 𝑥 is equal to one, 𝑢 is equal to one cubed minus five, which is negative four. And when 𝑥 is equal to negative one, 𝑢 is equal to negative one cubed minus five, which is equal to negative six. We can take the constant a third outside of the integral sign and rewrite one over 𝑢 squared as 𝑢 to the negative two. And we know that the integral of 𝑢 to the negative two is 𝑢 to the negative one divided by negative one, or negative 𝑢 to the negative one, which can then be written as negative one over 𝑢.
We substitute negative four and negative six. And we get a sixth of negative one over negative four minus negative one over negative six. Well, negative one over negative four is just a quarter. And negative one over negative six is a sixth. We subtract these fractions by creating a common denominator. And we see that the average value of our function is a sixth times a twelfth, which is one over 72.
