Question Video: Finding the Average Value of a Function on a Given Interval Involving Using Integration by Substitution

Find the average value of 𝑓(π‘₯) = π‘₯Β²/(π‘₯Β³ βˆ’ 5)Β² on the interval [βˆ’1, 1].

02:42

Video Transcript

Find the average value of 𝑓 of π‘₯ equals π‘₯ squared over π‘₯ 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 π‘Ž to 𝑏 is given as one over 𝑏 minus π‘Ž times the integral of 𝑓 of π‘₯ with respect to π‘₯ evaluated between π‘Ž and 𝑏. In this case, we can see that our function is equal to π‘₯ squared over π‘₯ cubed minus five all squared. And we see our closed interval is from negative one to one. So we let π‘Ž 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 π‘₯ squared over π‘₯ cubed minus five all squared with respect to π‘₯ 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 π‘₯ cubed minus five is three π‘₯ squared. And this tells us we can use integration by substitution. We let 𝑒 be equal to π‘₯ cubed minus five, and d𝑒 by dπ‘₯ is therefore equal to three π‘₯ squared. We know that d𝑒 by dπ‘₯ 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 d𝑒 equals π‘₯ squared dπ‘₯.

We’re now in a position to replace the various parts of our integral. We replace π‘₯ squared dπ‘₯ with a third d𝑒. And we replace π‘₯ cubed minus five with 𝑒. And we see that the average value of our function is equal to a half times the integral of a third times one over 𝑒 squared with respect to 𝑒. But what do we do about these limits?

Well, we use our definition of 𝑒. We said that 𝑒 was equal to π‘₯ cubed minus five. So for our upper limit, when π‘₯ is equal to one, 𝑒 is equal to one cubed minus five, which is negative four. And when π‘₯ is equal to negative one, 𝑒 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 𝑒 squared as 𝑒 to the negative two. And we know that the integral of 𝑒 to the negative two is 𝑒 to the negative one divided by negative one, or negative 𝑒 to the negative one, which can then be written as negative one over 𝑒.

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.

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