Video: Finding the Average Value of a Function

What is the average value of 8/(4 + π‘₯Β²) in the closed interval [0, 2]?

03:41

Video Transcript

What is the average value of eight over four plus π‘₯ squared in the closed interval zero to two?

The average value of a function 𝑓 of π‘₯ over some closed interval from π‘Ž to 𝑏 is one over 𝑏 minus π‘Ž times the definite integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯, assuming, of course, 𝑓 of π‘₯ is itself integrable. In this question, we can see that our 𝑓 of π‘₯ is defined as eight over four plus π‘₯ squared. We’re interested in finding the average value of this function over the closed interval from zero to two. So, we’re going to let π‘Ž be equal to zero and 𝑏 be equal to two.

And so, we calculate the average value of our function by substituting everything we know into the formula. And we get one over two minus zero times the definite integral between zero and two of eight over four plus π‘₯ squared with respect to π‘₯. Of course, two minus zero is simply two, so we can simplify this fraction a little. But how do we integrate eight over four plus π‘₯ squared? Well, here, it helps to be able to recognize the antiderivative.

We know that the derivative of the inverse tan or arctan of π‘₯ with respect to π‘₯ is one over one plus π‘₯ squared. And so, the antiderivative of one over one plus π‘₯ squared is the arctan or inverse tan of π‘₯. We’re going to manipulate the expression eight over four plus π‘₯ squared, so it looks a little more like this. To begin, we’ll divide both the numerator and denominator of our fraction by four. Eight divided by four is two, so our new numerator is two. Then, dividing our denominator by two, and we get one plus π‘₯ squared over four

But in fact, four is two squared, so we can write π‘₯ squared over four as π‘₯ over two all squared. And of course, we know that we can take any constant factors outside of the integral sign and focus on integrating the function itself. So, let’s take out a constant factor of two. And our average value becomes a half times two times the definite integral between zero and two of one over one plus π‘₯ over two all squared with respect to π‘₯.

Now, of course, a half times two is one. So, we have the definite integral between zero and two of one over one plus π‘₯ over two squared with respect to π‘₯. And now, we should notice this looks a little bit like the derivative of the inverse tan of π‘₯. There’s a difference, though; we have π‘₯ over two squared rather than π‘₯ squared. So, we’re going to perform a substitution. We’re going to let 𝑒 be equal to π‘₯ over two. And this means that d𝑒 by dπ‘₯, the derivative of 𝑒 with respect to π‘₯, is one-half.

Now, d𝑒 by dπ‘₯ isn’t a fraction, but in this process we do treat it a little like one. And we can say that this is equivalent to two d𝑒 equals dπ‘₯. We then replace π‘₯ over two with 𝑒 and dπ‘₯ with two d𝑒. But what about the limits on our integral? Well, we go back to our substitution. Our upper limit is π‘₯ equals two. And when π‘₯ equals two, 𝑒 is equal to two divided by two, which is one. Similarly, when π‘₯ is equal to zero, that’s our lower limit, 𝑒 is zero divided by two, which is zero. And so, the average value of our function is equal to the definite integral between zero and one of one over one plus 𝑒 squared two d𝑒.

Once again, we’ll take out this constant factor, so we get two times that definite integral. And so, by applying the antiderivative, we can see that the average value of our function is two times the inverse tan of 𝑒 between the limits of zero and one. Let’s substitute these limits in. We get two times the inverse tan of one minus the inverse tan of zero. That’s two times πœ‹ by four minus zero or simply two times πœ‹ by four, which gives us a value of πœ‹ by two. And so, the average value of eight over four plus π‘₯ squared in the closed interval zero to two is πœ‹ by two.

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