What is the average value of eight
over four plus 𝑥 squared in the closed interval zero to two?
The average value of a function 𝑓
of 𝑥 over some closed interval from 𝑎 to 𝑏 is one over 𝑏 minus 𝑎 times the
definite integral between 𝑎 and 𝑏 of 𝑓 of 𝑥 with respect to 𝑥, assuming, of
course, 𝑓 of 𝑥 is itself integrable. In this question, we can see that
our 𝑓 of 𝑥 is defined as eight over four plus 𝑥 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 𝑎 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 𝑥 squared
with respect to 𝑥. 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 𝑥 squared? Well, here, it helps to be able to
recognize the antiderivative.
We know that the derivative of the
inverse tan or arctan of 𝑥 with respect to 𝑥 is one over one plus 𝑥 squared. And so, the antiderivative of one
over one plus 𝑥 squared is the arctan or inverse tan of 𝑥. We’re going to manipulate the
expression eight over four plus 𝑥 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 𝑥 squared over four
But in fact, four is two squared,
so we can write 𝑥 squared over four as 𝑥 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
𝑥 over two all squared with respect to 𝑥.
Now, of course, a half times two is
one. So, we have the definite integral
between zero and two of one over one plus 𝑥 over two squared with respect to
𝑥. And now, we should notice this
looks a little bit like the derivative of the inverse tan of 𝑥. There’s a difference, though; we
have 𝑥 over two squared rather than 𝑥 squared. So, we’re going to perform a
substitution. We’re going to let 𝑢 be equal to
𝑥 over two. And this means that d𝑢 by d𝑥, the
derivative of 𝑢 with respect to 𝑥, is one-half.
Now, d𝑢 by d𝑥 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 d𝑢 equals d𝑥. We then replace 𝑥 over two with 𝑢
and d𝑥 with two d𝑢. But what about the limits on our
integral? Well, we go back to our
substitution. Our upper limit is 𝑥 equals
two. And when 𝑥 equals two, 𝑢 is equal
to two divided by two, which is one. Similarly, when 𝑥 is equal to
zero, that’s our lower limit, 𝑢 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
𝑢 squared two d𝑢.
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 𝑢 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 by four minus
zero or simply two times 𝜋 by four, which gives us a value of 𝜋 by two. And so, the average value of eight
over four plus 𝑥 squared in the closed interval zero to two is 𝜋 by two.