### Video Transcript

Find the average value of π of π‘
is equal to two π‘ divided by the square root of two π‘ squared plus two on the
closed interval from one to three.

Weβre given a function π of π‘,
and weβre asked to find the average value of this function, π of π‘, on the closed
interval from one to three. Letβs start by recalling what we
mean by the average value of a function π of π‘ on a closed interval from π to
π. We have π average is equal to one
over π minus π times the integral from π to π of π of π‘ with respect to
π‘. In our case, we want the average
value of π of π‘ on the closed interval from one to three. So, weβll set π equal to one, π
equal to three, and π of π‘ to be two π‘ divided by the square root of two π‘
squared plus two.

And of course, there is one thing
we need to check. We need to make sure that we can
actually evaluate the definite integral from π to π of π of π‘ with respect to
π‘. One way of doing this is to prove
that π of π‘ is continuous on the entire interval of integration. In this case, we need to show π of
π‘ is continuous on the closed interval from one to two. And the easiest way to do this is
to take a look at our function π of π‘. We can see we take the square root
of a polynomial in our denominator, and our numerator is a linear function. So, π of π‘ is the quotient and
composition of continuous functions. And we know this means π of π‘ is
continuous on its entire domain.

So, we should ask the question,
what is the domain of the function π of π‘? Well, π of π‘ will be defined
everywhere except where our denominator is equal to zero or where two π‘ squared
plus two is negative. In other words, π of π‘ wonβt be
defined when two π‘ squared plus two is less than or equal to zero because then
weβre either dividing by zero or taking the square root of a negative number. However, we can solve this
inequality. Weβll subtract two from both sides
and then divide through by two, giving us π‘ squared should be less than or equal to
negative one.

But we know π‘ squared is greater
than or equal to zero for all values of π‘. Therefore, π of π‘ is defined for
all real values of π‘, so π of π‘ is continuous for all real values of π‘. And that means itβs integrable on
the closed interval from one to three, so we can use this formula.

Now, all we need to do is
substitute π is equal to one, π is equal to three, and our expression for π of π‘
into our formula. This gives us π average is equal
to one over three minus one times the integral from one to three of two π‘ divided
by the square root of two π‘ squared plus two with respect to π‘. And we can simplify this expression
slightly. First, one divided by three minus
one is equal to one-half. All thatβs left to do is evaluate
our definite integral. Weβll do this by using
substitution. Weβre going to set π’ to be two π‘
squared plus two. We then differentiate both sides of
this expression with respect to π‘ by using the power rule for differentiation. This gives us dπ’ by dπ‘ is equal
to four π‘.

And of course, we know dπ’ by dπ‘
is not a fraction. However, when weβre integrating by
using substitution, we can treat it a little bit like a fraction. This gives us the equivalent
statement in terms of differentials, dπ’ is equal to four dπ‘. And we can see that four π‘ almost
appears in our integrand. Our numerator is two π‘. So, we can multiply our integrand
by two and then divide our entire integral by two. This wonβt change the value of our
integral; however, it may make our substitution easier.

So now, weβre multiplying our
entire integral by one-quarter, and our new numerator is four π‘. But weβre not yet done. Remember, weβre using integration
by substitution on a definite integral. So, we need to calculate the new
limits of integration. To find the new upper limit of
integration, we substitute π‘ is equal to three into our substitution for π’. This gives us π’ is equal to two
times three squared plus two, which we can calculate is equal to 20. We can do the same to find the new
lower limit of integration. We substitute π‘ is equal to one
into our expression for π’. This gives us π’ is equal to two
times one squared plus two, which we can calculate is equal to four.

Weβre now ready to use our
substitution to help us evaluate this integral. First, our new limits of
integration are four and 20. And then, by using the substitution
π’ is equal to two π‘ squared plus two, we can rewrite our integrand as one over the
square root of π’. And now, weβre integrating with
respect to π’. And we can evaluate this integral
by using the power rule for integration. Remember, one over the square root
of π’ is the same as π’ to the power of negative one-half. So to integrate this, we add one to
our exponent of π’ and then divide by this new exponent. This gives us the antiderivative π’
to the power of one-half divided by one-half.

And remember, this is a definite
integral, so we donβt need to add a constant of integration. And we need to add our limits of
integration, four and 20. And we can simplify our
antiderivative slightly. Instead of dividing by one-half, we
can multiply through by the reciprocal of one-half, which is two. This just gives us two π’ to the
power of one-half. And of course, we know π’ to the
power of one-half is the square root of π’. This gives us the following
expression.

And in fact, we can even simplify
this even further. We can take the factor of two
outside of our expression. And of course, two divided by four
is equal to one-half. So therefore, weβve shown π
average is equal to one-half times the square root of π’ evaluated at the limits of
integration, π’ is equal to four and π’ is equal to 20. So all thatβs left to do now is
evaluate this expression at the limits of integration. Evaluating this at the limits of
integration, we get one-half times the square root of 20 minus the square root of
four. And finally, all thatβs left to do
is evaluate this expression. If we do this, we get root five
minus one, which is our final answer.

Therefore, we were able to show the
average value of π of π‘ is equal to two π‘ divided by the square root of two π‘
squared plus two on the closed interval from one to three is given by the square
root of five minus one.