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

# Question Video: Finding the Average Value of a Function on a Given Interval Using Integration by Substitution Mathematics • Higher Education

Find the average value of π(π‘) = (2π‘)/β(2π‘Β² + 2) on the interval [1, 3].

05:20

### 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.

Weβre given a function π of π‘, and weβre asked to find the average value of this function, π of π‘, on the closed interval from one to three. Letβs 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, weβll 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‘ wonβt be defined when two π‘ squared plus two is less than or equal to zero because then weβre either dividing by zero or taking the square root of a negative number. However, we can solve this inequality. Weβll 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 itβs 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 thatβs left to do is evaluate our definite integral. Weβll do this by using substitution. Weβre 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 weβre 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 wonβt change the value of our integral; however, it may make our substitution easier.

So now, weβre multiplying our entire integral by one-quarter, and our new numerator is four π‘. But weβre not yet done. Remember, weβre 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.

Weβre 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, weβre 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 donβt 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, weβve 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 thatβs 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 thatβs 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.

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