Video: Finding the Average Value of a Root Function in a Given Interval

Find the average value of 𝑓(π‘₯) = √(2π‘₯) on the interval [0, 2].

04:01

Video Transcript

Find the average value of the function 𝑓 of π‘₯ is equal to the square root of two π‘₯ on the closed interval from zero to two.

The question gives us a function 𝑓 of π‘₯, which is root two π‘₯. And it wants us to find the average value of this function on the closed interval from zero to two. Let’s start by recalling what we mean by the average value of a function on a closed interval. If a function 𝑓 is continuous on the closed interval from π‘Ž to 𝑏, then the average value of that function on this closed interval, 𝑓 average, is equal to one divided by 𝑏 minus π‘Ž times the integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ with respect to π‘₯. In our case, we want the average value of the function 𝑓 of π‘₯ is equal to root two π‘₯ on the closed interval from zero to two. So we want our function 𝑓 of π‘₯ is root two π‘₯, π‘Ž is zero, and 𝑏 is two.

First, we need to check that our function 𝑓 is continuous on this closed interval. Well, we know that 𝑓 of π‘₯ is continuous on its entire domain. So in particular, it’s continuous on the closed interval from zero to two. So this tells us we can use our formula to find the average value of our function on this interval. 𝑓 average is equal to one divided by two minus zero times the integral from zero to two of the square root of two π‘₯ with respect to π‘₯. But we can’t yet evaluate our integrand. So we’ll need to rewrite it by using our laws of exponents. First, the square root of two π‘₯ is equal to root two times root π‘₯. Next, we’ll rewrite root π‘₯ as π‘₯ to the power of one-half. This we’ll let us use the power rule for integration.

So using this to rewrite our integrand, we have 𝑓 average is equal to one divided by two minus zero times the integral from zero to two of root two times π‘₯ to the power of one-half with respect to π‘₯. And we can now integrate this by using the power rule for differentiation. We recall this tells us, for constants π‘Ž and 𝑛, where 𝑛 is not equal to negative one, the integral of π‘Žπ‘₯ to the 𝑛th power with respect to π‘₯ is equal to π‘Ž times π‘₯ to the power of 𝑛 plus one divided by 𝑛 plus one plus 𝐢. We add one to our exponent of π‘₯ and then divide by this new exponent. In our case, our exponent of π‘₯ is equal to one-half.

We’re now ready to simplify this expression. First, we’ll simplify one over two minus zero to give us one-half. Next, we’ll evaluate our definite integral by using the power rule for integration, where our value of π‘Ž is root two and our value of 𝑛 is one-half. So we’ve shown 𝑓 average is equal to one-half times root two times π‘₯ to the power of one-half plus one divided by one-half plus one evaluated at the limits of our integral, zero and two. And we can simplify this expression slightly since one-half plus one is equal to three over two.

So we’re now ready to find the value of 𝑓 average. We just need to evaluate this at the limits of our integral. Evaluating this at the limits of our integral, we get one-half multiplied by root two times two to the power of three over two divided by three over two minus root two times zero to the power of three over two divided by three over two. And we can see the second term inside of our parentheses has a factor of zero. So this is equal to zero. And we can simplify this further. We can see in our denominator, we get two multiplied by three over two, which we can simplify to give us three.

So we’ve simplified our expression to show that 𝑓 average is equal to root two times two to the power of three over two divided by three. And we could leave our answer like this. However, we’ll simplify this further. By our laws of exponents, two to the power of three over two is equal to two to the power of one-half all cubed. But then, two to the power of one-half is the square root of two. So we can replace the two to the power of three over two in our numerator with the square root of two all cubed. Doing this, we can see our numerator is now root two times root two cubed. So our numerator is root two multiplied by itself four times. This is equal to four. And this gives us our final answer. 𝑓 average is equal to four over three.

Therefore, we’ve shown the average value of the function 𝑓 of π‘₯ is equal to the square root of two π‘₯ on the closed interval from zero to two is four over three.

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