Video: APCALC02AB-P1A-Q43-583184107896

Let 𝑓(π‘₯) = 2π‘₯Β² ln π‘₯. What is the average value of 𝑓 over the interval 2 ≀ π‘₯ ≀ 7?

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Video Transcript

Let 𝑓 of π‘₯ be equal to two π‘₯ squared times the natural log of π‘₯. What is the average value of 𝑓 over the interval π‘₯ is greater than or equal to two and less than or equal to seven?

Remember the formula we use for the average value of a function 𝑓 over a closed interval π‘Ž to 𝑏 is one over 𝑏 minus π‘Ž times the integral evaluated between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯. In our case, 𝑓 of π‘₯ is equal to two π‘₯ squared times the natural log of π‘₯. And we can let π‘Ž be equal to two and 𝑏 be equal to seven. Those are the limits of our closed interval. So the average value of 𝑓 is one over seven minus two times the integral evaluated between two and seven of two π‘₯ squared times the natural log of π‘₯ with respect to π‘₯.

Now since our function is the product of two functions, we can use integration by parts to evaluate it. This says that the integral of 𝑒 times d𝑣 by dπ‘₯ with respect to π‘₯ is equal to 𝑒 times 𝑣 minus the integral of 𝑣 times d𝑒 by dπ‘₯ with respect to π‘₯. We want to let 𝑒 be equal to the natural log of π‘₯. This means we will get d𝑒 by dπ‘₯ to be equal to one over π‘₯. And we know that some cancelling will occur in our second integral, making it much easier to integrate. This means we need to let d𝑣 by dπ‘₯ be equal to two π‘₯ squared. And we can find 𝑣 by integrating its two π‘₯ cubed over three. And we’ll deal with the limits in a moment.

But what this means is that our integral two π‘₯ squared times the natural log of π‘₯ with respect to π‘₯ is equal to the natural log of π‘₯ times two π‘₯ cubed over three minus the integral of two π‘₯ cubed over three times one over π‘₯ with respect to π‘₯. This simplifies to the integral of two π‘₯ squared over three with respect to π‘₯ which integrates to two π‘₯ cubed over nine. And usually, we would have a conserve integration here. But we are going to be evaluating this integral between the limits of two and seven. Let’s clear some space.

The average value of π‘₯ is now one-fifth of two π‘₯ cubed over three times the natural log of π‘₯ minus two π‘₯ cubed over nine evaluated between the limits two and seven. We substitute π‘₯ is equal to seven and π‘₯ is equal to two and then find the difference. So we get a fifth times two times seven cubed over three times the natural log of seven minus two times seven cubed over nine minus two times two cubed over three times the natural log of two minus two times two cubed over nine.

And whilst we could evaluate this in one go on our calculator, it’s always sensible to perform some intermediate steps. Here, I’ve worked out two times seven cubed over three times the natural of seven minus two times seven cubed over nine as 368.742. And then, I’ve also worked out the second half to be 1.919. So a fifth of the difference between these two values is 73.36459 and so on. Correct to three decimal places, the average value of 𝑓 over the closed interval two to seven is 73.365.

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