### Video Transcript

Determine, to the nearest thousandth, the area of the plane region bounded by the curve π¦ equals the square root of two π₯ minus two and the lines π₯ equals two, π₯ equals three, and π¦ equals zero.

Here is a graph containing the curve π¦ equals the square root of two π₯ minus two and the lines π₯ equals two and π₯ equals three. The line π¦ equals nought is simply the π₯-axis. And so, the area weβre trying to find is the area of the shaded region here. We could also define this region as the area underneath the curve π¦ equals the square root of two π₯ minus two between π₯ equals two and π₯ equals three. And so therefore, this area is equal to the integral from two to three of the square root of two π₯ minus two with respect to π₯.

Letβs now calculate the value of this integral. Although we could integrate this directly, we can make it easier for ourselves by substituting in something for the two π₯. So we donβt have to worry about the constant two. So letβs let two π₯ equal π’. And now we need to find dπ₯ in terms of dπ’. And we will use the formula that says dπ’ is equal to dπ’ by dπ₯ times dπ₯. And we can calculate dπ’ by dπ₯ using the fact that π’ is equal to two π₯. So differentiating this with respect to π₯ gives us just two. This means that dπ’ is equal to two dπ₯. Rearranging this for dπ₯ gives us dπ₯ is equal to one-half dπ’.

Since weβre integrating between π₯ equals two and π₯ equals three, we will need to find these in terms of π’. So we have two values of π’ which will give us an equivalent area when integrating with respect to π’. So when π₯ equals two, π’ is equal to two times π₯. So that means π’ is equal to four. Then, when π₯ is equal to three, π’ is equal to two times π₯. So thatβs six. So weβll be integrating between π’ equals four and π’ equals six.

And now weβre ready to make our substitution. So we put in π’ for every two π₯ term. And we substitute in one-half dπ’ for dπ₯. And now, weβre integrating between π’ equals four and π’ equals six. And this gives us the integral from four to six of the square root of π’ minus two times a half dπ’. And we can factor this constant of a half out of the integral.

Next, we note that we can write the square root of π’ minus two as π’ minus two to the power of a half. This will make it easier to integrate. At this stage, we are now ready to integrate. When integrating, the constant term of one-half will remain there. For the π’ minus two to the power of a half term, we need to increase the power by one and divide by the new power. Increasing the power by one gives us π’ minus two to the power of three over two. And then, we must divide by the new power. So we divide by three over two.

Now, we must be careful here since this is a function within a function. And so, we must divide by the derivative of the function inside the other function. So thatβs π’ minus two. And the derivative of π’ minus two is just one. So when we divide by one, nothing happens. So we do not need to worry about this here. Next, we mustnβt forget that we are integrating between six and four. This means that we substitute six into our formula and then subtract the formula again, but with four substituted in instead.

That leaves us with? Now, we have a common factor of one over three over two which we can factor out here. That leaves us with one over two timesed by one over three over two timesed by six minus two to the power of three over two minus four minus two to the power of three over two. And we can multiply one-half by one over three over two, which will leave us with one over three. And we can also simplify six minus two to four and four minus two to two.

Now, we can type this into our calculator to give us 1.723857625. And now, we note that the question asks us to find the answer to the nearest thousandth. And we note that one thousandth or one over one thousand is equal to 0.001. So therefore, we need to round our answer to three decimal places. This gives us a solution of 1.724.