### Video Transcript

Use a right Riemann sum to approximate the area under the curve of π of π₯ equals the square root of three minus π₯ in the interval zero, two. Use subintervals with π equals four. Approximate your answer to three decimal places.

So what does it mean by a right Riemann sum? Well, we use Riemann sums to approximate the area under a curve by using rectangles. For example, say we wanted to approximate the area under this curve in the interval between π and π using three rectangles. It would look something like this. And notice how the top right-hand corners of each rectangle touch the curve. So we call this a right Riemann sum. Instead, we couldβve drawn the rectangles in this way, and this wouldβve been a left Riemann sum. In both cases, we approximate the area under the curve by finding the total area of these rectangles. In mathematics, this can be helpful to approximate an integral because, remember, definite intervals represent the area under a curve of a function.

Letβs draw a quick sketch of this function so we can visualize what weβre doing here. Weβre wanting to estimate the area under this curve in the interval between zero and two. But instead of finding this exact area, weβre going to use a right Riemann sum. So we want to use rectangles with the top right-hand corner of each rectangle touching the curve. Weβve been asked to use subintervals with π equal to four. This means weβre using four rectangles, so letβs draw in those rectangles now. We want four rectangles between zero and two, so we can calculate that each rectangle must have a width of 0.5.

Alternatively, there is a handy formula Ξπ₯ equals π minus π over π, where π, π is the interval and π is the number of subintervals. So this is two minus zero over four, and that gives us 0.5. So thatβs the width of each rectangle. So letβs draw our four rectangles, each with a width of 0.5. Notice how itβs the top right-hand corner of each rectangle that touches the curve rather than the left. Then we can go ahead and find the area of each rectangle. We know that we find the area of a rectangle by multiplying the width by the height. In fact, the Riemann method has a formula for this. This says the area under the curve is approximately the sum from π equals one to π of Ξπ₯ multiplied by π of π₯π, where π is the number of subintervals which, in this case, thatβs four.

So if we find this sum with π equal to four, we have Ξπ₯ π of π₯ one add Ξπ₯ π of π₯ two add Ξπ₯ π of π₯ three add Ξπ₯ π of π₯ four. The π₯ one, π₯ two, π₯ three, and π₯ four are the right endpoints of each interval. In this case, thatβs 0.5, one, 1.5, and two. Note that these values would be different if we were doing a left Riemann sum. So Iβm going to replace π₯ one, π₯ two, π₯ three, and π₯ four with 0.5, one, 1.5, and two in our working. So then, π of 0.5, π of one, π of 1.5, and π of two simply give us these values at the top right-hand corner of each rectangle. In other words, π of 0.5, π of one, π of 1.5, and π of two are just the heights of each rectangle. And we know that Ξπ₯ is the width of each rectangle, so we can see how this formula works by finding the area of each rectangle and then summing them together.

So letβs go ahead and calculate the values of π of 0.5, π of one, π of 1.5, and π of two. Remember, π of π₯ is the square root of three minus π₯. So, π of 0.5 is the square root of three minus 0.5, which is the square root of 2.5. π of one is the square root of three minus one, which is the square root of two. π of 1.5 is the square root of three minus 1.5, which is the square root of 1.5. And π of two is the square root of three minus two, which is the square root of one, although we know that this is just one. So letβs go ahead and sum the areas of the rectangles together.

The area of the first rectangle is 0.5, which is the width, Ξπ₯, multiplied by the height, which is π of 0.5, which we found to be the square root of 2.5. The area of the second rectangle from the left is the width, which is 0.5 or Ξπ₯, multiplied by the height, which is π of one or root two. The area of the third rectangle from the left is the width, which is 0.5, thatβs Ξπ₯, multiplied by the height, which is π of 1.5, which we found to be the square root of 1.5. And for the fourth rectangle from the right, the width is 0.5, which is Ξπ₯, and the height is π of two, which we found to be one. And when we calculate these and add them together, we work this out to be 2.610 to three decimal places. Remember that this is just an estimate to the area under the curve.