Video: Finding the Riemann Sum of a Linear Function in a Given Interval by Dividing It into Subintervals and Using Their Right Endpoint

Given 𝑓(𝑥) = 2𝑥 − 5 and −6 ≤ 𝑥 ≤ 4, evaluate the Riemann sum for 𝑓 with five subintervals, taking sample points to be right endpoints.

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Video Transcript

Given 𝑓 of 𝑥 equals two 𝑥 minus five and 𝑥 is greater than or equal to negative six and less than or equal to four, evaluate the Riemann sum for 𝑓 with five subintervals, taking sample points to be right endpoints.

Riemann sums are a way of approximating the area between a curve and an axis by splitting it into rectangles. In this case, we’re told that our sample points are going to be right endpoints. So the height of each rectangle will be the value of the function at the right end of each subinterval. Let’s sketch this out. We have the function 𝑓 of 𝑥 equals two 𝑥 minus five. And it’s bounded by the lines 𝑥 equals negative six and 𝑥 equals four.

We’re going to split this region into five subintervals. And whilst it’s probably fairly intuitive, we can use the given formula. This is 𝛥𝑥, which is the width of each subinterval, is equal to 𝑏 minus 𝑎 over 𝑛, where 𝑏 and 𝑎 are the upper and lower limits of our region and 𝑛 is the number of subintervals. In this case, that’s four minus negative six over five. That’s 10 divided by five, which is equal to two. Each of our subintervals and therefore each of our rectangles has a width of two units. And so we add in a rectangle every two units, making sure that the height of the rectangle is at the right endpoint of each subinterval.

Our job is to calculate the area of each of these rectangles, remembering that when we have an area that lies below the 𝑥-axis, we subtract that area from the total. Let’s deal with the first rectangle. Let’s call the area 𝐴 sub one. Its area is its width multiplied by its height. Well, its width is two square units; we know that. But its height is the value of the function at the right end of its interval. We can see that’s at negative four. We therefore substitute 𝑥 equals negative four into our function. We get two times negative four minus five, which gives us negative 13. So the area is two times negative 13, which is negative 26.

Now, we don’t need to worry that we’ve got a negative value here. Remember, we said that we’re going to subtract the area. So technically, the area is 26 square units. But we’re going to subtract it. Let’s repeat this process for the second rectangle. The Riemann sum for our second rectangle is two times the value of the function at its right endpoint. This time, that’s at the point 𝑥 equals negative two. So we get two times negative two minus five. That’s 𝑓 of negative two. The Riemann sum becomes two times negative nine, which is negative 18. And of course, since we’re going to be subtracting the area, we were expecting a negative value.

Let’s look at our third rectangle. This time, the value of the function at the right end is 𝑓 of zero. 𝑓 of zero is two times zero minus five, which is negative five. So the Riemann sum for this rectangle is negative 10. We’ll now consider the fourth rectangle. This one still lies underneath the 𝑥-axis. So we’re expecting a negative value. This time, the height is equal to the value of the function when 𝑥 is equal to two. 𝑓 of two is two times two minus five, which is equal to negative one. So the Riemann sum is two times negative one, which is negative two.

Our next rectangle lies above the 𝑥-axis. So when we calculate the area, we should be expecting a positive value. At the right endpoint, the value of the function is 𝑓 of four. 𝑓 of four is two times four minus five, which is three. So we have two times three, which is equal to six. The Riemann sum is the sum of these values. It’s negative 26 plus negative 18 plus negative 10 plus negative two plus six, which is equal to negative 50.

The Riemann sum for our function with five subintervals taking sample points to be right endpoints, in other words, the right Riemann sum here is negative 50.

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