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

Let 𝑓(π‘₯) = 3/2π‘₯ over the interval 1 ≀ π‘₯ ≀ 5. Evaluate the Riemann sum of 𝑓 using four subintervals and right end point sample points, giving your answer to six decimal places.

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Video Transcript

Let 𝑓 of π‘₯ be equal to three divided by two π‘₯ over the interval where π‘₯ is greater than or equal to one and π‘₯ is less than or equal to five. Evaluate the Riemann sum of 𝑓 using four subintervals and right endpoint sample points, giving your answer to six decimal places.

The question gives us a function 𝑓 of π‘₯, which is only defined on the closed interval from one to five. The question wants us to evaluate the Riemann sum of this function 𝑓 by using four subintervals and right endpoints as our sample points for our Riemann sum. It wants us to give our answer to six decimal places.

Let’s start by sketching a graph of our function 𝑓 of π‘₯. Remember, since π‘₯ is greater than or equal to one and π‘₯ is less than or equal to five, we only need the positive π‘₯-axis. We see that 𝑓 of π‘₯ is equal to three divided by two π‘₯, which we can rewrite as three over two times one over π‘₯. In other words, this is a vertical stretch of factor three over two of our reciprocal function one over π‘₯. So our curve will have the same shape as one over π‘₯, stretched vertically by a factor of three over two.

Remember, we only sketch our graph for π‘₯ is greater than or equal to one and π‘₯ is less than or equal to five. We now want to set up our Riemann sum. We see the question wants us to use four subintervals where π‘₯ is between one and five. And we recall the width of our rectangles, Ξ”π‘₯, will be equal to 𝑏 minus π‘Ž divided by 𝑛, where 𝑏 and π‘Ž are the endpoints of our interval and 𝑛 is the number of subintervals. So we’ll set the value of 𝑏 equal to five, the value of π‘Ž equal to one, and the value of 𝑛 equal to four. This gives us Ξ”π‘₯ is equal to five minus one divided by four, which we can evaluate to just give us one.

So the widths of our rectangles will be increasing by one each time. This gives us the values of π‘₯ as one, two, three, four, and five. Next, the question tells us we want to use right endpoints as our sample points for our Riemann sum. We want to approximate the area of 𝑓 of π‘₯ by using four rectangles. The fact that the question tells us to use right endpoints as our sample points for our Riemann sum tells us that we should use the rightmost value of π‘₯ in each of our rectangles to find the heights of our rectangles to approximate the area under 𝑓 of π‘₯.

In other words, to find the height of our first rectangle, we go up from π‘₯ is equal to two all the way up to our curve 𝑓 of π‘₯. This gives us our first rectangle. Its width is Ξ”π‘₯, which is one, and its height is 𝑓 of two. We’ll label the area of this rectangle 𝐴 one. We do the same to find our second rectangle. Remember, we’re taking a right endpoint, so we want to go up from π‘₯ is equal to three. And, again, we see the height of this rectangle is 𝑓 evaluated at three and the width of this rectangle is Ξ”π‘₯.

We can do the same to find our remaining two rectangles. Our Riemann sum will be the sum of the four areas of these rectangles. We’ve already found the value of Ξ”π‘₯. So now we need to calculate the height of each of these rectangles to calculate their area. And the height of each of these rectangles is 𝑓 evaluated at two, 𝑓 evaluated at three, 𝑓 evaluated at four, and 𝑓 evaluated at five, respectively. Substituting π‘₯ is equal to two into our definition of 𝑓 of π‘₯, we get three divided by two times two, which is three-quarters.

We can do the same to find 𝑓 evaluated at three. We get three divided by two times three, which simplifies to give us one-half. And we can do the same to find 𝑓 evaluated at four, which is three-eighths, and 𝑓 evaluated at five, which is three-tenths. So our Riemann sum tells us that the area under 𝑓 of π‘₯ between π‘₯ is equal to one and π‘₯ is equal to five is approximately equal to the areas of our four rectangles added together. And we have the areas of each of these rectangles as their width multiplied by their height.

All of our rectangles have the same width of Ξ”π‘₯. And each of their heights are 𝑓 evaluated at two, 𝑓 evaluated at three, 𝑓 evaluated at four, and 𝑓 evaluated at five, respectively. So we now can calculate the areas of each of our rectangles. We know that Ξ”π‘₯ is equal to one, and we calculated 𝑓 evaluated at these points earlier. This gives us one times three-quarters plus one times one-half plus one times three-eighths plus one times three-tenths. Which if we calculate, we get 77 divided by 40.

But, remember, the question wants us to give our answer to six decimal places. And to six decimal places, this is equal to 1.925000. So we’ve shown the approximate area of our function 𝑓 of π‘₯ to six decimal places by using a Riemann sum with right endpoints and four subintervals is 1.925000.

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