### Video Transcript

The table gives the values of a function obtained from an experiment. Use them to estimate the integral from three to 27 of π of π₯ with respect to π₯ using three equal subintervals with right endpoints.

The question is asking us to estimate the integral from three to 27 of π of π₯ with respect to π₯. And it wants us to do this by using a right endpoint approximation by using three subintervals of equal width. To do this, weβre given a table obtained experimentally. It gives us some inputs of π₯ and their respective outputs of π of π₯.

Since the question wants us to approximate this integral by using three equal subintervals of right endpoints, weβll do this by using a Riemann sum. We recall we start by calling Ξπ₯ the width of our rectangles. This is equal to π minus π all divided by π, where π is the number of subintervals. Next, we need our sample points π₯ π. These will be equal to π plus π times Ξπ₯, where π ranges from one to π. These will be our right endpoints.

With these set up, our right Riemannβs sum tells us the integral from π to π of π of π₯ with respect to π₯ is approximately equal to the sum from π equals one to π of Ξπ₯ times π of π₯ π. We see from the question weβre asked to estimate the integral from three to 27 of π of π₯ with respect to π₯ by using three subintervals of equal width. π is the number of subintervals. So, in this case, itβs equal to three. And π and π are the endpoints for our integral. π is equal to three, and π is equal to 27.

Now, we can use this information to find the value of Ξπ₯ and the values of our sample points π₯ π. Letβs start with Ξπ₯ is equal to π minus π divided by π, which in this case is 27 minus three divided by three. And 27 minus three divided by three is 24 over three, which equals eight. Now, since weβre taking right endpoints, we want our sample points to be π₯ one all the way up to π₯ π.

Letβs start by finding π₯ one. Itβs equal to π plus one times Ξπ₯, which is three plus one times eight. Letβs now find π₯ two. This time, we set our value of π to be equal to two. This gives us three plus two times eight, which is equal to 19. And finally, we find the value of π₯ three by using π is equal to three. We get three plus three times eight, which is equal to 27. And remember, we only have three subintervals. So, this is our final sample point.

Weβre now ready to try and approximate our area. By using our formula for the right Riemannβs sum, we have the integral from three to 27 of π of π₯ with respect to π₯ is approximately equal to the sum from π equals one to three of eight times π evaluated at π₯ π. This is because our value of π is three and our value of Ξπ₯ is eight.

At this point, we can expand this series. We get eight times π of π₯ one plus eight times π of π₯ two plus eight times π of π₯ three. And this is π evaluated at our sample points. π₯ one is 11, π₯ two is 19, and π₯ three is 27. So, this gives us eight π of 11 plus eight π of 19 plus eight π of 27. Normally, we would substitute these values into our function π of π₯. But in this case, weβre not told our function π of π₯. Instead, weβre given a table, so we need to read these values from our table.

First, when π₯ is equal to 11, we can see that π of π₯ is equal to negative 0.7. Next, when π₯ is equal to 19, we can see that π of π₯ is equal to 2.3. Finally, when π₯ is equal to 27, we can see that π of π₯ is equal to 4.8. This gives us eight times negative 0.7 plus eight times 2.3 plus eight times 4.8. And if we evaluate this expression, we get our final answer of 51.2. Therefore, by using our table and a right Riemann sum with three equal subintervals, we were able to show the integral from three to 27 of π of π₯ with respect to π₯ is approximately equal to 51.2.