Video: APCALC02AB-P1B-Q36-585180842012

The following table shows values of π‘₯ and 𝑓(π‘₯). Use a right-hand Riemann sum with eight subintervals to approximate the integral from 1 to 16 of 𝑓(π‘₯) dπ‘₯.

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Video Transcript

The following table shows values of π‘₯ and 𝑓 of π‘₯. Use a right-hand Riemann sum with eight subintervals to approximate the integral from one to 16 of 𝑓 of π‘₯ with respect to π‘₯.

Let’s start by plotting our point on a graph. Using this, we can draw a rough sketch for what 𝑓 of π‘₯ may look like. Now the shape of 𝑓 does not particularly matter to this question since the question requires us to estimate the integral between one and 16 of 𝑓 of π‘₯ with respect to π‘₯ using a right-hand Riemann sum. We are told to use eight subintervals.

Using the information in the table, we can see that there are eight subintervals between the values given. And so, these will be the eight subintervals we’ll be using for our right-hand Riemann sum. A Riemann sum is used to approximate the area under a curve. For the right-hand Riemann sum, we approximate by using rectangles with a width equal to the width of each interval and a height which is equal to the 𝑓 value on the right of each interval. Let’s draw on our graph what the bars of each interval will look like.

The first interval goes between one and two. The 𝑓 value on the right of this interval is 𝑓 of two which is equal to five. Therefore, this bar will have a width equal to the difference of the π‘₯-values on either side, so two minus one which equals one and a height of five. It will look something like this. The next interval between two and four will have a width of two and a height of 𝑓 of four which is 11. We can continue this along as we have creating bars for each interval. This is what our bars will look like.

Now in order to find the value of our right Riemann sum and therefore approximate our integral, we simply add together the area of all of these bars. We can say that the integral from one to 16 of 𝑓 of π‘₯ with respect to π‘₯ is roughly equal to one times five plus two times 11 plus one times 13 plus two times nine plus three times eight plus one times 14 plus two times 16 plus three times 20. Now we can complete the multiplications within our calculation and then add the numbers in pairs.

Five plus 22 is 27. 13 plus 18 is 31. 24 plus 14 is 38. And 32 plus 60 is 92. And we can add these numbers in pairs again. 27 plus 31 is 58 and 38 plus 92 is 130. Adding these final two numbers together, we reach our answer. And that is, that using a right-hand Riemann sum with eight subintervals, we find that the integral between one and 16 of 𝑓 of π‘₯ with respect to π‘₯ is roughly equal to 188.

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