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

Given that 𝑓(π‘₯) = 4 cos π‘₯ and that 0 ≀ π‘₯ ≀ πœ‹/4, evaluate, to the nearest six decimal places, the Riemann sum for 𝑓 with six subintervals, taking the sample points to be left endpoints.

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Video Transcript

Given that 𝑓 of π‘₯ is equal to four cos π‘₯ and that π‘₯ is greater than or equal to zero and less than or equal to πœ‹ by four, evaluate, to the nearest six decimal places, the Riemann sum for 𝑓 with six subintervals, taking the sample points to be left endpoints.

Remember, we can approximate the definite integral of a function between the limits of, say, π‘Ž and 𝑏 by splitting the area between the curve and the π‘₯-axis into 𝑛 subintervals. Where the width of each rectangle, π›₯π‘₯, is equal to the difference between the upper and lower limits divided by 𝑛, the number of subintervals. In this question, we’re looking to split the area into six subintervals, so we let 𝑛 be equal to six. We let π‘Ž be equal to zero and 𝑏 be equal to πœ‹ by four. So π›₯π‘₯ here and the width of each of our rectangles is πœ‹ by four minus zero over six, which is πœ‹ over 24. Let’s now sketch out the curve of 𝑦 equals four cos π‘₯ between π‘₯ equals zero and πœ‹ by four radians and spilt it into these six subintervals.

This question requires us to use left endpoints. So the height of each of our rectangles will be equal to the value of the function at the left of each subinterval. The sketch is really useful here because we see that each rectangle sits above the π‘₯-axis. So we just need to find the positive area of each. Let’s consider our first rectangle. Its left endpoint is at π‘₯ equals zero. So we find the height of the rectangle by evaluating 𝑓 of zero. 𝑓 of zero is four times cos of zero, which is four. The area of this first rectangle then is therefore πœ‹ by 24, since that’s the width of each rectangle, times four. That’s πœ‹ by six square units.

We now need to find the next endpoint. We achieve this by adding πœ‹ by 24 to zero; that’s πœ‹ by 24. And we can therefore find the height of this rectangle by evaluating 𝑓 of πœ‹ by 24. That’s four cos of πœ‹ by 24, which is approximately 3.965779 and so on. The area of our second rectangle then is width times height or base times height. That’s πœ‹ by 24 times 3.965 and so on, which is roughly 0.51911931 and so on. Now, of course, this question is asking us to evaluate till six decimal places, which is why I’ve left the numbers as I have. We find the next endpoint by adding another πœ‹ by 24 to πœ‹ by 24 and we get πœ‹ by 12. The height of this third rectangle then is 𝑓 of πœ‹ by 12 which is four cos of πœ‹ by 12, which gives us root six plus root two. And then we see that the area of our third rectangle is πœ‹ by 24 multiplied by this value, which gives us an area of roughly 0.505 and so on square units.

We repeat this process by repeatedly adding πœ‹ by 24 to each of our endpoints. And when we do, we find that the left endpoint of each of our rectangles is πœ‹ by eight, πœ‹ by six, and π‘₯ equals five πœ‹ by 24. Their heights are respectively given by 𝑓 of πœ‹ by eight, 𝑓 of πœ‹ by six, and 𝑓 of five πœ‹ by 24. And when we multiply each of these values by the width, πœ‹ by 24, we find that the area of our fourth rectangle is roughly 0.4837. Our fifth rectangle is roughly 0.4534. And our sixth is approximately 0.4153. Remember, the Riemann sum is the total area of these six rectangles. It’s 2.901 and so on. That’s 2.901067 correct to six decimal places.

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