Video: Representing the Area under the Curve of a Given Function in Sigma Notation Using a Right Riemann Sum

Represent the area under the curve of the function 𝑓(π‘₯) = 1/(π‘₯ βˆ’ 2) in the interval [3, 5] in sigma notation using a right Riemann sum with 𝑛 subintervals.

02:25

Video Transcript

Represent the area under the curve of the function 𝑓 of π‘₯ equals one over π‘₯ minus two in the closed interval 3 to 5 in sigma notation using a right Riemann sum with 𝑛 subintervals.

Remember, when we’re writing a right Riemann sum, we take values of 𝑖 from one to 𝑛. The area is approximately equal to the sum of π›₯π‘₯ times 𝑓 of π‘₯ 𝑖 for values of 𝑖 between one and 𝑛. π›₯π‘₯ is 𝑏 minus π‘Ž divided by 𝑛, where π‘Ž and 𝑏 are the lower and upper endpoints of our interval, respectively. And 𝑛, of course, is the number of subintervals. Then π‘₯ 𝑖 is π‘Ž plus 𝑖 lots of π›₯π‘₯. We always begin by working out π›₯π‘₯. We see that our closed interval is from three to five. So we let π‘Ž be equal to three and 𝑏 be equal to five. π›₯π‘₯ is, therefore, five minus three over 𝑛 which is, of course, two over 𝑛.

We’re now ready to work out what π‘₯ 𝑖 is. It’s π‘Ž, which we know to be three, plus π›₯π‘₯, which is two over 𝑛, times 𝑖. Let’s just write that as three plus two 𝑖 over 𝑛. Now, obviously for our summation, we need to know 𝑓 of π‘₯ 𝑖. 𝑓 of π‘₯ 𝑖 must be 𝑓 of three plus two 𝑖 over 𝑛. So lets substitute three plus two 𝑖 over 𝑛 into our formula. This gives us one over three plus two 𝑖 over 𝑛 minus two which when we distribute our parentheses is simply one plus two 𝑖 over 𝑛. This still isn’t particularly nice. So what we’re going to do is simplify the denominator.

We’ll write it as one over one plus two 𝑖 over 𝑛. And then we’re going to multiply the numerator and denominator of one over one by 𝑛. This creates a common denominator of 𝑛 and then it means we can add the numerators. And we end up with 𝑛 plus two 𝑖 over 𝑛. Now over here, we’re working out one divided by 𝑛 plus two 𝑖 over 𝑛. Well, another way to think about that is think about its reciprocal. The reciprocal of 𝑛 plus two 𝑖 over 𝑛 is 𝑛 over 𝑛 plus two 𝑖. And we now have everything we need to write our right Riemann sum. It’s a right Riemann sum, so we start at 𝑖 equals one and we end at 𝑛. π›₯π‘₯ is two over 𝑛. And we multiply this by 𝑓 of π‘₯ 𝑖, which we just worked out to be 𝑛 over 𝑛 plus two 𝑖. We then see that these 𝑛s cancel. And we have our right Riemann sum using sigma notation. It’s the sum of two over 𝑛 plus two 𝑖 for values of 𝑖 between one and 𝑛.

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