Video Transcript
Represent the area under the curve
of the function 𝑓 of 𝑥 equals one over 𝑥 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 𝛥𝑥 times 𝑓 of 𝑥 𝑖 for values of 𝑖 between one and 𝑛. 𝛥𝑥 is 𝑏 minus 𝑎 divided by 𝑛,
where 𝑎 and 𝑏 are the lower and upper endpoints of our interval, respectively. And 𝑛, of course, is the number of
subintervals. Then 𝑥 𝑖 is 𝑎 plus 𝑖 lots of
𝛥𝑥. We always begin by working out
𝛥𝑥. We see that our closed interval is
from three to five. So we let 𝑎 be equal to three and
𝑏 be equal to five. 𝛥𝑥 is, therefore, five minus
three over 𝑛 which is, of course, two over 𝑛.
We’re now ready to work out what 𝑥
𝑖 is. It’s 𝑎, which we know to be three,
plus 𝛥𝑥, 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 𝑥 𝑖. 𝑓 of 𝑥 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. 𝛥𝑥 is two over 𝑛. And we multiply this by 𝑓 of 𝑥
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.
