Question Video: Using Sigma Notation to Represent a Right Riemann Sum Mathematics • Higher Education

Video Transcript

Represent the area under the curve of the function 𝑓 of 𝑥 equals 𝑥 squared plus four in the closed interval negative two to two 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 10. And when we’re writing a left Riemann sum, we take values of 𝑖 from zero to 𝑛 minus one. So to estimate the area under the curve of some function 𝑓 of 𝑥 in the closed interval of 𝑎 to 𝑏 using 𝑛 subintervals. We find the sum of Δ𝑥 times 𝑓 of 𝑥𝑖 for 𝑖 minus one to 𝑛 in a right Riemann sum and the sum of Δ𝑥 times 𝑓 of 𝑥𝑖 for 𝑥 equals zero to 𝑛 minus one in a left Riemann sum. And here Δ𝑥 is 𝑏 minus 𝑎 divided by 𝑛. And 𝑥𝑖 is equal to 𝑎 plus 𝑖 lots of Δ𝑥.

Now of course, in this question, we’re looking to find a right Riemann sum. So we will be taking values of 𝑖 from one to 𝑛. And we can calculate Δ𝑥 quite easily. We’ll let 𝑎 be equal to negative two and 𝑏 be equal to two. This means Δ𝑥 is equal to two minus negative two over 𝑛. And of course, two minus negative two is four. So Δ𝑥 is simply four over 𝑛.

Once we’ve calculated Δ𝑥, we can work out 𝑥 subscript 𝑖. It’s 𝑎, which we said was negative two plus 𝑖 lots of Δ𝑥. Here that’s 𝑖 times four over 𝑛 or four 𝑖 over 𝑛. So we know Δ𝑥 and we know 𝑥 subscript 𝑖. So next, we need to work out what 𝑓 of 𝑥𝑖 is.

Now in this question, 𝑓 of 𝑥 is equal to 𝑥 squared plus four. So we’re going to substitute our expression for 𝑥𝑖 into this function. And when we do, we find that 𝑓 of 𝑥𝑖 is negative two plus four 𝑖 over 𝑛 all squared plus four. We then distribute the parentheses and we obtain four minus 16𝑖 over 𝑛 plus 16𝑖 squared over 𝑛 squared. Combining like terms, we see that 𝑓 of 𝑥𝑖 is eight minus 16𝑖 over 𝑛 plus 16𝑖 squared over 𝑛 squared.

And actually, we’ve done enough now. We can substitute all of this into the summation formula. It’s the sum between 𝑖 equals one and 𝑛 of Δ𝑥 — that’s four over 𝑛 — times 𝑓 of 𝑥𝑖. And we found that to be eight minus 16𝑖 over 𝑛 plus 16𝑖 squared over 𝑛 squared. We’re going to add the values inside the parentheses. And we do that by multiplying the first one by 𝑛 squared, the second one by 𝑛, both the numerator and the denominator. And we create a common denominator of 𝑛 squared.

And so the part inside our parentheses becomes eight 𝑛 squared over 𝑛 squared minus 16𝑖𝑛 over 𝑛 squared plus 16𝑖 squared over 𝑛 squared. And that’s great because we can find the sum and difference of those numerators. Let’s clear some space for the final steps.

Now we take out a factor of eight over 𝑛 squared. And the bit outside our parenthesis becomes 32 over 𝑛 cubed. And we’re going to multiply that by 𝑛 squared minus two 𝑖𝑛 plus two 𝑖 squared. Now this might look a bit complicated. And you might wish to redistribute these parentheses and check that we have the same terms. And then this next step is a little special. 𝑛 cubed is independent of 𝑖, as of course is 32. This means we can take the fact of 32 over 𝑛 cubed outside of our summation. And we found the right Riemann sum. It’s 32 over 𝑛 cubed times the sum of 𝑛 squared plus two 𝑖 squared minus two 𝑖𝑛 for values of 𝑖 between one and 𝑛.