Video: Using Sigma Notation to Represent a Right Riemann Sum

Represent the area under the curve of the function 𝑓(π‘₯) = π‘₯Β² + 4 in the interval [βˆ’2, 2] in sigma notation using a right Riemann sum with 𝑛 subintervals.

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Video Transcript

Represent the area under the curve of the function 𝑓 of π‘₯ equals π‘₯ 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 π‘₯ in the closed interval of π‘Ž to 𝑏 using 𝑛 subintervals. We find the sum of Ξ”π‘₯ times 𝑓 of π‘₯𝑖 for 𝑖 minus one to 𝑛 in a right Riemann sum and the sum of Ξ”π‘₯ times 𝑓 of π‘₯𝑖 for π‘₯ equals zero to 𝑛 minus one in a left Riemann sum. And here Ξ”π‘₯ is 𝑏 minus π‘Ž divided by 𝑛. And π‘₯𝑖 is equal to π‘Ž plus 𝑖 lots of Ξ”π‘₯.

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 Ξ”π‘₯ quite easily. We’ll let π‘Ž be equal to negative two and 𝑏 be equal to two. This means Ξ”π‘₯ is equal to two minus negative two over 𝑛. And of course, two minus negative two is four. So Ξ”π‘₯ is simply four over 𝑛.

Once we’ve calculated Ξ”π‘₯, we can work out π‘₯ subscript 𝑖. It’s π‘Ž, which we said was negative two plus 𝑖 lots of Ξ”π‘₯. Here that’s 𝑖 times four over 𝑛 or four 𝑖 over 𝑛. So we know Ξ”π‘₯ and we know π‘₯ subscript 𝑖. So next, we need to work out what 𝑓 of π‘₯𝑖 is.

Now in this question, 𝑓 of π‘₯ is equal to π‘₯ squared plus four. So we’re going to substitute our expression for π‘₯𝑖 into this function. And when we do, we find that 𝑓 of π‘₯𝑖 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 π‘₯𝑖 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 Ξ”π‘₯ β€” that’s four over 𝑛 β€” times 𝑓 of π‘₯𝑖. 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.

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