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

Represent the area under the curve of the function π(π₯) = π₯Β² + 4 in the interval [β2, 2] in sigma notation using a right Riemann sum with π subintervals.

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### Video Transcript

Represent the area under the curve of the function π of π₯ equals π₯ squared plus four in the closed interval negative two to two in sigma notation using a right Riemann sum with π subintervals.

Remember, when weβre writing a right Riemann sum, we take values of π from one to 10. And when weβre writing a left Riemann sum, we take values of π from zero to π minus one. So to estimate the area under the curve of some function π of π₯ in the closed interval of π to π using π subintervals. We find the sum of Ξπ₯ times π of π₯π for π minus one to π in a right Riemann sum and the sum of Ξπ₯ times π of π₯π for π₯ equals zero to π minus one in a left Riemann sum. And here Ξπ₯ is π minus π divided by π. And π₯π is equal to π plus π lots of Ξπ₯.

Now of course, in this question, weβre looking to find a right Riemann sum. So we will be taking values of π from one to π. And we can calculate Ξπ₯ quite easily. Weβll let π be equal to negative two and π be equal to two. This means Ξπ₯ is equal to two minus negative two over π. And of course, two minus negative two is four. So Ξπ₯ is simply four over π.

Once weβve calculated Ξπ₯, we can work out π₯ subscript π. Itβs π, which we said was negative two plus π lots of Ξπ₯. Here thatβs π times four over π or four π over π. So we know Ξπ₯ and we know π₯ subscript π. So next, we need to work out what π of π₯π is.

Now in this question, π of π₯ is equal to π₯ squared plus four. So weβre going to substitute our expression for π₯π into this function. And when we do, we find that π of π₯π is negative two plus four π over π all squared plus four. We then distribute the parentheses and we obtain four minus 16π over π plus 16π squared over π squared. Combining like terms, we see that π of π₯π is eight minus 16π over π plus 16π squared over π squared.

And actually, weβve done enough now. We can substitute all of this into the summation formula. Itβs the sum between π equals one and π of Ξπ₯ β thatβs four over π β times π of π₯π. And we found that to be eight minus 16π over π plus 16π squared over π squared. Weβre going to add the values inside the parentheses. And we do that by multiplying the first one by π squared, the second one by π, both the numerator and the denominator. And we create a common denominator of π squared.

And so the part inside our parentheses becomes eight π squared over π squared minus 16ππ over π squared plus 16π squared over π squared. And thatβs great because we can find the sum and difference of those numerators. Letβs clear some space for the final steps.

Now we take out a factor of eight over π squared. And the bit outside our parenthesis becomes 32 over π cubed. And weβre going to multiply that by π squared minus two ππ plus two π 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. π cubed is independent of π, as of course is 32. This means we can take the fact of 32 over π cubed outside of our summation. And we found the right Riemann sum. Itβs 32 over π cubed times the sum of π squared plus two π squared minus two ππ for values of π between one and π.