# 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 π(π₯) = 1/(π₯ β 2) in the interval [3, 5] in sigma notation using a right Riemann sum with π subintervals.

02:25

### Video Transcript

Represent the area under the curve of the function π of π₯ equals one over π₯ minus two in the closed interval 3 to 5 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 π. The area is approximately equal to the sum of π₯π₯ times π of π₯ π for values of π between one and π. π₯π₯ is π minus π divided by π, where π and π are the lower and upper endpoints of our interval, respectively. And π, of course, is the number of subintervals. Then π₯ π is π plus π 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 be equal to three and π be equal to five. π₯π₯ is, therefore, five minus three over π which is, of course, two over π.

Weβre now ready to work out what π₯ π is. Itβs π, which we know to be three, plus π₯π₯, which is two over π, times π. Letβs just write that as three plus two π over π. Now, obviously for our summation, we need to know π of π₯ π. π of π₯ π must be π of three plus two π over π. So lets substitute three plus two π over π into our formula. This gives us one over three plus two π over π minus two which when we distribute our parentheses is simply one plus two π over π. This still isnβt particularly nice. So what weβre going to do is simplify the denominator.

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