### 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 weβre 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.

Weβre now ready to work out what π₯
π is. Itβs π, which we know to be three,
plus π₯π₯, which is two over π, times π. Letβs 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 isnβt particularly
nice. So what weβre going to do is
simplify the denominator.

Weβll write it as one over one plus
two π over π. And then weβre 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, weβre 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. Itβs 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 πs
cancel. And we have our right Riemann sum
using sigma notation. Itβs the sum of two over π plus
two π for values of π between one and π.