Video Transcript
Represent the area under the curve
of the function π of π₯ equals π₯ squared minus one on the close interval zero to
three in sigma notation using a right Riemann sum with π subintervals.
Remember when weβre finding a right
Riemann sum, we find the sum of π₯π₯ times π of π₯ π for values of π from one to
π. π₯π₯ is π minus π over π where
π and π are the lower and upper limits of our interval, respectively, and π is
the number of subintervals. π₯ π is π plus π lots of
π₯π₯. We always begin by working out what
π₯π₯ is. In our case, π is equal to zero,
π is equal to three, and, well, π is just π. This means π₯π₯ is three minus zero
over π or just three over π. Next, weβre going to work out what
π₯ π is. Itβs π, which we know to be zero,
plus π₯π₯, which is three over π, times π. Weβll write this as three π over
π.
Of course, we want to know what π
of π₯ π is. So it follows that to find π of π₯
π, we find π of three π over π. Letβs substitute three π over π
into our formula. Thatβs three π over π all squared
minus one which is nine π squared over π squared minus one. Weβre now ready to use the
summation formula. Weβre evaluating our sum for values
of π from one to π. Its π₯π₯, which is three over π,
multiplied by nine π squared over π squared minus one. We distribute our parentheses and
then weβre going to look to create a common denominator. We can do that by multiplying both
the numerator and denominator of our second fraction by π squared. That gives us three π squared over
π cubed, leaving us just to simply combine the numerators. We have 27 π squared minus three
π squared over π cubed.
Now, we can simplify this
somewhat. The numerators share a factor of 27
and three. And of course they have a common
denominator of π cubed. Both three and π cubed are
indepnent of π. This means we can take three over
π cubed outside of the sigma symbol, and that means weβre done. Weβve represented the area under
the curve of the function in sigma notation with a right Riemann sum. Itβs three over π cubed times the
sum of nine π squared minus π squared for values of π from one π.