### Video Transcript

Approximate the definite integral
between negative two and two of three π₯ squared minus five π₯ with respect to π₯
using a Riemann sum with right endpoints. Take π to be eight.

Remember, we can estimate the
solution to a definite integral by using Riemann sums, in this case is the definite
integral of three π₯ squared minus five π₯ between the limits of negative two and
two. And weβre going to be using right
endpoints. When writing Riemann sums with
right endpoints, we take values of π from one to π. Itβs the sum of Ξπ₯ times π of
π₯π for these values of π, where Ξπ₯ is π minus π divided by π. Bear in mind here that π is the
number of subintervals and π₯π is equal to π plus π lots of Ξπ₯. So letβs look at what we actually
have.

Our limits are from negative two to
two. So weβll let π be equal to
negative two and π be equal to two. Weβre told that we need to take π
to be eight. Geometrically, this tells us the
number of rectangles we have. And now, we can work out Ξπ₯. Thatβs the width of each
rectangle. According to our formula, Ξπ₯ is π
minus π over π. Thatβs two minus negative two over
eight, which is one-half. And once we have Ξπ₯, we can then
work out π₯π. Itβs π which we said is negative
two plus Ξπ₯. Thatβs half times π. In other words, π₯π is negative
two plus π over two.

Now, for our Riemann sum, we need
to work out π of π₯π. Thatβs clearly π of negative two
plus π over two. We achieve this by substituting
negative two plus π over two into our formula three π₯ squared minus five π₯. And if we distribute our
parentheses and simplify, we see that π of π₯π is three-quarters π squared minus
17 over two π plus 22. Letβs now substitute Ξπ₯ and π of
π₯π into our summation formula. We now see that an approximation to
our definite interval is the sum of a half times three-quarters π squared minus 17
over two π plus 22 for values of π from one to eight. Now, actually, this constant factor
a half is independent of π. So we could actually take it
outside of the sum. And now, whilst this step isnβt
entirely necessary, it canβt simplify things on occasion.

Weβre now going to substitute
values of π from one through to eight into three-quarters π squared minus 17 over
two π plus 22 and find their sum. When π is one, we get 0.75 minus
8.5 plus 22, which is 14.25. When π is two, we get eight. When π is three, we get 3.25. When π is four, we get zero. When π is five, itβs negative
1.75. When π is six, we get negative
two. When itβs seven, we get negative
0.75. And when π is eight, we get
two. Finding the sum of these values and
then multiplying it by one-half and we get 23 over two. And so, an approximation to our
definite integral weβre using a right Riemann sum with eight subintervals is 23 over
two.