### Video Transcript

Using the midpoint rule with π
equals five, round the definite integral from two to five of two π₯ over three π₯
plus two with respect to π₯ to four decimal places.

Remember, we can estimate a
definite integral by using Riemann sums. We split the region into
subintervals and create a rectangle in each. The total area of the rectangles
gives us an estimate to the integral. In a midpoint Riemann sum, the
height of each rectangle is equal to the value of the function at the midpoint of
its base. Now, working with midpoints isnβt
quite as nice as using the left or right Riemann sum. In this case, thereβs nothing to
stop us evaluating each of the areas in term. And letβs begin by working out the
width of each subinterval.

Geometrically, it tells us the
width of the rectangles. And itβs given by Ξπ₯ equals π
minus π over π, where π and π are the endpoints of the interval and π is the
number of subintervals. In our case, our lower limit is two
and our upper limit is five. We, therefore, let π be equal to
two and π be equal to five. And weβre told that π is equal to
five. Ξπ₯ is five minus two over five,
which is three-fifths or 0.6. And a table can make the next step
a little easier. In our table, weβre going to begin
by working out each of the subintervals.

We know the lower limit of our
definite integral to be two. To find the right endpoint of our
first rectangle on our first subinterval, we add 0.6 to two to get 2.6. This means our next rectangle
starts at π₯ equals 2.6. This time, we add 0.6 again. And we find the right endpoint to
be 3.2. The left endpoint of our next
rectangle must, therefore, be 3.2. And the right endpoint is 3.2 plus
Ξπ₯. Thatβs 3.8. Our next rectangle begins at
3.8. And adding 0.6, we find that it
ends at 4.4. And our fifth and final rectangle β
Remember, we wanted π to be equal to five β begins at 4.4 and then ends at 4.4 plus
0.6, which is five. And thatβs a really good start
because we know the upper limit of our interval is indeed five.

Next, weβre going to work out the
midpoint of each of these subintervals. Now, we can probably do this in our
head. But if weβre struggling, we add the
two values and divide by two. And when we do, we obtain the
midpoints to be 2.3, 2.9, 3.5, 4.1, and 4.7, respectively. To work out the height of each
rectangle, we need to work out the value of the function at these points. So, for example, in this first row,
weβll start by working out π of 2.3. To do that, we substitute 2.3 into
our function two π₯ over three π₯ plus two. And we get 46 over 89. Then, we repeat this process for π₯
equals 2.9. When we substitute 3.5 into our
function, we get 14 over 25. And the height of our final two
rectangles are 82 over 143 and 94 over 161 units, respectively.

And our very final step is to
calculate the area of each rectangle by multiplying its width by its height. Of course, the width of each
rectangle is Ξπ₯ at 0.6. So we multiply each of these
function values by 0.6 and then find their sum. Now, we could do this in turn or we
could find their sum and multiply by 0.6. Weβll get the same answer. When we do find the total of all
the values in our column titled Ξπ₯ times π of π₯π, we get 1.66571 and so on. And if we run this correct to four
decimal places, we find that an estimate that the definite integral between two and
five of two π₯ over three π₯ plus two with respect to π₯ is roughly 1.6657.