### Video Transcript

Estimate the definite integral
between one and nine of five sin of two times the square root of three π₯dπ₯ using
the midpoint rule with π equals four, giving your answer to four decimal
places.

Remember, we can estimate the value
of a definite integral by using Riemann sums. We split the region into π sub
intervals and create a rectangle in each. The total area of the rectangles
gives us an estimate to our 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. And whilst itβs not entirely
necessary, letβs sketch the graph of five sin of two root three π₯ out so we can see
whatβs going on.

The graph of π¦ equals five sin of
two root three π₯ looks a little something like this. Weβre going to split our region
into four sub intervals. So, before we add any rectangles to
our diagram, weβll work out the width of each sub interval. Geometrically, this tells us the
width of our rectangles. Itβs given by β³π₯ equals π minus
π over π, where π and π are the lower and upper limits of the definite integral,
and π is the number of sub intervals. In this case, weβre going to let π
be equal to one and π be equal to nine. And weβre told that π equals
four.

So, the width of each of our
rectangles is nine minus one over four, which is two units. This means our first rectangle will
be between one and two. The height of this rectangle is the
value of the function at the midpoint. So, thatβs at the point where π₯
equals two. In other words, the height of the
rectangle will be π of two. Now, of course, this rectangle sits
below the π₯-axis, so that will give us a negative value. Which is absolutely fine because
eventually weβre going to subtract the area of the rectangles that lie below the
π₯-axis from the area of the rectangles that lie above them.

Our next rectangle is going to go
from π₯ equals three to π₯ equals five. The height of this rectangle will
be the value of the function at the midpoint. Thatβs at the point where π₯ equals
four. So, the height of this rectangle
will be the value of π of four. Our third rectangle goes from π₯
equals five to π₯ equals seven. And, once again, the height of the
rectangle is the value of the function at the midpoint. So, thatβs when π₯ equals six. Its height will be the value of π
of six.

Our fourth and final rectangle will
go from π₯ equals seven to π₯ equals nine. We notice once again that we have a
rectangle that sits below the π₯-axis, so calculating the area will give us a
negative value. And thatβs fine because we want to
subtract the area from the area of the rectangles that sit above the π₯-axis. This time, the height of this
rectangle is the value of the function when π₯ equals eight. So, weβre going to need to work out
π of eight.

Now, we said that an estimate for
the definite integral is the sum of the areas of the rectangles. Take into account that weβre going
to subtract the areas of the rectangle that lie below the π₯-axis. And we know that the area of a
rectangle is given by base multiplied by height. And so, the negative area of the
first rectangle will be π of two multiplied by two. π of four multiplied by two will
give us the area of the second rectangle. π of six multiplied by two will
give us the positive area of the third rectangle. And π of eight multiplied by two
will give us the negative area of the fourth rectangle.

We can work out π of two, π of
four, π of six, and π of eight by substituting two, four, six, and eight into our
function, respectively. And so, an estimate for our
definite integral is five sin of two root six times two plus five sin of two root 12
times two plus five sin of two root 18 times two plus five sin of two root 24 times
two. Thatβs 0.6125207 and so on, which,
correct to four decimal places, is 0.6125. And so, using the midpoint rule,
weβve estimated the value of our definite integral. Itβs 0.6125.

Now, throughout the video, I
mentioned that we would subtract any areas of rectangles that lie below the
π₯-axis. And at this step, it may have seen
that we were adding these. However, if we work out the value
of five sin of two root six and five sin of two root 24, we get a negative value as
we would expect. And so, multiplying each of these
individually by two maintains a negative. And these end up getting subtracted
from the area of the other two rectangles. So, this does all of our work for
us. The definite integral is
approximately 0.6125.