### Video Transcript

Given that π of π₯ is equal to
four cos π₯ and that π₯ is greater than or equal to zero and less than or equal to
π by four, evaluate, to the nearest six decimal places, the Riemann sum for π with
six subintervals, taking the sample points to be left endpoints.

Remember, we can approximate the
definite integral of a function between the limits of, say, π and π by splitting
the area between the curve and the π₯-axis into π subintervals. Where the width of each rectangle,
π₯π₯, is equal to the difference between the upper and lower limits divided by π,
the number of subintervals. In this question, weβre looking to
split the area into six subintervals, so we let π be equal to six. We let π be equal to zero and π
be equal to π by four. So π₯π₯ here and the width of each
of our rectangles is π by four minus zero over six, which is π over 24. Letβs now sketch out the curve of
π¦ equals four cos π₯ between π₯ equals zero and π by four radians and spilt it
into these six subintervals.

This question requires us to use
left endpoints. So the height of each of our
rectangles will be equal to the value of the function at the left of each
subinterval. The sketch is really useful here
because we see that each rectangle sits above the π₯-axis. So we just need to find the
positive area of each. Letβs consider our first
rectangle. Its left endpoint is at π₯ equals
zero. So we find the height of the
rectangle by evaluating π of zero. π of zero is four times cos of
zero, which is four. The area of this first rectangle
then is therefore π by 24, since thatβs the width of each rectangle, times
four. Thatβs π by six square units.

We now need to find the next
endpoint. We achieve this by adding π by 24
to zero; thatβs π by 24. And we can therefore find the
height of this rectangle by evaluating π of π by 24. Thatβs four cos of π by 24, which
is approximately 3.965779 and so on. The area of our second rectangle
then is width times height or base times height. Thatβs π by 24 times 3.965 and so
on, which is roughly 0.51911931 and so on. Now, of course, this question is
asking us to evaluate till six decimal places, which is why Iβve left the numbers as
I have. We find the next endpoint by adding
another π by 24 to π by 24 and we get π by 12. The height of this third rectangle
then is π of π by 12 which is four cos of π by 12, which gives us root six plus
root two. And then we see that the area of
our third rectangle is π by 24 multiplied by this value, which gives us an area of
roughly 0.505 and so on square units.

We repeat this process by
repeatedly adding π by 24 to each of our endpoints. And when we do, we find that the
left endpoint of each of our rectangles is π by eight, π by six, and π₯ equals
five π by 24. Their heights are respectively
given by π of π by eight, π of π by six, and π of five π by 24. And when we multiply each of these
values by the width, π by 24, we find that the area of our fourth rectangle is
roughly 0.4837. Our fifth rectangle is roughly
0.4534. And our sixth is approximately
0.4153. Remember, the Riemann sum is the
total area of these six rectangles. Itβs 2.901 and so on. Thatβs 2.901067 correct to six
decimal places.