Video Transcript
Compute the right Riemann sum for π of π₯ is equal to one divided by π₯ times π₯
minus two on the closed interval from three to five, given that there are four
subintervals of equal width. Approximate your answer to the nearest three decimal places.
The question gives us a function π of π₯, and it wants us to compute the right
Riemann sum for this function on the closed interval from three to five with four
subintervals of equal width. We need to give our answer to three decimal places of accuracy.
Letβs start by recapping what we mean by a right Riemann sum. The right Riemann sum of a function π of π₯ on a closed interval from π to π with
π subintervals is given by the sum from π equals one to π of Ξπ₯ times π of π₯
π, where Ξπ₯ is the width of our subintervals, which will be π minus π divided by
π, and our π₯ π will be our sample points. Since weβre taking a right Riemann sum, these will be the right endpoints of our
subintervals. π₯ π will be equal to π plus π times Ξπ₯.
The Riemann sum gives us a method of approximating the area under our curve π of π₯
on the closed interval from π to π. First, since the question is asking us to compute the right Riemann sum of the
function π of π₯ is equal to one divided by π₯ times π₯ minus two, weβll set our
function π of π₯ equal to this. Next, since we want to calculate this on the closed interval from three to five,
weβll set our value of π equal to three and our value of π equal to five. Finally, since we want to use four subintervals of equal width, weβll set our value
of π equal to four.
We can now use this information to calculate the width of our subintervals. The width of our subintervals will be the length of our interval divided by the
number of subintervals, in other words, π minus π divided by π. We see this is equal to five minus three divided by four. And of course we can calculate this to give us one-half.
Next, we can find the values of each of our sample points. Remember, since our sum is going to go from π is equal to one to four, we need to
find π₯ one, π₯ two, π₯ three, and π₯ four. So letβs find π₯ one. We substitute π is equal to one, π is equal to three, and Ξπ₯ is equal to one-half
into our formula. This gives π₯ one is equal to three plus one times one-half. Weβll write this as 3.5.
We can do exactly the same to find π₯ two. We see itβs equal to three plus two times one-half. And if we calculate this expression, we see itβs equal to four. Next, we find π₯ three. We find that itβs equal to 4.5. And finally, we see that our last sample point, π₯ four, is equal to five.
Now that weβve found all of our sample points and the value of Ξπ₯, weβre ready to
find our right Riemann sum. Substituting in our value of Ξπ₯ is equal to one-half and our value of π is equal to
four and using our definition for our function π of π₯, we get that our right
Riemann sum is equal to the sum from π equals one to four of one-half times one
divided by π₯ π multiplied by π₯ π minus two.
Weβre almost ready to evaluate this series. Remember, one-half is a constant factor, so we can take this outside of our sum. So we now have one-half times the sum from π equals one to four of one divided by π₯
π times π₯ π minus two. We now want to expand our series term by term. Letβs start with the first term, which is when π is equal to one. We of course get one divided by π₯ one times π₯ one minus two. But remember, we found that π₯ one is equal to 3.5. So by using π₯ one is equal to 3.5, we get one divided by 3.5 times 3.5 minus
two. And if we evaluate this expression, we get four divided by 21.
We now want to find the second term in our series. This will be one divided by π₯ two times π₯ two minus two. And remember, we found that π₯ two is equal to four. So using this value of π₯ two, we get one divided by four times four minus two. And if we calculate this, we get one-eighth. For the third term in our series, we found π₯ three equal to 4.5. So we get the third term one divided by 4.5 times 4.5 minus two. And if we calculate this expression, we get four divided by 45. And finally, for our fourth term, we found π₯ four is equal to five. So we get one divided by five times five minus two, which we can calculate is equal
to one divided by 15.
So our Reimann sum is equal to one-half multiplied by four over 21 plus one over
eight plus four over 45 plus one over 15. And we can just evaluate this expression. To three decimal places, we get 0.236, which is our final answer.
Therefore, we were able to show if we compute the right Riemann sum for the function
π of π₯ is equal to one divided by π₯ times π₯ minus two on the closed interval
from three to five with four subintervals of equal width, then to three decimal
places we got 0.236.