Video Transcript
Compute the right Riemann sum for 𝑓 of 𝑥 is equal to the cos of two 𝜋𝑥 on the closed interval from zero to one-half, given that there are four subintervals of equal width.
We need to compute the right Riemann sum for the function 𝑓 of 𝑥 is equal to the cos of two 𝜋𝑥 on the closed interval from zero to one-half using four subintervals of equal width. Let’s start by recalling that a right Riemann sum for a function 𝑓 of 𝑥 on the closed interval from 𝑎 to 𝑏 with 𝑛 subintervals of equal width is given by the sum from 𝑖 equals one to 𝑛 of Δ𝑥 times 𝑓 of 𝑥 𝑖, where Δ𝑥 will be our subinterval width which will be the length of our internal divided by the number of subintervals, so in this case, 𝑏 minus 𝑎 divided by 𝑛. And since we’re using our right Riemann sum, our sample points 𝑥 𝑖 will be the right 𝑛 points of our subintervals. So we’ll have 𝑥 𝑖 is equal to 𝑎 plus 𝑖 times Δ𝑥.
Let’s now look at our question. We’re given that our function 𝑓 of 𝑥 is the cos of two 𝜋𝑥. We want to find the Riemann sum for this function on the closed interval from zero to one-half. So we set our value of 𝑎 equal to zero and our value of 𝑏 equal to one-half. Finally, we’re using four subintervals of equal width. Since 𝑛 is the number of subintervals, we’ll use 𝑛 equal to four. We’re now ready to find the width of our subintervals Δ𝑥. We have Δ𝑥 is equal to 𝑏 minus 𝑎 divided by 𝑛. We know 𝑏 is one-half, 𝑎 is zero, and 𝑛 is equal to four. So we get Δ𝑥 is one-half minus zero all divided by four. And we can calculate this is equal to one-eighth.
Before we find our values of 𝑥 𝑖, remember, our values of 𝑖 are going to go from one to 𝑛. Since 𝑛 is equal to four, we need to find values for 𝑥 one, 𝑥 two, 𝑥 three, and 𝑥 four. Let’s start with 𝑥 one. We know 𝑥 𝑖 is equal to 𝑎 plus 𝑖 times Δ𝑥. Our value of 𝑎 is equal to zero, our value of 𝑖 in this case is one, and Δ𝑥 is equal to one-eighth. So we get 𝑥 one is zero plus one times one-eighth. We can also find 𝑥 two; it’s equal to zero plus two times one-eighth, which we can evaluate is equal to one-quarter. We can do the same to find 𝑥 three. We get zero plus three times one-eighth, which is equal to three-eighths.
And if we do the same for 𝑥 four, we get zero plus four times one-eighth which is equal to one-half. Now that we found our values for 𝑥 𝑖, we can find an expression for our right Riemann sum. So by using 𝑛 is equal to four, Δ𝑥 is equal to one-eighth, and 𝑓 of 𝑥 is the cos of two 𝜋𝑥, we get that our right Riemann sum in this case is equal to the sum from 𝑖 equals one to four of one-eighth times the cos of two 𝜋 times 𝑥 𝑖. We could start evaluating the series. But remember, one-eighth is a constant, so we’ll take it outside. This gives us one-eighth times the sum from 𝑖 equals one to four of the cos of two 𝜋𝑥 𝑖.
We now want to expand this series out term by term. We’ll start with the first term when 𝑖 is equal to one. So we’ll get the cos of two 𝜋 times 𝑥 one. Remember, we found that 𝑥 one is equal to one-eighth. So we get the cos of two 𝜋 times one-eighth which we can simplify to be the cos of 𝜋 by four. And of course, we know the cos of 𝜋 by four is equal to the square root of two divided by two. Let’s now do the same for the second term in our expansion. We get the cos of two 𝜋 times 𝑥 two. Remember, we showed that 𝑥 two is equal to one-quarter. So substituting this in, we get the cos of two 𝜋 times one-quarter, which is the cos of 𝜋 by two. However, the cos of 𝜋 by two is just equal to zero. So this doesn’t affect the value of our sum.
Let’s now find the third term in our series when 𝑖 is equal to three. Remember, we showed that 𝑥 three is equal to three-eighths. This time, we get the cos of two 𝜋 times three over eight. We know this is equal to the cos of three 𝜋 by four. And we know the cos of three 𝜋 by four is equal to negative the square root of two divided by two. But now we see we have root two over two minus root two over two. So these two terms cancel. So we just need to find the last term in our series when 𝑖 is equal to four. We get the cos of two 𝜋 times 𝑥 four.
Remember, we found that 𝑥 four is equal to one-half. This gives us the cos of two 𝜋 times one-half, which is the cos of 𝜋. But we know the cos of 𝜋 is just equal to one. So our entire expression just simplifies to give us one-eighth. And since we use Riemann sums to approximate the area under a curve, this is an approximation of the area. So we’ll write this as a decimal 0.125.
This means when we calculated the right Riemann sum for our function 𝑓 of 𝑥 is equal to the cos of two 𝜋𝑥 on the closed interval from zero to one-half with four subintervals of equal width, we got the answer of 0.125.
