For the function given below, find a formula for the Riemann sum obtained by dividing the interval two to four into N equals subintervals and using the right hand endpoint for each C sub K. Then take a limit of the sum as N goes to Infinity to calculate the area under the curve over two to four. So F of X = 3 X. First of all, we need to figure out what the distance is between the interval they're giving us. So in between 2:00 and 4:00, we'd have 4 - 2, and we're going to subdivide it into N intervals. So 4 - 2 / n or 2 / n is all these various intervals. Now we need to pay attention that we're not starting until we actually hit two. So we're going to have 2 + 2 / n two plus 2 / n * 2, two plus 2 / n + 3 based on how many intervals we've gone out. So at some point we're going to have 2 + 2 / n * K for the K interval out. If we think about this, we'd have K going from one to N of 3X. But now we're at some interval out there, 2 + 2 K over N Remember, if we're at 2:00 to start, we have two. The next one out is going to be 2 + 2 / n, and the next one's out it's going to be 2 + 2 / n * 2. And the next one out is going to be 2 + 2 / n * 3. And we're going to go... And eventually we're going to have this two plus 2K over N. And then we're literally going to stick that into that F of X equation to come up with the height. So three times the 2 + 2 KN would be the height of that location. And then we have N intervals, each with a base of 2 / n So when we multiply this out, 3 * 2 * 2 / n would give us 12 / n. And three times two K / n * 2 / n is going to give us 12 KN squared. Now if K is going from one to N, there is no K here, so we have 12 / n + 12 / n + 12 / n + 12 over NN times, so 12 / n * n. This next piece, the 12 / n ^2 doesn't have AK involved, so we can pull it out and recall that the summation of any K equal 1 to north of just plain old K is a formula. It's n * n + 1 all over two. So that's where this piece down here came from. So then if we multiply that, we get the NS to cancel in the first term and 12 N times n + 1 / 2 N squared. We're going to look at the leading coefficient of the second term in just a moment. So when we look at the limit as N goes to Infinity, we get 12 + 12 N squared plus twelve n / 2 N squared leading coefficients. 12 / 2 is 6, so 12 + 6 is 18. So the limit of the sum as N goes to Infinity is 18.