Find the lower Riemann sum approximation for 𝑓 of 𝑥 equals five minus 𝑥 squared on the close interval one to two, given that 𝑛 equals four subintervals.
The lower Riemann sum approximation is going to be the smallest estimate for the area between the curve of 𝑓 of 𝑥 and the 𝑥-axis and bounded by the lines 𝑥 equals one on 𝑥 equals two that we’ll get. So we recall if 𝑓 of 𝑥 is increasing, that is, 𝑓 prime of 𝑥 is greater than zero over some interval, then a left Riemann sum will yield an underestimate for the area between the curve on the 𝑥-axis in that interval. And a right Riemann sum will yield an overestimate. For a decreasing function, the reverse is true. So let’s begin by finding the derivative of our function 𝑓 of 𝑥 and working out whether it’s greater than or less than zero over the closed interval one to two.
The first derivative of our function is negative two 𝑥. Now for values of 𝑥 that are greater than zero, in other words, positive values of 𝑥, negative two 𝑥 is going to be less than zero. This means that our function must be a decreasing function in the closed interval one to two. So we need to evaluate a right Riemann sum. To find a right Riemann sum, we take values of 𝑖 from one to 𝑛. It’s the sum of Δ𝑥 times 𝑓 of 𝑥𝑖 for these values of 𝑖, where Δ𝑥 is 𝑏 minus 𝑎 divided by 𝑛. And in our case, 𝑎 is equal to one and 𝑏 is equal to two. And of course, 𝑛 is the number of subintervals. And 𝑥𝑖 is 𝑎 plus 𝑖 lots of Δ𝑥.
Let’s begin by working out the value of Δ𝑥. It’s 𝑏 minus 𝑎, that’s two minus one, all divided by 𝑛, which is four. Two minus one divided by four is a quarter or 0.25. And now, we have enough information to be able to evaluate 𝑥𝑖. It’s 𝑎 which is one plus Δ𝑥 times 𝑖; that’s 0.25𝑖. Remember, to evaluate the summation, we’re going to need to find 𝑓 of 𝑥𝑖. That’s, of course, 𝑓 of one plus 0.25𝑖. So let’s swap 𝑥 in our function for one plus 0.25𝑖. When we do, we obtain 𝑓 of 𝑥𝑖 to be five minus one plus 0.25𝑖 all squared. Now whilst we could distribute these parentheses and simplify, there’s really no need. We have enough to substitute into the summation formula. We’re interested in values of 𝑖 from one to 𝑛, so that’s one to four. Then it’s Δ𝑥 which is 0.25 times five minus one plus 0.25𝑖 squared.
Now, it’s useful to know that 0.25 is independent of 𝑖. So we can take this out as a common factor. And we need to work out the sum of five minus one plus 0.25𝑖 squared between 𝑖 equals one and four and then multiply that by 0.25. To do this, we simply substitute values of 𝑖 from one to four into this formula and then add them all together. When 𝑖 is one, it’s five minus one plus 0.25 squared. When 𝑖 is two, we get five minus one plus 0.5 squared. When 𝑖 is three, it’s five minus one plus 0.75 squared. And when 𝑖 is four, its five minus one plus one all squared. Remember, we’re gonna add these together and then multiply it all by 0.25. When we do, we obtain an answer of 2.28125 which correct to two decimal places is 2.28. The lower Riemann sum approximation for our function is 2.28.