Question Video: Estimating the Definite Integration of a Polynomial Function in a Given Interval by Dividing It into Subintervals and Using Their Right Endpoint Mathematics • Higher Education

Video Transcript

Approximate the definite integral between negative two and two of three 𝑥 squared minus five 𝑥 with respect to 𝑥 using a Riemann sum with right endpoints. Take 𝑛 to be eight.

Remember, we can estimate the solution to a definite integral by using Riemann sums, in this case is the definite integral of three 𝑥 squared minus five 𝑥 between the limits of negative two and two. And we’re going to be using right endpoints. When writing Riemann sums with right endpoints, 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. Bear in mind here that 𝑛 is the number of subintervals and 𝑥𝑖 is equal to 𝑎 plus 𝑖 lots of Δ𝑥. So let’s look at what we actually have.

Our limits are from negative two to two. So we’ll let 𝑎 be equal to negative two and 𝑏 be equal to two. We’re told that we need to take 𝑛 to be eight. Geometrically, this tells us the number of rectangles we have. And now, we can work out Δ𝑥. That’s the width of each rectangle. According to our formula, Δ𝑥 is 𝑏 minus 𝑎 over 𝑛. That’s two minus negative two over eight, which is one-half. And once we have Δ𝑥, we can then work out 𝑥𝑖. It’s 𝑎 which we said is negative two plus Δ𝑥. That’s half times 𝑖. In other words, 𝑥𝑖 is negative two plus 𝑖 over two.

Now, for our Riemann sum, we need to work out 𝑓 of 𝑥𝑖. That’s clearly 𝑓 of negative two plus 𝑖 over two. We achieve this by substituting negative two plus 𝑖 over two into our formula three 𝑥 squared minus five 𝑥. And if we distribute our parentheses and simplify, we see that 𝑓 of 𝑥𝑖 is three-quarters 𝑖 squared minus 17 over two 𝑖 plus 22. Let’s now substitute Δ𝑥 and 𝑓 of 𝑥𝑖 into our summation formula. We now see that an approximation to our definite interval is the sum of a half times three-quarters 𝑖 squared minus 17 over two 𝑖 plus 22 for values of 𝑖 from one to eight. Now, actually, this constant factor a half is independent of 𝑖. So we could actually take it outside of the sum. And now, whilst this step isn’t entirely necessary, it can’t simplify things on occasion.

We’re now going to substitute values of 𝑖 from one through to eight into three-quarters 𝑖 squared minus 17 over two 𝑖 plus 22 and find their sum. When 𝑖 is one, we get 0.75 minus 8.5 plus 22, which is 14.25. When 𝑖 is two, we get eight. When 𝑖 is three, we get 3.25. When 𝑖 is four, we get zero. When 𝑖 is five, it’s negative 1.75. When 𝑖 is six, we get negative two. When it’s seven, we get negative 0.75. And when 𝑖 is eight, we get two. Finding the sum of these values and then multiplying it by one-half and we get 23 over two. And so, an approximation to our definite integral we’re using a right Riemann sum with eight subintervals is 23 over two.