# Question Video: Finding the Riemann Sum of a Quadratic Function in a Given Interval by Dividing It into Subintervals and Using Their Midpoint Mathematics • Higher Education

Given 𝑓(𝑥) = 𝑥² − 4 and −4 ≤ 𝑥 ≤ 2, evaluate the Riemann sum for 𝑓 with six subintervals, taking sample points to be midpoints.

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### Video Transcript

Given 𝑓 of 𝑥 is equal to 𝑥 squared minus four and 𝑥 is greater than or equal to negative four and less than or equal to two, evaluate the Riemann sum for 𝑓 with six subintervals, taking sample points to be midpoints.

Remember, we can approximate the definite integral of a function between the limits 𝑎 and 𝑏 by splitting the area between the curve and the 𝑥-axis into 𝑛 subintervals. The width of each rectangle is given by 𝛥𝑥, which is found by subtracting 𝑎 from 𝑏 and dividing this value by 𝑛. In this example, we want to split the area into six subintervals. So we let 𝑛 be equal to six, and then 𝑎 is negative four and 𝑏 is two. The width of each of our subintervals is given then by two minus negative four over six, which is equal to one. Next, we’re going to sketch the curve of 𝑦 equals 𝑥 squared minus four between the endpoints negative four and two. And we’re going to split this into six subintervals.

This question requires us to use midpoints. So the height of each rectangle will be equal to the value of the function at the middle of each subinterval. That’s going to look a little something like this. Now, we can work out the 𝑥-value at the end of each rectangle by repeatedly adding one to four. And that gives us each of these values. We can then see that the midpoints are going to be negative 3.5, 𝑥 equals negative 2.5, negative 1.5, and so on.

Now, we’re going to need to be extra careful here. We see that some of our rectangles sit below the 𝑥-axis. This means we’re going to find the negative value of their area. In other words, we’ll subtract the total area of the rectangles that sit below the 𝑥-axis from the total area of the rectangles that sit above the 𝑥-axis. Let’s begin by finding the function values at each midpoint. And that will tell us the height of each rectangle. That’s 𝑓 of negative 3.5, 𝑓 of negative 2.5, 𝑓 of negative 1.5, and so on. And, of course, our function is 𝑥 squared minus four. And when we substitute each of these values into that function, we get 8.25, 2.25, negative 1.75, negative 3.75, another negative 3.75, and another negative 1.75. We then obtain the area of our first rectangle here to be 8.25 times one.

Our second rectangle has an area of 2.25 times one. Our third rectangle has an area of 1.75 times one, not negative 1.75. Because we’re just dealing with areas at the moment. Our fourth rectangle has an area of 3.75 times one. Base times height with our fifth rectangle is 3.75 times one again. And for our last rectangle, it’s 1.75 times one. The Riemann sum is therefore 8.25 plus 2.25 minus the sum of 1.75, 3.75, 3.75, and another 1.75. And that gives us an approximation to the definite integral between the values of negative four and two of 𝑥 squared minus four. It’s negative 0.5. And, of course, 𝑥 squared minus four is a fairly simple function to integrate. So we can check our answer by evaluating that integral.

When we do, we get 𝑥 cubed over three minus four 𝑥 between the limits of negative four and two, which gives us a value of zero. And our estimate of negative 0.5 is pretty close. So we can assume we’ve probably done this correctly. It’s worth noting that a sketch of the curve won’t always be possible. So instead, we need to notice that when the value of 𝑓 of 𝑥 is less than zero, we subtract the area of the rectangle with that height.