# Video: Finding the Numerical Value of a Definite Integral of a Function Using the Right Endpoint Estimate

Calculate the right endpoint estimate of ∫_(0) ^(4) (𝑥² + 2) d𝑥 with 𝑛 = 2 subintervals. Is the result an overestimate or underestimate of the actual value?

02:35

### Video Transcript

Calculate the right endpoint estimate of the definite integral between zero and four of 𝑥 squared plus two d𝑥 with 𝑛 equals two subintervals. Is the result an overestimate or underestimate of the actual value?

The right endpoint estimate or the right Riemann sum is a way of estimating a definite integral by thinking about the area between the curve and the 𝑥-axis. In a right endpoint estimate, we split that area into a given number of rectangles. The total area of those rectangles gives us an estimate for the definite integral. So we can see what’s going on. Let’s begin by sketching the graph of our function out: 𝑓 of 𝑥 equals 𝑥 squared plus two.

We have the curve represented by 𝑓 of 𝑥 equals 𝑥 squared plus two. And the area we’re interested in is bounded by the vertical lines 𝑥 equals zero and 𝑥 equals four. Those are simply the upper and lower limits of our definite integral. Now, if we were to calculate the definite integral, it would tell us the exact area shaded. What we’re going to do, though, is split this region into two subintervals. And we’re going to form two rectangles. The height of each rectangle will be the value of the function at the right endpoint of each subinterval.

Now, it’s probably fairly obvious how we’re going to split this region up into two subintervals. But we can recall a formula, the width of each subinterval given by Δ𝑥 is equal to 𝑏 minus 𝑎 over 𝑛, where 𝑎 and 𝑏 are the vertical lines that bound our region and 𝑛 is the number of subintervals. In this case, the width of each subinterval will be four minus zero over two which is simply two. We count up a multiples of two. And actually, we see that we’re going to construct an extra vertical line at the point where 𝑥 equals two.

And so, we have our two rectangles. The height of the first rectangle will be the value of the function at the right endpoint, so 𝑓 of two. And the height of the second rectangle will be the value of the function when 𝑥 is equal to four. And of course, we know 𝑓 of 𝑥 to be equal to 𝑥 squared plus two. So, 𝑓 of two is two squared plus two, which is six. Similarly, 𝑓 of four is four squared plus two, which is 18.

Remember, we’re going to calculate the area of each of these rectangles. Now, since the area of a rectangle is simply its base multiplied by its height, we can say that an estimate to our definite integral is six multiplied by two plus 18 multiplied by two. That’s simply the height of each rectangle multiplied by its width, which we know to be two, which is equal to 48.

So we’ve calculated an estimate for the definite integral between zero and four of 𝑥 squared plus two d𝑥. But is this an overestimate or an underestimate? Well, if we simply go back to the diagram, we can see that we’ve actually calculated a little bit more area than we need. And so, we can say that 48 is an overestimate of the actual value.