Lean Calculus

Historically, the mathematician Bernhard Riemann is credited with the approach / solution to the area problem presented in this section. To see how it works, we’ll do a case study on something for which we know the answer. That way, when we apply Riemann’s approach, we’ll know whether it’s working or not. So, let’s look at \(f(x)=x\) on \([0,2]\text{.}\) Notice that this gives a triangle above the axis with leg lengths 2 and 2 so that the triangle has area 2. Riemann’s idea is to chop the interval \([0,2]\) into equally sized subintervals. For example, if we chop it into 4 subintervals, each subinterval would have size \((2-0)/4 = 1/2\text{.}\) The subintervals would be \([0,1/2], [1/2,2/2], [2/2,3/2],\) and \([3/2,4/2]\text{.}\) For each subinterval, Riemann evaluated the function (\(f(x)=x\) for us) somewhere in each subinterval and used that evaluation as the height of a rectangle and the subinterval as the base of the rectangle. For simplicity, let’s evaluate our function at the right endpoints of the subintervals. So, here’s what we have. Now, unlike the area under the curve, these rectangles *are* familiar geometric shapes so we can compute their areas. Even better, if we add up all of their areas, it will give us an *approximate* value of the area under the curve. Of course, it will be off by some amount since we can clearly see that the rectangles stick outside of the area in question. Still ... it gives some approximation. So, let’s compute it. Each rectangle has the same size base of \(1/2\) and the heights are \(f(1/2)=1/2, f(2/2)=2/2, f(3/2)=3/2, \) and \(f(4/2)=4/2\text{.}\) One quick note: Instead of writing things like \(4/2\) as \(2\text{,}\) I’m leaving it in terms of the subinterval size \(1/2\text{.}\) That will be helpful later so avoid reducing such things for now. So, the approximate area as given by these rectangular areas is

So? Not bad. We know the actual, exact, area is 2 but this process estimated it at 2.5. How could we improve? The answer is, hopefully, very apparent. Namely, we could just up the number of subintervals. When we do that, the size of the subintervals shrinks so that the rectangles are thinner (more like pins). So, the amount of the rectangle poking out of the actual area will be less. In other words, the estimate by using them *should* be more accurate. Let’s check! Let’s chop the interval into 8 equal subintervals this time. So each subinterval will have size \((2-0)/8=1/4\) this time. Now, let’s add up all those rectangular areas.

Using these 8 smaller subintervals gives the more accurate estimate 2.25 instead of 2.5 that we got with the 4 larger subintervals. So, we have every reason to expect if we continue to use more and more subintervals of equal size (so thinner and thinner rectangles) that we’ll get more and more accurate estimates of the area.