So far we have been using rectangles to approximate the area under a curve. We’ve determined the heights of these rectangles by evaluating the function at either the right or left endpoints of each subinterval [latex][x_{i-1},x_i][/latex].

However, we do not have to restrict the evaluation to just these points. We can evaluate the function at any point [latex]x_i^*[/latex] within the subinterval [latex][x_{i-1},x_i][/latex], and use [latex]f(x_i^*)[/latex] as the height of our rectangle. This gives us an estimate for the area of the form:

A sum of this form is called a Riemann sum, named after the 19th-century mathematician Bernhard Riemann.

When using left- and right-endpoint approximations, our estimates improve as we increase the number of subintervals [latex]n[/latex]. The same idea applies to Riemann sums: the more subintervals we use, the better our approximation. Now, let’s define the area under a curve using Riemann sums.

Taking the limit of a sum is a bit different from taking the limit of a function [latex]f(x)[/latex] as [latex]x[/latex] goes to infinity. We discuss limits of sums in more detail in the chapter on Sequences and Series in Calculus 2. For now, assume that the techniques we use to compute limits of functions also apply to sums.

We must also consider what happens if our sum converges to different limits for different choices of [latex]x_i^*[/latex]. If [latex]f(x)[/latex] is continuous on [latex][a,b][/latex], the limit:

[latex]\underset{n\to \infty }{\lim}\displaystyle\sum_{i=1}^{n} f(x_i^*)\Delta x[/latex]

is unique and does not depend on the choice of [latex]x_i^*[/latex].

Before we dive into examples, let’s discuss some specific choices for [latex]x_i^*[/latex]. Any choice for [latex]x_i^*[/latex] gives us an estimate of the area under the curve, but we might want to know if our estimate is too high or too low. We can choose [latex]x_i^*[/latex] to guarantee one result or the other.

• Overestimate: Choose [latex]x_i^*[/latex] so that [latex]f(x_i^*)[/latex] is the maximum value on [latex][x_{i-1},x_i][/latex]. This makes our Riemann sum an upper sum.
• Underestimate: Choose [latex]x_i^*[/latex] so that [latex]f(x_i^*)[/latex] is the minimum value on [latex][x_{i-1},x_i][/latex]. This makes our Riemann sum a lower sum.

Find a lower sum for [latex]f(x)=10-x^2[/latex] on [latex][1,2][/latex]; use [latex]n=4[/latex] subintervals.

Show Solution

With [latex]n=4[/latex] over the interval [latex][1,2], \, \Delta x=\frac{1}{4}[/latex]. We can list the intervals as [latex][1,1.25], \, [1.25,1.5], \, [1.5,1.75], \, [1.75,2][/latex]. Because the function is decreasing over the interval [latex][1,2][/latex], (Figure 14) shows that a lower sum is obtained by using the right endpoints.

The Riemann sum is

[latex]\begin{array}{ll}\displaystyle\sum_{k=1}^{4} (10-x^2)(0.25)& =0.25[10-(1.25)^2+10-(1.5)^2+10-(1.75)^2+10-(2)^2] \\ & =0.25[8.4375+7.75+6.9375+6] \\ & =7.28 \end{array}[/latex]

 

The area of [latex]7.28[/latex] is a lower sum and an underestimate.

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “5.1 Approximating Areas” here (opens in new window).