SUNY Geneseo Department of Mathematics

Areas Under Curves and Riemann Sums

Monday, November 16

Math 221 02
Fall 2020
Prof. Doug Baldwin

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Misc

SI

There’s a session tonight from 6:30 - 8:00, to replace yesterday’s.

The next one will be Wednesday from 6:30 - 8:00.

Remote Learning Plans

After Thanksgiving, this course will still revolve around readings, discussions, problem sets, and individual meetings, with periodic Zoom “classes” to collect answers to discussion questions. The Zoom “classes” shouldn’t introduce new material, but should rather serve to summarize questions and ideas from discussions and readings into coherent documents that will be posted in Canvas.

SI sessions will also continue after Thanksgiving, although on a different schedule. SI sessions continue to be reviews of what you learn through readings, discussion, and Zoom “classes,” not replacements for those things.

Flow diagrams: 'reading/discussion' leads to 'Zoom' and back, 'problem set' to 'meeting' and back, 'SI meetings' loop

Problem Set 9

Is now available through Canvas. Ideally, work on it this week and grade it late in the week or before you leave next week.

Area Under Curves

Based on “Approximating Area” and “Forming Riemann Sums” in section 5.1 of the textbook, and this discussion of areas under curves.

Riemann Sums

This is the key idea for finding areas under curves. So what is a Riemann sum?

A sum motivated by adding areas of rectangles under a graph in order to estimate the area between that graph and the x axis.

Region below graph divided into rectangles and a sum of their areas

Use a Riemann sum to estimate, by hand, the area under the line y = x from x = 0 to x = 1.

Triangle below graph of y equals x

There were several examples offered, differing in how many intervals they used and thus what Δx was, as well as in where the “xi*” was in each interval. In all cases though, the basic idea of multiplying an f(x) value by the width of the intervals and adding up the products was the same:

Sums of areas under y equals x using 2 or 5 intervals and left or right x values

The actual area is 1/2. None of the Riemann sums got exactly that answer, and they didn’t have to because they are estimates of the area.

Sketching the rectangles involved in some of these sums over the function’s graph shows that some are under-estimates, and some are over-estimates, which again is consistent with how the values of the Riemann sums compared to the actual area.

Region under y equals x divided into rectangles

Note that regardless of whether the error is to over-estimate or under-estimate, smaller intervals mean there will be less of it.

Area

The second key idea is how to use a Riemann sum to find the exact area under a curve. It’s based on the idea that smaller intervals mean less error. You can get smaller intervals by using more of them, and so taking the limit as the number of intervals goes to infinity gives the exact area:

A equals limit as n goes to infinity of sum from 1 to n of f of x sub i times delta x

Apply this to the area under y = x from x = 0 to x = 1.

Start by writing out the formula:

Area under graph y equals x from 0 to 1 and Riemann sum formula for area

This looks impenetrable, until you realize that with n equal-sized intervals covering a region 1 unit wide, each one must be 1/n units wide, i.e., Δx = 1/n. Then, if you use the right side of each rectangle as its xi*, the first right side is 1 Δx distance away from the origin, i.e., x1* = 1/n, the second is distance 2Δx from the origin, so x2* = 2/n, and so forth. Substituting these values into the sum makes it look a lot more tractable:

Riemann area formula with x sub i identified as i over n and delta x as 1 over n

Now factor 1/n2 out of the sum, since it doesn’t change as i changes (anything not dependent on the summation’s dummy variable is a “constant” for purposes of the constant multiple rule for summations):

Area equals limit as n goes to infinity of 1 over n squared times sum from 1 to n of i

Now the sum is simplified to one that has a known closed form, so replace it with that closed form and simplify:

Area simplifies to limit as n goes to infinity of 1 half plus 1 over 2 n

The limit is now a standard limit at infinity, with a constant term and a term that goes to 0:

Limit as n goes to infinity of 1 half plus 1 over 2 n equals 1 half

Notice that this value agrees with what we calculated geometrically, as we hoped it would.

Next

Integrals as Riemann sums.

We have informally defined a definite integral of a function over some interval as the area under the function’s graph over that interval. But we ought to allow for the possibility that functions are negative sometimes, and maybe even discontinuous.

Please read “Definition and Notation,” “Evaluating Definite Integrals,” “Area and the Definite Integral,” and “Properties of the Definite Integral” in section 5.2 of the textbook by class time Wednesday, and participate in this discussion of definite integrals.

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