## 20.5 Special Riemann Sums

From now on for our convenience we will consider Riemann sums in which all strips have the same width, d.

A general Riemann sum with fixed widths consists then of the sum of d multiplied by the values of f chosen in each strip.

We define the following four choices that are of special interest.

If we always evaluate f at the leftmost argument in each strip, call the value of the sum L(d).

If we always evaluate f at the rightmost argument in each strip, call the value of the sum R(d).

If we evaluate f at an argument that has maximum value for f in each strip, call the sum M(d).

If we evaluate f at an argument that has minimum value for f in each strip, call the sum m(d).

We can make the following observations. We assume that f is bounded so that m(d) and M(d) are both finite. Then the rightmost argument of one strip is the leftmost argument of the next.

Therefore the difference between L(d) and R(d) is only that R(d) gets a contribution of d * f(b) from the last strip that L(d) lacks and L(d) gets instead a contribution of d * f(a) that R(d) lacks.

The arguments in between at the endpoints of strips contribute to the L interval to the right and an identical amount to the R interval on the left.

Therefore we have

R(d) = L(d) + (f(b) - f(a) * d

which implies that R(d) and L(d) come together as d approaches 0.

Second, M(d) is greater than or equal to the true area and to any other Riemann sum for strips that can be obtained by subdividing the strips of width d. m(d), similarly is less than or equal to the true area or and to any other Riemann sum obtained by subdividing.

Finally, notice that if f is increasing between a and b then M(d) = R(d) and m(d) = L(d), which means that the true area and any Riemann sum obtained by subdividing gets sandwiched between bounds that come together as d goes to 0.

Which means that all Riemann sums obtained by subdividing must approach the same value, and the function f is integrable.

If f is not increasing between a and b but we can break up that interval into pieces in which f within each piece is increasing or decreasing, and f has bounded total variation between a and b, then f will be integrable by the same argument applied to all the pieces one at a time, and adding the resulting bound.
In the applet that follows, you can enter any standard function and limits and look at the the area computed by the left and right hand rules for different numbers of slices.

Exercises:

20.1 Compute for the function in the interval -1 to 2. If you add this to the left hand rule result for the integral of this function over this interval (get it from the applet below), what do you obtain? Try for different d values.

20.2 Compute the left hand rule and right hand rule for this integrand using a spreadsheet, for d = .01 and d = .02 and d = .005.