Build left and right endpoint sums

A cyclist's speed readings

A cyclist's speed is measured every 2 seconds, and the dots show the readings in meters per second. The distance is the area under the speed graph, but only five points of the graph are known.

To estimate the area, a rectangle stands on each 2-second piece of time, called a subinterval. A speed times a time is a distance, so each area is a distance in meters.

Use the left ends. Each rectangle takes its height from the reading at the left end of its subinterval: 3, 5, 8 and 12. The areas add up to \(2(3 + 5 + 8 + 12) = 56\) meters.

Use the right ends. Each height now comes from the right end: 5, 8, 12 and 14. The areas add up to \(2(5 + 8 + 12 + 14) = 78\) meters. The sums differ because the speed changes inside every subinterval.

One rectangle for each subinterval

This curve has no numbers, and its subintervals have different widths. A rectangle stands on each one.

The orange dot on each rectangle is its sample point, where its top meets the curve. Slide the sample point from the left end of each subinterval to the right end.

The heights change, but the widths do not. The button steps through the subintervals, and the green bracket shows the width of the one in focus.

Each area is the height at the chosen end times the width of its subinterval. The sum of the areas is the estimate.

Notation

1. Cut the interval from \(a\) to \(b\) at the points \(a = x_0 < x_1 < \dots < x_n = b\). Subinterval \(i\) runs from \(x_{i-1}\) to \(x_i\), so its width is the difference.

   \[\cg{\Delta x_i = x_i - x_{i-1}}\]

2. A left sum reads each height at the left end of its subinterval, \(\cy{f(x_{i-1})}\), and multiplies it by that subinterval's width.

   \[L_n = \sum_{i=1}^{n} \cy{f(x_{i-1})}\,\cg{\Delta x_i}\]

3. A right sum reads each height at the right end, \(\cy{f(x_i)}\), and uses the same widths.

   \[R_n = \sum_{i=1}^{n} \cy{f(x_i)}\,\cg{\Delta x_i}\]

4. Both sums use only the values of \(f\) at the cuts, so a table of readings is enough. If all widths are equal, each is \(\Delta x = \frac{b-a}{n}\), and it moves in front of the sum as a common factor.

   \[L_n = \cg{\Delta x}\left[\cy{f(x_0)} + \dots + \cy{f(x_{n-1})}\right]\]

When it fails

If a table gives readings at uneven times, the subintervals have different widths, and one common width gives the wrong areas.

The red lines show how far each height moves from its own time. Multiply each height by the width of its own subinterval instead.

Example 1: A tank filling

Water flows into a tank at the rate \(r(t) = 4 - t^2\) liters per minute for \(0 \le t \le 2\). Find the left sum and the right sum with 4 equal subintervals.

1. Four equal subintervals share the interval from 0 to 2, so each is 0.5 minutes wide and the ends are 0, 0.5, 1, 1.5 and 2.

   \[\cg{\Delta t = \frac{2 - 0}{4} = 0.5}\]

2. Substitute each end into \(r(t) = 4 - t^2\) to find the rate there.

   \[\cy{r} = 4,\ 3.75,\ 3,\ 1.75,\ 0\]

3. The left sum uses the rates at the left ends, the first four.

   \[\begin{aligned} L &= \cg{0.5}\,(\cy{4} + \cy{3.75} + \cy{3} + \cy{1.75}) \\ &= 6.25 \end{aligned}\]

4. The right sum uses the rates at the right ends, the last four. A rate in liters per minute times a time in minutes is an amount in liters.

   \[\begin{aligned} R &= \cg{0.5}\,(\cy{3.75} + \cy{3} + \cy{1.75} + \cy{0}) \\ &= 4.25 \end{aligned}\]

Example 2: Uneven readings

A car's speed is 4, 7, 9 and 10 meters per second at \(t = 0, 2, 5\) and 9 seconds. Find the left sum and the right sum for the distance.\[\begin{array}{c|ccccc} t & 0 & 2 & 5 & 9 \\ \hline v(t) & 4 & 7 & 9 & 10 \end{array}\]

1. The readings cut the interval from 0 to 9 into three subintervals, with widths 2, 3 and 4.

2. A tempting wrong move: Use the same width for every rectangle, \(9 \div 3 = 3\).

   \[L = 3\,(4 + 7 + 9) = 60\]

   The first subinterval is only 2 wide and the last is 4 wide. Rectangles 3 wide stand at the wrong times, and each red line shows how far a height moves from its own time.

3. Multiply each height by the width of its own subinterval. The left sum uses the heights at the left ends: 4, 7 and 9.

   \[L = \cy{4}\cdot\cg{2} + \cy{7}\cdot\cg{3} + \cy{9}\cdot\cg{4} = 65\]

4. The right sum uses the heights at the right ends: 7, 9 and 10, with the same widths.

   \[R = \cy{7}\cdot\cg{2} + \cy{9}\cdot\cg{3} + \cy{10}\cdot\cg{4} = 81\]

Practice

Let \(f(x) = \dfrac{1}{x}\). Find the left sum and the right sum on \([1, 3]\) with 4 equal subintervals.

The orange dots show where the heights are read. The left sum uses the first four of the five heights, and the right sum uses the last four.

Questions

The figure shows readings of a function \(g\): \(g(0) = 2\), \(g(2) = 5\), \(g(3) = 6\) and \(g(7) = 8\). The readings cut the interval from 0 to 7 into three subintervals.

1. The left rectangles are shown, and the third one is highlighted. What is its area?

   • 24
   • 32
   • 6
   Answer

   24

2. The right rectangles are shown, and the second one is highlighted. What is its area?

   • 6
   • 5
   • 12
   Answer

   6

3. A table gives a rate \(f\) at \(t = 0, 3, 4, 8\) and 10: the values are 5, 4, 6, 3 and 2. What is the right sum on the table's four subintervals?\[\begin{array}{c|ccccc} t & 0 & 3 & 4 & 8 & 10 \\ \hline f(t) & 5 & 4 & 6 & 3 & 2 \end{array}\]

   • 34
   • 49
   • 37.5
   Answer

   34

Summary

You build a left or right sum by multiplying each height at the chosen end by its subinterval's own width, and adding the products.

This works because each product is a rate times a time, which is a piece of the total quantity.

You use it to estimate an integral when the rate is known at only some points.

This is one of five skills in topic 6.2 of Which & Why: AP Calculus.