Find the width and sample points

Four equal pieces of an interval

The curve shows a function \(f\) on the interval from 2 to 10. The interval is cut into 4 pieces of equal width, and a teal rectangle stands on each piece.

The height of each rectangle is the value of \(f\) at one point of its piece, the sample point, marked by an orange dot.

The interval is \(10 - 2 = 8\) long and four equal pieces share it, so each piece is \(8 \div 4 = 2\) wide, as the green bracket shows. The pieces meet at the edges 2, 4, 6, 8 and 10.

Move the slider to the left end. The sample points become the left edges 2, 4, 6 and 8. At the right end they are 4, 6, 8 and 10, and in the middle 3, 5, 7 and 9.

Any interval, any number of pieces

The interval runs from \(a\) to \(b\) and has \(n\) equal pieces. The bracket above the curve spans the whole interval, which is \(b - a\) long.

Each piece takes an equal share of this length, and the green bracket marks the width of one piece.

Move the Pieces slider. More pieces are narrower, but in every case \(n\) widths add up to \(b - a\).

The first edge is \(a\) and each next edge is one width farther to the right, so after \(n\) widths the edge is \(b\).

Notation

1. The width of each piece is the length of the interval divided by the number of pieces. Its name is \(\cg{\Delta x}\), read "delta x".

   \[\cg{\Delta x = \dfrac{b-a}{n}}\]

2. Edge number \(i\) is \(a\) plus \(i\) widths, so the first edge, with \(i = 0\), is \(a\) itself.

   \[x_i = a + i\,\cg{\Delta x}, \qquad i = 0, 1, \dots, n\]

   The last edge is \(x_n = a + n\,\Delta x = b\), because \(n\) widths make the whole length.

3. Piece number \(i\) runs from \(x_{i-1}\) to \(x_i\), for \(i = 1\) to \(n\). Its sample point is its left edge, its right edge or its midpoint, the average of the two edges.

   \[\begin{aligned} \text{left: } & \cy{x_{i-1}} \\ \text{right: } & \cy{x_i} \\ \text{midpoint: } & \cy{\tfrac{x_{i-1} + x_i}{2}} \\ & \cy{{} = a + \left(i - \tfrac12\right)\Delta x} \end{aligned}\]

   The midpoint is the average of \(a + (i-1)\Delta x\) and \(a + i\,\Delta x\), so it is half a width past the left edge.

When it fails

The formula \(\frac{b-a}{n}\) gives every piece the same share of the interval, so it needs pieces of equal width. Make the pieces unequal.

With unequal pieces, the edges are 2, 3, 5, 8 and 10. Each width is the right edge of its piece minus the left edge, so the widths are 1, 2, 3 and 2. Together they still make 8.

Example 1: Left and right sample points

The interval from 1 to 13 has 4 equal pieces. Find the width, the left sample points and the right sample points.

1. The interval is \(13 - 1 = 12\) long and four equal pieces share it, so each piece is 3 wide.

   \[\cg{\Delta x = \dfrac{13 - 1}{4} = 3}\]

2. The edges start at 1 and each next edge is 3 farther to the right. The fifth edge is 13, the end of the interval.

   \[x_i = 1 + 3i\!:\quad 1,\ 4,\ 7,\ 10,\ 13\]

3. The left sample points are the first four edges, and the right sample points are the last four edges.

   \[\begin{aligned} \text{left: } & \cy{1,\ 4,\ 7,\ 10} \\ \text{right: } & \cy{4,\ 7,\ 10,\ 13} \end{aligned}\]

Example 2: Midpoints of five pieces

The interval from 3 to 7 has 5 equal pieces. Find the width and the midpoint sample points.

1. The bracket spans the interval from \(a = 3\) to \(b = 7\), which has \(n = 5\) pieces.

2. A tempting wrong move: Divide the end of the interval by the number of pieces.

   \[\Delta x = \dfrac{7}{5} = 1.4\]

   Five pieces of width 1.4 reach from 3 to \(3 + 5(1.4) = 10\). The red pieces run past the end of the interval at 7, so this width is too large.

3. The pieces must fill the interval from 3 to 7. It is \(7 - 3 = 4\) long, so each piece is 0.8 wide.

   \[\cg{\Delta x = \dfrac{7 - 3}{5} = 0.8}\]

4. The first midpoint is half a width after 3, at 3.4, and each next midpoint is one width farther. The last midpoint is 6.6, half a width before 7.

   \[\begin{gathered} \cy{3 + \left(i - \tfrac12\right)(0.8)}, \quad i = 1, \dots, 5 \\ \cy{3.4,\ 4.2,\ 5,\ 5.8,\ 6.6} \end{gathered}\]

Practice

The interval from \(-1\) to 5 has 4 equal pieces. Find the width and the left sample points.

On the figure, four equal pieces start at \(-1\). Move the Width slider until they end at 5.

Questions

Questions 1 and 2 use the figure, a curve on the interval from 1 to 9 cut into equal pieces. Each orange dot marks the sample point of one piece.

1. How wide is each piece?

   • 2
   • 8
   • 2.25
   Answer

   2

2. Which sample points does the figure use?

   • The midpoint of each piece
   • The left end of each piece
   • The right end of each piece
   Answer

   The midpoint of each piece

3. The interval from \(-3\) to 9 has 8 equal pieces. Each piece is read at its right end. What is the fifth sample point?

   • 4.5
   • 3
   • 7.5
   • 2.625
   Answer

   4.5

Summary

The width is the length of the interval divided by the number of pieces, and each edge is \(a\) plus a whole number of widths.

This works because \(n\) equal pieces fill the interval, so \(n\) widths add up to \(b - a\).

You find the width and the sample points at the start of every Riemann sum, because they decide every term.

This is one of four skills in topic 6.3 of Which & Why: AP Calculus.