Build a midpoint sum
A dragster's speed
A dragster's speed is \(v(t) = t^2\) meters per second, and the distance in the first 6 seconds is the area under the curve. Three rectangles, each standing on a 2-second piece of time called a subinterval, estimate this area.
Use the left ends. The rectangles lie under the curve, and the teal gaps are area that the sum misses. The areas add up to 40 meters, which is too small.
Use the right ends. The rectangles reach above the curve, and the red parts are area that the sum counts but that is not under the curve. The sum is 112 meters, which is too large.
Use the midpoints, the middle of each subinterval. The heights are \(v(1) = 1\), \(v(3) = 9\) and \(v(5) = 25\), so the sum is \(2(1 + 9 + 25) = 70\) meters. The area is exactly 72, so the midpoints give a much closer estimate.
The excess balances the gap
This curve has no numbers, and the top of each rectangle meets it above the middle of its subinterval.
The red excess is the part of a rectangle above the curve. The teal gap is the part under the curve that the rectangle misses.
On one side of the midpoint the top is above the curve, and on the other side it is below. The excess and the gap are about the same size, so they nearly cancel and each rectangle has about the correct area.
Move the sample point toward either end. One color grows while the other shrinks, so the balance is lost.
Notation
Subinterval \(i\) runs from \(x_{i-1}\) to \(x_i\). Its midpoint is halfway between its ends, so it is the average of the two.
\[\cy{m_i} = \frac{x_{i-1} + x_i}{2}\]The midpoint sum reads the height at each midpoint, \(\cy{f(m_i)}\), multiplies it by the width of its subinterval, and adds the products.
\[M_n = \sum_{i=1}^{n} \cy{f(m_i)}\,\cg{\Delta x_i}\]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.
\[M_n = \cg{\Delta x}\left[\cy{f(m_1)} + \dots + \cy{f(m_n)}\right]\]The sum needs a value of \(f\) at the midpoint of every subinterval.
When it fails
A table gives readings at the times 0, 2, 4, 6 and 8. Two subintervals, each 4 wide, have the midpoints 2 and 6, and the table has readings there.
Four subintervals, each 2 wide, have the midpoints 1, 3, 5 and 7, and the table has no readings there. Choose subintervals so that every midpoint is a time in the table.
Example 1: Rainfall
Rain falls at the rate \(R(t)\) millimeters per hour. A gauge records \(R\) every hour for 6 hours, as in the table. Use a midpoint sum with 3 subintervals to estimate the rainfall.\[\begin{array}{c|ccccccc} t & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline R(t) & 1 & 3 & 6 & 8 & 9 & 7 & 4 \end{array}\]
The three equal subintervals are from 0 to 2, from 2 to 4 and from 4 to 6, so their midpoints are 1, 3 and 5.
\[\cg{\Delta t = \frac{6 - 0}{3} = 2}\]Read the table at the midpoints.
\[R(1) = \cy{3},\ R(3) = \cy{8},\ R(5) = \cy{7}\]Multiply the sum of the heights by the width. A rate in millimeters per hour times a time in hours is an amount in millimeters.
\[M = \cg{2}\,(\cy{3} + \cy{8} + \cy{7}) = 36\]
Example 2: Bacteria
A culture grows at the rate \(g(t) = 2^t\) thousand cells per hour for \(0 \le t \le 4\). Use a midpoint sum with 2 subintervals to estimate the growth.
The two subintervals are from 0 to 2 and from 2 to 4, so the midpoints are 1 and 3.
A tempting wrong move: The midpoint is halfway between the ends, so take the height halfway between the end heights.
\[\frac{g(0) + g(2)}{2} = \frac{1 + 4}{2} = 2.5\]The height must be the value of the curve at the midpoint, \(g(1) = 2\). The curve bends upward, so at \(t = 1\) it is below the average of its end heights, as the red strip shows.
Substitute the midpoints into \(g\).
\[g(1) = \cy{2}, \quad g(3) = \cy{8}\]Multiply the sum of the heights by the width 2. The estimate is 20 thousand cells.
\[M = \cg{2}\,(\cy{2} + \cy{8}) = 20\]As a check, the exact area is about 21.6.
Practice
A river flows at the rate \(q(t) = \dfrac{20}{t + 1}\) liters per hour for \(0 \le t \le 4\). Use a midpoint sum with 2 subintervals to estimate the volume.
Each subinterval is 2 hours wide. Move the sample point to the middle of each one. Then find the height of the curve there.
Solution
The midpoints of the subintervals are 1 and 3. Substitute them into \(q\) to find the heights.
\[q(1) = \cy{10}, \quad q(3) = \cy{5}\]Multiply the sum of the heights by the width 2. The estimate is 30 liters.
\[M = \cg{2}\,(\cy{10} + \cy{5}) = 30\]
Questions
The figure shows \(k(x) = \tfrac12 x(8 - x)\) on the interval from 0 to 8, with four midpoint rectangles of equal width.
What is the height of the second rectangle?
- 7.5
- 6
- 8
Answer
7.5
What is the area of the first rectangle?
- 7
- 3.5
- 12
Answer
7
The table gives a rate \(f\) in liters per hour every 2 hours from \(t = 0\) to \(t = 12\). What is the midpoint sum with 3 equal subintervals?\[\begin{array}{c|ccccccc} t & 0 & 2 & 4 & 6 & 8 & 10 & 12 \\ \hline f(t) & 4 & 7 & 9 & 10 & 8 & 5 & 3 \end{array}\]
- 88 liters
- 44 liters
- 84 liters
Answer
88 liters
Summary
You build a midpoint sum by reading the height at the middle of each subinterval, multiplying it by the width, and adding the products.
This works because on each subinterval the excess above the curve nearly cancels the gap below it.
You use it when you want a closer estimate than a left or right sum and the function is known at the midpoints.
This is one of five skills in topic 6.2 of Which & Why: AP Calculus.