Determine the direction of error

Four estimates of one area

The curve \(f(x) = 8x - x^2\) rises from 0 to 4 and bends downward. The area under it is \(\frac{128}{3}\), about 42.7, and each estimate uses 4 subintervals.

Red marks area a shape counts that is not under the curve, and teal marks area under the curve that it misses.

Select the left sum. The curve rises, so every left rectangle lies under it, and the sum is 34, an underestimate. Every right rectangle reaches above the curve, so the right sum is 50, an overestimate.

Select the trapezoidal sum. Each top is a segment between two points of the curve, and the curve bends downward, so it lies above every segment. The sum is 42, an underestimate.

Select the midpoint sum. Each flat top can tilt onto the tangent line at its midpoint with no change of area, and the curve lies under its tangent lines. The sum is 43, an overestimate.

Two facts decide the direction

Two facts about the curve decide the direction of an error: whether it rises or falls, and which way it bends. Change each one and select the four sums.

If the curve rises, the left sum is too small and the right sum is too large. If the curve falls, these reverse.

If the curve bends downward, the trapezoidal sum is too small and the midpoint sum is too large. If it bends upward, these reverse.

Notation

1. Call the exact value of the integral \(\cb{I}\). If \(f\) is increasing on \([a, b]\), each left rectangle lies under the curve, and each right rectangle reaches above it.

   \[f' > 0:\quad L_n \le \cb{I} \le R_n\]

   If \(f\) is decreasing, the two sums change places.

2. If \(f\) is concave down, the graph bends downward. Each slanted top is a chord, a segment between two points of the graph, and it lies under the graph.

   \[f'' < 0:\quad T_n \le \cb{I}\]

3. Each midpoint top tilts onto the tangent line at its midpoint. The area does not change, because the top loses on one side what it gains on the other. The curve lies under its tangent lines, so the midpoint sum is too large.

   \[f'' < 0:\quad \cb{I} \le M_n\]

   If \(f\) is concave up, both inequalities reverse. Check \(f'\) for the left and right sums, and \(f''\) for the other two.

When it fails

The rules need \(f'\) or \(f''\) to keep one sign on the whole interval. From 0 to 8, the curve rises until \(x = 4\) and falls after it.

With 8 subintervals, the left and right sums are both 84, and the exact area is \(\tfrac{256}{3}\), about 85.3. Both are underestimates, and no rule here predicts that.

Example 1: A falling leak

The rate of a leak is decreasing. A table gives the rate in liters per hour at four times. Find the left sum for the first 6 hours. Is it an overestimate or an underestimate?\[\begin{array}{c|cccc} t & 0 & 2 & 4 & 6 \\ \hline r(t) & 9 & 6 & 4 & 3 \end{array}\]

1. The rate decreases, so the left end of each subinterval is its highest point. Each left rectangle reaches above the curve, so the left sum is an overestimate.

2. The width of each subinterval is 2.

   \[L = \cg{2}\,(\cy{9} + \cy{6} + \cy{4}) = 38\]

3. The right end of each subinterval is its lowest point, so the right sum is an underestimate. The leak is between 26 and 38 liters.

   \[R = \cg{2}\,(\cy{6} + \cy{4} + \cy{3}) = 26\]

Example 2: A rising rate

A rate is \(f(x) = x^2 + 1\) on the interval from 0 to 2. Is the trapezoidal sum with 2 subintervals an overestimate or an underestimate?

1. The first derivative is positive for \(x > 0\), so \(f\) is increasing.

   \[f'(x) = 2x > 0\]

2. A tempting wrong move: The function is increasing, so the trapezoidal sum is an underestimate, like the left sum.

   The tops are chords, and the curve bends upward, so the chords lie above it. The red parts show the excess. The rise decides only the left and right sums.

3. The second derivative is positive, so the curve is concave up and every chord lies above the curve.

   \[f''(x) = 2 > 0\]

4. The trapezoidal sum is an overestimate. As a check, \(T = 5\) and the exact area is \(\tfrac{14}{3}\), about 4.67.

   \[T = \tfrac12\left[1 + 2(2) + 5\right] = 5 > \tfrac{14}{3}\]

Practice

Let \(f(x) = e^{-x}\) on the interval from 0 to 2, with 2 subintervals. Which of the left, right, trapezoidal and midpoint sums are overestimates?

Find the signs of \(f'\) and \(f''\), or select each sum and read its colors.

Questions

The figure shows \(h(x) = 9 - x^2\) from 0 to 3. The curve falls and bends downward, and the first two questions use three equal subintervals.

1. The left rectangles are shown. Is the left sum an overestimate or an underestimate?

   • An overestimate, because the curve falls and the left end is the highest point of each subinterval.
   • An underestimate, because the left end of each subinterval is the lowest point.
   • An underestimate, because the curve bends downward.
   Answer

   An overestimate, because the curve falls and the left end is the highest point of each subinterval.

2. The trapezoids are shown. Is the trapezoidal sum an overestimate or an underestimate?

   • An underestimate, because the curve bends downward and each top lies under it.
   • An overestimate, because the curve falls, as for the left sum.
   • An overestimate, because the tops lie above the curve.
   Answer

   An underestimate, because the curve bends downward and each top lies under it.

3. Let \(f(x) = \dfrac{1}{x}\). The trapezoidal sum for \(\int_1^3 f(x)\,dx\) with 2 equal subintervals is which of these?

   • \(\tfrac{7}{6}\), an overestimate
   • \(\tfrac{7}{6}\), an underestimate
   • \(\tfrac{16}{15}\), an overestimate
   Answer

   \(\tfrac{7}{6}\), an overestimate

Summary

You decide the left and right sums by the rise or fall of the curve, and the trapezoidal and midpoint sums by its bend.

This works because a rising curve is highest at the right end, and a curve that bends up lies below its chords.

You use it when a problem asks whether an estimate from a graph or a table is too large or too small.

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