Concavity and Trapezoidal Over- or Underestimates

Why Concavity Predicts the Error Direction

A trapezoidal sum approximates the integral by replacing the graph on each subinterval with the straight segment joining its endpoint values. The key question is whether that segment lies above or below the graph. Concavity answers that question, so it lets us predict the direction of the approximation error without first calculating the exact integral.

The conclusion depends on the graph’s position relative to its endpoint chords, not on whether the function is positive or negative. A trapezoidal sum can be an overestimate even when the integral is negative. It is therefore important to compare the numerical values of the approximation and integral, rather than relying on a picture of positive area alone.

Concavity and the Second Derivative

More precisely, for \(a\leq x\leq b\), the chord from \((a,f(a))\) to \((b,f(b))\) has height \((1-t)f(a)+tf(b)\) at \(x=(1-t)a+tb\), where \(0\leq t\leq1\). Concavity up means \(f(x)\leq(1-t)f(a)+tf(b)\); concavity down means the reverse inequality.

When \(f\) is twice differentiable, the sign of \(f''\) is a standard way to check concavity. If \(f''(x)\geq0\) throughout an interval, then \(f\) is concave up there. If \(f''(x)\leq0\), then \(f\) is concave down there. These tests require the indicated derivative condition over the interval, not merely at one point. Concavity can also be established directly from the chord definition.

Trapezoidal Sums and the Main Result

Let \(a=x_0<x_1<\cdots<x_n=b\) be a partition of an interval, and let \(\Delta x_i=x_i-x_{i-1}\) be the width of its \(i\)th subinterval. The trapezoidal sum is

Each term is the area of the trapezoid whose parallel sides have heights \(f(x_{i-1})\) and \(f(x_i)\). The formula works with unequal as well as equal subinterval widths. For a continuous function, the comparison with the definite integral is:

The theorem applies to the signed definite integral, including when \(f\) is negative. If \(f\) is linear on the interval, its graph is its own endpoint chord, so the trapezoidal sum equals the integral exactly.

Why the Chord Determines the Comparison

Consider one subinterval \([u,v]\). Let \(L(x)\) be the linear function whose graph is the chord joining \((u,f(u))\) and \((v,f(v))\). The area of the trapezoid on this subinterval is the integral of that linear function:

If \(f\) is concave up, its graph lies on or below the chord, so \(f(x)\leq L(x)\) throughout \([u,v]\). Integrating this pointwise inequality gives

Thus the trapezoid’s contribution is at least the integral on that subinterval. Adding the inequalities for every subinterval gives \(T\geq\int_a^b f(x)\,dx\). For a concave-down function, \(f(x)\geq L(x)\), so the inequalities reverse and the trapezoidal sum is an underestimate.

The second-derivative test also has a direct justification. Suppose \(f\) is twice continuously differentiable and \(f''\geq0\) on \([u,v]\). For \(x<y\) in that interval, the Mean Value Theorem applied to \(f'\) gives \(f'(y)-f'(x)=f''(c)(y-x)\geq0\) for some \(c\) between \(x\) and \(y\). Hence \(f'\) is nondecreasing. If \(u<z<v\), apply the Mean Value Theorem to \(f\) on \([u,z]\) and \([z,v]\). The resulting points \(r\in(u,z)\) and \(s\in(z,v)\) satisfy \(r<s\), so

Multiplying by the positive widths and rearranging yields \(f(z)\leq\frac{v-z}{v-u}f(u)+\frac{z-u}{v-u}f(v)\), exactly the statement that the graph lies below its chord. If \(f''\leq0\), the same argument reverses the inequalities and gives concavity down.

Worked Examples

When Concavity Changes

The overestimate or underestimate conclusion requires the same concavity to hold throughout the interval being approximated. If \(f''\) is positive on some parts and negative on others, the chord can lie below the graph on some subintervals and above it on others. The errors may compete, so the sign of \(f''\) at just one point—or the function’s behavior on just one part of the interval—does not establish the direction of the total error.

A useful approach is to check concavity on every subinterval in the partition. If each is concave up, each trapezoid contributes at least the corresponding integral; if each is concave down, each contributes at most the corresponding integral. If the concavity differs across the partition, the simple global conclusion does not follow. The actual total comparison would require more information.

Common Pitfall

A frequent mistake is to reverse the chord comparison. For \(f(x)=x^2\) on \([0,2]\), the endpoint chord is \(L(x)=2x\), and \(x^2\leq2x\) on this interval. A mistaken conclusion might say, “The curve is below the chord, so the trapezoid is an underestimate.” The comparison actually shows the opposite: the chord’s integral is larger than the function’s integral. In fact, \(\int_0^2 2x\,dx=4\), while \(\int_0^2 x^2\,dx=\frac{8}{3}\). Thus the trapezoid overestimates.