Integral Bounds From Darboux Sums

Why Bound an Integral?

A definite integral can represent accumulated change or signed area, but its exact value is not always easy to calculate. A partition provides another approach: on each subinterval, enclose the function between a lower and an upper height, then add the resulting rectangle areas. The two totals give a guaranteed interval containing the integral.

The width of that interval is useful information in its own right. A narrow interval means the partition provides a precise estimate; a wide interval leaves greater uncertainty. The bounds do not require the integral’s exact value to be known in advance.

Partitions and Darboux Sums

Let \(f\) be a bounded function on the closed interval \([a,b]\), where \(a<b\). A partition \(P\) of this interval is a finite increasing list of points

The \(i\)th subinterval is \([x_{i-1},x_i]\), and its width is \(\Delta x_i=x_i-x_{i-1}\), which is positive. On that subinterval, let \(m_i\) be the infimum of \(f\), and let \(M_i\) be its supremum. Boundedness ensures these are finite. If \(f\) is continuous, it attains its minimum and maximum on each closed subinterval, so \(m_i\) and \(M_i\) are those values.

The heights may be negative. In that case, the corresponding rectangle contribution is a signed area: for example, a negative height times a positive width gives a negative contribution. The same formulas apply whether \(f\) stays above the horizontal axis, stays below it, or changes sign.

The Integral Lies Between the Sums

For a bounded Riemann-integrable function, the lower and upper sums for any fixed partition enclose its definite integral. The integral need not equal either sum. It is the total accumulated value that lies between these two estimates.

Here we use the monotonicity property of the definite integral: if integrable functions \(g\) and \(h\) satisfy \(g(x)\leq h(x)\) on an interval, then \(\int g\leq\int h\) over that interval. A continuous function on a closed bounded interval is bounded and Riemann integrable, so continuous functions satisfy the theorem’s hypotheses.

Why the Bounds Hold

On the \(i\)th subinterval, the definitions of infimum and supremum give \(m_i\leq f(x)\leq M_i\) at every point \(x\). Integrating these inequalities over that subinterval preserves their order. Adding the resulting inequalities over all subintervals gives the stated bounds. This uses additivity of the definite integral across adjacent intervals.

In equations, the full argument is

This reasoning also explains why the lower and upper sums are ordered. Since \(m_i\leq M_i\) and \(\Delta x_i>0\), each lower contribution is at most its corresponding upper contribution. Adding gives \(L(f,P)\leq U(f,P)\).

The Gap Measures Uncertainty

Subtracting the lower sum from the upper sum gives the width of the interval in which the integral is known to lie. Since both sums use the same partition, their difference can be calculated interval by interval.

Each \(M_i-m_i\) is the vertical range of the function on one subinterval. Multiplying this range by the subinterval’s width gives that subinterval’s contribution to the uncertainty. The total gap is nonnegative.

If we use the midpoint of the bounds as an estimate, define

Because the integral lies between the two endpoints, its distance from this midpoint cannot exceed half the gap:

This is a guaranteed error limit, not a claim that the midpoint is the exact integral. The integral might be anywhere within the bounds.

Worked Examples

A Common Mistake: Reversing the Bounds

A frequent error is to assume that the lower sum must be positive or that the upper sum must be the larger positive-looking area. The correct comparison is numerical: each interval’s infimum is at most its supremum, and the width is positive. This remains true when both values are negative.

For instance, on \([0,1]\), \(f(x)=-2\) is constant. With the one-interval partition, the lower and upper heights are both \(-2\), so

The integral is \(\int_0^1-2\,dx=-2\), so the correct relation is \(-2\leq-2\leq-2\). A calculation that changes the negative rectangle area to \(2\) has switched from signed accumulation to ordinary geometric area and does not give the definite integral.

Using the Bounds Responsibly

Darboux bounds are useful when the function’s range on each subinterval can be determined, even if the integral itself is inconvenient to evaluate. Before applying the theorem, check that the function is bounded and Riemann integrable on the interval; continuity on a closed, bounded interval is a common sufficient condition. Then determine the infimum and supremum on every subinterval, preserve the signs of the heights, and calculate both sums.

The final statement should include the whole interval of possible values, not just one selected estimate. For example, \(L\leq\int_a^b f(x)\,dx\leq U\) communicates the guarantee, while the gap \(U-L\) tells how much uncertainty remains for that partition.