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Integral Properties and Estimates · Tutorial 1002 of 1495

Monotone Functions And Darboux Sums

Use endpoint values to determine and calculate lower and upper Darboux sums for increasing and decreasing functions on equal-width partitions.

Intermediate 9 min read

What You'll Learn

  • Define increasing and decreasing functions using non-strict inequalities.
  • Identify the infimum and supremum on a subinterval from its endpoints.
  • Write lower- and upper-sum formulas for equal-width partitions.
  • Compute Darboux sums for increasing and decreasing functions.
  • Check endpoint choices, arithmetic, and the order of the two sums.

Why Monotonicity Simplifies Darboux Sums

A Darboux sum is built by dividing an interval into subintervals and using a function's infimum or supremum on each one as a rectangle height. For a general bounded function, finding those infima and suprema may require careful analysis. For a monotone function, the endpoint values determine them directly.

This lesson focuses on that endpoint choice and on computing the resulting sums when all subintervals have equal width. The direction of monotonicity matters: for an increasing function the left endpoint gives the lower value, while for a decreasing function the right endpoint gives the lower value.

Partitions and Darboux Sums

Let \(f:[a,b]\to\mathbb{R}\) be bounded, where \(a<b\). A partition \(P\) of \([a,b]\) is a finite increasing list of points

$$ a=x_0<x_1<\cdots<x_n=b. $$

The \(i\)-th subinterval is \([x_{i-1},x_i]\), and its width is \(\Delta x_i=x_i-x_{i-1}\). On this subinterval, let \(m_i\) be the infimum of \(f\)'s values and \(M_i\) be their supremum. The lower and upper Darboux sums are defined by

$$ L(f,P)=\sum_{i=1}^{n}m_i\Delta x_i, \qquad U(f,P)=\sum_{i=1}^{n}M_i\Delta x_i. $$
Definition: A function is increasing on an interval if \(x\le y\) implies \(f(x)\le f(y)\). It is decreasing if \(x\le y\) implies \(f(x)\ge f(y)\). These definitions allow equality, so constant functions are both increasing and decreasing.

The infimum and supremum are used in the definitions whether or not they are attained. For a monotone function on a closed subinterval, however, the endpoint values do attain them: monotonicity orders every function value between the two endpoint values.

Finding the Endpoint Values

Suppose \(f\) is increasing on \([a,b]\). For any subinterval \([x_{i-1},x_i]\), every \(x\) in that subinterval satisfies \(x_{i-1}\le x\le x_i\). Therefore

$$ f(x_{i-1})\le f(x)\le f(x_i). $$

Because both endpoints belong to the subinterval, the left endpoint value is its infimum and the right endpoint value is its supremum. For a decreasing function the inequalities reverse: the left endpoint value is the supremum and the right endpoint value is the infimum.

Endpoint rule: On each subinterval, an increasing function has its infimum at the left endpoint and its supremum at the right endpoint. A decreasing function has its supremum at the left endpoint and its infimum at the right endpoint.

These statements remain correct when the function is constant on a subinterval: the endpoint values then agree, so the infimum and supremum are equal.

Equal-Width Partitions

An equal-width partition divides \([a,b]\) into \(n\) subintervals of the same width. For a positive integer \(n\), define

$$ \Delta x=\frac{b-a}{n}, \qquad x_i=a+i\Delta x \quad (i=0,1,\ldots,n). $$

Since every subinterval has width \(\Delta x\), the endpoint rule gives compact formulas for both sums.

Formulas for equal-width partitions: If \(f\) is increasing, then $$ L(f,P)=\Delta x\sum_{i=1}^{n}f(x_{i-1}), \qquad U(f,P)=\Delta x\sum_{i=1}^{n}f(x_i). $$ If \(f\) is decreasing, then $$ L(f,P)=\Delta x\sum_{i=1}^{n}f(x_i), \qquad U(f,P)=\Delta x\sum_{i=1}^{n}f(x_{i-1}). $$

Each term is a function value multiplied by the width of its subinterval. The sum is not necessarily an ordinary geometric area if some function values are negative; it is the signed sum specified by the definition.

Worked Example: Increasing Linear Function

Find the lower and upper Darboux sums for \(f(x)=2x+1\) on \([0,4]\) using \(n=4\) equal subintervals.

First calculate the width:

$$ \Delta x=\frac{4-0}{4}=1. $$

The partition points are \(x_i=0+i\), so they are \(0,1,2,3,4\). The function values are

$$ f(0)=1,\quad f(1)=3,\quad f(2)=5,\quad f(3)=7,\quad f(4)=9. $$

Since \(f\) is increasing, the lower sum uses the left endpoint of each subinterval and the upper sum uses the right endpoint:

$$ L(f,P)=1(1+3+5+7)=1(16)=16, $$ $$ U(f,P)=1(3+5+7+9)=1(24)=24. $$

Check the endpoint choice: each successive value increases, so each left value is no greater than the corresponding right value. The calculated sums therefore satisfy \(16\le24\), as required.

Worked Example: Decreasing Linear Function

Find the lower and upper Darboux sums for \(g(x)=10-2x\) on \([1,5]\) using \(n=4\) equal subintervals.

The width and partition points are

$$ \Delta x=\frac{5-1}{4}=1, \qquad x_i=1+i, $$

so the points are \(1,2,3,4,5\). Evaluate the function at each point:

$$ g(1)=8,\quad g(2)=6,\quad g(3)=4,\quad g(4)=2,\quad g(5)=0. $$

The values decrease from left to right. Thus the right endpoints give the infima, and the left endpoints give the suprema:

$$ L(g,P)=1(6+4+2+0)=1(12)=12, $$ $$ U(g,P)=1(8+6+4+2)=1(20)=20. $$

Check: on every subinterval the right endpoint value is less than or equal to the left endpoint value. The lower sum is indeed no greater than the upper sum: \(12\le20\).

Worked Example: Increasing Function with a Nonlinear Formula

Find the lower and upper Darboux sums for \(h(x)=x^2\) on \([0,3]\) using \(n=3\) equal subintervals.

The width is

$$ \Delta x=\frac{3-0}{3}=1. $$

The partition points are \(0,1,2,3\), and the function values at them are

$$ h(0)=0^2=0,\quad h(1)=1^2=1,\quad h(2)=2^2=4,\quad h(3)=3^2=9. $$

The function \(x^2\) is increasing on \([0,3]\): if \(0\le x\le y\), then \(y^2-x^2=(y-x)(y+x)\ge0\). Therefore left endpoints give the infima and right endpoints give the suprema. The sums are

$$ L(h,P)=1(0+1+4)=5, \qquad U(h,P)=1(1+4+9)=14. $$

Check: the paired endpoint values on the three subintervals are \(0\le1\), \(1\le4\), and \(4\le9\). Adding the lower values gives \(5\), adding the upper values gives \(14\), and \(5\le14\).

Why the Formulas Work

The formulas follow directly from the definitions, and the argument also verifies the order of the sums. Consider the increasing case first. On the \(i\)-th subinterval, monotonicity gives

$$ f(x_{i-1})\le f(x)\le f(x_i) \quad\text{for every }x\in[x_{i-1},x_i]. $$

The endpoint values are included in the range on that subinterval, so its infimum is \(f(x_{i-1})\), and its supremum is \(f(x_i)\). Substituting those values into the definitions of \(L\) and \(U\), and using the common width \(\Delta x\), gives the increasing-function formulas.

For a decreasing function, monotonicity instead gives \(f(x_i)\le f(x)\le f(x_{i-1})\) throughout the same subinterval. Thus its infimum is \(f(x_i)\) and its supremum is \(f(x_{i-1})\), which gives the decreasing-function formulas. In either case, every subinterval's infimum is no greater than its supremum. Since every width is positive, adding the corresponding products yields

$$ L(f,P)\le U(f,P). $$

This justification uses monotonicity to locate the extrema; it does not assume that an arbitrary bounded function attains its infimum or supremum. For monotone functions on these closed subintervals, attainment follows from the endpoint values themselves.

A Common Endpoint Mistake

A frequent error is to choose the left endpoint for the lower sum regardless of whether the function is increasing or decreasing. For \(g(x)=10-2x\) on \([1,5]\), that incorrect choice would give

$$ 1(8+6+4+2)=20 $$

as the purported lower sum. But \(g\) is decreasing, so those left endpoint values are the suprema, not the infima. The value \(20\) is the upper sum. The correction is to use the right endpoint values for the lower sum:

$$ L(g,P)=1(6+4+2+0)=12. $$

A reliable check is to write down the endpoint values in order before summing. If they increase from left to right, the left values are lower; if they decrease, the right values are lower. Then check that the computed lower sum does not exceed the upper sum.

Why This Endpoint Rule Is Useful

Monotonicity replaces a separate search for the infimum and supremum on every subinterval with a simple comparison of endpoint values. This makes Darboux sums practical to compute even when the function is not linear, provided its direction of monotonicity is known on the whole interval. If monotonicity holds only on part of the interval, the endpoint rule can be used only on subintervals where that monotonicity condition holds.

The method also clarifies what information a sum uses: the partition determines the widths, monotonicity determines which endpoint is lower or upper, and the function formula supplies the endpoint values. Keeping these tasks separate helps prevent both endpoint errors and arithmetic errors.

Check Your Understanding

Use the definitions and endpoint rule to answer the following questions.

  1. For an increasing function on \([x_{i-1},x_i]\), which endpoint value is the infimum, and which is the supremum?
  2. For a decreasing function, which endpoint values are used in the lower and upper sums?
  3. For \(f(x)=5-x\) on \([0,2]\) with \(n=2\), find the common width and list the partition points.
  4. Why is it valid to use endpoint values for the infimum and supremum of a monotone function on a closed subinterval?
  5. A student obtains \(L(f,P)=18\) and \(U(f,P)=11\). What basic check shows that at least one calculation or endpoint choice must be wrong?

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