Why Monotonicity Gives Bounds
A rate describes how quickly some quantity accumulates. If the rate stays nonnegative, the accumulated quantity over a time interval is the area under the rate graph. Rectangle sums estimate that area by dividing time into subintervals and using a sampled rate as the height of each rectangle.
When the rate is monotone, its endpoint values tell us which way the graph can lie across each subinterval. A nondecreasing rate cannot fall below its value at the left endpoint; a nonincreasing rate cannot rise above it. This simple fact proves that endpoint sums bound the true area, and it determines which endpoint sum is larger.
Partitions and Endpoint Sums
Let \(f\) be a continuous, nonnegative rate on \([a,b]\), where \(a<b\). A partition divides this interval into finitely many subintervals by choosing points
The width of subinterval \(i\) is \(\Delta x_i=x_i-x_{i-1}\), which is positive. The left-endpoint sum uses \(f(x_{i-1})\) as the height on that subinterval, while the right-endpoint sum uses \(f(x_i)\). Write these sums as \(L\) and \(R\):
Let \(A\) denote the true area under the graph from \(a\) to \(b\). Because \(f\) is nonnegative, this area is also the accumulated amount represented by the rate over the interval; there is no below-axis area to subtract.
The Endpoint-Bound Theorem
- If \(f\) is nondecreasing, then \(L\le A\le R\).
- If \(f\) is nonincreasing, then \(R\le A\le L\).
- If \(f\) is nonconstant, both endpoint bounds are strict: for a nondecreasing rate, \(L<A<R\), and for a nonincreasing rate, \(R<A<L\). If \(f\) is constant, \(L=A=R\).
The bounds apply to any partition, including one whose subintervals have unequal widths. The key is that on each subinterval, monotonicity places every graph height between the two endpoint heights. Multiplying those height comparisons by the positive width preserves their order.
Why the Bounds Hold
If \(f\) is nondecreasing, then for every \(x\in[x_{i-1},x_i]\), \(f(x_{i-1})\le f(x)\le f(x_i)\). If \(f\) is nonincreasing, the inequalities reverse.
The rectangle has width \(\Delta x_i>0\). Its endpoint height multiplied by this width gives the rectangle area, so the height comparison gives the same ordering for the rectangle area and the graph's area over the subinterval.
Summing the subinterval areas gives the full left sum, right sum, and area. Adding inequalities in the same direction preserves them, which yields the stated bounds.
For instance, in the nondecreasing case, the pointwise height comparison gives a graph area on subinterval \(i\) between \(f(x_{i-1})\Delta x_i\) and \(f(x_i)\Delta x_i\). Summing over all subintervals gives the full inequality:
For a nonincreasing rate, \(f(x_i)\le f(x)\le f(x_{i-1})\) throughout each subinterval, so the same argument yields \(R\le A\le L\).
The strictness statement follows from continuity and monotonicity. If a monotone function has different endpoint values on a subinterval, continuity means it is strictly between those endpoint heights over a region of positive width. The graph's area there is strictly greater than the lower endpoint rectangle and strictly less than the upper one. If the function is nonconstant on \([a,b]\), at least one partition subinterval has different endpoint values: otherwise it would have equal values at the ends of every subinterval, and monotonicity would force it to be constant on each one. Thus both total inequalities are strict. If the function is constant, every rectangle has exactly the graph's height, so both sums equal the area.
Worked Examples
Worked Example: An Increasing Linear Rate
Let \(f(t)=2+3t\) on \([0,2]\), with partition points \(0,\frac12,1,\frac32,2\). The rate is nonnegative and increasing because its values rise as \(t\) increases. Each subinterval has width \(\frac12\). The sampled heights are \(2,\frac72,5,\frac{13}{2},8\).
The left sum uses the first four heights:
The right sum uses the last four heights:
The graph is a line segment from \((0,2)\) to \((2,8)\), so its area is the trapezoid area: average height times width. Hence \(A=\frac{2+8}{2}\cdot2=10\). Checking the result gives \(\frac{17}{2}<10<\frac{23}{2}\), as required for an increasing rate.
Worked Example: A Decreasing Rate on Unequal Subintervals
Let \(f(t)=10-2t\) on \([0,3]\), with partition points \(0,1,3\). The rate is nonnegative on this interval, falling from \(10\) to \(4\). The widths are \(1\) and \(2\), and the endpoint rates are \(f(0)=10\), \(f(1)=8\), and \(f(3)=4\).
The left sum is:
The right sum is:
The graph is a line segment from height \(10\) to height \(4\) across width \(3\). Its trapezoid area is \(A=\frac{10+4}{2}\cdot3=21\). Checking gives \(16<21<26\), so \(R<A<L\), as the theorem predicts for a decreasing rate.
Worked Example: An Increasing Piecewise-Linear Rate
Define \(f\) by joining, with straight line segments, the points \((0,1)\), \((1,3)\), \((3,3)\), and \((4,7)\). This gives a continuous nonnegative, nondecreasing rate. Use the partition \(0,1,3,4\), whose widths are \(1,2,1\), respectively. The endpoint heights at the partition points are \(1,3,3,7\).
The left sum is:
The right sum is:
Compute the true area one segment at a time. On \([0,1]\), the trapezoid area is \(\frac{1+3}{2}(1)=2\). On \([1,3]\), the rate is constant at \(3\), giving area \(3(2)=6\). On \([3,4]\), the trapezoid area is \(\frac{3+7}{2}(1)=5\). Therefore \(A=2+6+5=13\). Checking, \(10<13<16\), which confirms both bounds.
A Common Reversal
A frequent mistake is to assume that a left sum always underestimates, or that a right sum always overestimates. Neither claim is true without considering the direction of change. The endpoint comparison, not the name of the sum, determines the bound.
Worked Example: Correcting the Direction
For the decreasing rate \(f(t)=10-2t\) on \([0,3]\) with partition \(0,1,3\), one might incorrectly say, “The left sum is a lower bound because it uses left endpoints,” and conclude \(L<A\). But the actual left sum is \(26\) and the area is \(21\), so that proposed inequality is false: \(26<21\) does not hold.
Correct the reasoning by checking the rate direction. On each subinterval of a decreasing graph, the left endpoint height is at least every later height. Thus the left rectangles lie above the graph, and the right rectangles lie below it. The verified comparison is \(R=16<A=21<L=26\).
Why This Result Matters
These bounds turn a qualitative description of a rate into quantitative information about the total accumulation. If a rate is known to increase, the left sum supplies a lower bound and the right sum an upper bound; if it decreases, their roles reverse. The conclusion holds even when subinterval widths differ, because each comparison is made locally and then added.
The bounds also depend on the assumptions. Nonnegativity makes the area under the graph equal to the accumulated amount without cancellation from below the horizontal axis. Monotonicity orders the endpoint heights on each subinterval. Continuity ensures the graph's area behaves as described and gives the strictness conclusion. If a rate changes direction, this theorem alone does not give a single endpoint ordering over the whole interval.
Check Your Understanding
Use the endpoint-bound theorem and the definitions of the sums to answer these questions.
- A continuous nonnegative rate is nondecreasing on an interval. Which endpoint sum is a lower bound for its area, and which is an upper bound?
- For a nonincreasing rate, write the correct ordering of \(L\), \(A\), and \(R\).
- Why does the proof require each partition width to be positive when it multiplies endpoint-height inequalities?
- Under the theorem's assumptions, what happens to both endpoint sums if the rate is constant?
- Why can a left sum not be declared an underestimate without first checking whether the rate increases or decreases?
