Equal-Width Bounds For Monotone Functions

Why Equal Widths Simplify the Estimate

Upper and lower sums enclose the integral of a bounded function, but their usefulness as estimates depends on how far apart they are. For a monotone function, equal-width subintervals make this gap especially simple: the changes in function value across successive subintervals add up to the total change from one endpoint to the other.

This gives an error bound that can be calculated using only the interval length, the number of subintervals, and the function’s endpoint values. It can also tell us how many subintervals are sufficient to guarantee a chosen accuracy.

Equal-Width Sums for a Monotone Function

Let $f$ be bounded and monotone on $[a,b]$, where $a<b$. Here, monotone means either nondecreasing throughout the interval or nonincreasing throughout it. Divide $[a,b]$ into $n$ subintervals of equal width, with $n$ a positive integer. Write the partition points and width as

x_i=a+ih,\qquad h=\frac{b-a}{n},\qquad i=0,1,\ldots,n.

The lower sum $L_n$ uses the minimum value of $f$ on each subinterval, and the upper sum $U_n$ uses its maximum. Because $f$ is monotone, those extrema occur at the subinterval endpoints. For a nondecreasing function, the left endpoint supplies the minimum and the right endpoint the maximum. For a nonincreasing function, these roles are reversed.

Definition: For a monotone function on an equal-width partition, $L_n$ is the sum of the subinterval widths multiplied by their minimum function values, and $U_n$ is the sum of the widths multiplied by their maximum function values. The integral is enclosed by these sums: $L_n\leq \int_a^b f(x)\,dx\leq U_n$.

Monotone functions are integrable on a closed bounded interval, so this enclosure applies. The key result below quantifies its width.

The Equal-Width Gap

Theorem: If $f$ is monotone on $[a,b]$, and $L_n$ and $U_n$ are its lower and upper sums for $n$ equal-width subintervals, then $$ U_n-L_n=\frac{b-a}{n}\,|f(b)-f(a)|. $$ Consequently, using either sum as an estimate of the integral gives an absolute error no greater than $\frac{b-a}{n}|f(b)-f(a)|$.

Why the Gap Telescopes

First suppose $f$ is nondecreasing. On the subinterval from $x_{i-1}$ to $x_i$, the minimum is $f(x_{i-1})$ and the maximum is $f(x_i)$. Thus the sums are

L_n=h\sum_{i=1}^{n}f(x_{i-1}), \qquad U_n=h\sum_{i=1}^{n}f(x_i).

Subtract the lower sum from the upper sum. Every interior function value appears once with a positive sign and once with a negative sign, so those terms cancel:

\begin{aligned} U_n-L_n &=h\sum_{i=1}^{n}\bigl(f(x_i)-f(x_{i-1})\bigr)\\ &=h\bigl(f(x_n)-f(x_0)\bigr)\\ &=h\bigl(f(b)-f(a)\bigr). \end{aligned}

For a nonincreasing function, the upper sum uses left endpoints and the lower sum uses right endpoints. The same subtraction gives $U_n-L_n=h(f(a)-f(b))$. In either case, the result is $h|f(b)-f(a)|$. Substituting $h=(b-a)/n$ proves the formula.

Finally, the integral lies between $L_n$ and $U_n$. Therefore, its distance from either endpoint of that interval cannot exceed the interval’s width $U_n-L_n$. This proves the error guarantee for both endpoint sums.

Using the Bound to Set the Subinterval Count

Suppose the goal is to estimate the integral using either the upper sum or the lower sum, with absolute error at most a specified tolerance $\varepsilon>0$. The theorem guarantees this whenever the gap is at most $\varepsilon$. Since $n$ must be a positive integer, a sufficient choice is

n\geq \frac{(b-a)|f(b)-f(a)|}{\varepsilon}.

If the right-hand side is not an integer, choose the next integer above it. This is a sufficient condition, not a prediction of the actual error: a particular sum may be much closer to the integral than the bound requires. If $f(a)=f(b)$, monotonicity forces $f$ to be constant on the interval, so both sums equal the integral for every positive $n$.

Worked Example: Enclosing the Integral of an Increasing Function

Estimate $\int_0^2 x^2\,dx$ using four equal-width subintervals. Here, $f(x)=x^2$ is nondecreasing on $[0,2]$, and

h=\frac{2-0}{4}=\frac12, \qquad x_0=0,\ x_1=\frac12,\ x_2=1,\ x_3=\frac32,\ x_4=2.

The endpoint values are $f(x_0)=0$, $f(x_1)=\frac14$, $f(x_2)=1$, $f(x_3)=\frac94$, and $f(x_4)=4$. Since the function is increasing, the lower sum uses the first four values and the upper sum the last four:

\begin{aligned} L_4 &=\frac12\left(0+\frac14+1+\frac94\right) =\frac12\cdot\frac72 =\frac74,\\ U_4 &=\frac12\left(\frac14+1+\frac94+4\right) =\frac12\cdot\frac{15}{2} =\frac{15}{4}. \end{aligned}

The theorem predicts a gap of

\frac{2-0}{4}|f(2)-f(0)| =\frac12|4-0| =2.

Checking the computed sums, $\frac{15}{4}-\frac74=2$, as required. The integral is $\frac83$, and indeed $\frac74\leq\frac83\leq\frac{15}{4}$. Thus the endpoint sums enclose the integral, with the exact gap given by the theorem.

Worked Example: A Decreasing Function

Find equal-width upper and lower sums for $f(x)=5-2x$ on $[0,2]$ using four subintervals. The function is nonincreasing, and $h=(2-0)/4=\frac12$. The partition points are $0,\frac12,1,\frac32,2$, where the function values are $5,4,3,2,1$, respectively.

A decreasing function has its maximum at the left endpoint and minimum at the right endpoint of each subinterval. Therefore,

\begin{aligned} U_4&=\frac12(5+4+3+2)=\frac12\cdot14=7,\\ L_4&=\frac12(4+3+2+1)=\frac12\cdot10=5. \end{aligned}

The integral must lie between $5$ and $7$. The gap formula gives

\frac{2-0}{4}|f(2)-f(0)| =\frac12|1-5| =2,

and the calculated gap is $7-5=2$. As a check, direct integration gives $\int_0^2(5-2x)\,dx=[5x-x^2]_0^2=10-4=6$, which is in the interval $[5,7]$.

Worked Example: Choosing Enough Subintervals

How many equal-width subintervals are sufficient to estimate $\int_0^2(3x+2)\,dx$ using an upper or lower sum with error at most $0.1$? The function $f(x)=3x+2$ is increasing, and its endpoint values are $f(0)=2$ and $f(2)=8$. The gap bound is

U_n-L_n=\frac{2-0}{n}|8-2|=\frac{12}{n}.

To guarantee that either endpoint sum has error at most $0.1$, require the gap bound to be no greater than $0.1$:

\frac{12}{n}\leq 0.1 \quad\Longrightarrow\quad n\geq\frac{12}{0.1}=120.

Thus $n=120$ is sufficient. To check the guarantee directly, for this linear function the lower sum is

L_n=\frac{2}{n}\sum_{i=0}^{n-1}\left(2+\frac{6i}{n}\right) =4+\frac{12}{n^2}\cdot\frac{n(n-1)}{2} =10-\frac{6}{n}.

Since $U_n-L_n=12/n$, the upper sum is $U_n=10+6/n$. At $n=120$, these give $L_{120}=9.95$ and $U_{120}=10.05$. The exact integral is $[\,\frac32x^2+2x\,]_0^2=6+4=10$, which lies between the sums; each endpoint estimate is $0.05$ from the integral. The bound guaranteed error at most $0.1$, so this more accurate outcome is consistent with the guarantee.

A Common Pitfall: Confusing the Gap with the Actual Error

The gap $U_n-L_n$ is a bound on the error of either endpoint sum, not usually the exact error of either one. For the previous example with $n=120$, the gap is $0.1$, but the actual errors are both $0.05$. It would be incorrect to claim that an endpoint estimate must have error exactly $0.1$; the theorem says only that the error cannot exceed that amount.

Another important detail is the factor $1/n$. Increasing the number of equal-width subintervals reduces the stated gap bound in inverse proportion to $n$. The endpoint change $|f(b)-f(a)|$ and the interval length $b-a$ remain fixed; it is the width $h=(b-a)/n$ that shrinks.

Takeaway: For a monotone function, equal-width upper and lower sums differ by the subinterval width multiplied by the total endpoint change. This gives a direct, guaranteed accuracy test for either endpoint sum.

Check Your Understanding

Use the equal-width gap theorem and the distinction between a guaranteed bound and an actual error in answering these questions.

For a nondecreasing function, which endpoint of each subinterval supplies the minimum, and which supplies the maximum?
A monotone function on an interval of length $5$ has endpoint values differing by $3$. What is the upper-minus-lower gap for $n$ equal-width subintervals?
For the function in Question 2, what condition on $n$ guarantees that either endpoint sum has error at most $0.05$?
Does a gap of $0.2$ mean that the lower sum’s actual error must be $0.2$? Explain.
If a monotone function has equal endpoint values, what can be concluded about the function and its equal-width sums?

Equal-Width Bounds For Monotone Functions

What You'll Learn

Why Equal Widths Simplify the Estimate

Equal-Width Sums for a Monotone Function

The Equal-Width Gap

Why the Gap Telescopes

Using the Bound to Set the Subinterval Count

Worked Example: Enclosing the Integral of an Increasing Function

Worked Example: A Decreasing Function

Worked Example: Choosing Enough Subintervals

A Common Pitfall: Confusing the Gap with the Actual Error

Check Your Understanding

Test yourself: Calculus I final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

Equal-Width Bounds For Monotone Functions

What You'll Learn

Why Equal Widths Simplify the Estimate

Equal-Width Sums for a Monotone Function

The Equal-Width Gap

Why the Gap Telescopes

Using the Bound to Set the Subinterval Count

Worked Example: Enclosing the Integral of an Increasing Function

Worked Example: A Decreasing Function

Worked Example: Choosing Enough Subintervals

A Common Pitfall: Confusing the Gap with the Actual Error

Check Your Understanding

Test yourself: Calculus I final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

Previous

Next