Bounds for Integrals

Turning Pointwise Information into Integral Bounds

The previous tutorial, Integral Triangle Inequality, showed how pointwise inequalities can control integrals. Here we focus on a practical question: if the exact values of an integrable function are difficult to compute, what can be concluded from information about its size? The answer often comes from comparing the function with simpler upper and lower bounds, or with an approximation whose error is controlled.

We will use the established Theorem (Bounds for the Integral): if $f$ is Riemann integrable on $[a,b]$, where $a<b$, and $m\leq f(x)\leq M$ throughout the interval, then $m(b-a)\leq\int_a^b f(x)\,dx\leq M(b-a)$. We will also use order preservation, linearity, and additivity across adjacent intervals. These results let us turn pointwise comparisons into integral estimates without needing to evaluate the integral exactly.

The constant-bound theorem is one starting point. A more flexible approach is to find integrable functions $p$ and $q$ such that $p(x)\leq f(x)\leq q(x)$. Order preservation then gives $\int_a^b p(x)\,dx\leq\int_a^b f(x)\,dx\leq\int_a^b q(x)\,dx$. The bounds need not be constant; they can vary across the interval when that produces a tighter estimate.

Worked Example: Bounding an Integral with Constant Bounds

Suppose $f$ is Riemann integrable on $[1,4]$, and $2\leq f(x)\leq5$ at every point of the interval. The Bounds for the Integral theorem applies with $m=2$, $M=5$, and $b-a=4-1=3$. Therefore

2(3)\leq\int_1^4 f(x)\,dx\leq5(3), \qquad\text{so}\qquad 6\leq\int_1^4 f(x)\,dx\leq15.

This conclusion does not require knowing a formula for $f$. It uses only integrability and the pointwise bounds. If the range of $f$ varies substantially across the interval, however, one pair of global constants may give an unnecessarily wide estimate. Local bounds can improve it, as a later example shows.

Bounding the Error of an Approximation

A useful way to estimate an integral is to approximate the integrand by a simpler integrable function $g$. If $f$ stays within a known distance $\varepsilon$ of $g$, then the difference between their integrals is controlled by the interval length. This statement follows from the pointwise bounds $-\varepsilon\leq f-g\leq\varepsilon$, but its value is practical: it converts a uniform error estimate for functions into an error estimate for integrals.

Theorem (Uniform Approximation Bound for Integrals): Let $f,g:[a,b]\to\mathbb{R}$ be Riemann integrable, where $a<b$. If $|f(x)-g(x)|\leq\varepsilon$ for every $x\in[a,b]$, with $\varepsilon\geq0$, then $$ \left|\int_a^b f(x)\,dx-\int_a^b g(x)\,dx\right| \leq\varepsilon(b-a). $$

Proof. Put $h=f-g$. By linearity, $h$ is Riemann integrable. The hypothesis implies $-\varepsilon\leq h(x)\leq\varepsilon$ for all $x\in[a,b]$. Apply the Bounds for the Integral theorem to $h$, first with lower bound $-\varepsilon$ and then with upper bound $\varepsilon$:

-\varepsilon(b-a)\leq\int_a^b h(x)\,dx\leq\varepsilon(b-a).

Since $\int_a^b h(x)\,dx=\int_a^b f(x)\,dx-\int_a^b g(x)\,dx$, the two-sided inequality is exactly the claimed absolute-value bound. This proves the theorem. $\square$

The factor $b-a$ matters: the same pointwise error accumulates over a longer interval. Also, the theorem compares signed integrals. It does not say that the integrals of the absolute values differ by the same signed amount; that is a different quantity.

Worked Example: Estimating an Integral from a Step Approximation

On $[0,2]$, define a step function $g$ by $g(x)=1$ for $0\leq x<1$ and $g(x)=3$ for $1\leq x\leq2$. Suppose $f$ is Riemann integrable and satisfies $|f(x)-g(x)|\leq0.2$ throughout $[0,2]$. The integral of $g$ is obtained from its constant values on the two intervals:

\int_0^2 g(x)\,dx =1(1-0)+3(2-1) =1+3 =4.

The uniform approximation bound gives

\left|\int_0^2 f(x)\,dx-4\right| \leq0.2(2-0) =0.4.

Equivalently, $-0.4\leq\int_0^2 f(x)\,dx-4\leq0.4$, and hence

3.6\leq\int_0^2 f(x)\,dx\leq4.4.

The estimate does not require evaluating $f$. It uses a simple step function with a known integral and a uniform bound on the discrepancy.

Using Different Error Bounds on Different Intervals

A single uniform error tolerance can be wasteful. An approximation may be very accurate on one part of an interval and less accurate on another. A partition allows each part to have its own error bound, and the resulting estimates add according to the lengths of the subintervals.

Theorem (Piecewise Uniform Approximation Bound): Let $f,g:[a,b]\to\mathbb{R}$ be Riemann integrable, and let $a=x_0<x_1<\cdots<x_n=b$ be a partition. Suppose that for each $i\in\{1,\ldots,n\}$, there is an $\varepsilon_i\geq0$ such that $$ |f(x)-g(x)|\leq\varepsilon_i \quad\text{for every }x\in[x_{i-1},x_i]. $$ Then $$ \left|\int_a^b f(x)\,dx-\int_a^b g(x)\,dx\right| \leq\sum_{i=1}^{n}\varepsilon_i(x_i-x_{i-1}). $$

Proof. Let $h=f-g$, which is Riemann integrable by linearity. Restriction to subintervals and the stated pointwise bounds allow the Bounds for the Integral theorem to be applied on each $[x_{i-1},x_i]$. For each $i$, this gives

-\varepsilon_i(x_i-x_{i-1}) \leq\int_{x_{i-1}}^{x_i}h(x)\,dx \leq\varepsilon_i(x_i-x_{i-1}).

In particular, the absolute value of each subinterval integral is at most $\varepsilon_i(x_i-x_{i-1})$. By additivity across the partition, linearity, and the triangle inequality for real numbers,

\begin{aligned} \left|\int_a^b f(x)\,dx-\int_a^b g(x)\,dx\right| &=\left|\int_a^b h(x)\,dx\right|\\ &=\left|\sum_{i=1}^n\int_{x_{i-1}}^{x_i}h(x)\,dx\right|\\ &\leq\sum_{i=1}^n\left|\int_{x_{i-1}}^{x_i}h(x)\,dx\right|\\ &\leq\sum_{i=1}^n\varepsilon_i(x_i-x_{i-1}). \end{aligned}

This proves the theorem. $\square$

Worked Example: Improving a Bound by Splitting the Interval

Suppose $f$ and $g$ are Riemann integrable on $[0,3]$, and $g$ is a convenient approximation to $f$ whose integral is known. Assume the approximation error satisfies

|f(x)-g(x)|\leq \begin{cases} 0.1,&0\leq x\leq1,\\ 0.5,&1\leq x\leq2,\\ 0.2,&2\leq x\leq3. \end{cases}

The partition points are $0,1,2,3$, so all three subintervals have length $1$. The piecewise bound gives

\left|\int_0^3 f(x)\,dx-\int_0^3 g(x)\,dx\right| \leq0.1(1)+0.5(1)+0.2(1) =0.8.

If we instead used the largest error, $0.5$, as a single uniform bound on all of $[0,3]$, the uniform approximation theorem would give an error bound of $0.5(3)=1.5$. The local information improves the estimate from $1.5$ to $0.8$. If, for example, $\int_0^3 g(x)\,dx=6$, then the piecewise estimate yields $5.2\leq\int_0^3 f(x)\,dx\leq6.8$.

Choosing Bounds That Are Useful

A valid bound need not be a useful bound. If $f$ lies between two constants that are far apart, the resulting interval for its integral may be too wide to answer the question at hand. Whenever possible, choose bounds that reflect how $f$ behaves across the interval: use a variable lower or upper function, divide the interval into cells with separate bounds, or approximate $f$ by a simpler function with a controlled error.

For example, if $p$ and $q$ are integrable functions and $p(x)\leq f(x)\leq q(x)$ everywhere, order preservation gives an integral bracket using $\int p$ and $\int q$. This may be sharper than using the smallest constant below $f$ and the largest constant above it. In a partition-based estimate, local upper and lower bounds play the same role: each piece contributes its own bound multiplied by its length.

A common pitfall is to confuse a bound on the signed integral with a bound on the integral of the absolute value. If $m\leq f\leq M$, then the integral of $f$ is between $m(b-a)$ and $M(b-a)$. To bound $\int_a^b|f(x)|\,dx$, first bound $|f|$, or use an appropriate established absolute-value inequality. When $f$ changes sign, the signed integral can be small because of cancellation even when the integral of its absolute value is large.

Identify the quantity to estimate.
Decide whether the goal is a signed integral, an error from an approximation, or an integral of an absolute value.

Find a pointwise comparison.
Use constant bounds, integrable upper and lower functions, or a simpler approximation with a known error.

Use interval length or local lengths.
A uniform error contributes its size times the full interval length; separate errors contribute their sizes times the corresponding subinterval lengths.

Check what the estimate actually controls.
Keep the signed integral distinct from the integral of the absolute value, and verify that each function used in the comparison is integrable.

Check Your Understanding

Use the bounds and approximation estimates developed here to answer the following questions.

If $m\leq f(x)\leq M$ on $[a,b]$, what bounds does the Bounds for the Integral theorem give for $\int_a^b f(x)\,dx$?
If $|f-g|\leq\varepsilon$ throughout $[a,b]$, what controls the maximum error between the two integrals?
Why can separate error bounds on subintervals produce a tighter estimate than one global error bound?
What pointwise information is needed to compare $\int_a^b f$ with $\int_a^b g$ using upper and lower functions?
Why does a small signed integral not necessarily imply that $\int_a^b|f(x)|\,dx$ is small?

What You'll Learn