Why Bound an Integral?
A definite integral can be difficult to evaluate exactly, or an exact expression may be available but a numerical computation still needs checking. One useful check does not require finding an antiderivative: bound the values of the integrand, then use those bounds to constrain the integral.
The basic idea is simple. If a function never falls below \(m\) or rises above \(M\) on an interval, its accumulated integral over that interval cannot be less than the constant function \(m\) integrated there, or greater than the constant function \(M\) integrated there. The resulting interval of possible values can test a proposed exact answer or a numerical approximation.
The Integral Bound Theorem
Continuity of \(f\) on \([a,b]\) is a common sufficient condition for Riemann integrability. The inequalities must hold throughout the interval, not merely at a few sample points. The theorem also applies when the lower bound or upper bound is negative; it is the pointwise ordering that matters.
The factor \(b-a\) is the interval's length. For instance, if \(2\le f(x)\le 5\) throughout an interval of length \(3\), the integral lies between \(6\) and \(15\). An interval twice as long gives twice these constant-function bounds.
Why the Bounds Pass to the Integral
Here is a justification using the order property of the definite integral. If \(g\) and \(h\) are integrable on \([a,b]\) and \(g(x)\le h(x)\) throughout that interval, then \(\int_a^b g(x)\,dx\le\int_a^b h(x)\,dx\). To see why, \(h-g\) is nonnegative everywhere. Its integral is therefore nonnegative: every Riemann sum for \(h-g\) uses nonnegative function values and positive subinterval widths, so those sums are nonnegative, as is their limit. Linearity of the integral then gives \(\int_a^b h-\int_a^b g=\int_a^b(h-g)\ge0\).
Apply this order property twice. The constant functions with values \(m\) and \(M\) satisfy \(m\le f\le M\). Since the integral of a constant \(c\) over \([a,b]\) is \(c(b-a)\), it follows that
This proves the stated bounds. The reasoning also explains the requirement that the integrand be integrable and the pointwise inequalities hold over the full interval.
Worked Example: A Coarse Bound for a Reciprocal Integrand
Worked Example: A Coarse Bound for a Reciprocal Integrand
Bound \(I=\int_0^1\frac{1}{1+x^2}\,dx\), then check the proposed exact value \(\frac{\pi}{4}\).
For \(0\le x\le1\), squaring gives \(0\le x^2\le1\), so \(1\le1+x^2\le2\). All these quantities are positive. Taking reciprocals reverses the order, giving \(\frac12\le\frac{1}{1+x^2}\le1\).
The interval has length \(1-0=1\). The theorem therefore yields $$ \frac12(1-0)\le I\le1(1-0), \qquad\text{so}\qquad \frac12\le I\le1. $$ The proposed value \(\frac{\pi}{4}\) is about \(0.7854\), which lies between \(0.5\) and \(1\). The bound does not reject it. Since passing this check alone does not establish equality, this is a consistency check, not a proof that \(I=\frac{\pi}{4}\).
Check the numerical comparisons: \(\frac12=0.5<0.7854<1\). The proposed value is inside the bound, as required for it to remain plausible.
Sharpening a Bound with a Partition
A single pair of constant bounds can be too wide to be useful. Divide \([a,b]\) into smaller intervals and find separate bounds on each piece. If \(x_0=a<x_1<\cdots<x_n=b\), and \(m_i\le f(x)\le M_i\) on \([x_{i-1},x_i]\), then applying the integral bound theorem on each piece and adding gives
The reason addition is valid is additivity of the definite integral across adjacent intervals. A finer partition often improves the estimate because the bounds can adapt to local changes in the integrand. It need not improve every estimate automatically: the chosen bounds on each piece still need to be valid and sufficiently close.
For a monotone function, endpoint values provide convenient bounds. If \(f\) is increasing on a subinterval, its left endpoint value is a lower bound and its right endpoint value is an upper bound. If \(f\) is decreasing, the right endpoint value is a lower bound and the left endpoint value is an upper bound. These facts follow directly from the definition of increasing or decreasing: throughout the subinterval, the function values lie between the endpoint values.
Worked Example: Rejecting an Incorrect Value with Ten Subintervals
Worked Example: Rejecting an Incorrect Value with Ten Subintervals
Use a partition to test the proposed value \(\frac14\) for \(J=\int_0^1x^2\,dx\).
The function \(f(x)=x^2\) is increasing on \([0,1]\), since \(x^2\) increases as \(x\) increases there. Divide the interval into ten equal pieces, with endpoints \(x_i=\frac{i}{10}\) for \(i=0,1,\ldots,10\). Each width is \(x_i-x_{i-1}=\frac{1}{10}\).
On each piece, the left endpoint gives a lower bound and the right endpoint gives an upper bound. Thus $$ \frac{1}{10}\sum_{i=0}^{9}\left(\frac{i}{10}\right)^2 \le J\le \frac{1}{10}\sum_{i=1}^{10}\left(\frac{i}{10}\right)^2. $$ Using \(\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}\), the lower sum is $$ \frac{1}{1000}\sum_{i=0}^{9}i^2 = \frac{1}{1000}\left(\frac{9\cdot10\cdot19}{6}\right) = \frac{285}{1000} = \frac{57}{200}. $$ The upper sum is $$ \frac{1}{1000}\sum_{i=1}^{10}i^2 = \frac{1}{1000}\left(\frac{10\cdot11\cdot21}{6}\right) = \frac{385}{1000} = \frac{77}{200}. $$ Therefore \(\frac{57}{200}\le J\le\frac{77}{200}\), or \(0.285\le J\le0.385\). But \(\frac14=0.25<0.285\), so the proposed value is outside the rigorous bounds and is incorrect.
As a check on the interval itself, the Fundamental Theorem of Calculus gives \(J=[x^3/3]_0^1=\frac13\). Substitution at the endpoints gives \(1/3-0=1/3\), and \(0.285<1/3<0.385\), as required.
Algebraic Bounds Can Be More Effective
Partitioning is not the only way to find bounds. Sometimes an inequality involving simpler functions gives a tighter estimate directly. The important steps are to verify the inequality over the entire interval and then integrate both sides using the order property.
Worked Example: Testing a Numerical Approximation for \(\ln 2\)
Worked Example: Testing a Numerical Approximation for \(\ln 2\)
Show that \(\frac12\le\int_0^1\frac{1}{1+x}\,dx\le\frac34\), and use the bounds to assess the proposed approximation \(0.8\) for \(\ln 2\).
For \(0\le x\le1\), \(1+x>0\). To verify the lower bound, compare \(1-x\) with \(\frac{1}{1+x}\). Multiplying by the positive number \(1+x\), the required inequality is equivalent to $$ (1-x)(1+x)=1-x^2\le1. $$ This holds because \(x^2\ge0\). For the upper bound, compare \(\frac{1}{1+x}\) with \(1-\frac{x}{2}\). Again multiplying by \(1+x>0\), the required inequality is equivalent to $$ 1\le(1+x)\left(1-\frac{x}{2}\right) =1+\frac{x}{2}-\frac{x^2}{2}. $$ This holds because \(\frac{x}{2}-\frac{x^2}{2}=\frac{x(1-x)}{2}\ge0\) on \([0,1]\). Consequently, $$ 1-x\le\frac{1}{1+x}\le1-\frac{x}{2}. $$
Integrate all three expressions over \([0,1]\). The lower integral is \(\int_0^1(1-x)\,dx=[x-x^2/2]_0^1=1-\frac12=\frac12\). The upper integral is \(\int_0^1(1-x/2)\,dx=[x-x^2/4]_0^1=1-\frac14=\frac34\). Therefore $$ \frac12\le\int_0^1\frac{1}{1+x}\,dx\le\frac34. $$ The integral equals \(\ln 2\), since \([\ln(1+x)]_0^1=\ln2-\ln1=\ln2\). Thus \(0.5\le\ln2\le0.75\). The proposed approximation \(0.8\) exceeds the upper bound \(0.75\), so it cannot be correct.
For a final check, \(\ln2\) is approximately \(0.6931\), which lies within the derived interval. The integral bounds reject \(0.8\); the decimal value independently confirms the same conclusion.
How to Use a Bound as a Check
Suppose a calculation claims that an integral has exact value \(C\), while a rigorous estimate gives \(L\le\int_a^b f(x)\,dx\le U\). If \(C<L\) or \(C>U\), the claimed value is impossible. If \(L\le C\le U\), the estimate is consistent with the claim but does not prove it: many values may lie between \(L\) and \(U\).
The same logic applies to a numerical approximation. If a computation reports \(q\) with an error tolerance \(\varepsilon\), its claimed range is \([q-\varepsilon,q+\varepsilon]\). If this range does not overlap the rigorous integral interval \([L,U]\), the reported approximation and tolerance cannot both be correct. If the ranges overlap, the check does not by itself verify the approximation. A wide integral bound can be valid yet uninformative, so refine the partition or seek sharper pointwise inequalities when needed.
A reliable procedure is to identify the interval, establish pointwise lower and upper bounds there, integrate those bounds, and compare the resulting interval with the proposed value. Keep the direction of each inequality unchanged when passing to integrals, and distinguish a contradiction from mere consistency.
Check Your Understanding
Use the bounds and conditions in this tutorial to answer the following questions.
- If \(3\le f(x)\le 7\) on \([2,5]\) and \(f\) is integrable, what bounds follow for \(\int_2^5 f(x)\,dx\)?
- Why must a pointwise inequality used in an integral bound hold throughout the interval?
- For an increasing function on each piece of a partition, which endpoint values give the lower and upper bounds on that piece?
- A rigorous estimate gives \(1.2\le I\le1.8\). What can you conclude about proposed values \(1.1\) and \(1.5\)?
- A numerical result is reported as \(q=2.0\) with tolerance \(0.1\), while a rigorous bound is \(2.2\le I\le2.4\). Do the claimed approximation and tolerance pass the overlap check?
