Why Integral Estimates Need Care
Definite-integral estimates often look simple: bound the function, multiply by the interval length, and compare the result with the integral. But a small mistake about the bounds, the signs of function values, or the assumptions of a comparison can reverse or invalidate the conclusion.
Four questions help diagnose most errors:
- Are the limits written in increasing order, or has the integral's orientation been reversed?
- Does the function take negative values on part of the interval?
- Does the claimed pointwise comparison hold throughout the interval?
- Is the quantity being used an upper sum, or is it actually the integral?
The distinctions matter because estimates depend on hypotheses. A true numerical calculation on one example does not establish a general bound, and a graph that appears to sit below another graph is not a substitute for a valid comparison on the entire interval.
Integral Bounds and Their Conditions
Suppose \(f\) is Riemann integrable on \([a,b]\), where \(a<b\). If constants \(m\) and \(M\) satisfy \(m\leq f(x)\leq M\) for every \(x\in[a,b]\), then integrating the bounds gives:
The interval length \(b-a\) is positive, so multiplying by it preserves the inequalities. The requirement that the pointwise bounds hold across the whole interval is essential. A value at one point, or a comparison that fails elsewhere, does not justify this estimate.
To express Darboux bounds precisely, take a partition \(P\) of \([a,b]\), written \(a=x_0<x_1<\cdots<x_n=b\). On each subinterval \([x_{i-1},x_i]\), let \(m_i\) be the infimum of \(f\) and \(M_i\) its supremum. For bounded \(f\), these values exist even if the function does not attain them. The lower and upper Darboux sums are:
If \(f\) is Riemann integrable, every partition gives the bounds \(L(P,f)\leq\int_a^b f(x)\,dx\leq U(P,f)\). These are estimates, not a claim that either sum equals the integral.
Worked Example: Reversing the Bounds
Worked Example: Reversing the Bounds
Let \(f(x)=2x-1\) on \([0,2]\). Since \(-1\leq 2x-1\leq 3\) throughout this interval, estimate the integral in each orientation and check the exact values.
First integrate in the increasing direction. The interval length is \(2-0=2\), so the pointwise bounds give:
An antiderivative is \(x^2-x\). Therefore:
The value \(2\) lies between \(-2\) and \(6\), as required. Reversing the bounds changes the sign:
Negating the increasing-direction bounds also reverses their order:
The exact reversed value \(-2\) satisfies this estimate. A common error is to keep the same order of bounds after negating. The sign change applies to the whole inequality, so the order must change too.
Negative Values and Signed Cancellation
A definite integral combines positive and negative function values with their signs. A region below the horizontal axis contributes negatively to the integral; it is not automatically counted as positive area. Consequently, an estimate based only on the visible positive region can miss cancellation from a negative region.
Bounds such as \(m(b-a)\leq\int_a^b f\leq M(b-a)\) still apply when \(f\) changes sign, provided the bounds \(m\leq f(x)\leq M\) hold everywhere. But a positive geometric area is not, by itself, a lower bound for the signed integral.
Worked Example: A Positive Region Does Not Give a Lower Bound
Consider \(f(x)=x-1\) on \([0,2]\). The function is negative on \([0,1)\) and positive on \((1,2]\). Each triangular region has area \(1/2\), but their contributions to the definite integral have opposite signs.
Compute the integral directly:
The positive-region area is \(1/2\), so the claim \(\int_0^2(x-1)\,dx\geq 1/2\) is false: the integral is \(0\). Splitting at \(x=1\) shows exactly why:
As a check, the negative contribution is \(-1/2\), the positive contribution is \(1/2\), and their sum agrees with the direct calculation.
Pointwise Comparison Must Hold Throughout
If \(f\) and \(g\) are Riemann integrable on \([a,b]\) and \(f(x)\leq g(x)\) for every \(x\in[a,b]\), then their integrals satisfy \(\int_a^b f(x)\,dx\leq\int_a^b g(x)\,dx\). The comparison is pointwise: it must hold across the interval, not merely at one endpoint or at a few sample points.
To justify the result, set \(h=g-f\). The difference of two Riemann-integrable functions is Riemann integrable, and the pointwise hypothesis gives \(h(x)\geq0\) on the interval. For any partition \(P\), the infimum of \(h\) on each subinterval is therefore nonnegative. Since each subinterval has nonnegative width, the lower Darboux sum is nonnegative. The integral lies above that sum; linearity then gives the comparison.
Let \(h=g-f\). Then \(h\) is Riemann integrable and \(h(x)\geq0\) throughout \([a,b]\).
For each subinterval, the infimum \(m_i\) of \(h\) satisfies \(m_i\geq0\). Thus \(L(P,h)=\sum_{i=1}^n m_i(x_i-x_{i-1})\geq0\).
Because \(h\) is integrable, \(L(P,h)\leq\int_a^b h(x)\,dx\). Hence \(\int_a^b h(x)\,dx\geq0\).
Linearity gives \(\int_a^b h=\int_a^b g-\int_a^b f\geq0\), so \(\int_a^b f\leq\int_a^b g\).
Worked Example: A Comparison That Holds on the Whole Interval
Compare the integrals of \(x^2\) and \(x\) on \([0,1]\). Since \(0\leq x\leq1\), multiplying \(x\leq1\) by the nonnegative number \(x\) gives \(x^2\leq x\) at every point of the interval.
Both functions are continuous, hence Riemann integrable. The comparison theorem therefore applies. Calculate each integral:
Indeed, \(\frac13\leq\frac12\). The inequality follows both from the verified pointwise comparison and from the computed values.
For contrast, checking a comparison at just one point is not enough. On \([0,1]\), the functions \(f(x)=x\) and \(g(x)=0\) agree at \(x=0\), but \(x\leq0\) fails for every \(x>0\). In fact, \(\int_0^1x\,dx=1/2\), while \(\int_0^1 0\,dx=0\). Thus agreement at the endpoint cannot justify the claim that the first integral is no larger than the second.
An Upper Sum Is a Bound, Not Necessarily the Integral
The upper Darboux sum uses a supremum on each partition subinterval. It is constructed to bound the integral from above when the function is integrable. It is not defined as the exact integral, and it need not equal that integral for a given partition. Using an upper sum as though it were the integral can turn a valid estimate into an incorrect equality.
Worked Example: The Upper Sum Can Exceed the Integral
Let \(f(x)=x\) on \([0,1]\), and first use the partition \(P=\{0,1\}\). The supremum of \(f\) on \([0,1]\) is \(1\), the infimum is \(0\), and the interval width is \(1\). Thus:
The exact integral is:
So \(L(P,f)\leq\int_0^1f\leq U(P,f)\) reads \(0\leq1/2\leq1\), not \(1=\int_0^1x\,dx\).
For a further check, divide \([0,1]\) into \(n\) equal intervals, where \(n\) is a positive integer. Each width is \(1/n\). On interval \(i\), the supremum is \(i/n\) and the infimum is \((i-1)/n\), for \(i=1,\ldots,n\). Therefore:
For every positive integer \(n\), these satisfy \(\frac{n-1}{2n}\leq\frac12\leq\frac{n+1}{2n}\). The bounds enclose the exact integral; they are not themselves the integral in general.
A Reliable Diagnostic Procedure
When presented with an estimate, check the reasoning in a fixed order. This makes it easier to locate which assumption has failed rather than relying on a picture or a plausible-looking number.
If the lower limit exceeds the upper limit, rewrite the integral using \(\int_b^a f=-\int_a^b f\), and reverse inequality directions when negating an inequality.
Identify whether it is nonnegative, nonpositive, or changes sign. Do not substitute unsigned area for a signed integral.
Verify \(f(x)\leq g(x)\), or the claimed constant bounds, at every point in the interval before integrating them.
Distinguish the integral from a lower or upper Darboux sum. A sum provides a bound for an integrable function, not automatically its exact integral.
Whenever possible, calculate the integral independently and substitute its value into the proposed inequality or equality.
Common Mistake: Keeping the Inequality Direction After Negation
Suppose \(-2\leq I\leq6\). Negating all three expressions does not give \(-2\leq-I\leq6\). Multiplication by \(-1\) reverses each inequality, so the correct statement is \(-6\leq-I\leq2\). This is precisely the issue when an estimate for an integral with increasing limits is transferred to the reversed orientation.
For the first worked example, \(I=\int_0^2(2x-1)\,dx=2\), so \(-I=-2\). The correct reversed bounds give \(-6\leq-2\leq2\). The incorrect unchanged bounds would also happen to contain \(-2\) in this case, but that accidental success does not make the reasoning valid. Always reverse the inequality order when multiplying by a negative number.
Check Your Understanding
For each question, identify the relevant condition before deciding whether the estimate is valid.
- If \(-1\leq f(x)\leq4\) for all \(x\in[2,5]\), what bounds follow for \(\int_2^5 f(x)\,dx\)?
- If an integral over \([a,b]\) is known to lie between \(A\) and \(B\), with \(A\leq B\), what interval contains the same integral with its limits reversed?
- Why does the positive-region area of \(f(x)=x-1\) on \([0,2]\) fail to give a lower bound for its signed integral?
- What pointwise condition is needed before concluding \(\int_a^b f\leq\int_a^b g\) from a comparison of \(f\) and \(g\)?
- For \(f(x)=x\) on \([0,1]\) with partition \(\{0,1\}\), what are the upper sum and the exact integral, and why are they different?
