Bounding an Integral Using Extreme Values
A definite integral records accumulated signed values across an interval. Even when finding that integral exactly is difficult, it can be bounded if we know how small and how large the integrand can be. Continuity on a closed interval provides exactly the information needed: the function attains a minimum and a maximum there.
The resulting estimate has a simple interpretation. If every function value is at least \(m\), then the accumulated value cannot be smaller than a rectangle of height \(m\) and width equal to the interval length. If every function value is at most \(M\), the integral cannot exceed the corresponding rectangle of height \(M\).
Minimum and Maximum on a Closed Interval
A minimum value of \(f\) on \([a,b]\) is a value \(m\) such that \(f(x)\ge m\) for every \(x\in[a,b]\), with equality for at least one \(x\) in the interval. A maximum value \(M\) is defined in the corresponding way: \(f(x)\le M\) throughout the interval, with equality somewhere.
The requirement that these values are attained matters. A function can have a lower bound or an upper bound that it approaches without ever reaching, but such a bound is not necessarily its minimum or maximum. Here we use the extreme value theorem: a function continuous on a closed, bounded interval attains both a minimum and a maximum there.
For \(a<b\), the minimum-and-maximum bounds theorem gives the following estimate. The length of the interval is \(b-a\), which is positive.
If \(a=b\), the interval has length zero and the integral is zero, so the same displayed inequalities hold with both bounds equal to zero. In what follows, the examples use intervals of positive length.
Why the Bounds Hold
By the definitions of \(m\) and \(M\), every value of the function satisfies \(m\le f(x)\le M\) on the interval. The integral comparison principle says that pointwise inequalities between integrable functions are preserved when they are integrated over an interval in increasing order. Constants are integrable, and the integral of a constant is its value multiplied by the interval length. Combining these facts proves the estimate.
For each \(x\in[a,b]\), the definition of minimum and maximum gives \(m\le f(x)\le M\).
Since \(a<b\), integral comparison gives \(\int_a^b m\,dx\le \int_a^b f(x)\,dx\le\int_a^b M\,dx\).
Using \(\int_a^b c\,dx=c(b-a)\), the outer terms become \(m(b-a)\) and \(M(b-a)\).
This argument also explains why the interval length appears in the formula. If the same range of function values occurs over a wider interval, the bounding rectangles have greater area. The bounds are estimates; unless the function is constant, they need not equal the integral.
Worked Examples
Worked Example: A Quadratic on an Interval
Find minimum-and-maximum bounds for \(\int_{-1}^{2}(x^2+1)\,dx\), and check the exact integral against them.
The function \(f(x)=x^2+1\) is continuous on \([-1,2]\). Its smallest value occurs at \(x=0\), where \(f(0)=1\). To identify the largest value, compare the endpoint values and the interior minimum: \(f(-1)=2\), \(f(0)=1\), and \(f(2)=5\). Thus \(m=1\), \(M=5\), and the interval length is \(2-(-1)=3\).
For a check, an antiderivative is \(\frac{x^3}{3}+x\), since its derivative is \(x^2+1\). Evaluating at the endpoints gives
The exact value satisfies the estimate: \(3\le6\le15\).
Worked Example: A Function That Takes Positive and Negative Values
Bound \(\int_{0}^{\pi}\cos x\,dx\) using the minimum and maximum of cosine on \([0,\pi]\).
The function is continuous on this closed interval. Its maximum is \(1\), attained at \(x=0\), and its minimum is \(-1\), attained at \(x=\pi\). The interval length is \(\pi-0=\pi\). Therefore,
To check, an antiderivative is \(\sin x\), whose derivative is \(\cos x\). Hence
The value \(0\) lies between \(-\pi\) and \(\pi\). The example also shows that the integral can be zero even when the function is not zero: positive and negative contributions can offset one another.
Worked Example: A Concave-Down Quadratic
Use extreme values to bound \(\int_{0}^{4}\bigl(4-(x-2)^2\bigr)\,dx\), then verify the result exactly.
The function \(f(x)=4-(x-2)^2\) is continuous on \([0,4]\). Since a square is nonnegative, \(f(x)\le4\), with equality at \(x=2\); thus \(M=4\). On this interval, \((x-2)^2\le4\), so \(f(x)\ge0\), with equality at \(x=0\) and \(x=4\); thus \(m=0\). The interval length is \(4\), giving
Expanding the integrand gives \(4-(x^2-4x+4)=-x^2+4x\). An antiderivative is \(-\frac{x^3}{3}+2x^2\), whose derivative is \(-x^2+4x\). Therefore,
Since \(0\le\frac{32}{3}\le16\), the exact value meets the bounds.
Common Mistake: Assuming an Endpoint Value Is the Minimum
Checking only the endpoints does not necessarily identify the minimum or maximum on the whole interval. An interior point can have a more extreme value. Consider \(f(x)=(x-1)^2\) on \([0,2]\). The endpoint values are \(f(0)=1\) and \(f(2)=1\), but at the interior point \(x=1\), \(f(1)=0\). Thus the minimum is \(m=0\), not \(1\), while the maximum is \(M=1\).
If we incorrectly took the endpoint value \(1\) as the minimum, we would claim that the integral is at least \(1\cdot(2-0)=2\). But direct evaluation gives a smaller value:
So the incorrect lower bound \(2\le\frac{2}{3}\) is false. Using the actual minimum and maximum gives the valid estimate
When identifying extreme values, account for the entire closed interval, including its interior and its endpoints. Continuity guarantees that the extrema exist, but it does not say that they must occur at the endpoints.
Interpreting and Using the Estimate
The estimate is useful even when the exact integral is not known or is inconvenient to calculate. Once reliable values of \(m\) and \(M\) are available, the theorem immediately supplies an interval that must contain the integral. Its width is
This width depends on both the range of function values and the length of the interval. A narrow range or a short interval makes the bounds closer together. The estimate can also serve as a check: if an exact calculation lies outside the bounds, then at least one of the stated extrema, the interval length, or the integral calculation is wrong.
The theorem requires continuity on the whole closed interval so that the minimum and maximum are attained. Do not apply it as stated when that condition is absent. Also keep the interval length positive in the displayed order: the theorem is written for integration from \(a\) to \(b\) with \(a\le b\). Reversing the integration limits changes the sign of the integral, so the inequalities must be adjusted accordingly.
Check Your Understanding
Answer each question using the minimum-and-maximum bounds theorem.
- For a continuous function with minimum \(-2\) and maximum \(3\) on \([1,5]\), what bounds does the theorem give for its integral?
- Why does continuity on a closed interval matter when applying this theorem?
- For \(f(x)=(x-2)^2\) on \([0,4]\), identify the minimum and maximum, then write the resulting integral bounds.
- Can an integral be zero when the integrand has both positive and negative values? Give a reason based on the cosine example.
- What goes wrong if endpoint values are assumed to be the extrema without checking the interior?
