From Pointwise Information to an Integral Estimate
A definite integral combines the values of a function across an interval. Often, the exact values are difficult to calculate, but we know that the function stays between two simpler bounds. That information is already enough to estimate the integral.
The interval’s length matters: a constant bound contributes that constant multiplied by the length. The order of the limits matters too: reversing the limits changes the sign of the integral, and therefore changes how its lower and upper bounds are ordered.
Bounds on an Increasing Interval
Suppose \(a<b\), and \(f\) is Riemann integrable on \([a,b]\). If fixed constants \(m\) and \(M\) satisfy \(m\leq f(x)\leq M\) at every point \(x\) of the interval, then the integral is bounded by the same constants multiplied by the interval length.
The factor \(b-a\) is essential. A bound on the height of a function does not, by itself, determine the size of its integral: the same height bounds over a longer interval allow a larger range of possible integral values.
Why the Bound Holds
The result follows from the integral comparison principle and the integral of a constant. Those facts say, respectively, that pointwise order is preserved by integration from left to right, and that the integral of a constant is that constant times the interval length.
Since \(m\leq f(x)\) throughout \([a,b]\), integral comparison gives \(\int_a^b m\,dx\leq\int_a^b f(x)\,dx\).
Because \(a<b\), \(\int_a^b m\,dx=m(b-a)\). This gives the lower bound.
Since \(f(x)\leq M\) throughout \([a,b]\), comparison gives \(\int_a^b f(x)\,dx\leq\int_a^b M\,dx=M(b-a)\). This gives the upper bound.
Putting the two comparisons together proves the theorem. The assumption that \(f\) is integrable is needed for its definite integral to exist; continuity on \([a,b]\) is one familiar sufficient condition.
Worked Example: Bounding a Linear Function
Worked Example: Bounding a Linear Function
Suppose \(f(x)=2x+1\) on \([1,4]\). Find a bound for \(\int_1^4 f(x)\,dx\) using constant pointwise bounds.
First, \(1\leq x\leq4\). Multiplying by \(2\), which is positive, gives \(2\leq2x\leq8\). Adding \(1\) throughout gives \(3\leq2x+1\leq9\). Thus \(m=3\), \(M=9\), and the interval length is \(4-1=3\).
For a check, evaluate the integral exactly: an antiderivative is \(x^2+x\), so \(\int_1^4(2x+1)\,dx=(16+4)-(1+1)=20-2=18\). The value \(18\) lies between \(9\) and \(27\), as required.
When the Limits Are Reversed
The theorem above uses \(a<b\). If the limits are instead \(b\) and \(a\), with the same pointwise bounds on \([a,b]\), use the signed-orientation rule \(\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx\). Multiplication by \(-1\) reverses inequalities. Consequently, the upper bound on the forward integral becomes the lower bound on the reversed integral.
The interval length in these formulas is still \(b-a>0\), not \(a-b\). The sign change is accounted for separately by the reversed limits.
Worked Example: Bounding an Integral with Reversed Limits
On \([0,2]\), let \(f(x)=x+1\). Find bounds for \(\int_2^0 f(x)\,dx\).
Since \(0\leq x\leq2\), adding \(1\) gives \(1\leq x+1\leq3\). Thus \(m=1\), \(M=3\), and the interval length is \(2-0=2\). For the forward integral, the estimate is \(1(2)\leq\int_0^2f(x)\,dx\leq3(2)\), or \(2\leq\int_0^2f(x)\,dx\leq6\).
Reversing the limits negates the integral and reverses the inequality order:
Check by direct evaluation. An antiderivative is \(x^2/2+x\), so \(\int_2^0(x+1)\,dx=(0)-(2^2/2+2)=-(2+2)=-4\). The value \(-4\) lies between \(-6\) and \(-2\).
Zero-Length Intervals
If both limits are the same, the integral is zero. There is no interval over which the function accumulates, so the length factor is zero too. In particular, for any integrable \(f\) and any real number \(a\), \(\int_a^a f(x)\,dx=0\).
This case does not require choosing an orientation: the two limits coincide.
Bounds Given by Functions
The lower and upper bounds do not have to be constants. Suppose \(g\), \(f\), and \(h\) are Riemann integrable on \([a,b]\), where \(a<b\), and \(g(x)\leq f(x)\leq h(x)\) for every \(x\) in that interval. Applying integral comparison to each inequality gives \(\int_a^b g(x)\,dx\leq\int_a^b f(x)\,dx\leq\int_a^b h(x)\,dx\). These bounds are useful when the bounding functions have straightforward integrals.
Worked Example: Using Two Function Bounds
Suppose that \(x\leq f(x)\leq x+1\) for every \(x\in[0,2]\). Bound \(\int_0^2 f(x)\,dx\).
All three functions are integrable on \([0,2]\), and the given inequalities hold throughout that interval. Integrate the lower bound: \(\int_0^2 x\,dx=[x^2/2]_0^2=2/2-0=2\). Integrate the upper bound: \(\int_0^2(x+1)\,dx=[x^2/2+x]_0^2=(2+2)-0=4\). Therefore \(2\leq\int_0^2 f(x)\,dx\leq4\).
For a check, the function \(f(x)=x+\tfrac12\) satisfies the assumed bounds because \(x\leq x+\tfrac12\leq x+1\). Its integral is \([x^2/2+x/2]_0^2=2+1=3\), which lies between \(2\) and \(4\). This check illustrates the estimate for one function satisfying the bounds; the comparison argument applies to every such integrable \(f\).
A Common Mistake: Using the Wrong Length or Order
A frequent error is to treat an integral with reversed limits as though it were still positive. For example, take \(1\leq f(x)\leq3\) on \([0,2]\). The forward-integral estimate is \(2\leq\int_0^2 f(x)\,dx\leq6\). A mistaken calculation might claim \(2\leq\int_2^0 f(x)\,dx\leq6\), keeping the same bounds after reversing the limits.
That cannot be correct: \(\int_2^0f(x)\,dx=-\int_0^2f(x)\,dx\), so the correct estimate is \(-6\leq\int_2^0 f(x)\,dx\leq-2\). For instance, if \(f(x)=2\), then the reversed integral is \(2(0-2)=-4\), which satisfies the corrected estimate but not the mistaken one.
A second error is to omit the interval length. A pointwise bound of \(2\leq f(x)\leq5\) on an interval of length \(4\) gives \(8\leq\int f\leq20\) when the interval is traversed from left to right—not \(2\leq\int f\leq5\).
What the Estimate Tells You
An integral bound is a guarantee, not usually an exact value. Every integrable function that stays between the stated bounds has an integral in the resulting range. If that range is wide, the pointwise information may be too coarse to give a precise estimate; sharper pointwise bounds can narrow it.
For constant bounds, the calculation has a particularly direct interpretation: the integral is at least the lower constant times the interval length and at most the upper constant times that length. The same statement remains valid when either constant is negative. For example, if \(-4\leq f(x)\leq-1\) on an interval of length \(3\), then the forward integral satisfies \(-12\leq\int f\leq-3\). The inequalities are multiplied by the positive length, so their order does not change.
When applying the result, verify three things: the bounds hold at every point of the integration interval, the function is integrable there, and the orientation of the limits has been handled correctly. A correct pointwise inequality on only part of the interval is not enough to bound the integral over the whole interval.
Check Your Understanding
Use the pointwise information and interval orientation carefully in each question.
- If \(2\leq f(x)\leq5\) on \([1,4]\), what bounds follow for \(\int_1^4 f(x)\,dx\)?
- If \(-1\leq g(x)\leq2\) on \([-2,1]\), what bounds follow for \(\int_1^{-2} g(x)\,dx\)?
- If \(x^2\leq h(x)\leq x^2+1\) on \([0,2]\), how can you bound \(\int_0^2h(x)\,dx\)?
- Why does multiplying the bounds by the interval length preserve their order for a left-to-right integral?
- What happens to both the signs and the order of integral bounds when the limits are reversed?
