Why Bound the Integral of a Sum?
When several functions contribute to a total quantity, it is useful to estimate the size of their combined contribution without first determining its exact value. The integral triangle inequality provides such an estimate: it bounds the integral of the absolute value of a sum using the separate absolute-value integrals.
The key idea is familiar from real numbers. The magnitude of a sum cannot exceed the sum of the magnitudes. Applying that fact at each point of an interval, and then integrating, gives a bound for the entire interval.
Statement and Conditions
Let \(a<b\), and let \(f\) and \(g\) be Riemann integrable real-valued functions on \([a,b]\). Their absolute values and their sum are also Riemann integrable, so all the integrals below are defined.
In particular, the theorem applies when \(f\) and \(g\) are continuous on \([a,b]\), since continuous functions on a closed bounded interval are Riemann integrable. The same result holds when the endpoints coincide: each integral is then zero.
The absolute-value signs are essential. The theorem estimates the total size of the sum, not merely its signed integral. It also gives an upper bound, not in general an equality.
Why the Inequality Holds
For any real numbers \(u\) and \(v\), the ordinary triangle inequality says \(|u+v|\leq |u|+|v|\). Set \(u=f(x)\) and \(v=g(x)\). This gives a pointwise inequality for every \(x\) in the interval. Integral comparison says that if integrable functions \(p\) and \(q\) satisfy \(p(x)\leq q(x)\) throughout \([a,b]\), then \(\int_a^b p(x)\,dx\leq\int_a^b q(x)\,dx\). Applying that principle here, followed by linearity, proves the result.
At each \(x\in[a,b]\), \(|f(x)+g(x)|\leq |f(x)|+|g(x)|\).
Integral comparison gives \(\int_a^b |f(x)+g(x)|\,dx\leq\int_a^b (|f(x)|+|g(x)|)\,dx\).
The integral on the right is \(\int_a^b |f(x)|\,dx+\int_a^b |g(x)|\,dx\), as required.
This proof uses two earlier integral properties: comparison for pointwise ordered functions, and linearity for sums of integrable functions. The pointwise inequality supplies the order; integration preserves it.
Worked Examples
Worked Example: Two Nonnegative Functions
Use the integral triangle inequality for \(f(x)=x\) and \(g(x)=2\) on \([0,1]\), and compare the two sides.
On this interval, \(x\geq 0\) and \(2\geq 0\). Thus \(f+g=x+2\) is also nonnegative, so the absolute values can be evaluated directly:
For the right-hand side, compute each integral:
Therefore the right-hand side is \(\frac12+2=\frac52\), equal to the left-hand side. The inequality holds, and in this example it is an equality. As a check, differentiating the antiderivative \(\frac{x^2}{2}+2x\) gives \(x+2\), and differentiating \(\frac{x^2}{2}\) and \(2x\) gives \(x\) and \(2\), respectively.
Worked Example: A Strict Bound for Constant Functions
Let \(f(x)=2\) and \(g(x)=-1\) on \([0,1]\). The sum is \(f(x)+g(x)=1\), so
The separate absolute-value integrals are
Hence the bound reads \(1\leq 2+1=3\), which is true but not an equality. The antiderivatives used above differentiate to the respective constant integrands: \(1\), \(2\), and \(1\). This example confirms that the theorem guarantees an upper bound; it does not guarantee that the two sides match.
Worked Example: Estimating a Nonconstant Sum
Take \(f(x)=x-2\) and \(g(x)=3-2x\) on \([0,1]\). On this interval, \(x-2<0\) and \(3-2x>0\). Their sum is \(f(x)+g(x)=1-x\), which is nonnegative. Thus the left-hand side is
For the first term on the right, \(|x-2|=2-x\) on \([0,1]\). Therefore
For the second term, \(|3-2x|=3-2x\), so
The theorem gives \(\frac12\leq\frac32+2=\frac72\). This is a valid, though not especially tight, estimate. As a check, the derivatives of the three antiderivatives \(x-\frac{x^2}{2}\), \(2x-\frac{x^2}{2}\), and \(3x-x^2\) are \(1-x\), \(2-x\), and \(3-2x\), exactly the integrands used.
Finite Sums and Scalar Multiples
The same reasoning extends to any finite number of integrable functions. For real numbers \(u_1,\ldots,u_n\), repeated use of the ordinary triangle inequality gives \(\left|\sum_{j=1}^{n}u_j\right|\leq\sum_{j=1}^{n}|u_j|\). Applying this pointwise and then using integral comparison and linearity yields the finite-sum form.
Scalar multiples can be included because \(|c f(x)|=|c|\,|f(x)|\) for every real constant \(c\). In particular, for real constants \(c\) and \(d\),
To see this, apply the two-function result to \(c f\) and \(d g\). Then use \(\int_a^b |c f(x)|\,dx=|c|\int_a^b |f(x)|\,dx\) and the corresponding identity for \(d g\).
A Common Mistake: Replacing the Inequality With Equality
A tempting but incorrect step is to write \(|f+g|=|f|+|g|\) and then integrate both sides. The ordinary triangle inequality does not say that absolute values distribute over addition as an equality. For a concrete check, use the constant functions \(f(x)=2\) and \(g(x)=-1\) on \([0,1]\): the left integral is \(1\), while the sum of the separate absolute-value integrals is \(3\). They are not equal.
Another useful distinction is that this theorem does not claim \(\left|\int_a^b(f+g)\,dx\right|=\int_a^b|f+g|\,dx\). The absolute-value bound for a single integral concerns a different comparison. Here the integrand on the left is specifically the absolute value of a sum, and the right-hand side measures the separate absolute-value integrals.
Why the Estimate Is Useful
The integral triangle inequality can be applied before the exact integral of a sum is known. If each term has a manageable absolute-value integral, the theorem produces an upper bound for the combined integrand. It is especially useful when the sum itself is complicated but its components can be estimated separately.
The logic is also a model for many estimates in mathematics: first establish a pointwise bound, then use a principle that preserves the bound. Here, the real-number triangle inequality controls each value, and integral comparison transfers that control to the whole interval.
Check Your Understanding
Use the theorem and its proof to answer these questions.
- State the integral triangle inequality for two Riemann integrable functions on \([a,b]\).
- Which pointwise inequality is integrated in the proof?
- Which property of the integral allows the right-hand side to split into two integrals?
- For \(f(x)=3\) and \(g(x)=-2\) on \([0,1]\), compute both sides of the inequality and check their order.
- How does the theorem extend to a finite sum of integrable functions?
