Why Linearity Matters
A definite integral can describe accumulated change, signed area, or the limit of a sum of many small contributions. When the quantity being accumulated is a sum of simpler quantities, it is useful to compute the contributions separately and then combine them. The linearity properties of definite integrals make this possible.
For example, if a rate is the sum of two rates, its total accumulation over the same interval is the sum of the two separate accumulations. Likewise, multiplying a rate by a constant multiplies its accumulation by that constant. These are exact properties, not approximations.
In this tutorial, assume the interval has endpoints \(a<b\). The properties apply when the functions involved are Riemann integrable on \([a,b]\). Continuous functions on a closed interval are an important sufficient case.
The Linearity Properties
The difference rule follows from the sum and constant-multiple rules by taking the constant to be \(-1\):
These rules require the functions being combined to be integrable on the same interval. The coefficients \(c\) and \(d\) are constants with respect to \(x\); they do not vary across the interval.
Why the Rules Work: A Riemann-Sum Derivation
The definite integral is the limit of Riemann sums. Choose a partition of \([a,b]\), with subinterval widths \(\Delta x_i\), and choose a sample point \(\xi_i\) in each subinterval. For an integrable function \(f\), the corresponding Riemann sum is \(\sum_{i=1}^{n}f(\xi_i)\Delta x_i\), and suitable Riemann sums approach \(\int_a^b f(x)\,dx\) as the partition is refined.
The key fact is that sums for \(f\) and \(g\) can use the same partition and the same sample points. At each sample point, adding the function values before multiplying by the width gives exactly the same result as adding the two separate contributions.
For each subinterval, use the same width \(\Delta x_i\) and sample point \(\xi_i\) for both functions.
The equality \((f(\xi_i)+g(\xi_i))\Delta x_i=f(\xi_i)\Delta x_i+g(\xi_i)\Delta x_i\) holds for every \(i\).
Summing over all subintervals gives \(\sum_{i=1}^{n}(f(\xi_i)+g(\xi_i))\Delta x_i=\sum_{i=1}^{n}f(\xi_i)\Delta x_i+\sum_{i=1}^{n}g(\xi_i)\Delta x_i\).
As the partition is refined, each Riemann sum approaches its definite integral. The limit of the sum on the right is the sum of the two limits, giving the integral addition rule.
This proves the sum rule: the equality is true for each common-partition Riemann sum, and the functions are integrable, so the relevant limits exist. The constant-multiple rule follows by the same reasoning, because \(c f(\xi_i)\Delta x_i=c\bigl(f(\xi_i)\Delta x_i\bigr)\) for every subinterval. Taking limits gives \(\int_a^b cf(x)\,dx=c\int_a^b f(x)\,dx\).
Worked Example: Combining Known Integral Values
Worked Example: Combining Known Integral Values
Suppose \(f\) and \(g\) are integrable on \([0,3]\), with \(\int_0^3 f(x)\,dx=4\) and \(\int_0^3 g(x)\,dx=-1\). Find \(\int_0^3 (2f(x)-5g(x))\,dx\).
Apply linearity, taking the constants \(2\) and \(-5\) outside their integrals:
Check by combining the weighted contributions: \(2\) times the first integral contributes \(8\), while \(-5\) times the second integral contributes \(5\), because the second integral is \(-1\). Their sum is \(8+5=13\), as required.
Worked Example: Integrating a Polynomial by Separating Terms
Worked Example: Integrating a Polynomial by Separating Terms
Find \(\int_0^2 (3x^2+4x)\,dx\). To show the component values directly from the Riemann-sum definition, divide \([0,2]\) into \(n\) equal subintervals and use right endpoints \(x_i=2i/n\). The width is \(\Delta x=2/n\).
First calculate the integral of \(x^2\) from its right-endpoint sums:
As \(n\) increases without bound, \(3/n\) and \(1/n^2\) approach \(0\), so \(\int_0^2 x^2\,dx=8/3\).
For \(x\), the same right endpoints give:
Taking the limit gives \(\int_0^2 x\,dx=2\). Now apply linearity:
Check by recombining the two component contributions: \(3(8/3)=8\) and \(4(2)=8\), so the total is \(8+8=16\).
Worked Example: A Constant Term and a Variable Term
Worked Example: A Constant Term and a Variable Term
Find \(\int_1^5 (7-2x)\,dx\). The constant function \(7\) has integral \(7\) times the interval width, which is \(5-1=4\). Thus \(\int_1^5 7\,dx=7(4)=28\).
For the other term, use \(n\) equal subintervals of width \(4/n\) and right endpoints \(x_i=1+4i/n\). Its Riemann sum for \(x\) is
As \(n\) increases without bound, this approaches \(4+8=12\). Therefore \(\int_1^5 x\,dx=12\). Separate the terms and factor out \(-2\):
Check the two contributions: the constant term contributes \(28\), and the term \(-2x\) contributes \(-24\). Their sum is \(28+(-24)=4\).
A Common Mistake: Splitting a Product
Linearity allows an integrand that is a sum or difference to be separated. It does not allow a product to be split into the product of two integrals. For example, in general, \(\int_a^b f(x)g(x)\,dx\) is not equal to \(\left(\int_a^b f(x)\,dx\right)\left(\int_a^b g(x)\,dx\right)\). The Riemann-sum proof does not establish such a rule: distributing a sum over addition is valid, but multiplying separate accumulated totals does not reproduce the sum of the products \(f(\xi_i)g(\xi_i)\Delta x_i\).
A related error is to omit a term when expanding an expression before using linearity. Consider \(\int_0^1 (x+2)^2\,dx\). A wrong calculation might write \((x+2)^2=x^2+4\), then conclude that the integral is \(1/3+4=13/3\). The expansion is incorrect: the cross term \(4x\) is missing.
The correct expansion is \((x+2)^2=x^2+4x+4\). Using linearity and the component values \(\int_0^1 x^2\,dx=1/3\), \(\int_0^1 x\,dx=1/2\), and \(\int_0^1 1\,dx=1\), the correct result is
The correction is to expand accurately first, then apply linearity to the resulting sum.
How to Use Linearity Reliably
When simplifying a definite integral, identify the terms and their constant coefficients. Separate only sums and differences, keep every term on the same interval, and check that each coefficient is constant with respect to the variable of integration. The properties can simplify an integral, but they do not by themselves provide the value of every integral; each component still needs to be evaluated by an appropriate method.
Check Your Understanding
Use the linearity properties and show the main algebraic steps in each answer.
- If \(\int_0^2 f(x)\,dx=3\) and \(\int_0^2 g(x)\,dx=5\), what is \(\int_0^2 (4f(x)-g(x))\,dx\)?
- Rewrite \(\int_1^4 (6x^2+3)\,dx\) as a sum of constant multiples of integrals.
- In the Riemann-sum derivation, why must the same partition and sample points be used for the two functions?
- Is \(\int_a^b (f(x)+g(x))\,dx=\int_a^b f(x)\,dx+\int_a^b g(x)\,dx\) valid when both functions are integrable on \([a,b]\)? State the property that justifies your answer.
- Explain why linearity does not imply that the integral of a product is the product of the integrals.
