Why Linearity Is Useful
Many integrands are written as combinations of simpler expressions: a sum, a difference, or a constant times a function. If the antiderivatives of the component functions are known, linearity lets us combine them without starting over. The key is to apply the rules to the functions while remembering that an indefinite integral represents a whole family of antiderivatives.
For example, if \(F'(x)=f(x)\) and \(G'(x)=g(x)\), then differentiating \(F+G\) gives \(f+g\). This simple fact is the reason the sum rule works. The difference and constant-multiple rules follow from the same derivative rules.
Antiderivatives and Indefinite Integral Notation
Let \(I\) be an interval, and let \(f\) be a function defined on \(I\). A function \(F\) is an antiderivative of \(f\) on \(I\) if \(F\) is differentiable there and \(F'(x)=f(x)\) for every \(x\) in \(I\). The notation \(\int f(x)\,dx\) represents the family of all antiderivatives of \(f\) on the interval, rather than one particular function.
If \(F\) is one antiderivative of \(f\), then all antiderivatives on the interval have the form \(F(x)+C\), where \(C\) is a real constant. This follows because two antiderivatives of the same function have a difference whose derivative is zero; on an interval, a differentiable function with derivative zero is constant.
The Linearity Rules
Suppose \(f\) and \(g\) have antiderivatives \(F\) and \(G\), respectively, on the same interval \(I\). Let \(a\) and \(b\) be fixed real numbers. Then the function \(aF+bG\) has derivative \(af+bg\). Adding an arbitrary constant gives every antiderivative of that combination.
Choosing \(a=b=1\) gives the sum rule. Choosing \(a=1\) and \(b=-1\) gives the difference rule. Setting \(b=0\) gives the constant-multiple rule, with \(a\) as the constant factor. The same formula covers all three cases.
These formulas assume that the component functions have antiderivatives on the interval being considered. They do not require \(a\) or \(b\) to be nonzero. The arbitrary constant in the result is essential in every case, including when a coefficient is zero.
Why the Rules Work
The proof is short: differentiate the proposed antiderivative. The sum rule follows from the derivative rule for sums, and the constant-multiple rule follows from the derivative rule for a constant times a function. The difference rule is included by using a negative constant.
Assume \(F'=f\) and \(G'=g\) on the interval, with \(a\) and \(b\) fixed real constants.
Consider \(H(x)=aF(x)+bG(x)+C\), where \(C\) is any real constant.
By the sum and constant-multiple derivative rules, \(H'(x)=aF'(x)+bG'(x)+0=af(x)+bg(x)\).
Thus every such \(H\) is an antiderivative of \(af+bg\). Any two antiderivatives on the interval differ by a constant, so allowing every \(C\in\mathbb R\) gives the full family.
This argument also explains why the coefficients must be constants with respect to \(x\). If a coefficient depends on \(x\), differentiating its product with an antiderivative introduces an additional term, so the constant-multiple rule in this form does not apply.
Worked Examples
Worked Example: Integrating a Difference of Known Derivatives
Find \(\int \bigl(4(2x+3)-3(3x^2-2x)\bigr)\,dx\).
First identify antiderivatives of the component expressions. Since \(\frac{d}{dx}(x^2+3x)=2x+3\), an antiderivative of \(2x+3\) is \(x^2+3x\). Since \(\frac{d}{dx}(x^3-x^2)=3x^2-2x\), an antiderivative of \(3x^2-2x\) is \(x^3-x^2\).
Apply the constant-multiple and difference rules:
Check by differentiating the result:
The derivative matches the integrand, so the antiderivative is correct. The component antiderivatives here are used only as known functions whose derivatives can be checked; the point is how linearity combines them.
Worked Example: Including a Constant Term
Find \(\int \bigl(5(2x-4)+2(3x^2)-7\bigr)\,dx\). Use the known derivatives \(\frac{d}{dx}(x^2-4x)=2x-4\), \(\frac{d}{dx}(x^3)=3x^2\), and \(\frac{d}{dx}(-7x)=-7\).
Apply linearity to each term, carrying the coefficient along with its antiderivative:
For the simplification, \(5(x^2-4x)=5x^2-20x\), and \(-20x-7x=-27x\). Check by differentiation:
The derivative equals the original integrand. The constant term \(-7\) contributes the antiderivative \(-7x\); it does not disappear during integration.
Worked Example: Zero Coefficients and the Arbitrary Constant
Consider \(\int \bigl(0(2x+1)-0(3x^2-2)\bigr)\,dx\). The integrand simplifies to \(0\). Each component has an antiderivative: \(x^2+x\) differentiates to \(2x+1\), and \(x^3-2x\) differentiates to \(3x^2-2\).
The family formula gives
Check: for every real \(C\), \(\frac{d}{dx}C=0\), which is exactly the integrand. Thus the answer is the family of all constant functions, not just the single function zero.
What the Constant Does—and Does Not—Mean
It is common to write a compact expression such as \(\int (af+bg)\,dx=a\int f\,dx+b\int g\,dx\). This is useful shorthand, but it must be interpreted carefully: an indefinite integral denotes a family, and manipulating families as though they were ordinary single functions can hide the arbitrary constant.
To see the issue, suppose \(a=b=0\). The left side represents all antiderivatives of zero, namely all constant functions. But if the right side is interpreted as literal multiplication of every member of the two antiderivative families by zero, it produces only the zero function. The two families would not be equal under that interpretation.
Common Mistake: Dropping the Constant
A frequent error is to treat the displayed combination of antiderivatives as the only answer. For instance, one might write \(\int (2F'-2F')\,dx=2F-2F=0\), then conclude that zero is the only antiderivative. The calculation does show that zero is one antiderivative, but it does not give the full family.
The integrand simplifies to zero, and every constant function differentiates to zero. The complete answer is \(\int 0\,dx=C\), not just \(0\). More generally, after combining terms, include one arbitrary constant in the final result. Do not add a separate constant for every term unless you then correctly combine them; the full family of the sum is represented by one arbitrary constant.
When Linearity Matters
Linearity is useful whenever an integrand is expressed as a finite sum of terms whose antiderivatives are already known. It lets you preserve the structure of the expression, handle fixed numerical coefficients, and check each contribution independently. It also makes verification straightforward: differentiate the proposed result, distribute the derivative across the sum, and confirm that each term returns its corresponding integrand.
Linearity alone does not tell you an antiderivative for a component whose antiderivative is unknown. It provides a way to combine known antiderivatives; it is not a substitute for a rule that finds each one. In every application, check that the component functions share the interval under consideration and that the antiderivatives being used are valid there.
Check Your Understanding
For each question, keep in mind that an indefinite integral represents a family of functions.
- If \(F'=f\) and \(G'=g\), what function is an antiderivative of \(3f-2g\)? Write the full family.
- Why does the constant-multiple rule require a fixed constant rather than a coefficient that depends on \(x\)?
- Find an antiderivative of \(2(2x+1)-3\), and verify it by differentiation.
- What is the complete family of antiderivatives of the zero function?
- Explain why multiplying antiderivative families literally by zero can lose functions from the family of antiderivatives of zero.
