Why Linearity Matters
Differentiation respects sums and constant multiples: the derivative of a sum is the sum of the derivatives, and a constant factor can be taken outside a derivative. This has a useful consequence in reverse. If functions have antiderivatives, then sums and constant multiples of those functions also have antiderivatives.
The word indefinite is important here. An indefinite integral represents a family of antiderivatives, not just one function. Linearity describes how to construct an antiderivative of a combined function, while the arbitrary constant accounts for the fact that there are generally many antiderivatives.
Antiderivatives on an Interval
Let \(I\) be an interval, and let \(f:I\to\mathbb{R}\). A function \(F:I\to\mathbb{R}\) is an antiderivative of \(f\) on \(I\) if \(F\) is differentiable on \(I\) and \(F'(x)=f(x)\) for every \(x\) in \(I\). When \(F\) is one such function, the indefinite integral denotes the full family of antiderivatives:
where \(C\) is an arbitrary real constant. The earlier result that antiderivatives on an interval differ by a constant ensures that this notation includes all of them.
To state linearity precisely, take two functions \(f\) and \(g\) that each have antiderivatives on the same interval \(I\), and let \(a,b\in\mathbb{R}\). Choose particular antiderivatives \(F\) and \(G\), so \(F'=f\) and \(G'=g\). Then the function \(aF+bG\) is an antiderivative of \(af+bg\). The entire family is obtained by adding any real constant.
Why the Rule Works
The construction follows from the sum and constant-multiple rules for derivatives. The derivative rules establish that the proposed function has the required derivative. To know it gives the whole family, we also use the fact that any two antiderivatives of the same function on an interval differ by a constant.
Let \(F'=f\) and \(G'=g\) on the same interval \(I\).
Form \(H=aF+bG\), using the same real constants \(a\) and \(b\) as in the target function.
By the derivative rules, \(H'=aF'+bG'=af+bg\). Thus \(H\) is an antiderivative of the combined function.
Every other antiderivative of \(af+bg\) on \(I\) differs from \(H\) by a constant, so the full family is \(H+C\).
This proves both parts of the claim: the displayed combination works, and adding \(C\) accounts for every other antiderivative. The common interval matters because the statement about antiderivatives differing by a constant applies on an interval; the theorem does not assert that one constant describes a family across disconnected domains.
Worked Examples
Worked Example: An Antiderivative of a Sum
Find the full family of antiderivatives of \(2x+3\) by combining antiderivatives of its two terms. Work on \(I=\mathbb{R}\). From the earlier power rule, \(x^2\) is an antiderivative of \(2x\), and \(3x\) is an antiderivative of \(3\). Linearity therefore gives
Check by differentiating the proposed family:
The derivative is the original integrand, so the formula is correct. The constant is unrestricted, as required for the full family.
Worked Example: A Linear Combination
Find the full family of antiderivatives of \(4x-2\). View the function as \(4(x)-2(1)\). An antiderivative of \(x\) is \(\frac{x^2}{2}\), and an antiderivative of \(1\) is \(x\). Combining those particular antiderivatives with the same coefficients gives
Differentiate to check:
Thus \(2x^2-2x+C\) is the full family. The check also confirms that each coefficient has been carried through correctly.
Worked Example: A Constant Multiple and Its Arbitrary Constant
Find the full family of antiderivatives of \(-3(2x)\). Since \(x^2\) is an antiderivative of \(2x\), multiplying that particular antiderivative by \(-3\) gives one antiderivative of the scaled function:
The derivative check is
It is important that the final constant is still arbitrary. If one starts instead with the family \(x^2+C_1\), scaling it gives \(-3x^2-3C_1\). As \(C_1\) ranges over all real numbers, \(-3C_1\) also ranges over all real numbers. Therefore this scaled family is exactly \(-3x^2+C\), not a family with a fixed constant.
Worked Example: Combining Separate Families
Take the antiderivative families of \(2x\) and \(3\) separately:
Adding members of these families gives
The sum \(C_1+C_2\) is a real constant, so the result has the form \(x^2+3x+C\). Conversely, every real value of \(C\) can occur: choose \(C_1=C\) and \(C_2=0\). Thus combining the separate families gives exactly the full family of antiderivatives of \(2x+3\). Differentiation verifies the result:
How to Read the Linearity Notation
A compact way to express the rule is
This is standard shorthand for the theorem: choose antiderivatives \(F\) and \(G\), combine them as \(aF+bG\), and then include an arbitrary additive constant. Because indefinite integrals represent families, the shorthand should not be interpreted as ordinary equality between single functions. In particular, the arbitrary constants on separate expressions can be combined into one arbitrary constant in the resulting family.
The condition that \(f\) and \(g\) have antiderivatives is essential. Linearity tells us how to build an antiderivative when the component antiderivatives exist; it does not claim that every function has an antiderivative. The coefficients \(a\) and \(b\) must be constants, not functions of \(x\). If a coefficient depends on \(x\), the constant-multiple rule used in the proof does not apply as stated.
A Common Pitfall: The Zero Multiple
A careless reading of the shorthand can cause trouble when the coefficient is zero. For example, it is correct that
because every constant function has derivative zero. But multiplying a chosen antiderivative family \(F+C_1\) by zero gives only the zero function:
Its derivative is zero, but the expression \(0\) alone is not the full family of antiderivatives of zero. The missing constants must be included:
The safe statement of linearity handles this without exception: for any real coefficient, including zero, a chosen linear combination of particular antiderivatives is an antiderivative of the corresponding linear combination of functions, and the full answer then includes \(+C\).
Why the Arbitrary Constant Is Part of the Rule
Differentiation removes additive constants: the derivative of \(F+C\) is \(F'\). Consequently, a derivative alone cannot identify which constant was present in an antiderivative. Whenever a constructed function has the required derivative, every function differing from it by a constant has that same derivative.
For linearity, this means that \(aF+bG\) supplies a particular antiderivative of \(af+bg\), but it is not the complete answer by itself. Adding \(C\) gives the entire family on the interval. This distinction is useful throughout integration: a formula without the arbitrary constant may be a valid particular antiderivative, but it does not express the indefinite integral's full family.
Check Your Understanding
Answer these questions using the theorem and the meaning of an indefinite integral.
- If \(F'=f\) and \(G'=g\) on an interval, what is the derivative of \(aF+bG+C\), where \(a\) and \(b\) are constants?
- Why must \(f\) and \(g\) have antiderivatives before the construction \(aF+bG\) can be used?
- Why does the full family of antiderivatives include an arbitrary constant even after one antiderivative has been constructed?
- What is the full family of antiderivatives of the zero function, and why is the single function \(0\) not the full family?
- Why can two arbitrary constants from separate antiderivative families be represented by one arbitrary constant after the expressions are combined?
