Why Linearity Helps
An integrand can contain several terms that are easier to handle separately. For example, a sum of a polynomial term and a trigonometric term may have a straightforward antiderivative if each part is considered on its own. The reason this is legitimate is not just an algebraic convenience: it follows from how derivatives behave.
Differentiation respects sums and constant multiples. Since integration asks us to reverse differentiation, the corresponding antiderivative rules respect sums and constant multiples too. These rules let us reorganize an integral without changing the antiderivative problem.
Antiderivatives and Indefinite Integrals
A function \(F\) is an antiderivative of \(f\) on an interval \(I\) if \(F'(x)=f(x)\) for every \(x\in I\). The indefinite integral \(\int f(x)\,dx\) denotes the family of antiderivatives of \(f\), rather than one particular function. If \(F\) is one antiderivative, that family is written \(F(x)+C\), where \(C\) is an arbitrary constant.
The interval matters. For example, \(1/x\) is defined separately on the intervals \(x>0\) and \(x<0\), and an antiderivative must be considered on an interval where the integrand is defined. In the rules below, assume all functions and antiderivatives are defined on a common interval.
Where the Rules Come From
Let \(F\) and \(G\) be differentiable on an interval \(I\), and let \(a\) and \(b\) be constants. The derivative of \(aF+bG\) can be obtained directly from the definition of the derivative. For a point \(x\in I\), take \(h\) small enough that \(x+h\) is also in \(I\):
Because \(F\) and \(G\) are differentiable, the two difference quotients have limits as \(h\) approaches zero. Taking that limit gives
This proves the derivative rule for a linear combination: the derivative of a sum is the sum of the derivatives, and a constant multiple remains a constant multiple after differentiation. To obtain the integration rules, suppose \(F'=f\) and \(G'=g\). Then
Thus \(aF+bG\) is an antiderivative of \(af+bg\). Every other antiderivative on the same interval differs from it by an arbitrary constant. This gives the integration rule.
In the shorter notation \(\int f\,dx+\int g\,dx\), each indefinite integral represents a family. The displayed formulas mean that the resulting families agree, not that a particular choice of constants on separate terms must be preserved. When combining terms, one arbitrary constant at the end is enough.
Using Linearity to Split an Integrand
To use the rules, first identify terms that are joined by addition or subtraction. A subtraction is addition of a negative term. Then take any constant coefficients into account and find an antiderivative for each term. This is a change in organization, not a change to the function being integrated.
Separate terms joined by addition or subtraction, keeping each sign attached to its term.
Move a constant coefficient outside its term's integral, or leave it in place and integrate the term directly.
Use an antiderivative for each simpler term on the interval under consideration.
Write one arbitrary constant and differentiate the result to confirm that it returns the original integrand.
Worked Example: A Polynomial Sum
Find \(\int (6x^2-4x+5)\,dx\). The three terms are joined by addition and subtraction, so apply linearity to them separately:
Check by differentiating the answer:
The derivative is the original integrand, so the antiderivative is correct.
Worked Example: A Trigonometric Term and a Reciprocal Term
Find \(\int (3\cos x-2/x)\,dx\) on the interval \(x>0\). The term \(1/x\) is defined throughout this interval, and \(\sin x\) and \(\ln x\) have derivatives \(\cos x\) and \(1/x\), respectively. Split the expression and retain its coefficients:
Differentiate to check:
This matches the integrand on \(x>0\). On the separate interval \(x<0\), the corresponding logarithm can be written \(\ln|x|\).
More Than Two Terms
The same reasoning applies to any finite sum: repeatedly use the two-term sum rule. For example, an integrand with four terms can be split into four antiderivative problems, provided the terms have antiderivatives on the interval being used. Linearity does not tell us what those antiderivatives are; it only guarantees that finding them separately and combining them is valid.
Worked Example: Several Standard Antiderivatives
Find \(\int \left(2e^x+\frac{3}{1+x^2}-5\sin x\right)\,dx\). The expression is a sum of three terms with constant coefficients. Use antiderivatives \(e^x\), \(\arctan x\), and \(\cos x\) for \(e^x\), \(1/(1+x^2)\), and \(-\sin x\), respectively:
The term \(1/(1+x^2)\) is defined for all real \(x\), so this calculation holds on \(\mathbb R\). Check by differentiating each part:
The result is exactly the given integrand.
What Linearity Does Not Permit
The rules apply to sums and constant multiples. They do not allow an arbitrary factor that depends on \(x\) to be pulled outside the integral, and they do not turn a product into the product of two integrals. For instance, \(x\) is not a constant coefficient. The equation \(\int x f(x)\,dx=x\int f(x)\,dx\) is not a linearity rule; differentiating its right side generally produces an extra term.
A concrete example shows the danger. Since \(x^2=x\cdot x\), an invalid “split” might claim
But differentiating the proposed answer gives
which is not \(x^2\). The valid approach is to integrate the whole term \(x^2\), giving \(x^3/3+C\); its derivative is \(x^2\). The contrast is important: linearity splits a sum of terms, not a product of factors.
Why the Rule Matters
Linearity makes integration more manageable because an expression can be organized into pieces whose antiderivatives are familiar. It also helps preserve signs and coefficients: each term has a clear contribution to the final result. However, it is only a bookkeeping rule. It does not guarantee that every piece has an elementary antiderivative, nor does it choose a method for finding one. Each separated term still needs to be integrated correctly, on a suitable interval.
A reliable final check is to differentiate the proposed antiderivative. That check tests the entire result at once: coefficients, signs, and the arbitrary constant. Since the derivative of a constant is zero, the constant does not affect whether the result matches the integrand.
Check Your Understanding
Use the rules and conditions in this tutorial to answer the questions.
- Why does differentiating \(aF+bG\) lead to the integration rule for \(af+bg\), when \(F'=f\) and \(G'=g\)?
- Find \(\int (8x^3+6x-4)\,dx\), and differentiate your result to check it.
- Find \(\int (4\cos x+3\sin x)\,dx\), and verify the answer.
- Why can a constant factor be moved outside an integral, but \(x\) cannot be treated as a constant factor in \(\int x f(x)\,dx\)?
- What does the arbitrary constant represent in an indefinite integral?
