Mathematics · Ch 7 — Integrals

Some Properties of Indefinite Integral

7.2.1

Some Properties of Indefinite Integral

7.2.1 Some Properties of Indefinite Integral

This subsection establishes the fundamental properties of indefinite integrals — the rules you will use every time you break a complicated integral into simpler pieces.

Property (I): Differentiation and Integration are Inverse Processes

\frac{d}{dx}\left[\int f(x)\,dx\right] = f(x)

Proof. Let $F$ be any anti-derivative of $f$ , so $\frac{d}{dx}F(x) = f(x)$ and, by definition, $\int f(x)\,dx = F(x) + C$ . Differentiating both sides:

\frac{d}{dx}\left[\int f(x)\,dx\right] = \frac{d}{dx}[F(x) + C] = f(x) + 0 = f(x)

For the second statement, since $f'(x) = \frac{d}{dx}f(x)$ , integrating $f'(x)$ returns $f(x)$ up to a constant: $\int f'(x)\,dx = f(x) + C$ . The constant appears because the derivative of any constant is zero.

Note

The constant of integration $C$ is essential. Without it, $\int f'(x)\,dx = f(x)$ would be false, since $f(x) + 5$ also has derivative $f'(x)$ .

Property (II): Equivalence of Indefinite Integrals

Two indefinite integrals with the same derivative represent the same family of curves and are therefore equivalent.

Statement: If

\frac{d}{dx}\left[\int f(x)\,dx\right] = \frac{d}{dx}\left[\int g(x)\,dx\right]

then $\int f(x)\,dx$ and $\int g(x)\,dx$ differ only by a constant.

Proof. From the hypothesis,

\frac{d}{dx}\left[\int f(x)\,dx - \int g(x)\,dx\right] = 0

The only functions with zero derivative everywhere are constants, so $\int f(x)\,dx - \int g(x)\,dx = C$ , i.e. $\int f(x)\,dx = \int g(x)\,dx + C$ . The two families of curves are therefore the same set, each curve in one being a curve in the other shifted by a constant.

Important

This justifies writing $\int f(x)\,dx = \int g(x)\,dx$ even when the two sides differ by a constant — the equality is of families of anti-derivatives, not of individual functions.

Property (III): Integral of a Sum

\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx

Proof. By Property (I), the derivative of the left side is $f(x) + g(x)$ . The derivative of the right side is also $\frac{d}{dx}\int f(x)\,dx + \frac{d}{dx}\int g(x)\,dx = f(x) + g(x)$ . Since both sides have the same derivative, by Property (II) they are equivalent.

Property (IV): Constant Multiple

For any real number $k$ ,

\int k\,f(x)\,dx = k\int f(x)\,dx

Proof. By Property (I), the derivative of the left side is $k\,f(x)$ ; the derivative of the right side is $k\cdot\frac{d}{dx}\int f(x)\,dx = k\,f(x)$ . Equal derivatives, so by Property (II) the two sides are equivalent.

Watch out

This holds only when $k$ is a constant. You cannot pull a function of $x$ outside the integral sign.

Property (V): Generalised Linearity

Properties (III) and (IV) extend to any finite number of functions and constants:

\int [k_1 f_1(x) + k_2 f_2(x) + \cdots + k_n f_n(x)]\,dx = k_1\int f_1(x)\,dx + k_2\int f_2(x)\,dx + \cdots + k_n\int f_n(x)\,dx

This follows by repeated application of Properties (III) and (IV).

Tip

This is the workhorse property. Almost every integration problem begins by using Property (V) to break a complicated expression into a sum of simpler integrals from the standard table.

Integration by the Method of Inspection

Finding an anti-derivative is often done by inspection — you look at the given function and ask: "What function has this as its derivative?" This reverses differentiation and relies on your familiarity with derivative formulas.

Finding a Unique Anti-derivative with an Initial Condition …

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