Antiderivatives

The central idea

An antiderivative of a function $f$ on an interval $I$ is a differentiable function $F$ satisfying $F'(x)=f(x)$ for every $x\in I$. The notation $\int f(x)\,dx=F(x)+C$ represents the entire family, not one function. On an interval, any two antiderivatives of the same $f$ differ by a constant.

Definitions, hypotheses, and notation

The interval in the definition matters. The function $1/x$ has $\ln x+C$ as an antiderivative on $(0,\infty)$ and $\ln(-x)+C$ on $(-\infty,0)$; constants chosen on the two disconnected pieces need not agree. For a continuous integrand on one interval, however, the constant-difference theorem follows from the Mean Value Theorem: if $F'=G'$, then $(F-G)'=0$, so $F-G$ is constant. This explains both the $+C$ convention and its scope.

Antiderivatives also encode qualitative information. If $f\ge 0$, every antiderivative of $f$ is nondecreasing, and if $f>0$ throughout, it is strictly increasing. If $f$ is differentiable and nondecreasing, then $f'\ge 0$, so $F''=f'\ge 0$ and every antiderivative is convex. Initial-value problems use one point to set vertical position, while definite integrals use endpoint subtraction so the arbitrary constant disappears. These are two different ways of resolving the same family of vertically translated solutions.

A dependable method and decision rule

1. Rewrite roots and reciprocals as powers when that makes the power rule visible.
2. Integrate term by term, using $\int x^n dx=x^{n+1}/(n+1)$ only when $n\ne-1$.
3. Use $\int x^{-1}dx=\ln|x|+C$ for the exceptional exponent $-1$.
4. Attach one arbitrary constant $C$ after integrating the whole expression.
5. Differentiate the result; then use any initial value to determine $C$.

Fully worked example

Common mistakes and why they fail

A shape check before solving

An antiderivative carries geometric information as well as algebra. If $F'=f$ and $f\ge0$ on an interval, then $F$ is nondecreasing there; if $f>0$ throughout, $F$ is strictly increasing. If $f$ is differentiable and $f'\ge0$, then $F''=f'\ge0$, so $F$ is convex. These checks do not determine the additive constant: an initial value such as $F(a)=b$ still selects one curve from the family $F+C$. Predict sign and rough shape before integrating, then differentiate the candidate and compare those predictions. This catches sign errors, especially when $f$ changes sign or has a zero where strict increase cannot be asserted.