Calculus I · Adding Up the Infinite

The other face: the antiderivative

Now set area aside entirely and ask a question that sounds unrelated. Differentiation turns $F$ into $F'$. Can we run it backward — given a function $f$, find a function $F$ whose derivative is $f$? Such an $F$ is called an antiderivative of $f$.

For example, an antiderivative of $2x$ is $x^2$, because $\tfrac{d}{dx}x^2 = 2x$. But so is $x^2 + 1$, and $x^2 - 7$ — adding any constant leaves the derivative unchanged. This is not sloppiness; it is the whole truth, and Chapter 6 already proved it: if two functions have the same derivative, they differ by a constant (because $F' = G'$ means $(F-G)' = 0$, and a function with zero derivative everywhere is constant — a consequence of the Mean Value Theorem). So antiderivatives come in families, all parallel, separated by a constant:

The indefinite integral. $\displaystyle\int f(x)\,dx = F(x) + C$, where $F' = f$ and $C$ is an arbitrary constant — the whole family of antiderivatives at once.

(Notice the same symbol $\int$ now appears without limits $a$ and $b$. That collision of notation is a deliberate hint: the area problem and the antiderivative problem are about to turn out to be the same problem. For now, treat them as separate — definite integral with limits is a number, indefinite integral without limits is a family of functions.)

The standard antiderivative library

Finding antiderivatives is just running the Chapter 4 rules backward. Every derivative you learned becomes an antiderivative read right to left:

• Power rule, reversed: $\displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ (for $n \ne -1$). Check by differentiating the answer. (Raise the power by one and divide by the new power — the opposite of “drop the power and subtract one.”)
• The exception, $n = -1$: $\displaystyle\int \frac1x\,dx = \ln|x| + C$. The power rule would divide by zero here, so this one stands apart — and it is no accident, as Chapter 11’s capstone will reveal.
• Exponential: $\displaystyle\int e^x\,dx = e^x + C$ (it is its own antiderivative, just as it’s its own derivative).
• Trigonometric — now within reach, because Chapter 4 derived the matching derivatives: since $\tfrac{d}{dx}\sin x = \cos x$, $$\int \cos x\,dx = \sin x + C, \qquad \int \sin x\,dx = -\cos x + C, \qquad \int \sec^2 x\,dx = \tan x + C.$$ (Mind the minus sign on $\int\sin$: the antiderivative of sine is negative cosine, because the derivative of cosine carries the minus.)
• Linearity, as for definite integrals: antiderivative of a sum is the sum of antiderivatives, and constants pull out.

So $\displaystyle\int (3x^2 - 4\cos x + e^x)\,dx = x^3 - 4\sin x + e^x + C$ — reverse each term, attach one $C$ at the end. Differentiate it back to check, and you recover the integrand exactly.

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