Integration as inverse process of differentiation

Anti-derivatives undo derivatives. If $\dfrac{d}{dx}F(x) = f(x)$ for every $x$ in some interval, $F$ is called an anti-derivative (or primitive) of $f$. Finding $F$ from $f$ is integration — the operation exactly reverses differentiation, the way "add 5" reverses "subtract 5."

The answer is never one function — it is a family. If $F(x)$ is an anti-derivative of $f(x)$, then so is $F(x) + C$ for any constant $C$, since $\dfrac{d}{dx}[F(x)+C] = f(x) + 0 = f(x)$; differentiation always erases an additive constant, so integration can never recover it. This gives the standard notation for the indefinite integral:

where: $f(x)$ is the integrand, $x$ is the variable of integration, $dx$ marks integration with respect to $x$, $F$ is any one anti-derivative, and $C$ is the arbitrary constant of integration.

Geometrically, $F(x) + C$ is a stack of parallel curves. Every value of $C$ shifts the graph of $F$ straight up or down without changing its shape, so at any fixed $x$ every curve in the family has the identical slope $f(x)$ — they are vertical translates of one another, never crossing.

Every derivative you know reads backwards into an integral. Since $\dfrac{d}{dx}\left(\dfrac{x^{n+1}}{n+1}\right) = x^n$ for $n \neq -1$, reversing gives $\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$. Likewise $\dfrac{d}{dx}(\sin x) = \cos x$ reversed gives $\int \cos x\,dx = \sin x + C$, and $\dfrac{d}{dx}(e^x) = e^x$ gives $\int e^x\,dx = e^x + C$. This backwards habit rebuilds the entire standard-integrals table on the spot, without rote memorisation, whenever a result is forgotten.

The "same interval" fine print. The uniqueness-up-to-a-constant result needs a connected domain. For $f(x) = 1/x^2$ (defined for all $x \neq 0$, really two separate pieces $(-\infty,0)$ and $(0,\infty)$), the constant is allowed to differ across the two pieces — the theorem is stated "on an interval," not "on a domain," for exactly this reason, though it rarely changes a Class 12 exam answer.

Verify that $F(x) = x^3 - 4x + 7$ is an anti-derivative of $f(x) = 3x^2 - 4$, write the general indefinite integral, and then find the one particular curve in that family passing through the point $(0, 7)$.

Step 1. Differentiate $F(x)$ to check it against $f(x)$: $F'(x) = \dfrac{d}{dx}(x^3 - 4x + 7) = 3x^2 - 4 - 0 = 3x^2 - 4 = f(x)$. Since $F'(x) = f(x)$, $F$ is indeed an anti-derivative of $f$.

Step 2. Write the general family. The constant of integration is arbitrary, so the full indefinite integral is $\int (3x^2 - 4)\,dx = x^3 - 4x + C$, where $C$ is any real number — the specific value $7$ used in $F(x)$ above is just one member of this family, chosen for a reason that Step 3 makes explicit.

Step 3. Pin down the one curve through $(0,7)$. Substitute $x=0, y=7$ into $y = x^3 - 4x + C$: $7 = 0^3 - 4(0) + C$, so $C = 7$. Answer: the general integral is $x^3 - 4x + C$, and the particular anti-derivative through $(0,7)$ is $x^3 - 4x + 7$ — recovering the very $F(x)$ we started with, and showing exactly how one extra condition (a point on the curve) selects a single member from an infinite family.

Competitive stretch

Find $\int \dfrac{d}{dx}\big(\sin^2 x\big)\,dx$ and explain why the answer is $\sin^2 x + C$, not merely $\sin^2 x$. By the very definition of integration as the inverse of differentiation, integrating the derivative of a function returns the function itself up to an arbitrary constant: $\int \frac{d}{dx}(\sin^2 x)\,dx = \sin^2 x + C$. As a check, $\frac{d}{dx}(\sin^2 x) = 2\sin x \cos x = \sin 2x$, so the question is also asking for $\int \sin 2x\,dx = -\frac{\cos 2x}{2} + C$; using $\cos 2x = 1 - 2\sin^2 x$ this equals $\sin^2 x - \frac12 + C = \sin^2 x + C'$, the same family with the constant simply relabelled.

• Writing an anti-derivative without the "+C" — for an indefinite integral this is treated as an incomplete, penalised answer.
• Assuming there is only one correct-looking anti-derivative; $\sin^2 x + C$ and $-\cos^2 x + C'$ can both be correct if they differ by a constant — check by differentiating, not by eyeballing the algebra.
• Confusing "an anti-derivative" (one specific $F$) with "the indefinite integral" (the whole family $F(x)+C$) — the symbol $\int f(x)\,dx$ always denotes the family.
• Forgetting the anti-derivative relationship must hold on a single interval; the "differ only by a constant" theorem assumes a connected domain, so a broken domain can carry different constants on each piece.

Ask the tutor: if you are still unsure why "+C" is unavoidable or want to practice turning more differentiation formulas backwards into integrals, ask the AI tutor to walk through extra examples with you.