Antiderivatives

Differentiation runs a function through a machine and hands you its slope. An antiderivative runs that machine in reverse: given f, we hunt for a function F whose derivative is f.

F \text{ is an antiderivative of } f \quad\Longleftrightarrow\quad F'(x) = f(x).

For example F(x) = x^2 is an antiderivative of f(x) = 2x, because differentiating x^2 gives 2x. But so is x^2 + 7, and x^2 - \tfrac12 — the constant differentiates away. That little ambiguity is the whole story of this page.

Antiderivatives differ by a constant — line by line

Suppose two functions F and G are both antiderivatives of the same f, so F' = G' = f everywhere on an interval. We will prove they can differ by no more than a constant. The engine is the Mean Value Theorem: on [a, b] there is some interior point c with

\frac{H(b) - H(a)}{b - a} = H'(c).

Step 1 — name the difference. Let H(x) = F(x) - G(x). This new function packages "how far apart the two antiderivatives are."

H(x) = F(x) - G(x).

Step 2 — differentiate it. The derivative is linear (the difference of derivatives is the derivative of the difference), so

H'(x) = F'(x) - G'(x) = f(x) - f(x) = 0 \quad\text{for every } x.

Step 3 — a zero-slope function cannot move. Pick any two points a < b. By the Mean Value Theorem there is a c between them with

\frac{H(b) - H(a)}{b - a} = H'(c) = 0.

Step 4 — solve for the gap. Multiplying both sides by (b - a) \ne 0,

H(b) - H(a) = 0 \cdot (b - a) = 0 \quad\Longrightarrow\quad H(b) = H(a).

Step 5 — read off the conclusion. The points a and b were arbitrary, so H takes the same value everywhere — it is a constant, call it C. Hence

F(x) - G(x) = C \quad\Longrightarrow\quad F(x) = G(x) + C.

So once you know one antiderivative F, every other is F(x) + C for some constant. We call F(x) + C the general antiderivative, and C the constant of integration.

Step 3 quietly assumed F and G live on a single connected interval. Drop that and the theorem fails. The function F(x) = \operatorname{sgn}(x) on the broken domain x \ne 0 has F'(x) = 0 everywhere it is defined, yet it is -1 on the left and +1 on the right — not a single constant. The Mean Value Theorem needs a connecting path between a and b; across a gap there is none.

Let f be defined on an interval I. Then:

Reversing the power rule — line by line

The power rule says differentiating drops the exponent by one and multiplies by the old exponent. To antidifferentiate x^n we run that backwards: raise the exponent by one, and divide by the new exponent to cancel the factor the power rule will produce.

Step 1 — guess the shape. To land on x^n after differentiating, we need an x^{n+1} upstairs. Try F(x) = \dfrac{x^{n+1}}{n+1} (legal only when n \ne -1, so the denominator is non-zero).

Step 2 — differentiate the guess. The constant \tfrac{1}{n+1} rides along, and the power rule handles x^{n+1}:

F'(x) = \frac{1}{n+1}\,\frac{d}{dx}\big(x^{n+1}\big) = \frac{1}{n+1}\,(n+1)\,x^{(n+1)-1}.

Step 3 — cancel. The (n+1) in front meets the \tfrac{1}{n+1} and they annihilate:

F'(x) = \frac{n+1}{n+1}\,x^{n} = x^{n}. \checkmark

Step 4 — restore the constant. By the theorem above, every antiderivative is this one plus a constant. So the reverse power rule reads

\boxed{\;\int x^{n}\,dx = \frac{x^{n+1}}{n+1} + C \qquad (n \ne -1).\;}

For instance x^2 \mapsto \tfrac{x^3}{3} + C and x^{5} \mapsto \tfrac{x^{6}}{6} + C. The lone forbidden case n = -1 — antidifferentiating 1/x — needs a logarithm, which we meet in the indefinite integral.

The family of antiderivatives

Here is f(x) = 2x (faint, straight) together with its antiderivatives F(x) = x^2 + C for several constants C. Slide C and watch the bold parabola ride up and down. Crucially, every member of the family has exactly the same slope at each x — that shared slope is f(x) = 2x. Shifting a curve vertically never changes its steepness, which is precisely why the constant is invisible to differentiation.