Why Antiderivatives Differ By A Constant

Why the Constant Is the Only Difference

When a function has an antiderivative, reverse differentiation does not usually produce just one function. If \(F'(x)=f(x)\), then adding a constant to \(F\) leaves its derivative unchanged:

So every function \(F+C\), where \(C\) is constant, is also an antiderivative of \(f\). But is this the whole story? Could there be two antiderivatives that differ by something more complicated than a constant?

On an interval, the answer is no. The reason is a basic fact about derivatives: if a differentiable function has derivative zero everywhere on an interval, then it is constant there. Subtracting two antiderivatives produces exactly such a function.

Antiderivatives on an Interval

The interval is the domain on which we compare the functions. The definition says that the derivative of \(F\) agrees with \(f\) throughout that domain. The same definition applies to another function \(G\).

If \(F\) and \(G\) are both antiderivatives of \(f\), their derivatives are equal. Subtraction then gives

To conclude that \(F-G\) is constant, we need the zero-derivative property. Its proof uses the Mean Value Theorem.

The Zero-Derivative Property

Here is why the theorem holds. Choose any two points \(a\) and \(b\) in the interval with \(a<b\). The segment \([a,b]\) lies in the interval. The Mean Value Theorem says that if \(H\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then for some \(c\in(a,b)\),

Since \(H'(c)=0\), the right-hand side is zero, so \(H(b)=H(a)\). The two points were arbitrary; therefore, \(H\) takes the same value at every pair of points in the interval and is constant there. If the interval has endpoints, continuity at those endpoints is included in the hypotheses of the Mean Value Theorem.

Why Antiderivatives Differ by a Constant

The conclusion follows by applying the zero-derivative property to their difference. Because \(F\) and \(G\) are differentiable, \(H=F-G\) is differentiable, hence continuous. At every interior point of \(I\), both derivatives equal \(f\), so \(H'(x)=0\). The zero-derivative property then says that \(H\) is constant.

The converse is also important. If \(G\) is an antiderivative of \(f\), then \(G+C\) is another one for every constant \(C\), because the derivative of a constant is zero. Together, the two directions say that all antiderivatives on an interval have the form one antiderivative plus a constant.

Worked Examples

Why the Interval Condition Matters

The theorem does not say that antiderivatives must differ by one constant on every possible domain. The domain being an interval matters because the Mean Value Theorem applies on the segment between any two points in an interval.

For example, let the domain be \(D=(-\infty,0)\cup(0,\infty)\), which is not an interval. Define \(F(x)=0\) for every \(x\in D\), and define \(G(x)=0\) when \(x<0\) and \(G(x)=1\) when \(x>0\). On each part of \(D\), both derivatives are zero. But \(F-G\) equals \(0\) on the negative part and \(-1\) on the positive part, so there is no single constant difference across all of \(D\).

This does not contradict the theorem: \(D\) is not an interval, and there is no segment within \(D\) joining a negative point to a positive point. On each separate interval, the functions differ by a constant. The theorem guarantees one constant across an interval, not across a domain with separated pieces.

A Common Mistake

A frequent error is to claim that because \(H'(x)=0\) at every point, \(H\) must have value zero. Zero derivative means constant, not necessarily the zero function. For example, if \(H(x)=5\), then \(H'(x)=0\) everywhere, but \(H(x)\) is not zero.

The same distinction matters when subtracting antiderivatives. From \((F-G)'=0\), the valid conclusion is \(F-G=C\), where \(C\) may be any real constant. The conclusion \(F-G=0\) would require additional information that the two functions have the same value at some point.