Why an Indefinite Integral Has More Than One Answer
An antiderivative is not usually a single function. If \(F'(x)=f(x)\), then adding any constant to \(F\) produces another function with the same derivative. The notation for an indefinite integral reflects this: it describes the whole family of antiderivatives, not just one representative.
There is also a converse. On an interval, if two differentiable functions have the same derivative, their difference must be constant. This fact explains both why the constant appears and why one constant accounts for every antiderivative on that interval.
Adding a Constant Leaves the Derivative Unchanged
Let \(F\) be an antiderivative of \(f\) on an interval \(I\), and let \(C\) be any real number. The derivative of the constant function \(C\) is zero, so
Thus \(F+C\) is also an antiderivative of \(f\). The value of \(C\) can be positive, negative, or zero; it does not depend on \(x\). All these functions have the same rate of change, even though their output values may differ.
Worked Example: A Family of Antiderivatives
Find a family of functions whose derivative is \(2x\), and check the result.
We know that the derivative of \(x^2\) is \(2x\). Adding a real constant \(C\) gives the candidate family
Differentiate term by term. Since \(C\) is constant, its derivative is zero:
Therefore every function \(x^2+C\), for real \(C\), is an antiderivative of \(2x\). The check is direct: differentiating any member of the family returns \(2x\).
Why the Difference Must Be Constant
Showing that \(F+C\) always works establishes that constants can be added. To show that these are the only differences between antiderivatives, suppose \(F\) and \(G\) both have derivative \(f\) on the same interval \(I\). Consider their difference \(H=F-G\). Its derivative is
Why does a zero derivative force a function to be constant? The Mean Value Theorem supplies the reason. For any two distinct points \(a,b\) in \(I\), the theorem says that a function continuous on \([a,b]\) and differentiable on \((a,b)\) has some \(c\in(a,b)\) such that its change over the interval equals its derivative at \(c\) times the interval length. Here \(H\) meets those conditions because it is differentiable on \(I\).
Let \(a,b\in I\), with \(a<b\). Because \(I\) is an interval, every point between \(a\) and \(b\) is also in \(I\).
There is a point \(c\in(a,b)\) such that \(H(b)-H(a)=H'(c)(b-a)\).
Since \(H'(c)=0\), the equation gives \(H(b)-H(a)=0\), so \(H(b)=H(a)\).
The two points were arbitrary, so \(H\) has the same value at every pair of points in \(I\). Hence \(H\) is constant on \(I\).
This proves the full description of the antiderivatives. If \(F'=f\) on \(I\), adding any constant gives another antiderivative. Conversely, if \(G'=f\) on \(I\), then \(G'=F'\), so the theorem says \(G=F+C\) for one constant \(C\) on that interval.
What the Indefinite Integral Notation Means
When \(f\) has an antiderivative \(F\) on an interval \(I\), the indefinite integral records all antiderivatives there:
This notation has no bounds of integration. It represents a family of functions, and the expression \(+C\) indicates that every real vertical shift is included. It is not an extra operation to perform after finding an antiderivative. It is part of a complete description of all the possible answers.
Worked Example: Comparing Two Antiderivatives
Consider \(F(x)=x^3-2x\) and \(G(x)=x^3-2x+9\). Determine whether they are antiderivatives of the same function, and find their difference.
Differentiate \(F\):
Differentiate \(G\), showing the derivative of each term:
The derivatives match, so both functions are antiderivatives of \(3x^2-2\). Their difference is
The difference is the constant \(9\), as the theorem predicts. Checking by differentiation, \(\frac{d}{dx}(G-F)=\frac{d}{dx}(9)=0\), consistent with \(G'=F'\).
Why the Interval Requirement Matters
The theorem guarantees one constant across an interval. If a domain has separate pieces with a gap between them, the values on those pieces need not be linked by one constant. There is no path within the domain connecting points across the gap, and the Mean Value Theorem cannot be applied across it.
Worked Example: Separate Constants on Separate Intervals
Let the domain be \(D=(-2,-1)\cup(1,2)\). Define \(H\) by
On the first interval, \(H\) is constant at \(0\), so \(H'(x)=0\). On the second interval, \(H\) is constant at \(5\), so \(H'(x)=0\) there as well. Thus \(H'(x)=0\) everywhere in its domain.
But \(H\) is not one constant on all of \(D\): for example, \(H(-\frac{3}{2})=0\), while \(H(\frac{3}{2})=5\). There is no contradiction with the theorem, because \(D\) is not an interval. Instead, \(H\) is constant separately on each component interval, and those constants may differ.
The check is direct: each chosen point belongs to the stated component, and the definition gives the two different values. The derivative is zero at every point of \(D\), but no single constant describes \(H\) on the entire disconnected domain.
A Common Mistake: Leaving Out the Constant
A common incomplete answer is to write \(\int 2x\,dx=x^2\). This does give one antiderivative, since differentiating \(x^2\) returns \(2x\). However, it leaves out every other antiderivative of \(2x\), such as \(x^2+4\) and \(x^2-11\).
The correction is to include the arbitrary constant:
Differentiating confirms the whole family:
Another error is to treat \(C\) as a quantity that changes with \(x\). If it depended on \(x\), it would not be a constant, and its derivative would not generally be zero. The notation \(C\) means one fixed real number for the entire interval under consideration.
Why the Constant Matters
Antiderivatives describe functions from their rates of change. A derivative records how a function changes, but it does not record the function's vertical position. Shifting a graph vertically changes its output values while leaving every derivative value unchanged. The constant of integration accounts exactly for that information the derivative does not determine.
Consequently, derivative information alone identifies an antiderivative only up to a constant on an interval. Additional information about a function's value at a point can select one member of the family, but without such information all members must be represented.
Check Your Understanding
Use the definition of an antiderivative and the relationship between equal derivatives and constant differences.
- If \(F'(x)=f(x)\), explain why \(F(x)-6\) is also an antiderivative of \(f\).
- Two differentiable functions \(P\) and \(Q\) have equal derivatives on an interval. What can you conclude about \(P-Q\), and why?
- Why does the Mean Value Theorem apply to the difference of two antiderivatives between any two points of the same interval?
- Can a function have derivative zero everywhere on a disconnected domain without being one constant on the whole domain? Explain.
- What information does the \(+C\) represent in \(\int f(x)\,dx=F(x)+C\)?
