Why the Choice of Antiderivative Does Not Matter
An antiderivative is not unique: if \(F'(x)=f(x)\), then adding a constant to \(F\) leaves its derivative unchanged. This freedom may seem to create a problem when using an antiderivative to evaluate a definite integral. The reason it does not is that the same constant appears at both endpoints and cancels in the difference.
To make that reasoning precise, we need to be careful about the interval on which the functions are defined. In particular, differentiability on the interior of a closed interval alone says nothing about continuity at its endpoints.
Antiderivatives and Their Domains
The open-interval convention avoids endpoint qualifications: every point of \(I\) is an interior point where the derivative is defined. If we instead work on a closed interval \([a,b]\), the appropriate conditions are that \(F\) is continuous on \([a,b]\), differentiable on \((a,b)\), and satisfies \(F'(x)=f(x)\) for every \(x\in(a,b)\). These conditions allow endpoint evaluation and ensure the hypotheses needed for the Mean Value Theorem.
In the Fundamental Theorem of Calculus setting, \(f\) is continuous on \([a,b]\). The uniqueness argument below actually needs less: it only needs two functions with the same derivative. Continuity of \(f\) is important for the standard definite-integral evaluation formula.
Two Antiderivatives Differ by a Constant
The proof uses the Mean Value Theorem, which states that if a function \(H\) is continuous on \([u,v]\) and differentiable on \((u,v)\), then for some \(c\in(u,v)\), \(H(v)-H(u)=H'(c)(v-u)\). The continuity and differentiability conditions are essential; this is why the endpoint conditions have been stated explicitly.
Define \(H(x)=G(x)-F(x)\). On the open interval, \(H\) is differentiable, and \(H'(x)=G'(x)-F'(x)=f(x)-f(x)=0\).
Choose \(u,v\in I\) with \(u<v\). Since \(I\) is an interval, \([u,v]\subset I\). The function \(H\) is continuous on \([u,v]\) and differentiable on \((u,v)\), so the Mean Value Theorem applies.
For some \(c\in(u,v)\), \(H(v)-H(u)=H'(c)(v-u)=0\). Thus \(H(v)=H(u)\). Since the points were arbitrary, \(H\) has one constant value throughout \(I\).
Let that constant value be \(C\). Then \(G(x)-F(x)=C\), or equivalently \(G(x)=F(x)+C\), for every \(x\in I\). The closed-interval case follows by applying the same argument on the segment between any two points of \([a,b]\), using continuity at included endpoints.
This proves the theorem: the difference of two antiderivatives has derivative zero, and the Mean Value Theorem shows that a function with zero derivative on an interval must have the same value at every pair of points in that interval.
Why the Constant Cancels at the Endpoints
Suppose \(f\) is continuous on \([a,b]\), and \(F\) is an antiderivative on an open interval containing \([a,b]\). The Fundamental Theorem of Calculus, Part II, gives \(\int_a^b f(x)\,dx=F(b)-F(a)\). If \(G\) is another antiderivative, the theorem just proved gives \(G(x)=F(x)+C\) throughout the interval. Consequently,
Thus both antiderivatives produce the same endpoint difference and therefore the same definite integral. The cancellation depends on using the same constant at both endpoints. It does not mean that the individual values \(F(a)\) and \(G(a)\), or \(F(b)\) and \(G(b)\), must agree.
Worked Examples
Worked Example: A Polynomial Antiderivative with an Added Constant
Evaluate \(\int_{-1}^{2}(2x+1)\,dx\) using two antiderivatives.
First, \(F(x)=x^2+x\) is an antiderivative because \(F'(x)=2x+1\). Another is \(G(x)=x^2+x+8\), since \(G'(x)=2x+1\) as well. Their difference is \(G(x)-F(x)=8\), the same constant for every \(x\).
Using \(F\), evaluate at the endpoints: \(F(2)=2^2+2=4+2=6\), and \(F(-1)=(-1)^2+(-1)=1-1=0\). Therefore \(F(2)-F(-1)=6-0=6\).
Using \(G\), \(G(2)=4+2+8=14\), and \(G(-1)=1-1+8=8\). Hence \(G(2)-G(-1)=14-8=6\), agreeing with the first calculation.
Check: \(F'(x)=2x+1\) and \(G'(x)=2x+1\), so both are antiderivatives of the integrand. Their endpoint differences both equal \(6\), and the Fundamental Theorem gives \(\int_{-1}^{2}(2x+1)\,dx=6\).
Worked Example: Different Forms of an Antiderivative
Evaluate \(\int_{1}^{e^2}\frac{1}{x}\,dx\) using \(F(x)=\ln x\) and \(G(x)=\ln(3x)\). The interval is positive, so both logarithms are defined throughout it.
Differentiate \(F\): \(F'(x)=1/x\). For \(G\), the chain rule gives \(G'(x)=(1/(3x))\cdot 3=1/x\). Also, \(G(x)=\ln 3+\ln x=F(x)+\ln 3\), so the difference is constant.
Using \(F\), \(F(e^2)=\ln(e^2)=2\) and \(F(1)=\ln 1=0\). Thus \(F(e^2)-F(1)=2-0=2\).
Using \(G\), \(G(e^2)=\ln(3e^2)=\ln 3+2\) and \(G(1)=\ln 3\). Thus \(G(e^2)-G(1)=(\ln 3+2)-\ln 3=2\).
Check: both derivatives equal \(1/x\), and both endpoint differences equal \(2\). Since \(1/x\) is continuous on \([1,e^2]\), the Fundamental Theorem applies and the integral is \(2\).
Worked Example: A Trigonometric Antiderivative
Evaluate \(\int_{0}^{\pi/2}\cos x\,dx\) using \(F(x)=\sin x\) and \(G(x)=\sin x-4\).
Differentiate each function: \(F'(x)=\cos x\), and \(G'(x)=\cos x-0=\cos x\). Their difference is \(G(x)-F(x)=-4\).
Using \(F\), \(F(\pi/2)=\sin(\pi/2)=1\), while \(F(0)=\sin 0=0\). The endpoint difference is \(1-0=1\).
Using \(G\), \(G(\pi/2)=1-4=-3\) and \(G(0)=0-4=-4\). The endpoint difference is \((-3)-(-4)=-3+4=1\).
Check: both derivatives equal the integrand, and both endpoint differences equal \(1\). The integrand is continuous on the interval, so \(\int_{0}^{\pi/2}\cos x\,dx=1\).
A Necessary Endpoint Caution
For an antiderivative defined on an open interval, every point in its domain is an interior point, so the uniqueness proof applies directly. On a closed interval, however, one must not define an antiderivative by requiring only that the derivative equal \(f\) on the interior. That weaker condition permits endpoint values that do not match the limiting values from inside the interval.
Worked Example: Why Endpoint Continuity Is Needed
Consider \([0,1]\), let \(f(x)=0\), and define \(F(x)=0\) for all \(x\in[0,1]\). Define another function by \(G(x)=0\) for \(0\leq x<1\), but \(G(1)=1\).
For every \(x\in(0,1)\), \(F'(x)=0\) and \(G'(x)=0\), so both derivatives equal \(f(x)\) on the interior. But \(G\) is not continuous at \(1\): its values approach \(0\) as \(x\) approaches \(1\) from the left, while \(G(1)=1\).
If one incorrectly used only the interior derivative condition, one might call both functions antiderivatives and conclude their endpoint differences should agree. In fact, \(F(1)-F(0)=0-0=0\), while \(G(1)-G(0)=1-0=1\). They disagree because \(G\) violates the endpoint-continuity condition.
Check: the integral of \(f=0\) over \([0,1]\) is \(0\), and \(F\) gives the correct endpoint difference \(0\). The calculation using \(G\) fails because \(G\) is not continuous on the closed interval and therefore does not satisfy the stated antiderivative conditions there.
A Common Mistake: Cancelling the Constant Incorrectly
A frequent algebra error is to subtract the endpoint values but change the sign of the constant at only one endpoint. If \(G(x)=F(x)+C\), the incorrect calculation might be written as
The error is the replacement of \(G(a)=F(a)+C\) by \(F(a)-C\). The same relation holds at both endpoints. The correct substitution is
Keeping the entire second endpoint expression inside parentheses helps prevent this sign error.
Why This Result Matters
The definite integral is determined by the integrand and the endpoints, not by which antiderivative happens to be easiest to write down. The uniqueness result explains why the Fundamental Theorem's endpoint evaluation is well-defined even though an integrand has infinitely many antiderivatives: once one antiderivative is known, every other one differs from it by a constant, and that constant vanishes in the endpoint difference.
When evaluating an integral, it is therefore reasonable to choose any convenient antiderivative, provided it is defined and satisfies the required derivative condition throughout an interval containing the integration interval. If endpoint values are included in the antiderivative's domain, continuity there must also be in place. These conditions protect the evaluation formula from the boundary failure illustrated above.
Check Your Understanding
Use the definitions and reasoning in this tutorial to answer the following questions.
- If \(F'(x)=f(x)\) and \(G'(x)=f(x)\) on an open interval, what is the derivative of \(G-F\), and what does the Mean Value Theorem imply?
- If \(G(x)=F(x)-6\), explain why \(G(b)-G(a)=F(b)-F(a)\).
- What continuity and differentiability conditions make the uniqueness result valid when working on a closed interval \([a,b]\)?
- Why is differentiability only on \((a,b)\), without continuity at the endpoints, insufficient for endpoint evaluation?
- For \(f(x)=3x^2\), compare the endpoint differences from \(F(x)=x^3\) and \(G(x)=x^3+5\) on \([0,2]\).
