Checking FTC Results by Differentiation

Why Check an FTC Result?

A result involving the Fundamental Theorem of Calculus can often be checked with two direct tests. For an accumulation function, differentiate the proposed formula and compare the result with the integrand. For a definite integral, differentiate the proposed antiderivative and check that its endpoint values have been substituted in the correct order.

These checks are useful because the most common errors are small but consequential: a missing constant in an accumulation function, an incorrectly differentiated expression, or subtraction of the endpoint values in the wrong order. A check does not replace the theorem or its hypotheses. It helps confirm that the stated calculation is consistent with them.

Checking an Accumulation Function by Differentiation

Let \(f\) be continuous on an interval, fix a basepoint \(a\), and define the accumulation function \(A\) by \(A(x)=\int_a^x f(t)\,dt\). FTC Part I says that, at an interior point \(x\), \(A'(x)=f(x)\). The variable \(t\) is the integration variable; \(x\) is the endpoint and the input of \(A\).

Suppose a formula \(G(x)\) is proposed for \(A(x)\). Differentiating \(G\) tests whether its rate of change is the integrand. There is one more check: \(A(a)=0\), because an integral with equal endpoints is zero. Thus the proposed formula should also satisfy \(G(a)=0\).

Why check both? If two functions have equal derivatives on an interval, their difference has derivative zero there and so is constant, by the Mean Value Theorem. Derivative agreement alone does not determine that constant. Matching the basepoint value fixes it.

Why FTC Part I Gives the Derivative

The derivative rule for an accumulation function follows from the definition of derivative and continuity of the integrand. This short argument explains why differentiating a proposed formula is a meaningful check.

Fix an interior point \(x\), and let \(A(u)=\int_a^u f(t)\,dt\). For nonzero \(h\) sufficiently small, additivity of definite integrals gives \(A(x+h)-A(x)=\int_x^{x+h}f(t)\,dt\). Therefore,

Taking absolute values, the absolute value of the right-hand side is at most the largest value of \(|f(t)-f(x)|\) between \(x\) and \(x+h\). More precisely,

This bound also holds when \(h<0\): reversing the integral's orientation changes its sign, while taking the absolute value removes that sign. Since \(f\) is continuous at \(x\), the supremum on the right tends to zero as \(h\) tends to zero. The difference quotient therefore tends to \(f(x)\), proving \(A'(x)=f(x)\).

Checking a Definite-Integral Evaluation

FTC Part II evaluates a definite integral using an antiderivative, but its conditions matter. If \(f\) is continuous on \([a,b]\), and \(F\) is continuous on \([a,b]\), differentiable on \((a,b)\), and satisfies \(F'(x)=f(x)\) for every \(x\) in \((a,b)\), then

In particular, it is sufficient that \(F\) be differentiable on an open interval containing \([a,b]\) and satisfy \(F'=f\) there. Continuity of \(F\) at the endpoints is part of the conditions when only differentiability on the interior is specified.

To check an evaluation, first differentiate the stated \(F\) and verify that it gives \(f\). Then evaluate \(F\) separately at \(b\) and \(a\), and subtract in that order. The notation \([F(x)]_a^b\) means \(F(b)-F(a)\), not \(F(a)-F(b)\).

Constants and Endpoint Signs

An antiderivative is not unique: if \(F'=f\), then \((F+C)'=f\) for any constant \(C\). For a definite integral, this constant cancels because \((F(b)+C)-(F(a)+C)=F(b)-F(a)\). For an accumulation formula, however, the basepoint value must still be checked. The formula with the wrong constant can have the correct derivative and still fail to equal the accumulation function.

A Common Subtraction Error

A frequent mistake is to reverse the endpoint subtraction while keeping the bounds in their original order. For the integral from \(1\) to \(3\) in the earlier example, an incorrect calculation would be

This is not the value of \(\int_1^3(2x-1)\,dx\), because the upper endpoint is \(3\), so the formula requires \(F(3)-F(1)\). The corrected calculation is \(6-0=6\). The negative value \(-6\) would instead correspond to the reversed integral \(\int_3^1(2x-1)\,dx\), since reversing bounds changes the sign.

Another error is to check only the derivative of a proposed accumulation formula. For example, \(G(x)=x^3-2x^2+x\) has the same derivative as the correct formula in the first worked example, but \(G(2)=2\), not \(0\). It is therefore not the accumulation function based at \(2\). Subtracting \(2\) gives the correct formula and the required basepoint value.

A Reliable Checking Routine

Use these checks in order. Keeping derivative verification separate from endpoint arithmetic makes it easier to locate an error.

What the Checks Establish

Differentiation checks the local rate of change, while endpoint substitution checks the net change over an interval. For an accumulation function, the derivative test and basepoint value together identify the proposed formula. For a definite integral, a correct antiderivative and correctly ordered endpoint values verify the FTC calculation.

These checks are not a substitute for verifying that the theorem applies. Nor do they justify a proposed formula if its derivative or endpoint values have not been checked. Their value is that they turn a result into a short sequence of specific, inspectable claims.