Why Checking the Hypotheses Matters
The Fundamental Theorem of Calculus (FTC) connects derivatives and definite integrals. Its formulas are powerful, but each has conditions. Before using one, identify which part you need and check that its hypotheses hold on the relevant interval or at the relevant point.
The two parts address different questions. Part I concerns the derivative of an accumulated integral. Part II evaluates an integral using an antiderivative. Continuity of the integrand is a standard sufficient condition for both, but the precise requirements are not identical.
The Two Standard Forms
For Part I, let \(f\) be continuous at an interior point \(x\) of an interval, and define the accumulation function \(G\) by integrating from a fixed point \(a\) to a variable endpoint. Then \(G\) is differentiable at \(x\), and its derivative is the value of the integrand there.
Continuity of \(f\) throughout an interval ensures this derivative formula at every interior point of that interval. At an endpoint, the corresponding conclusion is a one-sided derivative. The integration variable \(t\) is a dummy variable; \(x\) is the endpoint and the input of \(G\).
For Part II, a standard version says that if \(f\) is continuous on \([a,b]\), and \(F\) is an antiderivative of \(f\) with the required endpoint continuity, then the definite integral is the difference of endpoint values.
Here \(F\in C([a,b])\) means that \(F\) is continuous on the closed interval, including at both endpoints. This condition matters: differentiability on the interior and merely assigning values at the endpoints are not enough. The usual convenient hypothesis is that \(F\) is differentiable on an interval containing \([a,b]\); then its continuity on \([a,b]\) is automatic.
Why Continuity Gives the Part I Formula
The key idea is that the average value of a continuous function over a sufficiently short interval near \(x\) is close to \(f(x)\). The difference quotient for \(G\) is exactly such an average.
For \(h\ne 0\), additivity of definite integrals gives \(G(x+h)-G(x)=\int_x^{x+h} f(t)\,dt\). Thus the quotient is this integral divided by \(h\).
Since \(\int_x^{x+h} f(x)\,dt=h f(x)\), the difference between the quotient and \(f(x)\) is \(\frac{1}{h}\int_x^{x+h}(f(t)-f(x))\,dt\).
Its absolute value is at most \(\sup_{t\text{ between }x\text{ and }x+h}|f(t)-f(x)|\). As \(h\) tends to zero, continuity at \(x\) makes this bound tend to zero.
Therefore the difference quotient tends to \(f(x)\), which proves \(G'(x)=f(x)\). The argument works for positive and negative \(h\): the absolute value of the interval length divided by \(|h|\) is \(1\). Continuity is sufficient for the result, though it is not necessary in every individual example.
Worked Example: A Continuous Integrand
Worked Example: A Continuous Integrand
Let \(f(t)=t^2+1\) on \([0,2]\), and define \(G(x)=\int_0^x(t^2+1)\,dt\). Check the hypotheses and find \(G'(1)\) and \(\int_0^2 f(t)\,dt\).
The polynomial \(t^2+1\) is continuous on \([0,2]\). Part I therefore gives \(G'(x)=f(x)\) for interior \(x\), so \(G'(1)=1^2+1=2\).
For Part II, an antiderivative is \(F(x)=\frac{x^3}{3}+x\). It is continuous on \([0,2]\) and differentiable in the interior, with \(F'(x)=x^2+1=f(x)\). Hence
Check: differentiating \(F\) gives \(F'(x)=x^2+1\), the integrand. Thus both the Part I derivative and the Part II evaluation are justified.
Piecewise-Continuous Integrands
A function is piecewise continuous on a closed interval if the interval can be divided into finitely many subintervals on whose interiors the function is continuous, with finite one-sided limits at the division points. Such functions are Riemann integrable. At points where \(f\) is continuous, Part I applies. At a jump, it need not apply: the accumulated integral can have different left- and right-hand derivatives.
For Part II, a jump means that the standard continuous-integrand version cannot be applied directly across the whole interval. Often the right approach is to split the integral at the jump. On each subinterval, use the continuous extension of the relevant piece and its antiderivative. A single point has no effect on a Riemann integral, but the one-sided pieces and their endpoint values must still be handled correctly.
Worked Example: A Jump in the Integrand
Worked Example: A Jump in the Integrand
Define \(f(x)=1\) for \(-1\le x<0\) and \(f(x)=3\) for \(0\le x\le1\). Let \(G(x)=\int_{-1}^x f(t)\,dt\). Check Part I at \(x=-\frac12\) and \(x=0\), then find \(\int_{-1}^1 f(x)\,dx\).
The function is continuous at \(-\frac12\), so Part I applies there and \(G'(-\frac12)=f(-\frac12)=1\). At \(0\), the left-hand values of \(f\) are \(1\) and the right-hand values are \(3\). The difference quotient for \(G\) from the left tends to \(1\), while the one from the right tends to \(3\). Since these are unequal, \(G'(0)\) does not exist.
For the integral, split at the jump. The value assigned at the single point \(0\) does not change either integral. The continuous extensions on the two pieces are constant functions, with antiderivatives \(F_1(x)=x\) and \(F_2(x)=3x\), respectively. Apply the endpoint formula to each piece:
Check: the first interval has length \(1\) and height \(1\), giving \(1\); the second has length \(1\) and height \(3\), giving \(3\). Their sum is \(4\). The computation used the continuous pieces separately, not Part II as if \(f\) were continuous across the jump.
Discontinuous Functions: Check What Fails
A discontinuous integrand requires more careful diagnosis. A discontinuity at one point does not automatically prevent integrability, and an integral may still be computed. But Part I’s continuity hypothesis does not identify the derivative at that point, and Part II cannot be invoked using a function that is not an antiderivative there.
Worked Example: A Removable Discontinuity
On \([0,1]\), let \(f(x)=1\) when \(x\ne\frac12\), and let \(f(\frac12)=5\). Define \(G(x)=\int_0^x f(t)\,dt\). Does Part I give \(G'(\frac12)=f(\frac12)\)? Can \(F(x)=x\) be used directly as an antiderivative in Part II?
The function is discontinuous at \(\frac12\), so the continuity hypothesis for Part I is absent there. Directly, for small \(h\) with \(\frac12+h\in[0,1]\), changing the value at one point does not change the integral, and \(G(\frac12+h)-G(\frac12)=h\). Therefore \(G'(\frac12)=1\), not \(f(\frac12)=5\).
The integral over the full interval is \(1\), since \(f=1\) except at one point. Although \(F(x)=x\) has the correct derivative wherever \(x\ne\frac12\), it is not an antiderivative of \(f\) on the whole interior: \(F'(\frac12)=1\ne5\). So the stated Part II hypotheses do not hold for this \(F\) and \(f\). One may instead integrate the continuous representative that equals \(1\) everywhere, since changing a bounded function at one point does not change its Riemann integral. That gives \([x]_0^1=1\), as required.
Check: the derivative of \(F(x)=x\) is \(1\), and its endpoint difference is \(1-0=1\). The numerical answer is correct, but the direct claim that \(F'=f\) throughout the interior would be incorrect.
Not every discontinuous function is Riemann integrable. For example, the function on \([0,1]\) that equals \(1\) at rational inputs and \(0\) at irrational inputs is discontinuous everywhere and is not Riemann integrable. In the usual Riemann-integral form of the FTC, there is no definite integral to which the theorem can be applied. This is a different issue from a finite number of removable discontinuities or jumps.
A Common Antiderivative Pitfall
For Part II, it is not enough to check \(F'=f\) only in the open interval and then assign arbitrary endpoint values to \(F\). The values at the endpoints determine the difference \(F(b)-F(a)\), so they must agree with the continuous extension of the antiderivative.
Worked Example: Endpoint Continuity Is Essential
Let \(f(x)=1\) on \([0,1]\). Define \(F(x)=x\) for \(0<x\le1\), but set \(F(0)=7\). Then \(F'(x)=1=f(x)\) for every interior \(x\), but \(F\) is not continuous at \(0\). The proposed endpoint calculation gives
Thus the endpoint formula fails for this \(F\). Correct the antiderivative by using the continuous extension \(F(x)=x\) on all of \([0,1]\). Now \(F(0)=0\), so \(F(1)-F(0)=1\), matching the integral. Differentiation confirms \(F'(x)=1\). The correction is to require continuity on the closed interval, not merely defined endpoint values.
A Practical Hypothesis Check
Before using either part, work through the appropriate checks. These questions help prevent applying a valid formula outside its stated conditions.
Are you differentiating an accumulated integral (Part I), or evaluating an integral using an antiderivative (Part II)?
For Part I, is \(f\) continuous at the endpoint value where the derivative is claimed? For Part II’s standard form, is \(f\) continuous on the full closed interval?
For Part II, verify \(F'=f\) in the interior and ensure that \(F\) is continuous on the closed interval, including the endpoints.
If \(f\) is piecewise continuous, inspect the jump points and split the integral where needed. If it is more severely discontinuous, first check whether the integral exists in the framework being used.
These conditions are practical safeguards, not a claim that every theorem has only one possible hypothesis set. Some broader versions exist. For a first application, however, use the standard hypotheses stated here unless a stronger result has been established and its conditions checked.
Check Your Understanding
For each question, identify which FTC hypotheses are satisfied and explain what conclusion is justified.
- If \(f\) is continuous at \(c\), what does Part I say about the derivative of \(G(x)=\int_a^x f(t)\,dt\) at \(c\)?
- A piecewise-constant function has a jump at \(c\). What are the left- and right-hand derivative values of its accumulation function when the two constant values differ?
- Why must an antiderivative used in Part II be continuous at both endpoints?
- A bounded integrand differs from a continuous function at only one point. Does that change its Riemann integral? Does it ensure Part I gives the derivative at that point?
- What must be checked before applying the standard Part II formula to a piecewise-continuous integrand with a jump?
