Questions: Fundamental Theorem of Calculus (Rigorous)

Question 1

Let f be Riemann integrable but discontinuous on [0,1], with a jump at x = 1/2 (f(x) = 0 for x < 1/2, f(x) = 1 for x ≥ 1/2). Define F(x) = ∫₀ˣ f(t) dt. Which statement is correct?

• A. F'(x) = f(x) for all x ∈ [0,1] by FTC Part 1
• B. F is differentiable at every point where f is continuous; at x = 1/2, F may not be differentiable (or F'(1/2) may not equal f(1/2))
• C. F is not well-defined because f is not continuous
• D. F'(x) = f(x) everywhere because integration smooths out the discontinuity

Answer: (B)

FTC Part 1 requires f to be *continuous* at x to conclude F'(x) = f(x). The proof works by showing f stays near f(x) on a short interval — exactly what continuity guarantees. At a jump discontinuity x = 1/2, f makes a sudden jump no matter how small the interval, breaking the proof. F is still well-defined (f is integrable) and in fact continuous — but need not be differentiable at the jump point, or if it is, the derivative need not equal f(1/2). Option D is the common misconception: continuity is required, not bypassed by integration.

Question 2

FTC Part 2 states that ∫ₐᵇ F'(x) dx = F(b) − F(a). Which hypothesis is required by Part 2 but NOT by Part 1?

• A. f = F' must be continuous on [a,b]
• B. F must be a known antiderivative — Part 2 starts from a given F with F' = f, rather than constructing F from the integral
• C. The interval [a,b] must be bounded
• D. F must be differentiable at the endpoints a and b

Answer: (B)

Parts 1 and 2 answer different questions. Part 1 constructs an antiderivative: given integrable f, define F(x) = ∫ₐˣ f(t)dt and prove F' = f when f is continuous. Part 2 evaluates an integral: given a function F that you already know is an antiderivative of f (with F' integrable), conclude ∫ₐᵇ f = F(b) − F(a). Part 2 has *weaker* hypotheses for f (integrability suffices; continuity is not required) but presupposes you have already found an antiderivative F. The two parts together link the analytic construction of antiderivatives to the geometric computation of area.

Question 3

FTC Part 1 proves that every continuous function on [a,b] has an antiderivative, constructed explicitly as the accumulation function F(x) = ∫ₐˣ f(t) dt.

• T. True
• F. False

Answer: True

This is precisely what Part 1 establishes: if f is continuous, then F(x) = ∫_a^x f(t)dt is differentiable and F'(x) = f(x). In other words, F is an explicit antiderivative. This is a non-trivial existence theorem: it guarantees that continuous functions always have antiderivatives, even when no closed-form expression exists. The antiderivative need not be expressible in elementary functions — but existence is guaranteed by the integral construction.

Question 4

FTC Part 2 requires f to be continuous — the same hypothesis as Part 1 — because the evaluation formula F(b) − F(a) breaks down without continuity.

• T. True
• F. False

Answer: False

Part 2 has strictly weaker hypotheses than Part 1. Part 1 requires continuity of f (to ensure the accumulation function is differentiable at each point). Part 2 only requires that F is a continuous antiderivative of f and that F' is Riemann integrable — f itself need not be continuous. The proof of Part 2 uses the MVT applied to F and the accumulation function G, not the continuity of f. The two parts have different hypotheses, different proofs, and serve different purposes.

Question 5

Explain intuitively why Part 1 of the FTC requires continuity of f, and what breaks down in the proof if f has a jump discontinuity at x.

Model answer: Part 1 proves (F(x+h)−F(x))/h → f(x) by showing the average value of f on [x, x+h] approaches f(x) as h → 0. This relies on f staying near f(x) on a shrinking interval — precisely what continuity guarantees. If f has a jump discontinuity at x, then on any interval [x, x+h], f takes values far from f(x) on a portion that doesn't shrink away. The mean value of f on [x, x+h] does not converge to f(x), so the difference quotient fails to converge to f(x). Continuity is not just a convenient assumption — it is exactly the hypothesis that makes the average-value argument work.

The precise statement is: for any ε > 0, continuity gives δ such that |f(t) − f(x)| < ε for |t − x| < δ. This bounds |(F(x+h)−F(x))/h − f(x)| < ε. At a jump, no such δ exists, so no such bound can be made, and the limit may fail or yield the wrong value.