Integral properties and the fundamental theorem

Try

Let F(x)=∫₀ˣ(1+t)dt. From elementary sum calculations, F(x)=x+x²/2, so F′(x)=1+x. The derivative of accumulated quantity recovers the original density. The fundamental theorem generalizes this relationship, but it has two logically distinct parts: differentiating an accumulation function and evaluating an integral using an already known primitive. Neither part should be treated as the definition of integration after Darboux sums have supplied that definition.

Predict

Predict what happens if the density jumps. For f(t)=0 when t<0 and f(t)=1 when t≥0, accumulation from −1 is zero to the left of zero and equals x to the right. It is continuous at zero but has different one-sided derivatives there. Integrability alone guarantees useful accumulation control, but not the derivative equality at every point. Continuity of the integrand at the point supplies the missing condition.

Experiment

Keep the roles separate.

Object	Definition	Relevant conclusion
f	Integrable density	Has an integral
F(x)	Accumulation from a fixed endpoint	Continuous when f is bounded
G	A primitive with G′=f	Endpoint difference evaluates the integral under suitable hypotheses

For bounded |f|≤M, derive |F(y)−F(x)|≤M|y−x|. This bound proves accumulation is Lipschitz continuous even at discontinuities of f. Explain why it does not assert differentiability everywhere.

Understand

Integral monotonicity follows from lower and upper sums: if f≤g and both are integrable, then ∫f≤∫g. Linearity follows by comparing sums or approximating both functions on a common refinement; additivity across a split point follows by including that point in the partition. These properties give |∫ₓʸf|≤M|y−x| for bounded f, with orientation handled when y<x.

Now suppose f is continuous at an interior point c. Subtract f(c) from the accumulation difference quotient: (F(c+h)−F(c))/h−f(c)=(1/h)∫ from c to c+h of (f(t)−f(c)). For sufficiently small |h|, continuity makes the integrand’s absolute value less than ε throughout the joining interval. The integral bound makes the absolute quotient error at most ε. The same estimate works for negative h after reversing integration limits. Hence F′(c)=f(c).

Apply

For continuous f on [a,b], suppose G is continuous there and differentiable inside with G′=f. The accumulation F has the same derivative inside, so G−F has zero derivative. MVT makes it constant on the interval, including endpoints by continuity. Since F(a)=0 and F(b)=∫ₐᵇf, subtraction gives ∫ₐᵇf=G(b)−G(a).

Repair “an antiderivative formula permits integration across any singularity.” The theorem’s hypotheses concern the entire interval. The primitive −1/x for 1/x² cannot be evaluated across zero to produce a finite ordinary integral. That singular endpoint requires a separate limit process, and the next lesson will show it diverges.

Substitution: orientation is part of the theorem

A path can visit the same density twice while travelling in opposite directions. Before calculating, predict whether these visits cancel. In this inspector f(u)=1+a u² is positive, and all variables are dimensionless teaching quantities.

Which integral did the substitution preserve?

Parameter map φ(t) 2t+t² on[0,1]t² on[−1,1]t³−3t on[−2,2] Density f(u)=1+a u², choose a 012

Predict the signed result from the map's first and last values, then compare absolute traversal and image-once integration. This density is positive on every branch.

Parameter interval	Image traversal	Direction	Signed contribution	With |φ′|
-1 to 0	1 to 0	decreasing	-4/3	4/3
0 to 1	0 to 1	increasing	4/3	4/3

Signed ∫f(φ(t))φ′(t)dt: 0. With |φ′|: 8/3. Image once, from 0 to 1: 4/3.

Omitting the derivative factor gives ∫f(φ(t))dt=12/5, a different quantity.

Reset

All displayed fractions come from exact polynomial primitives on at most three analytically verified monotone branches. Critical endpoints have derivative zero but cause no singularity in this forward formula. Controls are temporary and do not award mastery.

For the default map φ(t)=t² on [−1,1], with a=1, the signed integral is zero: the image travels from 1 to 0 and back to 1. Integrating with |φ′| instead gives 8/3, twice the image-once integral 4/3. Omitting the derivative gives 12/5, a different integral with respect to parameter length.

Prove the forward identity

Suppose f is continuous on an open interval J containing the image of a continuously differentiable map φ on [A,B]. Fix u₀ in J and define F(u)=∫ from u₀ to u of f(v)dv. The differentiation part of FTC gives F′=f. The chain rule gives (F∘φ)′(t)=f(φ(t))φ′(t); this derivative is continuous. The evaluation part of FTC therefore gives

∫ from A to B of f(φ(t))φ′(t)dt = F(φ(B))−F(φ(A)) = ∫ from φ(A) to φ(B) of f(u)du.

Every step has a hypothesis: continuity supplies the primitive derivative, the chain rule supplies the derivative factor, and continuous differentiability supplies an integrable transformed derivative. No inverse map is used, so injectivity and a strictly positive derivative are unnecessary for this forward signed identity.

The MIT 18.100B problem-set 9 substitution problem assumes f continuous on [a,b], φ′ continuous and positive, and endpoint images a and b. MVT makes φ increasing and keeps its image in [a,b]. Extending f constantly beyond each endpoint gives a continuous function on an open interval, so the preceding proof applies without an endpoint-domain ambiguity. This is the same chain-rule mechanism introduced in calculus substitution, now with its hypotheses and FTC dependencies explicit.

Repair an absolute-value substitution

For φ(t)=t³−3t on [−2,2], φ′=3(t²−1). Split at −1 and 1. The image traversals are −2 to 2, then 2 to −2, then −2 to 2. For f(u)=1+u², each forward integral is 28/3. Signed contributions add to 28/3; absolute-derivative contributions add to 28. Almost every image point is visited three times. Thus |φ′| removes orientation but retains multiplicity; it does not generally produce an image-once integral.

Because this inspector uses positive f, each branch’s absolute-derivative integral is the absolute value of its signed integral. For a sign-changing density that shortcut is invalid: the correct branch formula is the oriented image integral with its limits arranged increasingly, which can itself be negative. With a=0 in the cubic example, omitting φ′ happens to give the same value 4 as signed substitution. Change a to 1 to expose the error: the omitted-factor value becomes 412/35. One numerical coincidence does not validate a substitution rule.

Continue

Optional depth extends the evaluation theorem to an integrable derivative under appropriate assumptions; that stronger version is not obtained merely by dropping continuity from every line of the proof. Likewise, differentiability of an accumulation at some discontinuity points can occur, but continuity gives a clean sufficient condition rather than a necessary one.

Write the local derivative proof with both signs of h and identify the integral property controlling its error. Then write the endpoint-evaluation proof using MVT. Keeping these arguments separate prevents the common circular move of proving integration from antiderivatives while simultaneously proving antiderivatives from integration.