The fundamental theorem of calculus

Compare a signed rate with its accumulated change

Suppose a signed rate is f(t)=2t−1 for 0≤t≤2. From time zero to time x, the accumulated change is A(x)=x²−x. Its exact values at x=0,1/2,1,3/2,2 are 0,−1/4,0,3/4,2. Initially the rate is negative and accumulation falls; after one-half the rate is positive and accumulation rises.

At x=1, the accumulated change is zero but its instantaneous rate is f(1)=1. At x=2, the accumulated change is 2 and its inward rate is 3. A value of an accumulated quantity and the rate of that quantity answer different questions.

The fundamental theorem explains this connection in two directions. An integral with a moving endpoint produces an antiderivative where the integrand is continuous. Conversely, a suitable antiderivative evaluates an integral by its endpoint difference. Neither direction removes its hypotheses.

Name the variable, the interval and the units

Use the published preparation links for function domains, signed arithmetic and physical quantities. The integral, continuity and derivative definitions needed here are supplied locally. First follow the area and rate comparisons; the final proof sections justify the interval facts used by the theorems.

Let f be a bounded real-valued function on a finite closed interval [a,b], with a<b. A partition a=t₀<t₁<⋯<tₙ=b has positive widths Δtᵢ=tᵢ−tᵢ₋₁ and mesh equal to the largest width. Choose any tag ξᵢ in each closed cell and form S=Σf(ξᵢ)Δtᵢ.

Riemann integrability means that one real number I satisfies this all-input promise: for every ε>0 there is δ>0 such that every partition of mesh less than δ, with every permitted choice of tags, has |S−I|<ε. Write I=∫_a^b f(t)dt; a is the lower bound and b the upper bound. One favourite sequence of rectangles is not that full promise.

Here t is a bound variable: ∫_a^b f(t)dt=∫_a^b f(s)ds. With fixed bounds the integral is a number. With x as an endpoint, A(x)=∫_a^x f(t)dt is a function of x. The symbol dt accompanies the input width in the sums; it is not an instruction to substitute the upper bound into the integrand.

A rate in litres per minute multiplied by widths in minutes gives accumulated litres. A velocity in metres per second gives signed displacement in metres. Differentiating that accumulated quantity with respect to time returns the rate units. Units belong to the original variables, not to the letters chosen for them.

Signed integrals keep track of direction

Set ∫_a^a f=0 and define ∫_b^a f=−∫_a^b f. Integrating from a larger input to a smaller one reverses the sign. This convention is useful even when the integrand is nonnegative; a reversed integral is not a negative geometric area.

For integrable f, splitting an interval gives ∫_a^c f+∫_c^b f=∫_a^b f. Integrals add and scale with their integrands. If m≤f≤M on [a,b], every tagged sum lies between m(b−a) and M(b−a), so its integral does too. In particular |∫_u^v f|≤K|v−u| when |f|≤K on the intervening interval.

Linearity follows by adding or scaling tagged sums and making their errors small. To justify restriction and splitting, insert a split point into any fine partition. Only one cell is divided, so with |f|≤K its tagged sum changes by at most 2K times the mesh. Conversely join fine partitions on the two pieces. The all-tag limits on those pieces exist by the same Cauchy comparison of joined sums, and their sum is the full limit. Orientation extends the identity to every order of a,c,b.

For f(t)=t−1 on [0,2], the integral is zero: the negative triangle on [0,1] cancels the positive triangle on [1,2]. The geometric area is instead ∫_0^2 |t−1|dt=1. Signed displacement can be zero while travelled distance is positive.

Continuity concerns the actual integrand value

Continuity at c in a domain D means: for every ε>0 there is δ>0 such that t∈D and |t−c|<δ imply |f(t)−f(c)|<ε. The assigned value f(c) matters. Continuity on a closed interval uses only permitted inputs, including inward approaches at its endpoints.

The finite derivative A′(c)=m at an interior input is the all-small-increment limit of [A(c+h)−A(c)]/h: for every ε>0, every sufficiently small nonzero h of either sign gives error less than ε from m. At an interval endpoint use only inward increments and name the derivative one-sided. A singleton interval has no approaching derivative under this convention.

If f is Riemann integrable and |f|≤K, its accumulator A(x)=∫_a^x f satisfies |A(y)−A(x)|≤K|y−x|. Thus A is continuous everywhere on [a,b], even when f is not. Continuity of A alone does not guarantee its differentiability.

The first direction: the rate of accumulation

FTC1: let f be Riemann integrable on [a,b] and continuous at an interior c. Then A(x)=∫_a^x f(t)dt is differentiable at c, and A′(c)=f(c). If f is continuous throughout [a,b], this holds throughout (a,b). At a or b, continuity there gives the corresponding inward one-sided derivative, not a two-sided derivative of this restricted function.

The proof begins with additivity: A(c+h)−A(c)=∫_c^{c+h} f(t)dt. Subtract f(c)h and divide by h. The error is [1/h]∫_c^{c+h} [f(t)−f(c)]dt. This is an average signed rate error over a shrinking interval, not the value of a single sampled rectangle.

Given ε>0, continuity supplies a radius within which |f(t)−f(c)|<ε/2. For every permitted nonzero h smaller than that radius, the integral bound gives an error at most (ε/2)|h|/|h|=ε/2<ε. This works for h<0 as well: both integral orientation and the divisor change sign. The two-sided quotient therefore approaches f(c).

The same estimate proves the endpoint claim using inward h only. It also shows why continuity only at the target suffices: discontinuities elsewhere do not enter the shrinking interval once h is small. Integrability on the full interval is still required to define A.

A corner and a discontinuity have different effects

For continuous f(t)=|t|, the integrand has a corner at zero but A(x)=∫_0^x |t|dt=x|x|/2 has derivative |x|, including A′(0)=0. Directly A(h)/h=|h|/2→0. FTC1 requires continuity of f, not differentiability of f. The accumulator has no second derivative at zero because the left and right slopes of |x| disagree.

For f(t)=−1 when t<0, +1 when t>0, and f(0)=0, the accumulator is A(x)=|x|. At zero its left quotient is −1 and its right quotient is 1, so A′(0) does not exist. Changing only f(0) cannot remove this corner.

For f(t)=0 except f(0)=5, the integral on every interval is zero. With mesh d, at most two tagged cells can select zero, and their total contribution has magnitude at most 10d. Hence the all-tag limit is zero. Its accumulator is constant, with A′(0)=0≠f(0). A discontinuity can therefore leave a derivative that exists but does not equal the assigned integrand value.

More generally, changing a bounded integrand at finitely many points changes any tagged sum by at most twice the number of changed points times the largest change times the mesh. Integrability and the integral are preserved. This does not preserve the continuity needed for FTC1 at a changed point.

The step is integrable too: away from the cell or two cells meeting zero it is constant. Only those cells can disagree with the exact signed lengths, and their error tends to zero with the mesh. These finite exceptions are not a licence to integrate an unbounded function across a missing pole.

The second direction: evaluate by an endpoint difference

An antiderivative of f on an open interval I is a differentiable function F with F′(x)=f(x) for every x∈I. FTC2 in a useful precise form is: f is Riemann integrable on [a,b], F is continuous on [a,b], and F′=f throughout (a,b). Then ∫_a^b f(t)dt=F(b)−F(a). No derivative of F at an endpoint is required.

For each cell [tᵢ₋₁,tᵢ], the mean value theorem supplies an interior ξᵢ with F(tᵢ)−F(tᵢ₋₁)=F′(ξᵢ)Δtᵢ=f(ξᵢ)Δtᵢ. Summing telescopes to F(b)−F(a). This is a permitted tagged sum for every partition. As the mesh tends to zero, integrability forces these sums to the integral, proving the equality. The mean value theorem and its interval hypotheses are justified in the final proof bridge.

In the familiar sufficient case, continuous f on [a,b] is integrable, and FTC1 produces a continuous antiderivative A. Any other antiderivative F, continuous at the endpoints, has the same endpoint difference. Continuity of f is sufficient here; the more general evaluation statement only requires its integrability and the stated derivative identity inside.

For f=t², direct expansion gives [(x+h)³−x³]/(3h)=x²+xh+h²/3→x², so F(x)=x³/3 is an antiderivative. On [−1,2], ∫_{−1}^2 t²dt=[t³/3]₋₁²=8/3−(−1/3)=3. Reverse the limits and the answer is −3. A definite integral is this number, without an extra arbitrary constant.

Antiderivatives differ by one constant on an interval

If F′=G′ on an open interval I, then H=F−G has derivative zero there. For any x<y in I, the mean value theorem on [x,y] gives H(y)−H(x)=H′(ξ)(y−x)=0. Thus F−G is one constant on that interval. The derivative of a constant is zero by its quotient; conversely the interval argument is needed to infer constancy.

On a disconnected domain, each component can have a different constant. The function that is zero for x<0 and seven for x>0 has derivative zero on ℝ\{0}, yet is not one constant on that whole set. Nor may an antiderivative defined only off a pole be used as if it were continuous across that pole.

Endpoint continuity of F matters. On [0,1], set F(x)=x for x<1 but F(1)=8. Then F′=1 throughout (0,1), while ∫_0^1 1dt=1 and F(1)−F(0)=8. The missing endpoint hypothesis prevents FTC2 from claiming that false equality.

Values of f at endpoints do not enter the interior derivative identity. For example f(t)=t for 0<t≤2 but f(0)=5 is integrable with ∫_0^2 f=2, and continuous F(t)=t²/2 satisfies F′=f inside. Its accumulator’s inward derivative at zero is 0, not f(0)=5; FTC2 still applies, while FTC1’s endpoint continuity condition fails.

A moving endpoint contributes its own speed

Let f be continuous on an open interval J, and let differentiable g map a neighbourhood of c into J. For B(x)=∫_r^{g(x)} f(t)dt with fixed r∈J, FTC1 and the chain rule give B′(c)=f(g(c))g′(c). The factor g′ describes how fast the endpoint moves; replacing only t by g(x) misses it.

Here is the local chain proof, including the case g(c+h)=g(c). Differentiability of A at v=g(c) means A(v+k)=A(v)+A′(v)k+ρ(k)k, with ρ(k)→0; set the last product to zero at k=0. With k=g(c+h)−g(c), divide by h. The quotient becomes [A′(v)+ρ(k)]k/h. Differentiability of g makes k→0 and k/h bounded with limit g′(c), so the error product tends to zero. Thus B′=f(g)g′, even when g′=0.

For B(x)=∫_0^{x²} t dt, B=x⁴/2 and B′=2x³, because f(x²)·2x=x²·2x. At x=−1 its slope is −2 although the integrand at the moving endpoint is positive. The endpoint is moving left as x increases there.

A variable lower endpoint changes the sign: C(x)=∫_{g(x)}^r f=−∫_r^{g(x)} f, so C′=−f(g)g′. With two differentiable endpoints u,v taking values in J, write H(x)=A(v(x))−A(u(x)). Then H′=f(v)v′−f(u)u′. These identities remain true when the endpoint order reverses or when u=v.

For H(x)=∫_{x²}^{2x} t dt, H=2x²−x⁴/2 and H′=4x−2x³. At x=1 the two contributions are 4 and −2, giving slope 2. At x=0 both limits agree and both contributions vanish. Do not silently extend these open-neighbourhood hypotheses to a target where the integrand is discontinuous or a moving bound leaves its domain.

Accumulation gives change; an initial value gives the quantity

If a continuously differentiable quantity Q has rate q=Q′ on [a,b], with endpoint derivatives named inward, FTC2 gives Q(b)−Q(a)=∫_a^b q(t)dt. Hence Q(x)=Q(a)+∫_a^x q(t)dt. The initial quantity supplies the constant that a rate alone cannot determine.

With velocity v(t)=t−1 metres per second for 0≤t≤2 seconds, displacement is [t²/2−t]₀²=0 metres. Distance is ∫_0^2 |t−1|dt=1 metre. If the initial position is 6 metres, the final position is also 6 metres; that does not mean no motion occurred.

If t has minutes and a moving endpoint g(x) has minutes while x has seconds, then f(g(x)) in litres per minute multiplied by g′(x) in minutes per second gives B′ in litres per second. The chain factor carries both the magnitude and the unit conversion.

Check the whole interval before subtracting endpoint values

For 1/t² across zero, the original rule is undefined there and unbounded nearby. It is not a bounded Riemann integrand on [−1,1]. The formal antiderivative −1/t exists on each separate component, but subtracting its endpoint values would give −2 for a positive integrand. That calculation violates FTC2’s interval and continuity hypotheses; it is not a definite Riemann integral.

A discontinuity need not prevent integrability, as the step and spike show. It does prevent using FTC1 at that target without further reasoning. A corner in a continuous integrand does not prevent FTC1. Check integrability, actual target continuity, interior versus endpoint derivative, and the domain of any moving bounds separately.

The Explorer supplies original integrands, bounds and accumulated readings. Predict the accumulator’s slope before revealing its exact result. Finite diagrams and quotient tables check the arithmetic; the all-increment estimates and interval proofs establish the theorem.

Why the continuous case really defines an integral

This optional proof uses completeness of the real numbers. Every bounded sequence in [a,b] has a convergent subsequence: keep a closed half containing infinitely many indexed terms, repeat, and select increasing indices from those nested halves. If their left endpoints have supremum c, nestedness puts c in every half; widths (b−a)/2ⁿ→0 make the selected inputs approach c∈[a,b].

A continuous f on [a,b] is bounded: otherwise choose inputs with |f|>n, take the subsequence just constructed, and continuity at its limit contradicts unbounded outputs. It is uniformly continuous: if some ε₀>0 had input pairs sₙ,tₙ with |sₙ−tₙ|<1/n but |f(sₙ)−f(tₙ)|≥ε₀, take a convergent subsequence of sₙ. The corresponding tₙ approach the same limit, and continuity of both output sequences is a contradiction.

For bounded f, define L(P)=ΣmᵢΔtᵢ and U(P)=ΣMᵢΔtᵢ using the infimum and supremum in each cell. Refinement increases L and decreases U. A common refinement therefore shows L(P)≤U(Q) for any P,Q. Completeness gives ℓ=supₚL(P)≤u=infₚU(P), and L(P)≤ℓ≤u≤U(P) for every partition.

Uniform continuity makes each cell’s oscillation Mᵢ−mᵢ at most ε/[2(b−a)] when the mesh is small. Hence U(P)−L(P)≤ε/2, forcing ℓ=u=I. Every tagged sum lies between L(P) and U(P), so every fine tagged sum has |S−I|≤ε/2<ε. This proves the all-tag definition for continuous f, rather than assuming that an integral sign alone guarantees existence.

The same lower/upper argument gives a useful existence test: if for every ε>0 some partition has U−L<ε, then ℓ=u. For all sufficiently fine partitions, refining by that fixed partition changes either bound by at most 2K times the number of its interior breakpoints times the mesh. Thus their gap, and every tagged error from I, become arbitrarily small. This justifies piecewise continuous bounded examples with finitely many exceptional points as well.

Why the mean value step is valid

For a continuous F on [a,b], boundedness was proved above. Its supremum S is attained: choose inputs with F>S−1/n, take a convergent subsequence in [a,b], and use continuity to obtain F(c)=S. Apply this to −F for an attained minimum. A singleton has its own value as both extrema; the mean value theorem below requires a<b.

Differentiability implies continuity because F(c+h)−F(c)=hQ(h) and a convergent finite quotient Q is bounded nearby. At an interior differentiable extremum, right and left quotient signs force the common limit to be zero. If F is continuous on [a,b], differentiable inside and has equal endpoint values, either it is constant or an attained non-endpoint extremum gives F′(c)=0. This is Rolle’s theorem.

Subtract the secant line with slope s=[F(b)−F(a)]/(b−a). The resulting continuous function has equal endpoint values and derivative F′−s inside. Rolle gives F′(c)=s for some c∈(a,b), proving F(b)−F(a)=F′(c)(b−a). This is the exact interval statement used for FTC2 and antiderivative uniqueness; continuity at every cell endpoint is part of it.

Return to the concrete accumulation examples after this proof. The two directions now rest on explicit all-tag integrability, local continuity estimates and a justified mean value step. Later integration methods can use these conclusions without dropping their domains or endpoint conditions.

Worked examples

Try it yourself

1. For A(x)=∫_0^x 3dt, find A(−2) and A′(−2).

   Show the answer and reasoning

   A(−2)=−6 and A′(−2)=3.

   The reversed integral is signed, while the local derivative equals the continuous constant rate.

2. Does replacing t by s in ∫_0^2 (t²+1)dt change its value?

   Show the answer and reasoning

   No; consistent renaming gives the same integral.

   The bound variable labels inputs within the sum. Fixed endpoints make this integral a number.

3. Evaluate ∫_0^3 t²dt using a verified antiderivative.

   Show the answer and reasoning

   9.

   F=t³/3 is continuous on [0,3] and has derivative t² inside, so F(3)−F(0)=27/3=9.

4. Differentiate B(x)=∫_0^{x²} t dt at x=−1.

   Show the answer and reasoning

   B′(−1)=−2.

   The derivative is f(x²)·2x=2x³. The positive endpoint rate is multiplied by the negative endpoint speed.

5. For C(x)=∫_{x²}^1 t dt, find C′(1).

   Show the answer and reasoning

   −2.

   A variable lower limit contributes −f(x²)·2x=−2x³.

6. If H(x)=∫_{x²}^{x²} t dt for every x, what are H and H′?

   Show the answer and reasoning

   Both are zero.

   The limits coincide identically, and the two equal chain contributions cancel. Coinciding only at one input would not alone force the derivative to vanish there.

7. For A(x)=∫_0^x t²dt restricted to [0,2], what derivative claim is justified at zero?

   Show the answer and reasoning

   The inward right derivative exists and equals zero.

   The restricted domain has no left input approach. A two-sided derivative requires an extension and its own conditions.

8. Let f=−1 left of zero and +1 right of zero, with f(0)=0. Does A(x)=∫_0^x f have a derivative at zero?

   Show the answer and reasoning

   No finite two-sided derivative exists.

   A=|x| has left quotient −1 and right quotient 1. The assigned value at one input cannot remove the side disagreement.

9. Does every Riemann-integrable function satisfy A′(x)=f(x) everywhere?

   Show the answer and reasoning

   No; a single assigned spike gives a counterexample.

   For f=0 except f(0)=5, A is identically zero, so A′(0)=0≠f(0). Target continuity is sufficient for the identity.

10. Must f be differentiable at c to use FTC1 there?

    Show the answer and reasoning

    No; integrability and continuity at c suffice.

    For f=|t|, the integrand has a corner at zero while its accumulator has derivative zero there.

11. F(x)=x for x<1 but F(1)=8 on [0,1]. Why is F(1)−F(0) not ∫_0^1 1dt?

    Show the answer and reasoning

    F is not continuous at the upper endpoint.

    Its interior derivative equals one, but FTC2 also requires endpoint continuity; the integral equals one, not eight.

12. May you evaluate ∫_{−1}^1 1/t²dt by [−1/t]₋₁¹=−2 as a bounded Riemann integral?

    Show the answer and reasoning

    No; the rule is undefined and unbounded near zero.

    The antiderivative is not continuous on the entire interval, and the bounded Riemann integral is not defined. Separate improper integration is a later topic.

13. If a derivative is zero on ℝ\{0}, must the function be one constant across both components?

    Show the answer and reasoning

    No; the constants may differ on the two components.

    Values zero for x<0 and seven for x>0 give derivative zero at every allowed input, while the interval hypothesis for one constant is absent.

14. A rate f is L/min and an upper bound g is measured in minutes as a function of seconds. What units does f(g)g′ have?

    Show the answer and reasoning

    Litres per second.

    (L/min)·(min/s)=L/s. The endpoint speed contributes the conversion factor as well as its sign.

What to carry forward

• A definite integral has signed orientation; accumulated value and current rate are different quantities.
• FTC1 needs integrability and continuity of the integrand at the target; endpoint claims are explicitly one-sided.
• FTC2 needs an integrable derivative record, a continuous antiderivative on the closed interval and the derivative identity inside.
• Antiderivatives differ by one constant on an interval; disconnected components can carry different constants.
• A moving upper limit contributes f(v)v′ and a moving lower limit contributes −f(u)u′.
• Corners, single-point values and jumps affect continuity and differentiability in different ways.
• Check original domains, endpoints, signs, initial values and units before evaluating or differentiating an integral.