Lecture 31 | Source: Calculus I Lecture Notes, Texas A&M University
Tags: fundamental theorem of calculus, FTC part 1, FTC1, antiderivative, accumulation function, definite integral, derivative of integral, continuous function, Extreme Value Theorem, Squeeze Theorem, calculus 1, TAMU MATH
The Fundamental Theorem of Calculus, Part 1 says that if you define a function by integrating a continuous function from a fixed starting point up to x, the derivative of that new function simply gives you back the original function. In short, differentiation undoes integration. The proof uses the limit definition of the derivative, a bounding property of integrals, and the Squeeze Theorem.
Fundamental Theorem of Calculus, Part 1 (FTC1)
If f is continuous on an interval and g(x) = ∫ₐˣ f(t) dt, then g′(x) = f(x). Differentiation and integration are inverse operations.
Accumulation function
The function g(x) = ∫ₐˣ f(t) dt, which gives the net signed area under f from a fixed lower limit a up to the variable upper limit x. Sometimes called an "area-so-far" function.
Extreme Value Theorem (EVT)
If f is continuous on a closed interval [a, b], then f attains both an absolute minimum value m and an absolute maximum value M somewhere on that interval. Used in the FTC1 proof to bound the integral.
Bounding property of integrals
For a continuous function f on [x, x + h], if m is the minimum value and M is the maximum value on that interval, then m·h ≤ ∫ₓˣ⁺ʰ f(t) dt ≤ M·h.
Squeeze Theorem (Sandwich Theorem)
If a quantity is trapped between two expressions that both tend to the same limit, the quantity itself must tend to that limit. This is the final step that clinches the FTC1 proof.
Let f be continuous on an open interval containing a.
Define g(x) = ∫ₐˣ f(t) dt.
Then g is differentiable, and g′(x) = f(x).
In words: the derivative of the integral (with a variable upper limit) returns the original integrand.
Step 1 – Write g′(x) using the limit definition
g′(x) = lim(h→0) [g(x + h) − g(x)] / h
Substituting the integral definition of g:
g(x + h) = ∫ₐˣ⁺ʰ f(t) dt
g(x) = ∫ₐˣ f(t) dt
So the difference quotient becomes: lim(h→0) (1/h) [∫ₐˣ⁺ʰ f(t) dt − ∫ₐˣ f(t) dt]
Step 2 – Simplify using the integral splitting property
By the additive property of definite integrals:
∫ₐˣ⁺ʰ f(t) dt − ∫ₐˣ f(t) dt = ∫ₓˣ⁺ʰ f(t) dt
The difference quotient reduces to: lim(h→0) (1/h) ∫ₓˣ⁺ʰ f(t) dt
Step 3 – Bound the integral using the Extreme Value Theorem
f is continuous on the closed interval [x, x + h], so by the EVT there exist points u and v in [x, x + h] where:
f(u) = m (the minimum value on the interval)
f(v) = M (the maximum value on the interval)
The bounding property gives: m·h ≤ ∫ₓˣ⁺ʰ f(t) dt ≤ M·h
Dividing through by h: f(u) ≤ (1/h) ∫ₓˣ⁺ʰ f(t) dt ≤ f(v)
Step 4 – Apply the Squeeze Theorem
As h → 0, the interval [x, x + h] shrinks to the single point x.
Both u and v are squeezed toward x, so u → x and v → x.
Since f is continuous: f(u) → f(x) and f(v) → f(x).
The expression (1/h) ∫ₓˣ⁺ʰ f(t) dt is trapped between f(u) and f(v), both tending to f(x).
By the Squeeze Theorem: g′(x) = f(x). ∎
The theorem itself:
If f continuous: d/dx [∫ₐˣ f(t) dt] = f(x)
The difference quotient form used in the proof:
g′(x) = lim(h→0) (1/h) ∫ₓˣ⁺ʰ f(t) dt
The bounding inequality (the heart of the argument):
f(u) ≤ (1/h) ∫ₓˣ⁺ʰ f(t) dt ≤ f(v)
where f(u) = min on [x, x+h] and f(v) = max on [x, x+h].
Diagram note: The lecture sketch shows a curve f on the interval [x, x + h] with horizontal dashed lines at the min value m and max value M. The area under f on that small interval is trapped between the two rectangles m·h (below) and M·h (above). As h shrinks to 0, both rectangles collapse and the average value of f on the interval is forced to equal f(x).
⚠️ FTC1 is one of the most commonly tested proofs in Calculus I. Be ready to state it precisely and reproduce the proof.
⚠️ The continuity of f is essential. Without it, the Extreme Value Theorem cannot be applied and the bounding step fails.
⚠️ Common mistake: confusing FTC1 with FTC2. FTC1 is about differentiating an integral (gives back the integrand). FTC2 is about evaluating a definite integral using an antiderivative (∫ₐᵇ f(x) dx = F(b) − F(a)).
⚠️ Watch the variable of integration. The integrand uses a dummy variable t, while the upper limit is the "live" variable x. Mixing these up is a frequent error.
⚠️ If the upper limit is not simply x but some function u(x), you need the chain rule: d/dx [∫ₐᵘ⁽ˣ⁾ f(t) dt] = f(u(x)) · u′(x). This extension is a very common exam question.
Q: State the Fundamental Theorem of Calculus, Part 1.
A: If f is continuous on an interval containing a, and g(x) = ∫ₐˣ f(t) dt, then g is differentiable and g′(x) = f(x).
Q: In the proof of FTC1, why is the Extreme Value Theorem needed?
A: Because f is continuous on the closed interval [x, x + h], the EVT guarantees that f attains a minimum m and a maximum M on that interval. These values are used to bound the integral ∫ₓˣ⁺ʰ f(t) dt between m·h and M·h, which sets up the Squeeze Theorem.
Q: What integral property is used to simplify g(x + h) − g(x)?
A: The additive (splitting) property: ∫ₐˣ⁺ʰ f(t) dt − ∫ₐˣ f(t) dt = ∫ₓˣ⁺ʰ f(t) dt. This isolates the integral over the small interval [x, x + h].
Q: Find d/dx [∫₂ˣ cos(t²) dt].
A: By FTC1, the answer is cos(x²). The integrand evaluated at the upper limit.
Q: Find d/dx [∫₀ˣ³ eᵗ dt].
A: By FTC1 combined with the chain rule: e^(x³) · 3x². You evaluate the integrand at the upper limit u(x) = x³, then multiply by u′(x) = 3x².
Q: Why does the proof break down if f is not continuous?
A: Continuity is required at two points. First, it allows the EVT to guarantee the existence of min and max values on [x, x + h]. Second, it ensures that f(u) → f(x) and f(v) → f(x) as h → 0, which is what makes the Squeeze Theorem work.
fundamental theorem of calculus, FTC part 1, FTC1, FTC part 2, accumulation function, area function, integral as a function of its upper limit, derivative of an integral, antiderivative, definite integral, Extreme Value Theorem, EVT, bounding property, Squeeze Theorem, Sandwich Theorem, chain rule with integrals, Leibniz integral rule, calculus 1, MATH 151, MATH 171, Cal31, Texas A&M calculus