Lecture practice | Source: Cal1ptac lecture notes, Texas A&M University
Tags: fundamental theorem of calculus, FTC part 1, FTC1, derivative of an integral, antiderivative, accumulation function, Extreme Value Theorem, Squeeze Theorem, calculus proof, MATH 151, MATH 171
The Fundamental Theorem of Calculus Part 1 says that if you define a function by integrating a continuous function from a fixed point to a variable upper limit, then the derivative of that new function just gives you back the original function. In short, differentiation undoes integration. The proof relies on the definition of the derivative, properties of integrals, the Extreme Value Theorem, and a squeeze argument.
Fundamental Theorem of Calculus, Part 1 (FTC1)
If f is continuous on [a, b] and g(x) = ∫ from a to x of f(t) dt, then g'(x) = f(x). Differentiation and integration are inverse operations.
Accumulation function
A function defined as g(x) = ∫ from a to x of f(t) dt. It measures the net signed area under f from a fixed starting point a up to the variable endpoint x. Sometimes called an "area-so-far" function.
Extreme Value Theorem (EVT)
If a function is continuous on a closed interval, it attains both an absolute minimum and an absolute maximum on that interval. Used in the FTC1 proof to guarantee the existence of min and max values on [x, x + h].
Squeeze Theorem (Sandwich Theorem)
If a quantity is trapped between two bounds that converge to the same limit, the quantity itself converges to that limit. The closing step of the FTC1 proof uses this to pin down g'(x) = f(x).
Continuity
A function f is continuous on [a, b] if, loosely, you can draw its graph without lifting the pen. More precisely, the limit of f at every point in the interval equals the function value. Continuity is the key hypothesis that makes FTC1 work.
Let f be continuous on [a, b].
Define g(x) = ∫ from a to x of f(t) dt.
Then g is differentiable on (a, b) and g'(x) = f(x).
The variable x is the upper limit of integration. The dummy variable t is just a placeholder inside the integral.
Step 1: Apply the definition of the derivative
Start from first principles:
g'(x) = lim (h → 0) of [g(x + h) − g(x)] / h
Substitute the definition of g:
g'(x) = lim (h → 0) of (1/h) [ ∫ from a to (x+h) of f(t) dt − ∫ from a to x of f(t) dt ]
Step 2: Use the additive property of integrals
The difference of two integrals with the same lower limit simplifies:
∫ from a to (x+h) of f(t) dt − ∫ from a to x of f(t) dt = ∫ from x to (x+h) of f(t) dt
So:
g'(x) = lim (h → 0) of (1/h) ∫ from x to (x+h) of f(t) dt
This is the average value of f on the small interval [x, x + h].
Step 3: Bound the integral using EVT
Since f is continuous on [x, x + h] (a closed interval), the Extreme Value Theorem guarantees:
A minimum value m = f(u) for some u in [x, x + h]
A maximum value M = f(v) for some v in [x, x + h]
By the comparison property of integrals:
m · h ≤ ∫ from x to (x+h) of f(t) dt ≤ M · h
Dividing through by h (taking h → 0⁺):
f(u) ≤ (1/h) ∫ from x to (x+h) of f(t) dt ≤ f(v)
Step 4: Squeeze to the result
As h → 0⁺, both u → x and v → x (they are trapped in the shrinking interval [x, x + h]).
Since f is continuous:
lim (h → 0⁺) f(u) = f(x)
lim (h → 0⁺) f(v) = f(x)
By the Squeeze Theorem, the expression in the middle also converges to f(x):
g'(x) = lim (h → 0) of (1/h) ∫ from x to (x+h) of f(t) dt = f(x)
Core formula:
If g(x) = ∫ from a to x of f(t) dt, then g'(x) = f(x)
Chain rule extension (common exam variant):
If g(x) = ∫ from a to u(x) of f(t) dt, then g'(x) = f(u(x)) · u'(x)
This follows from FTC1 combined with the chain rule. Appears frequently in exam problems where the upper limit is something like x², sin(x), or 3x.
Bounding inequality used in the proof:
m · h ≤ ∫ from x to (x+h) of f(t) dt ≤ M · h
where m and M are the minimum and maximum of f on [x, x + h].
⚠️ FTC1 is one of the most commonly tested proofs in Calculus I. Be prepared to reproduce the full argument, not just state the theorem.
⚠️ The continuity of f is essential. Without it, the Extreme Value Theorem does not apply and the proof breaks down. If an exam question changes the hypothesis (e.g. f has a jump discontinuity), the conclusion can fail.
⚠️ Do not confuse FTC Part 1 with FTC Part 2. Part 1 says differentiation undoes integration (g'(x) = f(x)). Part 2 says ∫ from a to b of f(x) dx = F(b) − F(a) where F is any antiderivative of f.
⚠️ Watch for chain rule variants. If the upper limit is not simply x but some function u(x), you must multiply by u'(x). This is the single most common FTC1 exam trick.
⚠️ The dummy variable (t in the integral) is not the same as the variable of differentiation (x). Mixing these up is a frequent source of errors.
Q: State the Fundamental Theorem of Calculus, Part 1.
A: If f is continuous on [a, b] and g(x) = ∫ from a to x of f(t) dt, then g'(x) = f(x) for all x in (a, b).
Q: In the proof, why is the Extreme Value Theorem needed?
A: Because f is continuous on the closed interval [x, x + h], the EVT guarantees there exist points u and v in that interval where f attains its minimum m = f(u) and maximum M = f(v). These bounds are used to squeeze (1/h) ∫ from x to (x+h) f(t) dt between f(u) and f(v).
Q: What happens to u and v as h → 0⁺, and why does that matter?
A: Both u and v are trapped in [x, x + h], so as h → 0⁺, both u → x and v → x. By continuity of f, f(u) → f(x) and f(v) → f(x). The Squeeze Theorem then forces the middle expression to converge to f(x).
Q: Find d/dx of ∫ from 2 to x³ of sin(t²) dt.
A: By FTC1 with the chain rule: the answer is sin((x³)²) · 3x² = 3x² sin(x⁶).
Q: Why does the proof consider h → 0⁺ rather than h → 0?
A: The argument is shown for h → 0⁺ for clarity (so the interval [x, x + h] makes sense with h positive). A symmetric argument works for h → 0⁻. Both one-sided limits equal f(x), so the two-sided limit exists and equals f(x).
Q: What hypothesis of FTC1 would fail if f had a jump discontinuity at some point c in (a, b)?
A: Continuity of f on [a, b]. Without continuity at c, the EVT cannot guarantee min/max values on intervals containing c, and the squeeze argument breaks down. The accumulation function g would still be continuous but would not be differentiable at c.
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