Chapter 17: The Fundamental Theorems of Calculus

Why this mattersThe Fundamental Theorem of Calculus is the reason you will never again add up infinitely many rectangles by hand. It says differentiation and integration are inverse operations — so the hardest sum in the world collapses into evaluating two numbers.

1 · Functions defined by integrals

Fix a constant a and let x move. Then F(x) = ∫ₐˣ f(t)dt is a function in its own right: it answers how much area has accumulated from a to x. The dummy variable t is the integration variable; x is the frontier. Note F(a) = 0 — no width, no area. If f is positive, F is the area under f so far, and it grows whenever f is positive and shrinks whenever f goes negative.

Why is this worth a name? Because the accumulated-area function is exactly what connects a velocity curve to a displacement, a rate to a total, a graph to its running total — and its derivative turns out to be f itself.

2 · The First Fundamental Theorem (F′ = f)

First FTC: if f is continuous on an interval containing a and x, and F(x) = ∫ₐˣ f(t)dt, then F′(x) = f(x).

Why it works: [F(x+h) − F(x)]/h = (1/h)∫ₓ^{x+h} f(t)dt, which is the average value of f over a skinny sliver of width h. As h → 0 the sliver shrinks to the point x and the average collapses to f(x). Integrating then differentiating hands you back the integrand — every time.

3 · Antiderivatives and the simple rules

An antiderivative of f is any function F with F′ = f. Antiderivatives come in families — differing by a constant, since constants vanish under differentiation — so we write the indefinite integral as

∫ f(x)dx = F(x) + C

The +C is not decoration; without it you are writing one member of an infinite family as if it were the whole family. The starter kit:

∫ xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1)
∫ eˣ dx = eˣ + C · ∫ (1/x) dx = ln|x| + C
∫ sin x dx = −cos x + C · ∫ cos x dx = sin x + C
∫ sec²x dx = tan x + C · ∫ 1/(1+x²) dx = arctan x + C

4 · The Second Fundamental Theorem

Second FTC: if F is any antiderivative of a continuous f, then

∫ₐᵇ f(x)dx = F(b) − F(a)

Find one antiderivative, plug in the top limit, plug in the bottom, subtract. The limit of Riemann sums is now a two-line computation. Why the two halves are one theorem: apply the Second FTC to F(x) = ∫ₐˣ f. The First FTC is what licenses F(b) − F(a) = ∫ₐᵇ f.

Worked example Example 1 — Evaluate ∫₀² (3x² − 2x)dx.
Step 1: an antiderivative is F(x) = x³ − x² (check: F′ = 3x² − 2x ✓).
Step 2: F(2) = 8 − 4 = 4; F(0) = 0.
Step 3: ∫₀² (3x² − 2x)dx = 4 − 0 = 4.

5 · The four classic F′ variations

Exams remix F′(x) = f(x) into four disguises. Learn all four and none will surprise you:

Simple: F(x) = ∫₀ˣ √(1+t³) dt ⇒ F′(x) = √(1+x³). Just strip the integral sign and rename the variable.
Variable lower limit: d/dx ∫ₓᵃ f(t)dt = −f(x). Swap the limits, pick up a minus sign.
Tricky limit (chain rule): d/dx ∫₀^{x²} f(t)dt = f(x²)·(x²)′ = 2x·f(x²). The upper limit is a function, so multiply by its derivative.
Limit that is a derivative in disguise: lim (h→0) (1/h)∫ₐ^{a+h} f(t)dt = f(a). That is the definition of F′(a) with the integral sign still attached — a derivative wearing a fake mustache.

Worked example Example 2 — Differentiate F(x) = ∫₃^{x³} cos t dt.
Step 1: upper limit u = x³, so by the chain rule F′(x) = cos(x³)·(x³)′.
Step 2: (x³)′ = 3x².
Step 3: F′(x) = 3x²cos(x³). The answer is a product — that is the point of variation 3.

6 · Definite integrals and unsigned area

The Second FTC computes net signed area. For genuine area between f and the x-axis, find the zeros of f in (a, b), split the integral there, and add the absolute values of the pieces. Antiderivatives make each piece trivial, so the only real work is locating the zeros.

Worked example Example 3 — Signed vs unsigned area of f(x) = x³ on [−1, 1].
Step 1: net: ∫₋₁¹ x³ dx = [x⁴/4]₋₁¹ = 1/4 − 1/4 = 0 (integrand is odd).
Step 2: the curve crosses the axis at x = 0, so split: ∫₋₁⁰ x³ dx = −1/4 and ∫₀¹ x³ dx = 1/4.
Step 3: unsigned area = |−1/4| + |1/4| = 1/2. Net 0, area ½ — never confuse the two.

Watch outF(b) − F(a) is the net integral, not the area. If f changes sign on [a, b], evaluate F at each zero, take absolute values of the pieces, and add. And in variation 2 above, the minus sign is mandatory: differentiating ∫ₓᵃ is not f(x).

Chapter formula sheet F(x) = ∫ₐˣ f(t)dt ⇒ F′(x) = f(x) (First FTC)
d/dx ∫_{u(x)}^{v(x)} f(t)dt = f(v(x))·v′(x) − f(u(x))·u′(x)
∫ₐᵇ f(x)dx = F(b) − F(a), F′ = f (Second FTC)
∫ f(x)dx = F(x) + C · ∫ xⁿdx = xⁿ⁺¹/(n+1) + C (n ≠ −1)
∫ eˣ dx = eˣ + C · ∫ (1/x)dx = ln|x| + C · ∫ cos x dx = sin x + C
∫ sin x dx = −cos x + C · ∫ sec²x dx = tan x + C
lim (h→0) (1/h)∫ₐ^{a+h} f(t)dt = f(a) · Unsigned area = Σ|∫ over each piece|

Common mistakes

Writing F′(x) = f(c) for some mysterious c — no: F′(x) = f(x), same input.
Dropping the minus sign when the variable sits in the lower limit.
Forgetting the chain-rule factor in variation 3 — f(x²)·2x, not f(x²).
Adding +C inside a definite integral — it cancels, but write F(b) − F(a) cleanly.
Answering net integral when the question asked for area — split at zeros, use |f|.
Using F(a) − F(b) — it is top limit minus bottom limit, always.