# Chapter 17: The Fundamental Theorems of Calculus

> The bridge between differentiation and integration: accumulation functions F(x) = ∫ₐˣ f(t)dt, the First FTC (F′ = f), antiderivatives, the Second FTC (top minus bottom), the four classic F′ variations, and unsigned area via splitting at zeros.

**Canonical:** https://calculus.kitzo.work/chapter/ftc  
**Level:** Core  
**Estimated time:** 45 minutes  
**Exercises:** 8 (auto-marked; answers stay server-side)

## Topics
- Antiderivatives
- First FTC
- Second FTC
- Indefinite integrals
- Variable limits
- Unsigned area & |f|
- F′ = f

**Why this matters**

The 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 out**

F(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.

## Practice exercises

1. If F(x) = ∫₀ˣ √(1 + t³) dt, then F′(x) equals:

- A. √(1 + x³)
- B. √(1 + t³)
- C. (1 + x³) 3/2 /3
- D. 1/√(1 + x³)

2. Let F(x) = ∫₂ˣ (cos t + 4) dt. Compute F′(π/2).

3. Evaluate ∫₀¹ 4x³ dx.

4. Compute d/dx ∫₀ x² sin(t²) dt.

- A. sin(x⁴)
- B. 2x·sin(x⁴)
- C. x²·sin(t²)
- D. 2x·sin(x²)

5. Evaluate ∫₁³ 2x dx using the Second FTC.

6. Compute d/dx ∫ x² ⁵ t³ dt evaluated at x = 1.

7. If ∫₀² f(x)dx = 5 and ∫₂³ f(x)dx = −3, what is the total UNSIGNED area between f and the axis on [0, 3]?

8. Write the indefinite integral ∫ 6x² dx. (Include + C.)

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