# Fundamental theorem of calculus

> IB Mathematics: Analysis and Approaches HL · IB AA HL
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u5-fundamental-theorem-of-calculus/

The Fundamental Theorem of Calculus (FTC) connects differentiation and integration, the two core operations of calculus. We cover both key parts of the theorem, their core applications, and common exam question patterns.

**Prerequisites:** [Basic differentiation rules](https://www.owlsprep.com/study/ib-math-aa-hl-basic-differentiation-rules/); [Definite integrals as Riemann sums](https://www.owlsprep.com/study/ib-math-aa-hl-definite-integral-riemann-sum/)

## Learning objectives

- State both parts of the Fundamental Theorem of Calculus (FTC)
- Apply FTC Part 1 to differentiate integrals with variable bounds
- Apply FTC Part 2 to evaluate definite integrals accurately

## Part 1 of the Fundamental Theorem of Calculus

**FTC Part 1** — $F$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $F'(x) = f(x)$ for all $x \in (a,b)$. This proves differentiation and integration are inverse operations.

*Notation:* If $f$ is continuous on $[a,b]$, and $F(x) = \int_a^x f(t) dt$ for $x \in [a,b]$

*Example:* For $F(x) = \int_2^x t^2 dt$, $F'(x) = x^2$ directly by FTC Part 1.

When the upper bound is a function of $x$ rather than $x$ itself, we apply the chain rule to find the derivative. For an upper bound $u(x)$, the derivative becomes $\frac{d}{dx} \int_a^{u(x)} f(t) dt = f(u(x)) \cdot u'(x)$.

**Worked example:** Differentiate $G(x) = \int_0^{x^2} \cos(t) dt$ with respect to $x$.

1. Recognize $G(x)$ uses a composite upper bound $u(x) = x^2$. By chain rule: $G'(x) = \frac{dG}{du} \cdot \frac{du}{dx}$.
2. By FTC Part 1, $\frac{dG}{du} = \cos(u)$.
3. Calculate the derivative of the upper bound: $\frac{du}{dx} = 2x$.
4. Substitute back $u = x^2$ to get the final result:
5. $$G'(x) = \cos(x^2) \cdot 2x = 2x \cos(x^2)$$

> **tip**
>
> If both bounds are variable, split the integral at a constant $c$: $\int_{v(x)}^{u(x)} f(t) dt = \int_c^{u(x)} f(t) dt - \int_c^{v(x)} f(t) dt$, then differentiate each term separately.

> **Exam tip:** Most IB exam questions on FTC Part 1 use variable composite bounds, never forget the chain rule derivative of the bound.

*Calculator:* forbidden

## Part 2 of the Fundamental Theorem of Calculus

**FTC Part 2** — The definite integral of $f$ over $[a,b]$ equals $F(b) - F(a)$. This lets us calculate exact definite integral values without approximating Riemann sums.

*Notation:* If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$ ($F' = f$)

*Example:* $\int_1^2 x dx = \left[\frac{1}{2}x^2\right]_1^2 = 2 - \frac{1}{2} = \frac{3}{2}$

The constant of integration from indefinite integrals always cancels out in $F(b) - F(a)$, so you never need to include it when evaluating definite integrals.

**Worked example:** Evaluate the definite integral $\int_0^{\frac{\pi}{2}} (2 \sin x + 3x^2) dx$.

1. Find the antiderivative of each term: antiderivative of $2 \sin x$ is $-2 \cos x$, antiderivative of $3x^2$ is $x^3$.
2. Combine to get $F(x) = -2 \cos x + x^3$ (ignore the constant of integration).
3. Evaluate at the upper bound $\frac{\pi}{2}$:
4. $$F\left(\frac{\pi}{2}\right) = -2 \cos\left(\frac{\pi}{2}\right) + \left(\frac{\pi}{2}\right)^3 = 0 + \frac{\pi^3}{8} = \frac{\pi^3}{8}$$
5. Evaluate at the lower bound $0$:
6. $$F(0) = -2 \cos(0) + 0^3 = -2(1) = -2$$
7. Subtract lower bound from upper bound to get the final result:
8. $$\int_0^{\frac{\pi}{2}} (2 \sin x + 3x^2) dx = F\left(\frac{\pi}{2}\right) - F(0) = \frac{\pi^3}{8} - (-2) = \frac{\pi^3}{8} + 2 \approx 5.87$$

> **Exam tip:** Always show $F(b) - F(a)$ explicitly in your working, even if you use a calculator for the final value, to earn full method marks.

*Calculator:* allowed

## Core Proof and Connections

The FTC is the unifying result of calculus, formalizing the inverse relationship between differentiation (finding rates of change) and integration (finding net area). This connection underpins all applications of calculus from kinematics to optimization.

**Derivation:** Prove FTC Part 1: $\frac{d}{dx} \int_a^x f(t) dt = f(x)$

*Starting from:* Definition of the derivative: $F'(x) = \lim_{h \to 0} \frac{F(x+h) - F(x)}{h}$

1. Substitute $F(x) = \int_a^x f(t) dt$: $$\frac{F(x+h) - F(x)}{h} = \frac{1}{h} \left(\int_a^{x+h} f(t) dt - \int_a^x f(t) dt\right) = \frac{1}{h} \int_x^{x+h} f(t) dt$$
2. By the Mean Value Theorem for integrals, there exists some $c$ between $x$ and $x+h$ such that $\int_x^{x+h} f(t) dt = f(c) h$.
3. Substitute back to get $\frac{1}{h} \cdot f(c) h = f(c)$.
4. As $h \to 0$, $c \to x$, and since $f$ is continuous, $\lim_{h \to 0} f(c) = f(x)$.

*Conclusion:* Therefore, $F'(x) = f(x)$, which completes the proof.

**Check your understanding**

Test your understanding

1. If $F(x) = \int_5^x \frac{1}{1 + t^3} dt$, what is $F'(2)$?

   - A) $\frac{1}{9}$
   - B) $\frac{1}{1+8}$
   - C) $\frac{1}{2}$
   - D) $0$

   *Why:* By FTC Part 1, $F'(x) = \frac{1}{1+x^3}$, so $F'(2) = \frac{1}{1+8} = \frac{1}{9}$.

*Calculator:* forbidden

## Common pitfalls

- **Wrong:** Forgetting the chain rule when differentiating integrals with variable bounds
  - Why it fails: FTC Part 1 is stated for an upper bound of $x$, not a function of $x$, so the chain rule is required for composite bounds
  - Correct: Always apply: $\frac{d}{dx} \int_a^{u(x)} f(t) dt = f(u(x)) \cdot u'(x)$
- **Wrong:** Calculating $F(a) - F(b)$ instead of $F(b) - F(a)$ for definite integrals
  - Why it fails: Mixing up the order of lower and upper bounds, leading to a negative result for positive area
  - Correct: Always subtract the value at the lower bound from the value at the upper bound
- **Wrong:** Retaining the dummy variable of integration after evaluation
  - Why it fails: Confusing the bound variable with the integration variable inside the integral
  - Correct: The dummy variable is eliminated after substitution of the bound values, so the final result will never contain the integration variable
- **Wrong:** Applying FTC to discontinuous functions over intervals containing discontinuities
  - Why it fails: The theorem requires continuity of $f$ on the entire closed interval $[a,b]$
  - Correct: Split the integral at the discontinuity and evaluate each part separately, checking for convergence

## Cheatsheet

| Result Type | Formula | Use Case |
| --- | --- | --- |
| FTC Part 1 | $\frac{d}{dx} \int_a^{u(x)} f(t) dt = f(u(x))u'(x)$ | Differentiate variable-bound integrals |
| FTC Part 2 | $\int_a^b f(x) dx = F(b) - F(a), F'=f$ | Evaluate exact definite integrals |
| Two variable bounds | $\frac{d}{dx} \int_{v(x)}^{u(x)} f(t) dt = f(u)u' - f(v)v'$ | Differentiate integrals with two variable bounds |
| Constant of integration | $C$ cancels in $F(b)-F(a)$ | Definite integral evaluation |

## What's next

Now that you've mastered the Fundamental Theorem of Calculus, you hold the core tool for all advanced work in integral calculus. The FTC enables you to move beyond approximations to solve applied problems from kinematics to volumes of revolution, and to tackle more complex integration techniques. The inverse relationship between differentiation and integration it establishes is also foundational for solving differential equations, a major exam-weighted topic in IB AA HL that appears across both papers. This topic opens the door to all further calculus work you will do for the exam and in future university study.

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