# Definite integration and fundamental theorem of calculus

> IB Mathematics AA SL · IB AA SL
> Source: https://www.owlsprep.com/study/ib-math-aa-sl-u5-definite-integration-and-fundamental-theorem/

This subtopic introduces definite integration, explains how the Fundamental Theorem of Calculus connects differentiation and integration, and teaches how to evaluate definite integrals and interpret their results for IB exams.

**Prerequisites:** [Indefinite integration and antiderivatives](https://www.owlsprep.com/study/ib-math-aa-sl-u5-indefinite-integration-antiderivatives/); [Basic differentiation of common functions](https://www.owlsprep.com/study/ib-math-aa-sl-u4-rules-of-differentiation/)

## Learning objectives

- Interpret definite integrals as net area under a curve
- Apply the First and Second Fundamental Theorems of Calculus correctly
- Evaluate definite integrals of common functions
- Recognize and avoid common exam pitfalls with definite integrals

## Definite Integrals as Net Area

A definite integral of $f(x)$ over $[a, b]$ is the net area between $y=f(x)$, the x-axis, $x=a$ and $x=b$. Area below the x-axis contributes a negative value, while area above contributes a positive value.

**Definite Integral (Riemann Sum Definition)** — The limit of the sum of areas of infinitely many rectangles approximating the area under the curve

*Notation:* \int_a^b f(x) dx = \lim_{n\to\infty} \sum_{i=1}^n f(x_i^*) \Delta x, \Delta x = \frac{b-a}{n}

*Example:* For $f(x)=2$ over $[0,3]$, the sum converges to $6$, the area of the rectangle.

**Worked example:** Find the net area represented by $\int_{-1}^2 2x dx$ using area interpretation.

1. Sketch $y=2x$ over $[-1,2]$. The line crosses the x-axis at $x=0$, negative for $x<0$ and positive for $x>0$.
2. Calculate the area of the triangle below the x-axis:
3. $$\frac{1}{2} \times 1 \times 2 = 1 \implies \text{contribution} = -1$$
4. Calculate the area of the triangle above the x-axis:
5. $$\frac{1}{2} \times 2 \times 4 = 4 \implies \text{contribution} = +4$$
6. Sum to get the final net area:
7. $$\int_{-1}^2 2x dx = -1 + 4 = 3$$

## First Fundamental Theorem of Calculus

The First FTC eliminates the need to calculate Riemann sums, giving a simple way to evaluate definite integrals using antiderivatives.

**First Fundamental Theorem of Calculus** — If $f(x)$ is continuous on $[a,b]$ and $F(x)$ is any antiderivative of $f(x)$ ($F'(x) = f(x)$), then $\int_a^b f(x) dx = F(b) - F(a)$

*Example:* The constant of integration $C$ from the antiderivative cancels out when calculating the difference.

**Worked example:** Evaluate $\int_1^4 (3x^2 + 2e^x) dx$ using the First FTC.

1. First find the general antiderivative of the integrand:
2. $$\int (3x^2 + 2e^x) dx = x^3 + 2e^x + C$$
3. Apply the First FTC by evaluating at the upper and lower limits, then subtracting:
4. $$\left[ x^3 + 2e^x \right]_1^4 = (4^3 + 2e^4) - (1^3 + 2e^1)$$
5. Simplify to get the final result:
6. $$64 + 2e^4 - 1 - 2e = 63 + 2(e^4 - e)$$

> **tip**
>
> Always write your antiderivative in square brackets with limits to show you are following the correct process, even for simple calculations.

## Second Fundamental Theorem of Calculus

The Second FTC proves that differentiation and integration are inverse operations. It is a very common exam question for derivatives of integrals with variable limits.

**Second Fundamental Theorem of Calculus** — If $f(t)$ is continuous on interval $I$, and $a$ is a constant in $I$, then for any $x \in I$: $\frac{d}{dx} \left( \int_a^x f(t) dt \right) = f(x)$

*Example:* Differentiation reverses the effect of integration for a constant lower bound.

**Worked example:** Find $\frac{d}{dx} \left( \int_2^{x^3} \sin(t) dt \right)$.

1. Use the chain rule: let $u = x^3$, so we calculate $\frac{d}{du} \left( \int_2^u \sin(t) dt \right) \times \frac{du}{dx}$.
2. Apply FTC 2 to the first term: the derivative equals the integrand evaluated at $u$:
3. $$\frac{d}{du} \int_2^u \sin(t) dt = \sin(u)$$
4. Calculate the derivative of $u$ with respect to $x$:
5. $$\frac{du}{dx} = \frac{d}{dx}(x^3) = 3x^2$$
6. Multiply and substitute back $u=x^3$:
7. $$\frac{d}{dx} \int_2^{x^3} \sin(t) dt = 3x^2 \sin(x^3)$$

## Key Properties of Definite Integrals

- $\int_a^a f(x) dx = 0$ (no area over an interval of zero width)
- $\int_a^b f(x) dx = - \int_b^a f(x) dx$ (swapping limits changes sign)
- $\int_a^b k f(x) dx = k \int_a^b f(x) dx$ (constant multiple rule)
- $\int_a^b [f(x) \pm g(x)] dx = \int_a^b f(x) dx \pm \int_a^b g(x) dx$ (sum/difference rule)
- $\int_a^c f(x) dx = \int_a^b f(x) dx + \int_b^c f(x) dx$ (additivity over intervals)

**Worked example:** Given $\int_2^6 f(x) dx = 10$ and $\int_4^6 f(x) dx = 3$, find $\int_2^4 f(x) dx$.

1. Apply the additivity property of definite integrals:
2. $$\int_2^6 f(x) dx = \int_2^4 f(x) dx + \int_4^6 f(x) dx$$
3. Rearrange to solve for the required integral:
4. $$\int_2^4 f(x) dx = 10 - 3 = 7$$

## Common pitfalls

- **Wrong:** Forgetting the negative sign for area below the x-axis when calculating net area.
  - Why it fails: Definite integrals calculate net area, not total geometric area, so negative contributions change the result.
  - Correct: Always check for x-intercepts in the interval of integration, and assign the correct sign to each area segment.
- **Wrong:** Only substituting the upper limit of integration, omitting the lower limit substitution in FTC 1.
  - Why it fails: The definite integral is the difference $F(b)-F(a)$, omitting $F(a)$ gives an incorrect result.
  - Correct: Always write the antiderivative in square brackets with limits, e.g. $[F(x)]_a^b$, before computing the difference.
- **Wrong:** Forgetting the chain rule when differentiating an integral with a variable upper limit function.
  - Why it fails: FTC 2 directly applies only when the upper limit is $x$, not a function of $x$.
  - Correct: Always use the chain rule: $\frac{d}{dx} \int_a^{g(x)} f(t) dt = f(g(x))g'(x)$.
- **Wrong:** Calculating net area when the question asks for total geometric area.
  - Why it fails: Students often default to direct definite integral evaluation, leading to cancelled positive and negative areas.
  - Correct: When asked for total area, split the integral at x-intercepts, take the absolute value of each segment, then add.

## Cheatsheet

| Concept | Formula/Rule |
| --- | --- |
| Definite integral via FTC 1 | $\int_a^b f(x) dx = F(b)-F(a), F'=f$ |
| FTC 2 (base case) | $\frac{d}{dx} \int_a^x f(t) dt = f(x)$ |
| FTC 2 + chain rule | $\frac{d}{dx} \int_a^{g(x)} f(t) dt = f(g(x))g'(x)$ |
| Swap limits rule | $\int_a^b f dx = -\int_b^a f dx$ |
| Additivity over intervals | $\int_a^c f dx = \int_a^b f dx + \int_b^c f dx$ |
| Net area | Area above axis minus area below axis |

## What's next

Definite integration is the foundation for all remaining calculus topics in IB AA SL, from substitution to volumes of revolution and differential equations. The Fundamental Theorem of Calculus is a core concept that appears regularly across both Paper 1 and Paper 2 exams, so mastering it now will save you time and marks later. Next, you will learn how to evaluate more complex definite integrals using substitution, before moving on to core applications like finding areas between two curves and volumes of revolution. A strong understanding of the rules and common pitfalls here will make these more advanced topics far easier to master.

- [Integration by substitution, area and volume of revolution](https://www.owlsprep.com/study/ib-math-aa-sl-u5-integration-by-substitution-area-and/)

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