Definite integration and fundamental theorem of calculus

IB Mathematics AA SL· 45 min read

1. Definite Integrals as Net Area★★☆☆☆⏱ 15 min

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.

📘 Definition

Definite Integral (Riemann Sum Definition)

\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}

The limit of the sum of areas of infinitely many rectangles approximating the area under the curve

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 $\int_{- 1}^{2} 2 x d x$ using area interpretation.

1
Sketch $y = 2 x$ 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
$\int_{-1}^2 2x dx = -1 + 4 = 3$

2. First Fundamental Theorem of Calculus★★★☆☆⏱ 20 min

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

📘 Definition

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 $\int_{a}^{b} f (x) d x = F (b) - F (a)$

Example:

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

📐 Worked Example

Evaluate $\int_{1}^{4} (3 x^{2} + 2 e^{x}) d x$ 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
$64 + 2e^4 - 1 - 2e = 63 + 2(e^4 - e)$

3. Second Fundamental Theorem of Calculus★★★★☆⏱ 20 min

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.

📘 Definition

Second Fundamental Theorem of Calculus

If $f (t)$ is continuous on interval $I$ , and $a$ is a constant in $I$ , then for any $x \in I$ : $\frac{d}{d x} (\int_{a}^{x} f (t) d t) = f (x)$

Example:

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

📐 Worked Example

Find $\frac{d}{d x} (\int_{2}^{x^{3}} sin (t) d t)$ .

1
Use the chain rule: let $u = x^{3}$ , so we calculate $\frac{d}{d u} (\int_{2}^{u} sin (t) d t) \times \frac{d u}{d x}$ .
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
$\frac{d}{dx} \int_2^{x^3} \sin(t) dt = 3x^2 \sin(x^3)$

4. Key Properties of Definite Integrals★★☆☆☆⏱ 15 min

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

📐 Worked Example

Given $\int_{2}^{6} f (x) d x = 10$ and $\int_{4}^{6} f (x) d x = 3$ , find $\int_{2}^{4} f (x) d x$ .

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
$\int_2^4 f(x) dx = 10 - 3 = 7$

5. Common Pitfalls

Wrong move:

Forgetting the negative sign for area below the x-axis when calculating net area.

Why:

Definite integrals calculate net area, not total geometric area, so negative contributions change the result.

Correct move:

Always check for x-intercepts in the interval of integration, and assign the correct sign to each area segment.

Wrong move:

Only substituting the upper limit of integration, omitting the lower limit substitution in FTC 1.

Why:

The definite integral is the difference $F (b) - F (a)$ , omitting $F (a)$ gives an incorrect result.

Correct move:

Always write the antiderivative in square brackets with limits, e.g. $[F (x)]_{a}^{b}$ , before computing the difference.

Wrong move:

Forgetting the chain rule when differentiating an integral with a variable upper limit function.

Why:

FTC 2 directly applies only when the upper limit is $x$ , not a function of $x$ .

Correct move:

Always use the chain rule: $\frac{d}{d x} \int_{a}^{g (x)} f (t) d t = f (g (x)) g^{'} (x)$ .

Wrong move:

Calculating net area when the question asks for total geometric area.

Why:

Students often default to direct definite integral evaluation, leading to cancelled positive and negative areas.

Correct move:

When asked for total area, split the integral at x-intercepts, take the absolute value of each segment, then add.

6. Quick Reference Cheatsheet

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

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Evaluate definite integral of exponential
2023 · 1
Differentiate integral with variable limit

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