# Definite integrals and area under curves

> IB Mathematics Applications and Interpretation HL · Unit 4: Calculus
> Source: https://www.owlsprep.com/study/ib-math-ai-hl-u4-definite-integrals-and-area-under/

This module covers evaluating definite integrals via the Fundamental Theorem of Calculus, and using definite integrals to find net and total area under a curve, including cases where the curve crosses the x-axis, with real-world applications common in IB exams.

**Prerequisites:** [Antiderivatives and indefinite integration](https://www.owlsprep.com/study/ib-math-ai-hl-u4-antiderivatives-indefinite-integration/); [Basic differentiation rules](https://www.owlsprep.com/study/ib-math-ai-hl-u4-basic-differentiation-rules/)

## Learning objectives

- Evaluate definite integrals using the Fundamental Theorem of Calculus
- Distinguish between net area and total geometric area under a curve
- Calculate total area between a curve and the x-axis when the curve crosses the axis
- Apply area calculations to real-world contexts like kinematics

## Definite Integrals: Definition and Evaluation

**Definite Integral** — The definite integral of a function f(x) from lower limit a to upper limit b is computed via the Fundamental Theorem of Calculus, producing a numerical result rather than a function.

*Notation:* \int_a^b f(x) \, dx

*Example:* \int_2^4 2x \, dx = [x^2]_2^4 = 16 - 4 = 12

The Fundamental Theorem of Calculus (FTC) Part 2 tells us that if $F(x)$ is an antiderivative of $f(x)$, meaning $F'(x) = f(x)$, then:

$$\int_a^b f(x) \, dx = F(b) - F(a)$$

**Worked example:** Evaluate the definite integral $\int_1^3 (3x^2 + 2x) \, dx$

1. First find the general antiderivative of the integrand. The constant of integration cancels out in definite integrals, so we can omit it.
2. $$\int (3x^2 + 2x) dx = x^3 + x^2$$
3. Write the antiderivative in square brackets with the upper and lower bounds marked:
4. $$[x^3 + x^2]_1^3$$
5. Evaluate at the upper bound, subtract the value at the lower bound:
6. $$(3^3 + 3^2) - (1^3 + 1^2) = (27 + 9) - (1 + 1) = 34$$

## Net Area vs Total Area

A common source of exam confusion is the difference between the value of a definite integral (net area) and the total geometric area between a curve and the x-axis.

**Net Area** — The value of the definite integral counts area above the x-axis as positive and area below the x-axis as negative, resulting in a net signed area.

**Total Area** — Total geometric area is the sum of the absolute values of each signed region, so all area counts as positive regardless of position relative to the x-axis.

> **warning**
>
> IB exam questions always specify whether they want the value of the integral (net area) or total geometric area. Read the question wording very carefully.

**Check your understanding**

Test your understanding of the difference between net and total area:

1. What is the value of $\int_{-2}^2 x dx$ (net area)?

   - 0
   - 2
   - 4
   - -4

   *Why:* Correct! The area above the axis from 0 to 2 cancels the negative area below from -2 to 0, so net area is 0, while total geometric area is 4.

## Calculating Total Area Between a Curve and the x-axis

To find the total geometric area between $y=f(x)$, the x-axis, $x=a$ and $x=b$, follow this 3-step process:
1. Find all x-intercepts of $f(x)$ between $a$ and $b$
2. Split the interval into sub-intervals between consecutive intercepts
3. Take the absolute value of each sub-integral and add them together

**Worked example:** Find the total geometric area between $y = x$ and the x-axis from $x=-2$ to $x=2$.

1. Find the x-intercept to split the interval: $y=x$ crosses the x-axis at $x=0$, so we split into $[-2, 0]$ and $[0, 2]$.
2. Evaluate the integral over each sub-interval:
3. $$\int_{-2}^0 x dx = \left[\frac{1}{2}x^2\right]_{-2}^0 = 0 - 2 = -2$$
4. $$\int_0^2 x dx = \left[\frac{1}{2}x^2\right]_0^2 = 2 - 0 = 2$$
5. Take absolute values of each result and add for total area:
6. $$|-2| + |2| = 2 + 2 = 4$$

## Real-World Applications

Definite integrals and area calculations are very common in real-world modelling problems in IB AI HL. For example, the net area under a velocity-time graph gives total displacement, while total area gives total distance travelled.

**Worked example:** A ball is thrown straight up, with velocity $v(t) = 20 - 10t$ m/s, where $t$ is time in seconds after release. Find the total distance travelled by the ball between $t=0$ and $t=3$ seconds.

1. Find where velocity changes sign: set $v(t)=0$ → $20 - 10t = 0$ → $t=2$ seconds. Split into $[0,2]$ (upward movement, positive velocity) and $[2,3]$ (downward movement, negative velocity).
2. Integrate over each interval:
3. $$\int_0^2 (20 - 10t) dt = [20t - 5t^2]_0^2 = 20$$
4. $$\int_2^3 (20 - 10t) dt = [20t - 5t^2]_2^3 = -5$$
5. Add absolute values for total distance:
6. $$|20| + |-5| = 25 \text{ meters}$$

## The Trapezoidal Rule for Numerical Integration

When a function is hard or impossible to integrate exactly, or when you only have a table of data values, you can still estimate a definite integral numerically. The trapezoidal rule is the signature numerical method in IB AI, approximating the area under a curve with a series of trapezoids rather than rectangles.

**Trapezoidal Rule** — Divide $[a,b]$ into $n$ strips of equal width $h = \frac{b-a}{n}$, giving ordinates $y_0, y_1, \ldots, y_n$. Each strip is approximated by a trapezoid; the end ordinates are counted once and every interior ordinate is counted twice.

*Notation:* \int_a^b y \, dx \approx \frac{h}{2}\left(y_0 + y_n + 2(y_1 + y_2 + \cdots + y_{n-1})\right)

*Example:* With 4 strips you need 5 ordinates $y_0, y_1, y_2, y_3, y_4$, and the two interior-pair ordinates $y_1, y_2, y_3$ are doubled.

> **note**
>
> The trapezoidal rule gives an approximation, not an exact value. For a curve that is concave up (bending upwards) the trapezoids sit above the curve, so the rule overestimates the true area; for a concave-down curve it underestimates.

**Worked example:** Use the trapezoidal rule with 4 strips to estimate $\int_0^2 \sqrt{1 + x^3} \, dx$, giving your answer to 3 significant figures.

1. Find the strip width: $h = \frac{2 - 0}{4} = 0.5$, so the x-values are $0, 0.5, 1, 1.5, 2$.
2. Evaluate the ordinates $y = \sqrt{1 + x^3}$ at each x-value:
3. $$y_0 = 1, \ y_1 = 1.0607, \ y_2 = 1.4142, \ y_3 = 2.0917, \ y_4 = 3$$
4. Apply the formula, doubling the interior ordinates $y_1, y_2, y_3$:
5. $$\int_0^2 \sqrt{1+x^3}\,dx \approx \frac{0.5}{2}\left(1 + 3 + 2(1.0607 + 1.4142 + 2.0917)\right)$$
6. $$= 0.25\left(4 + 2(4.5666)\right) = 0.25(13.133) \approx 3.28$$

**Check your understanding**

Test your understanding of the trapezoidal rule:

1. To estimate an integral with the trapezoidal rule using 5 strips of equal width, how many ordinates (y-values) do you need?

   - 5
   - 6
   - 10

   *Why:* Correct! With $n$ strips you need $n+1$ ordinates, so 5 strips require 6 y-values: $y_0$ through $y_5$.

## Common pitfalls

- **Wrong:** Forgetting to split the integral at x-intercepts when calculating total area
  - Why it fails: Integrating over the entire interval lets negative area cancel positive area, resulting in net area instead of total area
  - Correct: Always find all x-intercepts in the interval, split the integral into sub-regions, then add absolute values of each part
- **Wrong:** Assuming the value of the definite integral is always equal to geometric area
  - Why it fails: The definite integral gives signed net area, which can be zero or negative even when total area is positive
  - Correct: Check the question wording: if it asks for area, it means total positive geometric area
- **Wrong:** Evaluating $F(a) - F(b)$ instead of $F(b) - F(a)$ for the definite integral
  - Why it fails: Swapping the bounds flips the sign of the result, leading to an incorrect answer, often negative area
  - Correct: Always subtract the antiderivative evaluated at the lower limit from the value at the upper limit
- **Wrong:** Forgetting to take the absolute value of negative sub-integrals for total area
  - Why it fails: Leaving negative area as negative reduces the final total, leading to an incorrect result
  - Correct: Take the absolute value of each sub-integral before adding to get total area

## Cheatsheet

| Concept | Formula / Method |
| --- | --- |
| Definite integral (FTC) | $\int_a^b f(x) dx = F(b) - F(a), F'(x)=f(x)$ |
| Net area (integral value) | Area above x-axis minus area below |
| Total geometric area | Sum of $\left\|\int_{x_i}^{x_{i+1}} f(x) dx\right\|$ for each interval between intercepts |
| Area above x-axis | $\int_a^b f(x) dx, f(x) \geq 0$ |
| Area below x-axis | $-\int_a^b f(x) dx, f(x) \leq 0$ |

## What's next

Now that you have mastered definite integrals and area under curves, you have built a core foundational skill for all advanced integration applications in IB AI HL. Definite integrals are used to calculate volumes of revolution, the average value of a function, and accumulated quantities like population growth or energy use in modelling problems. Understanding the difference between net and total area is also critical for interpreting integration results when working with real-world data and functions in your internal assessment and exams. You can now extend your skills to more complex integration scenarios that appear regularly in IB assessments.

- [Integration by substitution](https://www.owlsprep.com/study/ib-math-ai-hl-u4-integration-techniques-substitution-and-by/)
- [Area between curves, volumes of revolution](https://www.owlsprep.com/study/ib-math-ai-hl-u4-area-between-curves-volumes-of/)
- [First order differential equations: Euler's method](https://www.owlsprep.com/study/ib-math-ai-hl-u4-first-order-differential-equations-euler/)

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