# Definite Integrals and Plane Areas

> CIE IGCSE Additional Mathematics · 0606 (2025-2027)
> Source: https://www.owlsprep.com/study/cie-0606-u14-definite-integrals-and-plane-areas/

This guide covers evaluating definite integrals and calculating plane areas for CIE IGCSE Additional Mathematics 0606. You will learn to find areas under curves, between lines and curves, and between two curves, including composite split areas.

**Prerequisites:** Indefinite integration of polynomials, trigonometric, exponential and logarithmic functions; Solving equations to find intersection points of lines and curves

## Learning objectives

- Evaluate definite integrals as an antiderivative evaluated between the limits, $F(b) - F(a)$
- Calculate the area bounded by a curve and the x-axis, including regions above and below the axis
- Find the area between a line and a curve, and between two non-linear curves
- Split composite plane areas into simpler regions to calculate total area correctly

## Evaluating Definite Integrals

To evaluate a definite integral, first find an antiderivative $F(x)$ of $f(x)$ (integration is the reverse of differentiation), then take its value at the upper limit minus its value at the lower limit. So the definite integral of $f(x)$ between the limits $a$ and $b$ is equal to $F(b) - F(a)$.

**Evaluating a Definite Integral** — The definite integral of a continuous function $f(x)$ over interval $[a,b]$ is equal to the difference between the antiderivative of $f(x)$ evaluated at the upper bound $b$ and the lower bound $a$.

*Notation:* \int_a^b f(x) dx = F(b) - F(a)

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

1. Find the antiderivative of the integrand:
2. $$\int (2x + 3x^2) dx = x^2 + x^3$$
3. Evaluate the antiderivative at the upper limit $x=3$:
4. $$F(3) = 3^2 + 3^3 = 9 + 27 = 36$$
5. Evaluate the antiderivative at the lower limit $x=1$:
6. $$F(1) = 1^2 + 1^3 = 1 + 1 = 2$$
7. Subtract the lower limit value from the upper limit value to get the final result:
8. $$F(3) - F(1) = 36 - 2 = 34$$

> **Exam tip:** Always omit the constant of integration $C$ when evaluating definite integrals, as it cancels out during calculation and will lead to errors if included.

*Calculator:* allowed

## Area Between a Curve and the X-axis

The area between a curve $y=f(x)$ and the x-axis over interval $[a,b]$ is the sum of the absolute values of definite integrals for regions above and below the x-axis. If the curve crosses the x-axis between $a$ and $b$, you must split the integral at each x-intercept to avoid negative area values.

> **warning**
>
> A negative integral result only indicates the region is below the x-axis, not a negative area. Always take the absolute value of integrals for regions below the axis.

**Worked example:** Find the total area between $y = x^2 - 4$ and the x-axis from $x=0$ to $x=3$

1. Find the x-intercept of the curve in the interval $[0,3]$:
2. $$x^2 - 4 = 0 \implies x = 2$$
3. Calculate the area of the region below the x-axis from $x=0$ to $x=2$, taking the absolute value of the integral:
4. $$\left| \int_0^2 (x^2 -4) dx \right| = \left| \left[ \frac{x^3}{3} - 4x \right]_0^2 \right| = \left| \frac{8}{3} - 8 \right| = \frac{16}{3}$$
5. Calculate the area of the region above the x-axis from $x=2$ to $x=3$:
6. $$\int_2^3 (x^2 -4) dx = \left[ \frac{x^3}{3} - 4x \right]_2^3 = (9 - 12) - \left( \frac{8}{3} - 8 \right) = \frac{7}{3}$$
7. Add the two areas to get the total area:
8. $$\text{Total Area} = \frac{16}{3} + \frac{7}{3} = \frac{23}{3} \text{ units}^2$$

## Area Between a Line and a Curve

To find the area between a line $y=g(x)$ and a curve $y=f(x)$, first find their intersection points to get the bounds of integration. If $f(x) \geq g(x)$ across the interval, the area is the integral of the upper function minus the lower function over the interval.

**Worked example:** Find the area enclosed between the curve $y = x^2 + 1$ and the line $y = x + 3$

1. Find the intersection points of the two functions:
2. $$x^2 + 1 = x + 3 \implies x^2 - x - 2 = 0 \implies x = -1, x = 2$$
3. Test a value between $x=-1$ and $x=2$ to find the upper function: at $x=0$, line $y=3 >$ curve $y=1$, so upper function is $y=x+3$, lower function is $y=x^2+1$
4. Set up the integral of upper minus lower function:
5. $$\int_{-1}^2 \left[(x+3) - (x^2 + 1)\right] dx = \int_{-1}^2 (-x^2 + x + 2) dx$$
6. Evaluate the integral:
7. $$\left[ -\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{-1}^2 = \left(-\frac{8}{3} + 2 + 4\right) - \left(\frac{1}{3} + \frac{1}{2} - 2\right) = \frac{9}{2} \text{ units}^2$$

> **Exam tip:** Sketch a quick graph of the line and curve to easily identify which function is the upper function across the interval.

## Area Between Two Curves and Composite Areas

The same upper minus lower function rule applies for areas between two non-linear curves. For composite areas, split the region at points where the upper function changes, calculate the area of each separate region, then sum the results to get the total area.

**Worked example:** Find the total area enclosed by $y = x^2$, $y = 2 - x$, and the x-axis $y=0$

1. Find the key intersection points: $x^2=0 \implies x=0$, $2-x=0 \implies x=2$, $x^2 = 2 - x \implies x=1$ (positive root)
2. Split the region into two parts: $[0,1]$ where upper function is $y=x^2$, lower function $y=0$; and $[1,2]$ where upper function is $y=2-x$, lower function $y=0$
3. Calculate area of first region:
4. $$\int_0^1 x^2 dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3}$$
5. Calculate area of second region:
6. $$\int_1^2 (2 - x) dx = \left[ 2x - \frac{x^2}{2} \right]_1^2 = (4 - 2) - (2 - 0.5) = \frac{1}{2}$$
7. Sum the two areas for total area:
8. $$\text{Total Area} = \frac{1}{3} + \frac{1}{2} = \frac{5}{6} \text{ units}^2$$

## Common pitfalls

- **Wrong:** Forgetting to split the integral when a curve crosses the x-axis, taking the direct integral and getting a lower value than the actual area.
  - Why it fails: Regions below the x-axis give negative integral values, so net signed area is not equal to total absolute area.
  - Correct: Find all x-intercepts between the bounds, split the integral at each intercept, take the absolute value of each integral, then sum the results.
- **Wrong:** Subtracting the upper function from the lower function when calculating area between two curves, getting a negative result.
  - Why it fails: Area is a positive quantity, so subtracting the larger function from the smaller one gives an incorrect negative value.
  - Correct: Test a value between the limits to identify the upper function, then compute upper minus lower, or take the absolute value of the integral result.
- **Wrong:** Including the constant of integration $C$ when evaluating definite integrals.
  - Why it fails: The constant cancels out when you subtract $F(a)$ from $F(b)$, so including it is redundant and can lead to arithmetic errors.
  - Correct: Omit $C$ entirely when working with definite integrals.
- **Wrong:** Using intersection points outside the bounded region as limits of integration.
  - Why it fails: Using incorrect bounds leads to calculating extra area that is not part of the enclosed shape.
  - Correct: Solve for all intersection points of the boundary functions, then confirm they form the bounds of the enclosed region with a rough sketch.
- **Wrong:** Attempting to calculate volumes of revolution when asked for plane areas.
  - Why it fails: Volumes of revolution are out of scope for the 0606 syllabus, and you will waste time working on an incorrect answer.
  - Correct: Only calculate 2D plane areas using the definite integral rules covered in this guide.

## Cheatsheet

| Scenario | Formula | Key Note |
| --- | --- | --- |
| Evaluate definite integral | $\int_a^b f(x) dx = F(b) - F(a)$ | Omit constant of integration $C$ |
| Area between curve and x-axis (curve above axis) | $\int_a^b f(x) dx$ | Take absolute value if curve is below x-axis |
| Area between two functions ($f(x) \geq g(x)$) | $\int_a^b [f(x) - g(x)] dx$ | $a, b$ are intersection points of the two functions |
| Composite area | Sum of areas of individual split regions | Split at points where upper function changes or curve crosses x-axis |

## What's next

Now that you have mastered definite integrals and plane areas, you can move on to applying integration to kinematics problems, the final major calculus topic in the CIE IGCSE Additional Mathematics 0606 syllabus. These skills are frequently tested in both Paper 1 and Paper 2, so make sure you practice both calculator and non-calculator past paper questions to build speed and accuracy. Sketching quick graphs for every area question will help you avoid common mistakes with upper/lower functions and split points, even if the question does not require a diagram.

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