Definite Integrals and Plane Areas

Q: Do I need to include the constant of integration for definite integrals?

No, the constant cancels out when you subtract the value of the antiderivative at the lower limit from the upper limit, so it can be omitted entirely for definite integral calculations.

CIE IGCSE Additional Mathematics· 25 min read

1. Evaluating Definite Integrals★★☆☆☆⏱ 6 min

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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 $F (b) - F (a)$ .

📘 Definition

Evaluating a Definite Integral

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

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 $a$ .

📐 Worked Example

Evaluate $\int_{1}^{3} (2 x + 3 x^{2}) d x$

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.

2. Area Between a Curve and the X-axis★★★☆☆⏱ 7 min

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.

📐 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
$\text{Total Area} = \frac{16}{3} + \frac{7}{3} = \frac{23}{3} \text{ units}^2$

3. Area Between a Line and a Curve★★★☆☆⏱ 6 min

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.

4. Area Between Two Curves and Composite Areas★★★★☆⏱ 6 min

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: $x^{2} = 0 ⟹ x = 0$ , $2 - x = 0 ⟹ x = 2$ , $x^{2} = 2 - x ⟹ x = 1$ (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
$\text{Total Area} = \frac{1}{3} + \frac{1}{2} = \frac{5}{6} \text{ units}^2$

5. Common Pitfalls

Wrong move:

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:

Regions below the x-axis give negative integral values, so net signed area is not equal to total absolute area.

Correct move:

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 move:

Subtracting the upper function from the lower function when calculating area between two curves, getting a negative result.

Why:

Area is a positive quantity, so subtracting the larger function from the smaller one gives an incorrect negative value.

Correct move:

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 move:

Including the constant of integration $C$ when evaluating definite integrals.

Why:

The constant cancels out when you subtract $F (a)$ from $F (b)$ , so including it is redundant and can lead to arithmetic errors.

Correct move:

Omit $C$ entirely when working with definite integrals.

Wrong move:

Using intersection points outside the bounded region as limits of integration.

Why:

Using incorrect bounds leads to calculating extra area that is not part of the enclosed shape.

Correct move:

Solve for all intersection points of the boundary functions, then confirm they form the bounds of the enclosed region with a rough sketch.

Wrong move:

Attempting to calculate volumes of revolution when asked for plane areas.

Why:

Volumes of revolution are out of scope for the 0606 syllabus, and you will waste time working on an incorrect answer.

Correct move:

Only calculate 2D plane areas using the definite integral rules covered in this guide.

6. Quick Reference Cheatsheet

Scenario	Formula	Key Note
Evaluate definite integral	$\int_{a}^{b} f (x) d x = F (b) - F (a)$	Omit constant of integration $C$
Area between curve and x-axis (curve above axis)	$\int_{a}^{b} f (x) d x$	Take absolute value if curve is below x-axis
Area between two functions ( $f (x) \geq g (x)$ )	$\int_{a}^{b} [f (x) - g (x)] d x$	$a, b$ 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

7. Frequently Asked

Do I take the absolute value of negative integral results for area calculations?

Yes, area is a positive scalar quantity. Always take the absolute value of integrals for regions below the x-axis, or subtract the lower function from the upper function when finding areas between curves to avoid negative results.

Can I use a calculator to evaluate definite integrals?

Only for Paper 2 of the 0606 exam. For Paper 1 (non-calculator), you must evaluate integrals algebraically, including simplifying exact values of logs, exponentials and trigonometric functions.

Do I need to include the constant of integration $C$ for definite integrals?

No, the constant $C$ cancels out when you subtract the value of the antiderivative at the lower limit from the upper limit, so it can be omitted entirely for definite integral calculations.

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.