Integration

CIE A-Level Mathematics· 9709 2025-2026 Syllabus: Section 5. Integration· 20 min read

1. Indefinite Integration and the Power Rule★★☆☆☆⏱ 5 min

📘 Definition

Indefinite Integral

\int f(x) dx

The general family of antiderivatives of $f (x)$ , which always includes an arbitrary constant of integration $C$ to account for all possible antiderivatives.

Example:

$\int 2 x d x = x^{2} + C$

Integration reverses the process of differentiation. For any power function $x^{n}$ where $n \neq = - 1$ , the power rule for integration follows a simple two-step process: add 1 to the exponent, then divide by the new exponent, before adding the constant of integration.

\int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1

📐 Worked Example

Find the indefinite integral of $3 x^{2} - 4 x + 5$ with respect to $x$ .

1
Integrate the first term $3 x^{2}$ using the power rule:
$3 \times \frac{x^{2+1}}{2+1} = \frac{3x^3}{3} = x^3$
2
Integrate the second term $- 4 x$ using the power rule:
$-4 \times \frac{x^{1+1}}{1+1} = \frac{-4x^2}{2} = -2x^2$
3
Integrate the constant term $5 = 5 x^{0}$ :
$5 \times \frac{x^{0+1}}{0+1} = 5x$
4
Add the constant of integration and combine all terms:
$x^3 - 2x^2 + 5x + C$

Exam tip:

Examiners always allocate one full mark for the constant of integration in indefinite integral questions — never leave it out.

2. Definite Integrals and the Fundamental Theorem★★★☆☆⏱ 6 min

📘 Definition

Definite Integral

\int_a^b f(x) dx

A numerical value equal to the net area between $y = f (x)$ , the x-axis, the lower bound $x = a$ , and the upper bound $x = b$ , calculated using the Fundamental Theorem of Calculus.

Example:

$\int_{0}^{2} 2 x d x = [x^{2}]_{0}^{2} = 4 - 0 = 4$

The Fundamental Theorem of Calculus connects antiderivatives to definite integrals. If $F (x)$ is any antiderivative of $f (x)$ , the definite integral between $a$ and $b$ is calculated as:

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

📐 Worked Example

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

1
First find the indefinite integral of the integrand:
$\int (2x + 1) dx = x^2 + x + C$
2
Write the antiderivative in standard bracket notation (the constant cancels so we omit it):
$[x^2 + x]_1^3$
3
Evaluate the antiderivative at the upper bound $x = 3$ :
$F(3) = 3^2 + 3 = 12$
4
Evaluate the antiderivative at the lower bound $x = 1$ :
$F(1) = 1^2 + 1 = 2$
5
Subtract the lower bound result from the upper bound result:
$12 - 2 = 10$

3. Calculating Areas Under Curves★★★☆☆⏱ 6 min

Integration gives the area between a curve and the x-axis between two bounds. If the curve lies entirely above the x-axis, the integral equals the area directly. If the curve crosses the x-axis between your bounds, you must split the integral into separate regions, because the integral will be negative for regions below the axis, and area is always positive.

📐 Worked Example

Find the total area bounded by $y = x^{2} - 4$ , the x-axis, $x = 0$ and $x = 3$ .

1
Find where the curve crosses the x-axis between 0 and 3: $x^{2} - 4 = 0 ⟹ x = 2$ . $y < 0$ for $0 \leq x < 2$ , $y > 0$ for $2 < x \leq 3$ .
2
Calculate the integral for the first (negative) region:
$\int_0^2 (x^2 - 4) dx = \left[ \frac{x^3}{3} - 4x \right]_0^2 = -\frac{16}{3}$
3
Take the absolute value to get the area of the negative region:
$\text{Area}_1 = \left| -\frac{16}{3} \right| = \frac{16}{3}$
4
Calculate the integral for the second (positive) region:
$\int_2^3 (x^2 - 4) dx = \left[ \frac{x^3}{3} - 4x \right]_2^3 = \frac{7}{3}$
5
Add the two areas to get total area:
$\text{Total Area} = \frac{16}{3} + \frac{7}{3} = \frac{23}{3} = 7\frac{2}{3}$

4. Integration of Negative and Fractional Exponents★★★★☆⏱ 5 min

The power rule works for all exponents except $n = - 1$ , including negative and fractional exponents. You just need to rewrite roots and reciprocal terms as power functions first before applying the rule.

📐 Worked Example

Find $\int (2 x + \frac{3}{x ^{2}}) d x$ .

1
Rewrite the integrand terms with exponents:
$2\sqrt{x} = 2x^{1/2}, \quad \frac{3}{x^2} = 3x^{-2}$
2
Integrate the first term: add 1 to the exponent, divide by the new exponent:
$2 \times \frac{x^{3/2}}{3/2} = \frac{4}{3}x^{3/2}$
3
Integrate the second term:
$3 \times \frac{x^{-1}}{-1} = -\frac{3}{x}$
4
Add the constant of integration:
$\frac{4}{3}x^{3/2} - \frac{3}{x} + C$

✓ Quick check

Test your understanding of the power rule for negative exponents:

What is the correct result for $\int x^{- 3} d x$ ?
- $- \frac{1}{3} x^{- 2} + C$
- $- \frac{1}{2} x^{- 2} + C$
- $- 2 x^{- 2} + C$
- $\frac{1}{- 2} x^{- 2}$
Reveal answer
1 —
Add 1 to the exponent: $- 3 + 1 = - 2$ , then divide by the new exponent $- 2$ . Don't forget the constant of integration $C$ !

5. Common Pitfalls

Wrong move:

Forgetting to add the constant of integration for indefinite integrals

Why:

Examiners explicitly allocate one mark for $C$ , so this is an easy mark to lose

Correct move:

Always add $+ C$ at the end of any indefinite integral result

Wrong move:

Not splitting the integral when the curve crosses the x-axis for area problems

Why:

Negative areas from below the axis cancel positive areas, giving a total area smaller than the true value

Correct move:

Find all x-intercepts between your bounds, split the integral by interval, take absolute value of each integral, then add

Wrong move:

Dividing by the original exponent instead of the new exponent

Why:

This is a common arithmetic slip when integrating negative or fractional powers

Correct move:

Always remember: you add 1 to the exponent first, then divide by this new exponent

Wrong move:

Subtracting the upper bound result from the lower bound result

Why:

This gives the wrong sign for positive integrals, leading to lost marks for area problems

Correct move:

Always calculate $F (upper bound) - F (lower bound)$

Wrong move:

Trying to integrate $x^{- 1}$ using the power rule in P1

Why:

This gives division by zero, and integration of $x^{- 1}$ is only covered in P2

Correct move:

If you encounter $x^{- 1}$ in a P1 integration question, check your simplification of the integrand

6. Quick Reference Cheatsheet

Rule Type	Formula	Key Note
Indefinite Power Rule	$\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C, n \neq = - 1$	Always add $+ C$
Definite Integral	$\int_{a}^{b} f (x) d x = F (b) - F (a)$	$F (x)$ is antiderivative of $f (x)$
Area (above x-axis)	Area = $\int_{a}^{b} f (x) d x$	No absolute value needed
Area (below x-axis)	Area = $\int_{a}^{b} f (x) d x$	Always take absolute value

7. Frequently Asked

Do I always need to add the constant of integration?

Always add $+ C$ for indefinite integrals. It cancels out in definite integral calculations, so you can omit it there.

Can area be negative?

Area is always positive. If the curve is below the x-axis, the integral will be negative, so you must take the absolute value to get area.

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 · 12
Definite integral area calculation
2023 · 11
Indefinite integral of polynomial
2024 · 13
Area between curve and axis

Going deeper

syllabusCIE 9709 2025-2026 Official Syllabus

What's Next

Integration is a core foundation for all further calculus topics in A-Level Mathematics. Mastering the basic power rule and area calculation here will prepare you for more advanced concepts, including areas between two curves, volumes of revolution, and integration of transcendental functions like exponentials and logarithms. These topics appear in both Pure Mathematics 2 and 3, so building a strong understanding of basic integration now will make harder problems much easier to tackle later.

→Differentiation
→Pure Mathematics 3