# Integration

> CIE A-Level Mathematics · Pure Mathematics 1 (P1)
> Source: https://www.owlsprep.com/study/cie-9709-u1-integration/

This module covers core integration concepts for CIE A-Level P1, including indefinite integrals, definite integrals, the Fundamental Theorem of Calculus, and area calculation. You will work through step-by-step exam-style examples.

**Prerequisites:** [Differentiation of polynomials](https://www.owlsprep.com/study/cie-9709-u1-differentiation/); Exponent rules for negative and fractional powers

## Learning objectives

- Find indefinite integrals of polynomials, power functions, and root functions
- Integrate expressions of the form $(ax+b)^n$ using the reverse of the chain rule
- Evaluate definite integrals using the Fundamental Theorem of Calculus
- Calculate total areas bounded by curves and the x-axis
- Find the area between a curve and a line or between two curves
- Calculate volumes of revolution about the x-axis and the y-axis
- Avoid common exam errors in integration problems

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 P1 past papers (2016–2025): **Integration** appears **130 times** in the last 10 years — **17.5%** of all P1 questions (130 of 692).

Most-tested forms: Indefinite integration (59), Volume of revolution (29), Area under a curve (28), Area between two curves (17), Definite integration (6).

## Indefinite Integration and the Power Rule

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

*Notation:* \int f(x) dx

*Example:* $\int 2x dx = x^2 + C$

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 $3x^2 - 4x + 5$ with respect to $x$.

1. Integrate the first term $3x^2$ using the power rule:

   $$3 \times \frac{x^{2+1}}{2+1} = \frac{3x^3}{3} = x^3$$
2. Integrate the second term $-4x$ using the power rule:

   $$-4 \times \frac{x^{1+1}}{1+1} = \frac{-4x^2}{2} = -2x^2$$
3. Integrate the constant term $5 = 5x^0$:

   $$5 \times \frac{x^{0+1}}{0+1} = 5x$$
4. Add the constant of integration and combine all terms:

   $$x^3 - 2x^2 + 5x + C$$

A linear expression raised to a power, $(ax+b)^n$, can be integrated directly by reversing the chain rule: apply the power rule to the bracket, then divide by the derivative of the bracket, $a$. This is one of the most frequently tested integration skills.

$$\int (ax+b)^n \, dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C, \quad n \neq -1$$

**Worked example:** Find $\int (2x-3)^4 \, dx$.

1. Raise the power of the bracket by 1 and divide by the new power, exactly as in the basic power rule:

   $$\frac{(2x-3)^{5}}{5}$$
2. Now divide by the derivative of the bracket, $a = 2$, because the inside is linear. This is the step students most often forget:

   $$\frac{(2x-3)^{5}}{5 \times 2} = \frac{(2x-3)^{5}}{10}$$
3. Add the constant of integration:

   $$\frac{(2x-3)^{5}}{10} + C$$

When you are given the gradient function $\frac{dy}{dx}$ and a point the curve passes through, integrate to recover $y$ (remembering $+C$), then substitute the point to find the value of $C$. This picks out the one specific curve rather than the whole family.

**Worked example:** A curve has gradient $\frac{dy}{dx} = 2x + 1$ and passes through the point $(1, -2)$. Find the equation of the curve.

1. Integrate the gradient function to find the general equation, including the constant of integration:

   $$y = \int (2x + 1) \, dx = x^2 + x + C$$
2. Substitute the known values $x=1$, $y=-2$ to form an equation in $C$:

   $$-2 = 1^2 + 1 + C = 2 + C$$
3. Solve for the constant of integration:

   $$C = -4$$
4. Write the equation of the specific curve:

   $$y = x^2 + x - 4$$

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

## Definite Integrals and the Fundamental Theorem

**Definite Integral** — 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.

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

*Example:* $\int_0^2 2x dx = [x^2]_0^2 = 4 - 0 = 4$

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 $\int_1^3 (2x + 1) dx$.

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:

   $$12 - 2 = 10$$

**Worked example:** Evaluate the improper integral $\int_1^{\infty} x^{-2} \, dx$.

1. An infinite upper limit cannot be substituted directly, so replace $\infty$ with a finite value $X$ and integrate as usual:

   $$\int_1^{X} x^{-2} \, dx = \left[ -\frac{1}{x} \right]_1^{X} = 1 - \frac{1}{X}$$
2. Now let $X \to \infty$; the term $\frac{1}{X}$ tends to $0$:

   $$\lim_{X \to \infty} \left( 1 - \frac{1}{X} \right) = 1$$
3. The value is finite, so the integral converges to $1$. If this limit had instead been infinite, the integral would diverge and have no finite value.

## Calculating Areas Under Curves

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: $x^2 - 4 = 0 \implies x=2$. $y<0$ 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:

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

To find the area enclosed between two curves (or a curve and a line), first find the x-coordinates where they intersect by setting the two expressions equal. Between those limits, integrate the upper function minus the lower function. Subtracting first means you never have to split for sign — the result is already the positive enclosed area.

$$\text{Area} = \int_a^b \left( y_{\text{top}} - y_{\text{bottom}} \right) dx$$

**Worked example:** Find the area of the region enclosed between the curve $y = x^2$ and the line $y = x + 2$.

1. Find the intersection points by setting the curve equal to the line:

   $$x^2 = x + 2 \implies x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0$$
2. So the curve and line meet at $x = -1$ and $x = 2$; these are the limits of integration.

   $$x = -1, \quad x = 2$$
3. Between these limits the line lies above the curve, so integrate (line minus curve):

   $$\int_{-1}^{2} \left[ (x + 2) - x^2 \right] dx = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{2}$$
4. Evaluate at the upper limit and the lower limit:

   $$\left( 2 + 4 - \tfrac{8}{3} \right) - \left( \tfrac{1}{2} - 2 + \tfrac{1}{3} \right) = \frac{10}{3} - \left( -\frac{7}{6} \right)$$
5. Combine to get the enclosed area:

   $$\text{Area} = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$$

## Integration of Negative and Fractional Exponents

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 $\int \left( 2\sqrt{x} + \frac{3}{x^2} \right) dx$.

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:

   $$\frac{4}{3}x^{3/2} - \frac{3}{x} + C$$

**Check your understanding**

Test your understanding of the power rule for negative exponents:

1. What is the correct result for $\int x^{-3} dx$?

   - $-\frac{1}{3}x^{-2} + C$
   - $-\frac{1}{2}x^{-2} + C$
   - $-2x^{-2} + C$
   - $\frac{1}{-2}x^{-2}$

   *Answer:* $-\frac{1}{2}x^{-2} + C$

   *Why:* Add 1 to the exponent: $-3 + 1 = -2$, then divide by the new exponent $-2$. Don't forget the constant of integration $C$!

## Volumes of Revolution

When a region bounded by a curve and an axis is rotated through $360^\circ$ about that axis, it sweeps out a solid. Slicing the solid into thin disks of radius $y$ and thickness $dx$, each disk has area $\pi y^2$, and summing them by integration gives the volume. The single most common error is forgetting to square the radius.

$$\text{About the x-axis:} \quad V = \pi \int_a^b y^2 \, dx$$

$$\text{About the y-axis:} \quad V = \pi \int_a^b x^2 \, dy$$

**Worked example:** The region under $y = x^2$ from $x = 0$ to $x = 2$ is rotated $360^\circ$ about the x-axis. Find the volume of the solid formed.

1. Use the x-axis formula and square $y$ first — square the whole expression, do not just copy $y$:

   $$V = \pi \int_0^2 y^2 \, dx = \pi \int_0^2 (x^2)^2 \, dx = \pi \int_0^2 x^4 \, dx$$
2. Integrate using the power rule:

   $$\pi \left[ \frac{x^5}{5} \right]_0^2$$
3. Evaluate at the limits:

   $$\pi \left( \frac{32}{5} - 0 \right) = \frac{32\pi}{5}$$

If the rotated region does not touch the axis, the solid has a hole through it — a washer. Take the volume swept by the outer boundary and subtract the volume swept by the inner boundary. Crucially this is the difference of the squares, $y_{\text{outer}}^2 - y_{\text{inner}}^2$, not the square of the difference.

$$V = \pi \int_a^b \left( y_{\text{outer}}^2 - y_{\text{inner}}^2 \right) dx$$

**Worked example:** The region between the curve $y = 9 - x^2$ and the line $y = 5$ is rotated $360^\circ$ about the x-axis. Find the volume of the solid formed.

1. Find where the curve meets the line to get the limits:

   $$9 - x^2 = 5 \implies x^2 = 4 \implies x = \pm 2$$
2. The outer radius is the curve $y = 9 - x^2$ and the inner radius is the line $y = 5$. Set up the washer integral as a difference of squares:

   $$V = \pi \int_{-2}^{2} \left[ (9 - x^2)^2 - 5^2 \right] dx$$
3. Expand the squares before integrating:

   $$(9 - x^2)^2 - 25 = 81 - 18x^2 + x^4 - 25 = 56 - 18x^2 + x^4$$
4. Integrate term by term:

   $$V = \pi \left[ 56x - 6x^3 + \frac{x^5}{5} \right]_{-2}^{2}$$
5. The integrand is even, so the value equals twice the result from $0$ to $2$:

   $$V = 2\pi \left( 112 - 48 + \frac{32}{5} \right) = 2\pi \cdot \frac{352}{5} = \frac{704\pi}{5}$$

> **Exam tip:** Whenever you rotate about an axis, square the radius before integrating and keep the factor of $\pi$ outside. For a region that does not reach the axis, use outer squared minus inner squared, never the square of the difference.

## Common pitfalls

- **Wrong:** Forgetting to add the constant of integration for indefinite integrals
  - Why it fails: Examiners explicitly allocate one mark for $C$, so this is an easy mark to lose
  - Correct: Always add $+ C$ at the end of any indefinite integral result
- **Wrong:** Not splitting the integral when the curve crosses the x-axis for area problems
  - Why it fails: Negative areas from below the axis cancel positive areas, giving a total area smaller than the true value
  - Correct: Find all x-intercepts between your bounds, split the integral by interval, take absolute value of each integral, then add
- **Wrong:** Dividing by the original exponent instead of the new exponent
  - Why it fails: This is a common arithmetic slip when integrating negative or fractional powers
  - Correct: Always remember: you add 1 to the exponent first, then divide by this new exponent
- **Wrong:** Subtracting the upper bound result from the lower bound result
  - Why it fails: This gives the wrong sign for positive integrals, leading to lost marks for area problems
  - Correct: Always calculate $F(\text{upper bound}) - F(\text{lower bound})$
- **Wrong:** Trying to integrate $x^{-1}$ using the power rule in P1
  - Why it fails: This gives division by zero, and integration of $x^{-1}$ is only covered in P2
  - Correct: If you encounter $x^{-1}$ in a P1 integration question, check your simplification of the integrand

## Cheatsheet

| Rule Type | Formula | Key Note |
| --- | --- | --- |
| Indefinite Power Rule | $\int x^n dx = \frac{x^{n+1}}{n+1} + C, n \neq -1$ | Always add $+C$ |
| Definite Integral | $\int_a^b f(x) dx = F(b) - F(a)$ | $F(x)$ is antiderivative of $f(x)$ |
| Area (above x-axis) | Area = $\int_a^b f(x) dx$ | No absolute value needed |
| Area (below x-axis) | Area = $\left\| \int_a^b f(x) dx \right\|$ | Always take absolute value |
| Linear Power Rule | $\int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C$ | Divide by $a(n+1)$; $n \neq -1$ |
| Area Between Curves | Area = $\int_a^b (y_{\text{top}} - y_{\text{bottom}}) dx$ | Upper minus lower |
| Volume (about x-axis) | $V = \pi \int_a^b y^2 dx$ | Square $y$; keep $\pi$ outside |
| Volume (about y-axis) | $V = \pi \int_a^b x^2 dy$ | Square $x$; keep $\pi$ outside |

## What's next

Integration is a core foundation for all further calculus topics in A-Level Mathematics. The area-between-curves and volume-of-revolution techniques in this topic are themselves part of Pure Mathematics 1 — you have already met them above. What is deferred to Pure Mathematics 2 and 3 is the integration of transcendental functions: exponentials such as $e^x$, the reciprocal $\frac{1}{x}$ (which integrates to a logarithm), and trigonometric functions, together with methods such as integration by substitution and by parts. Building a strong command of the power rule, definite integrals, areas, and volumes now will make those later topics much easier to tackle.

- [Differentiation](https://www.owlsprep.com/study/cie-9709-u1-differentiation/)
- [Pure Mathematics 3](https://www.owlsprep.com/study/cie-9709-u2-overview/)

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