Integration (Pure 1) — A-Level Mathematics Pure 1 Study Guide

For: A-Level Mathematics candidates sitting Paper 1 (Pure Mathematics 1).

Covers: All integration topics tested in A-Level Mathematics Paper 1: antiderivatives, constant of integration, definite integrals, area under/between curves, signed area, volume of revolution about the x-axis, and reverse chain rule for (ax+b)^n functions.

You should already know: IGCSE / Add-Maths algebra, sketching basic curves, solving linear and simple quadratic equations.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Mathematics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Integration?

Integration is the reverse process of differentiation, used to calculate quantities like total area, volume, and total change from a rate of change function. Also called antidifferentiation, it makes up 15-20% of marks on A-Level Mathematics Paper 1, appearing in both short 1-2 mark questions and longer 5-7 mark problem-solving questions. The core notation uses the $\int$ symbol to denote integration, with $dx$ indicating the variable being integrated (almost always $x$ in Paper 1).

2. Indefinite integral as antiderivative — power rule reversed

If differentiating $F(x)$ gives you $f(x)$, then $F(x)$ is the antiderivative of $f(x)$, written as: $$\int f(x)dx=F(x)$$ The power rule for differentiation states $\frac{d}{dx}[x^n]=n{x}^{n-1}$, so reversing this rule gives the basic integration power rule: you add 1 to the exponent of $x$, then divide by the new exponent. The full formula is: $$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C,n\ne -1$$ This works for all integer, fractional, positive, and negative values of $n$ except $n=-1$ (which you will cover in Pure 3). You can integrate polynomial functions term by term, just like you differentiate term by term.

Worked Example: Integrate $f(x)=4x^3+2x^2-5x+1$

1. Integrate each term separately:

• $\int 4x^{3}dx=4\times \frac{x^4}{4}=x^4$
• $\int 2x^{2}dx=2\times \frac{x^3}{3}=\frac{2x^3}{3}$
• $\int -5xdx=-5\times \frac{x^2}{2}=-\frac{5x^2}{2}$
• $\int 1dx=\int x^{0}dx=x$

2. Combine terms and add the constant of integration: $\int f(x)dx=x^4+\frac{2x^3}{3}-\frac{5x^2}{2}+x+C$ You can verify this by differentiating the result: you will get exactly the original $f(x)$.

Exam Tip: Examiners test this rule with negative and fractional exponents regularly, so rewrite terms like $\frac{1}{x^2}$ as $x^{-2}$ and $\sqrt{x}$ as $x^{1/2}$ before integrating.

3. Constant of integration — using a point to fix it

When you reverse differentiation, the constant term from the original function disappears (since the derivative of a constant is 0). This means there are infinitely many antiderivatives for any function, differing only by a constant value. We account for this by adding the arbitrary constant of integration $C$ to all indefinite integral results.

If you are given a point $(x,y)$ that lies on the curve $y=F(x)$, you can substitute these values into the integrated function to solve for $C$, giving you the unique particular integral for that curve.

Worked Example: The gradient function of a curve is $\frac{dy}{dx}=6x^2-4x$, and the curve passes through the point $(2,10)$. Find the equation of the curve.

1. Integrate the gradient function: $$y=\int (6x^2-4x)dx=2x^3-2x^2+C$$
2. Substitute $x=2,y=10$ to solve for $C$: $$10=2(2{)}^3-2(2{)}^2+C$$ $$10=16-8+C⟹C=2$$
3. Final equation: $y=2x^3-2x^2+2$

Exam Tip: Always write $+C$ for every indefinite integral answer. You will lose 1 mark per question for omitting this, even if all other integration steps are correct.

4. Definite integrals — ${\int }_a^{b}f(x)dx=F(b)-F(a)$

A definite integral has fixed lower limit $a$ and upper limit $b$, and it returns a numerical value instead of a function with an arbitrary constant. The result is calculated using the Fundamental Theorem of Calculus, Part 2: $${\int }_a^{b}f(x)dx=[F(x){]}_a^b=F(b)-F(a)$$ where $F(x)$ is the antiderivative of $f(x)$. The $[F(x){]}_a^b$ notation means you evaluate $F(x)$ at the upper limit $b$, subtract the value of $F(x)$ at the lower limit $a$. The constant $C$ cancels out, so you do not need to write it for definite integrals.

Worked Example: Evaluate ${\int }_1^3(2x+3)dx$

1. Find the antiderivative: $F(x)=x^2+3x$
2. Evaluate at upper limit: $F(3)=9+9=18$
3. Evaluate at lower limit: $F(1)=1+3=4$
4. Subtract: ${\int }_1^3(2x+3)dx=18-4=14$

Exam Tip: Always write the antiderivative in square brackets with limits before substituting values. This makes it easier to spot arithmetic errors, and examiners award partial marks for correct antiderivatives even if your final calculation is wrong.

5. Area under a curve and area between curves

For a non-negative function $f(x)\ge 0$ on the interval $[a,b]$, the definite integral ${\int }_a^{b}f(x)dx$ gives the exact area bounded by the curve $y=f(x)$, the x-axis, and the vertical lines $x=a$ and $x=b$.

For two curves $y=f(x)$ and $y=g(x)$ where $f(x)\ge g(x)$ (f(x) is the upper curve) on the interval $[a,b]$, the area between the two curves is the integral of the difference between the upper and lower function: $$A={\int }_a^b[f(x)-g(x)]dx$$

Worked Example: Find the area between the curves $y=x^2$ and $y=4x$ from $x=0$ to $x=2$, where $4x\ge x^2$ on this interval.

1. Subtract the lower function from the upper function: $4x-x^2$
2. Set up the integral: $A={\int }_0^2(4x-x^2)dx$
3. Evaluate: $[2x^2-\frac{x^3}{3}{]}_0^2=(8-\frac{8}{3})-0=\frac{16}{3}$ square units.

Exam Tip: If the two curves cross at a point between $a$ and $b$, split the integral at the intersection point to avoid signed values canceling each other out.

6. Area below the $x$-axis (signed area) — interpreting negatives

The definite integral calculates signed area: if $f(x)<0$ (the curve is below the x-axis) on the interval $[a,b]$, the integral result will be negative. Since actual area is always positive, you take the absolute value of the integral for regions below the x-axis before adding to get total area.

Worked Example: Find the total area between $y=x^2-4$, the x-axis, $x=0$ and $x=3$.

1. Find x-intercepts in the interval: $x^2-4=0⟹x=2$
2. From $0\le x\le 2$, $f(x)\le 0$, from $2<x\le 3$, $f(x)\ge 0$
3. Calculate integral for lower region: ${\int }_0^2(x^2-4)dx=[\frac{x^3}{3}-4x{]}_0^2=\frac{8}{3}-8=-\frac{16}{3}$, absolute value = $\frac{16}{3}$
4. Calculate integral for upper region: ${\int }_2^3(x^2-4)dx=[\frac{x^3}{3}-4x{]}_2^3=(9-12)-(\frac{8}{3}-8)=\frac{7}{3}$
5. Total area: $\frac{16}{3}+\frac{7}{3}=\frac{23}{3}$ square units.

Exam Trap: Never integrate directly across the x-intercept for area questions. If you just calculate ${\int }_0^3(x^2-4)dx=-3$, you will lose at least 2 marks for ignoring signed area.

7. Volume of revolution about the $x$-axis — $V=\pi {\int }_a^b{y}^{2}dx$

When you rotate the region under the curve $y=f(x)$ from $x=a$ to $x=b$ 360° around the x-axis, you form a solid of revolution. The volume of this solid is derived by summing the volume of thin circular slices (each with area $\pi y^2$ and thickness $dx$) across the interval, giving the formula: $$V=\pi {\int }_a^b{y}^{2}dx$$

Worked Example: Find the volume generated when $y=x^2+1$ is rotated around the x-axis from $x=0$ to $x=1$.

1. Calculate $y^2$: $(x^2+1{)}^2=x^4+2x^2+1$
2. Set up the integral: $V=\pi {\int }_0^1(x^4+2x^2+1)dx$
3. Evaluate the integral: $[\frac{x^5}{5}+\frac{2x^3}{3}+x{]}_0^1=\frac{1}{5}+\frac{2}{3}+1=\frac{28}{15}$
4. Multiply by $\pi$: $V=\frac{28\pi }{15}$ cubic units.

Exam Tip: Always expand $y^2$ fully before integrating. A common mistake is writing $(x^2+1{)}^2=x^4+1$, which will cost you 2+ marks. Write out $(y)(y)$ explicitly to avoid this error.

8. Reverse chain rule for $(ax+b{)}^n$ type integrands

For functions of the form $(ax+b{)}^n$, where $a$, $b$, and $n$ are constants and $n\ne -1$, you use the reverse of the chain rule for differentiation. Recall that $\frac{d}{dx}[(ax+b{)}^{n+1}]=a(n+1)(ax+b{)}^n$, so rearranging gives the integration formula: $$\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+C$$

Worked Example: Integrate $\sqrt{2x-3}=(2x-3{)}^{1/2}$

1. Identify values: $a=2$, $n=1/2$
2. Apply formula: $\frac{(2x-3{)}^{3/2}}{2\times \frac{3}{2}}+C=\frac{(2x-3{)}^{3/2}}{3}+C$
3. Verify by differentiating: $\frac{3}{2}\times \frac{(2x-3{)}^{1/2}\times 2}{3}=(2x-3{)}^{1/2}$, which matches the original integrand.

Exam Tip: The most common mistake here is forgetting to divide by the coefficient $a$ of $x$ inside the bracket. Always differentiate your result to check that you get the original function.

9. Common Pitfalls (and how to avoid them)

• Wrong move: Omitting the $+C$ for indefinite integrals. Why it happens: You focus on integrating terms and forget the arbitrary constant. Correct move: Write $+C$ immediately after integrating all terms, before doing any substitution steps, to lock in the easy mark.
• Wrong move: Integrating $(ax+b{)}^n$ as $\frac{(ax+b{)}^{n+1}}{n+1}$ without dividing by $a$. Why it happens: You confuse the basic power rule with the linear function adjustment. Correct move: Always divide by the coefficient of $x$ inside the bracket, then differentiate your answer to confirm.
• Wrong move: Using the raw signed integral result for total area questions where the curve dips below the x-axis. Why it happens: You mix up the integral's signed value with absolute area. Correct move: First find all x-intercepts in the interval, split the integral at those points, take the absolute value of each integral result before adding.
• Wrong move: Squaring individual terms instead of expanding the full bracket for $y^2$ in volume of revolution questions. Why it happens: You rush the algebra step and make the $(a+b{)}^2=a^2+b^2$ mistake. Correct move: Write out $(y)(y)$ explicitly and expand term by term, cross-checking your expansion before integrating.
• Wrong move: Swapping upper and lower limits in definite integrals, leading to negative results. Why it happens: You mix up the order when substituting values. Correct move: Always subtract the lower limit value from the upper limit value, remember ${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$ if you do swap them by accident.

10. Practice Questions (A-Level Mathematics Paper 1 Style)

Question 1

A curve has gradient function $\frac{dy}{dx}=3x^2-2x+4$, and passes through the point $(1,6)$. (a) Find the equation of the curve. [3 marks] (b) Calculate the area bounded by the curve, the x-axis, $x=0$ and $x=2$. [4 marks]

Worked Solution

(a) Integrate the gradient function: $$y=\int (3x^2-2x+4)dx=x^3-x^2+4x+C$$ Substitute $x=1,y=6$: $$6=1-1+4+C⟹C=2$$ Final equation: $y=x^3-x^2+4x+2$

(b) The function is positive for all $x\ge 0$, so area is: $$A={\int }_0^2(x^3-x^2+4x+2)dx=[\frac{x^4}{4}-\frac{x^3}{3}+2x^2+2x{]}_0^2$$ Evaluate at $x=2$: $\frac{16}{4}-\frac{8}{3}+8+4=4-\frac{8}{3}+12=\frac{40}{3}$ Area = $\frac{40}{3}$ square units.

Question 2

(a) Evaluate $\int (4x-1{)}^{3}dx$ [2 marks] (b) Find the volume of the solid formed when $y=\sqrt{3x+2}$ is rotated 360° about the x-axis from $x=0$ to $x=4$. [4 marks]

Worked Solution

(a) Apply reverse chain rule, $a=4,n=3$: $$\int (4x-1{)}^{3}dx=\frac{(4x-1{)}^4}{4\times 4}+C=\frac{(4x-1{)}^4}{16}+C$$

(b) First calculate $y^2=3x+2$: $$V=\pi {\int }_0^4(3x+2)dx=\pi [\frac{3x^2}{2}+2x{]}_0^4$$ Evaluate integral: $\frac{3(16)}{2}+8=24+8=32$ Volume = $32\pi$ cubic units.

Question 3

Find the total area between the curve $y=x^2-3x$, the x-axis, $x=0$ and $x=4$. [5 marks]

Worked Solution

Find x-intercepts: $x^2-3x=0⟹x=0,x=3$ From $0\le x\le 3$, $y\le 0$; from $3<x\le 4$, $y\ge 0$ Integral for lower region: $${\int }_0^3(x^2-3x)dx=[\frac{x^3}{3}-\frac{3x^2}{2}{]}_0^3=9-\frac{27}{2}=-\frac{9}{2}$$ Absolute value = $\frac{9}{2}$ Integral for upper region: $${\int }_3^4(x^2-3x)dx=[\frac{x^3}{3}-\frac{3x^2}{2}{]}_3^4=(\frac{64}{3}-24)-(-\frac{9}{2})=\frac{5}{6}$$ Total area: $\frac{9}{2}+\frac{5}{6}=\frac{32}{6}=\frac{16}{3}$ square units.

11. Quick Reference Cheatsheet

12. What's Next

Integration is a foundational skill for the rest of the A-Level Mathematics syllabus. In Paper 3 (Pure Mathematics 3), you will build on these basics to learn integration by substitution, integration by parts, and integration of trigonometric, exponential, and logarithmic functions, as well as volumes of revolution about the y-axis. It also directly connects to kinematics in Paper 4 (Mechanics), where you will use integration to find displacement and velocity from acceleration functions.

If you struggle with any of the concepts, rules, or practice questions in this guide, you can ask Ollie for step-by-step explanations, extra practice problems, or clarification of mark scheme conventions at any time. Head to the homepage] to access personalized support, topic quizzes, and full past paper walkthroughs tailored to A-Level Mathematics Paper 1.

Aligned with the Cambridge International AS & A Level Mathematics 9709 syllabus. OwlsAi is not affiliated with Cambridge Assessment International Education.