Integration

1. Overview

Integration is a core topic in IGCSE Additional Mathematics (0606). It's the reverse process of differentiation, allowing you to find functions from their derivatives, calculate areas under curves, and solve problems in kinematics and other applications. Mastering integration is crucial for success in both Paper 1 (non-calculator) and Paper 2. Expect to see integration problems in various contexts, often linked to differentiation questions via the word "hence". This revision note covers the key integration techniques, common mistakes, and exam strategies to help you excel.

Key Definitions

• Indefinite Integral: An integral without limits, representing a family of functions. It must always include an arbitrary constant ($+ c$).
• Definite Integral: An integral evaluated between two specific values (limits), resulting in a fixed numerical value.
• Constant of Integration ($c$): A constant added to the end of an indefinite integral to account for the constant term lost during differentiation.
• Integrand: The function being integrated (the expression following the $\int$ symbol).
• Limits of Integration: The values $a$ and $b$ in $\int_a^b f(x) dx$ that define the interval of integration.

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Core Content

A. Basic Rules and the Power Rule

To integrate $x^n$, increase the power by 1 and divide by the new power. This is a fundamental rule you'll use constantly.

Formula (memorise): $$\int x^n dx = \frac{x^{n+1}}{n+1} + c \quad (\text{where } n \neq -1)$$

For the special case where $n = -1$:

Formula (memorise): $$\int \frac{1}{x} dx = \ln|x| + c$$

B. Integrating Linear Composites: $(ax + b)^n$

When integrating a function of a linear expression $(ax+b)$, you must divide by the coefficient of $x$ (the $a$ value). This is effectively the reverse of the chain rule.

Formula (memorise): $$\int (ax + b)^n dx = \frac{(ax + b)^{n+1}}{a(n+1)} + c \quad (n \neq -1)$$

Formula (memorise): $$\int \frac{1}{ax + b} dx = \frac{1}{a} \ln|ax + b| + c$$

C. Trigonometric and Exponential Functions

Note: All trigonometric integration in Additional Maths is performed in radians. Make sure your calculator is in radian mode for Paper 2.

Formulas (memorise):

• $\int \sin(ax + b) dx = -\frac{1}{a} \cos(ax + b) + c$
• $\int \cos(ax + b) dx = \frac{1}{a} \sin(ax + b) + c$
• $\int \sec^2(ax + b) dx = \frac{1}{a} \tan(ax + b) + c$
• $\int e^{ax + b} dx = \frac{1}{a} e^{ax + b} + c$

D. Definite Integrals and Area

The definite integral $\int_a^b f(x) dx$ calculates the signed area between the curve and the $x$-axis. Remember to show your substitution of limits clearly.

1. Area between a curve and the x-axis: $A = \int_a^b y dx$
2. Area between a curve and a line (or two curves): $A = \int_a^b (y_{top} - y_{bottom}) dx$

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Worked Examples

Worked example 1 — Indefinite Integration (Power Rule and Linear Form)

Find $\int \left( 6x^5 + \frac{4}{(3x+2)^3} \right) dx$.

Step 1: Rewrite the expression in index form. $$\int \left( 6x^5 + 4(3x+2)^{-3} \right) dx$$ Step 2: Integrate term by term. For $6x^5$: $$\int 6x^5 dx = \frac{6x^6}{6} = x^6$$ Apply the power rule. For $4(3x+2)^{-3}$: $$\int 4(3x+2)^{-3} dx = \frac{4(3x+2)^{-2}}{-2 \times 3} = -\frac{2}{3(3x+2)^2}$$ Apply the power rule for linear composites, dividing by the coefficient of x (which is 3). Step 3: Combine and add the constant $c$. $$x^6 - \frac{2}{3(3x+2)^2} + c$$

Final Answer: $$x^6 - \frac{2}{3(3x+2)^2} + c$$

Worked example 2 — Definite Integration with Trigonometry

Evaluate $\int_{\pi/6}^{\pi/3} 2\sin(3x) dx$, leaving your answer in exact form.

Step 1: Integrate the function. $$\int 2\sin(3x) dx = 2 \left[ -\frac{1}{3} \cos(3x) \right]{\pi/6}^{\pi/3} = \left[ -\frac{2}{3} \cos(3x) \right]{\pi/6}^{\pi/3}$$ Apply the integral of sin(ax+b). Step 2: Substitute the upper and lower limits. $$\left( -\frac{2}{3} \cos(3 \cdot \frac{\pi}{3}) \right) - \left( -\frac{2}{3} \cos(3 \cdot \frac{\pi}{6}) \right)$$ Substitute the limits of integration. Step 3: Simplify using exact trig values. $$-\frac{2}{3} \cos(\pi) + \frac{2}{3} \cos(\frac{\pi}{2})$$ $$-\frac{2}{3}(-1) + \frac{2}{3}(0) = \frac{2}{3}$$ Evaluate the cosine terms.

Final Answer: $$\frac{2}{3}$$

Worked example 3 — Area between a line and a curve

Find the area of the region bounded by $y = x^2 + 1$ and $y = 3x + 1$.

Step 1: Find intersection points to determine limits. $$x^2 + 1 = 3x + 1 \implies x^2 - 3x = 0 \implies x(x-3) = 0$$ Limits are $x=0$ and $x=3$. Equate the two functions and solve for x. Step 2: Identify the "top" function. Between 0 and 3, $3x + 1 > x^2 + 1$. Choose a test value between 0 and 3, such as x=1, to determine which function is greater. Step 3: Set up the integral. $$Area = \int_0^3 ((3x + 1) - (x^2 + 1)) dx = \int_0^3 (3x - x^2) dx$$ Subtract the bottom function from the top function. Step 4: Integrate and evaluate. $$\left[ \frac{3x^2}{2} - \frac{x^3}{3} \right]_0^3 = \left( \frac{3(3)^2}{2} - \frac{(3)^3}{3} \right) - \left( \frac{3(0)^2}{2} - \frac{(0)^3}{3} \right)$$ $$\frac{27}{2} - \frac{27}{3} = \frac{27}{2} - 9 = \frac{27 - 18}{2} = \frac{9}{2}$$ Apply the power rule and substitute the limits.

Final Answer: $$\frac{9}{2} \text{ square units.}$$

Worked example 4 — Integrating $e^{ax+b}$

Evaluate $\int_0^1 4e^{2x} dx$, leaving your answer in exact form.

Step 1: Integrate the function. $$\int 4e^{2x} dx = \left[ \frac{4}{2} e^{2x} \right]_0^1 = \left[ 2e^{2x} \right]_0^1$$ Apply the integral of $e^{ax+b}$. Step 2: Substitute the upper and lower limits. $$2e^{2(1)} - 2e^{2(0)}$$ Substitute the limits of integration. Step 3: Simplify. $$2e^2 - 2e^0 = 2e^2 - 2(1) = 2e^2 - 2$$ Evaluate the exponential terms.

Final Answer: $$2e^2 - 2$$

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Extended Content (Extended Only)

Additional Mathematics is a single-tier syllabus — all content above applies to all students.

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Key Equations

Function $f(x)$	Integral $\int f(x) dx$	Notes
$x^n$	$\frac{x^{n+1}}{n+1} + c$	$n \neq -1$
$\frac{1}{x}$	$\ln	x
$e^{ax+b}$	$\frac{1}{a}e^{ax+b} + c$	$a$ is the coefficient of $x$
$\sin(ax+b)$	$-\frac{1}{a}\cos(ax+b) + c$	Radians only
$\cos(ax+b)$	$\frac{1}{a}\sin(ax+b) + c$	Radians only
$\sec^2(ax+b)$	$\frac{1}{a}\tan(ax+b) + c$	Radians only

Note: These formulas must be memorised, as they are not all provided on the 0606 formula sheet.

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Common Mistakes to Avoid

• ❌ Wrong: $\int \cos(3x) dx = 3\sin(3x) + c$ ✓ Right: $\int \cos(3x) dx = \frac{1}{3}\sin(3x) + c$ (Divide by $a$ in integration; multiply in differentiation).
• ❌ Wrong: $\int \frac{1}{x^2} dx = \ln(x^2) + c$ ✓ Right: $\int x^{-2} dx = -x^{-1} + c$ (Only $1/x$ results in a natural log).
• ❌ Wrong: Forgetting $+ c$ on indefinite integrals. (This will cost 1 mark every time).
• ❌ Wrong: Using $0.707$ for $\sin(\pi/4)$. ✓ Right: Use $\frac{\sqrt{2}}{2}$ or $\frac{1}{\sqrt{2}}$. Always use exact values unless decimals are requested.
• ❌ Wrong: $\int e^{5x} dx = 5e^{5x} + c$ ✓ Right: $\int e^{5x} dx = \frac{1}{5}e^{5x} + c$ (Remember to divide by the coefficient of $x$).
• ❌ Wrong: $\int \frac{5}{2x+1} dx = 5\ln|2x+1| + c$ ✓ Right: $\int \frac{5}{2x+1} dx = \frac{5}{2}\ln|2x+1| + c$ (Divide by the coefficient of $x$).
• ❌ Wrong: Not showing substitution of limits in definite integrals, especially when one limit is zero. ✓ Right: Show $[F(b)] - [F(a)]$ even if $F(a) = 0$.
• ❌ Wrong: Forgetting brackets when substituting limits, leading to sign errors. ✓ Right: Use brackets carefully: $[F(b)] - [F(a)]$.
• ❌ Wrong: $\ln(-2)$ is defined. ✓ Right: $\ln(x)$ is only defined for $x > 0$. Check for domain errors when dealing with logarithms.

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Exam Tips

• The "Hence" Rule: If Part (a) asks you to differentiate $x \ln x$ and Part (b) asks you to integrate $\ln x$, use your answer from (a). Integration is the reverse of differentiation; look for the link!
• Show Limit Substitutions: Even if the lower limit is 0, write out the substitution step. Examiners need to see $( \dots ) - ( 0 )$. For functions like $\cos(0)$ or $e^0$, the result is not zero.
• Command Words:
  • "Find the exact value": Use surds, $\pi$, $e$, and fractions. No calculator decimals unless explicitly allowed.
  • "Hence, evaluate": You must use the result from the previous part of the question.
• Calculators: In Paper 2, use the integral function on your calculator to verify your definite integral answer, but you must still show the algebraic integration steps to earn full marks.
• Check Domain: $\ln(x)$ is only defined for $x > 0$. If your area calculation involves negative $x$ values, re-check your boundaries.
• Paper 1 Focus: Paper 1 (non-calculator) will test your ability to integrate algebraically and evaluate definite integrals with exact values. Practice trigonometric integrals and simplifying surds.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0606 papers.

Exam-Style Question 1 — Paper 1 (No Calculator Allowed) [7 marks]

Question:

(a) Find $\int (4e^{2x} - \frac{6}{x+1}) dx$. [4]

(b) Given that $y = 5$ when $x = 0$, find the particular solution for $y$ in terms of $x$. [3]

Worked Solution:

(a)

1. Integrate $4e^{2x}$ $\int 4e^{2x} dx = 2e^{2x}$ [Using the standard integral of $e^{ax}$]

2. Integrate $-\frac{6}{x+1}$ $\int -\frac{6}{x+1} dx = -6\ln|x+1|$ [Using the standard integral of $\frac{1}{x}$]

3. Combine and add the constant of integration. $\int (4e^{2x} - \frac{6}{x+1}) dx = 2e^{2x} - 6\ln|x+1| + c$ [Putting the two integrals together]

(b)

1. Substitute $x=0$ and $y=5$ into the general solution. $5 = 2e^{2(0)} - 6\ln|0+1| + c$ [Substituting the given values to find c]

2. Simplify to find c. $5 = 2(1) - 6(0) + c \implies c = 3$ [Evaluating $e^0$ and $\ln 1$]

3. State the particular solution. $y = 2e^{2x} - 6\ln|x+1| + 3$ [Substituting the value of c back into the general solution]

Final Answer: $\boxed{y = 2e^{2x} - 6\ln|x+1| + 3}$

How to earn full marks: For part (a), remember to include the constant of integration, "+ c". For part (b), show your substitution of the initial values clearly.

Exam-Style Question 2 — Paper 1 (No Calculator Allowed) [8 marks]

Question:

(a) Given that $\frac{5x-1}{(x-1)(x+2)}$ can be written in the form $\frac{A}{x-1} + \frac{B}{x+2}$, find the values of $A$ and $B$. [3]

(b) Hence, find $\int \frac{5x-1}{(x-1)(x+2)} dx$. [5]

Worked Solution:

(a)

1. Express the right-hand side with a common denominator. $\frac{A}{x-1} + \frac{B}{x+2} = \frac{A(x+2) + B(x-1)}{(x-1)(x+2)}$ [Combining the fractions]

2. Equate the numerators. $5x - 1 = A(x+2) + B(x-1)$ [Equating numerators of the two fractions]

3. Solve for $A$ and $B$. Let $x = 1$: $5(1) - 1 = A(1+2) + B(1-1) \implies 4 = 3A \implies A = \frac{4}{3}$. Let $x = -2$: $5(-2) - 1 = A(-2+2) + B(-2-1) \implies -11 = -3B \implies B = \frac{11}{3}$. [Solving for A and B by substituting values of x]

(b)

1. Rewrite the integral using the partial fractions found in part (a). $\int \frac{5x-1}{(x-1)(x+2)} dx = \int (\frac{4/3}{x-1} + \frac{11/3}{x+2}) dx$ [Using the result from part (a)]

2. Integrate $\frac{4/3}{x-1}$. $\int \frac{4/3}{x-1} dx = \frac{4}{3} \ln|x-1|$ [Integrating the first term]

3. Integrate $\frac{11/3}{x+2}$. $\int \frac{11/3}{x+2} dx = \frac{11}{3} \ln|x+2|$ [Integrating the second term]

4. Combine and add the constant of integration. $\int \frac{5x-1}{(x-1)(x+2)} dx = \frac{4}{3} \ln|x-1| + \frac{11}{3} \ln|x+2| + c$ [Combining the two integrals]

Final Answer: $\boxed{\frac{4}{3} \ln|x-1| + \frac{11}{3} \ln|x+2| + c}$

How to earn full marks: In part (a), show your working clearly when solving for A and B. In part (b), don't forget the constant of integration and the absolute value signs within the logarithms.

Exam-Style Question 3 — Paper 2 (Calculator Allowed) [7 marks]

Question:

The diagram shows the curve $y = x^2 \sqrt{4-x}$, for $0 \le x \le 4$.

(a) Find the $x$-coordinate of the point on the curve where $y$ is a maximum. Give your answer to 3 significant figures. [3]

(b) Calculate the area of the region enclosed by the curve and the $x$-axis. Give your answer to 3 significant figures. [4]

Worked Solution:

(a)

1. Differentiate $y = x^2 \sqrt{4-x}$ using the product rule. $\frac{dy}{dx} = x^2 \cdot \frac{1}{2}(4-x)^{-1/2} \cdot (-1) + 2x \sqrt{4-x} = -\frac{x^2}{2\sqrt{4-x}} + 2x\sqrt{4-x}$ [Applying the product rule]

2. Set $\frac{dy}{dx} = 0$ and solve for $x$. $0 = -\frac{x^2}{2\sqrt{4-x}} + 2x\sqrt{4-x} \implies \frac{x^2}{2\sqrt{4-x}} = 2x\sqrt{4-x} \implies x^2 = 4x(4-x) \implies x^2 = 16x - 4x^2 \implies 5x^2 - 16x = 0 \implies x(5x-16) = 0$ [Setting the derivative equal to zero to find the maximum]

3. Solve for $x$. $x = 0$ or $x = \frac{16}{5} = 3.2$ [Solving the quadratic equation] Since $x=0$ is the minimum, $x = 3.2$.

(b)

1. Set up the definite integral for the area. $Area = \int_0^4 x^2 \sqrt{4-x} dx$ [Setting up the definite integral]

2. Evaluate the definite integral using a calculator. $Area = \int_0^4 x^2 \sqrt{4-x} dx \approx 8.5333...$ [Using a calculator to evaluate the definite integral]

3. Round to 3 significant figures. $Area \approx 8.53$

Final Answer: (a) $\boxed{3.20}$ (b) $\boxed{8.53}$

How to earn full marks: In part (a), show the differentiation process and the steps to solve for x. In part (b), state the definite integral before using your calculator, and round your final answer to 3 significant figures.

Exam-Style Question 4 — Paper 2 (Calculator Allowed) [8 marks]

Question:

The gradient of a curve is given by $\frac{dy}{dx} = \frac{10x}{(2x^2 + 1)^2}$. The curve passes through the point $(1, 2)$.

(a) Find the equation of the curve. [5]

(b) Find the $y$-coordinate of the point where $x = 2$, giving your answer to 3 significant figures. [3]

Worked Solution:

(a)

1. Integrate $\frac{10x}{(2x^2 + 1)^2}$ with respect to $x$. Let $u = 2x^2 + 1$, then $\frac{du}{dx} = 4x$, so $dx = \frac{du}{4x}$. $\int \frac{10x}{(2x^2 + 1)^2} dx = \int \frac{10x}{u^2} \cdot \frac{du}{4x} = \int \frac{5}{2u^2} du = \frac{5}{2} \int u^{-2} du = \frac{5}{2} \cdot \frac{u^{-1}}{-1} + c = -\frac{5}{2u} + c = -\frac{5}{2(2x^2+1)} + c$ [Using substitution to integrate]

2. Substitute the point $(1, 2)$ to find $c$. $2 = -\frac{5}{2(2(1)^2+1)} + c \implies 2 = -\frac{5}{6} + c \implies c = 2 + \frac{5}{6} = \frac{17}{6}$ [Substituting to find the constant of integration]

3. State the equation of the curve. $y = -\frac{5}{2(2x^2+1)} + \frac{17}{6}$ [Putting it all together]

(b)

1. Substitute $x = 2$ into the equation of the curve. $y = -\frac{5}{2(2(2)^2+1)} + \frac{17}{6} = -\frac{5}{2(9)} + \frac{17}{6} = -\frac{5}{18} + \frac{17}{6} = -\frac{5}{18} + \frac{51}{18} = \frac{46}{18} = \frac{23}{9}$ [Substituting x=2 into the equation]

2. Calculate the value of y. $y = \frac{23}{9} \approx 2.5555...$

3. Round to 3 significant figures. $y \approx 2.56$

Final Answer: (a) $\boxed{y = -\frac{5}{2(2x^2+1)} + \frac{17}{6}}$ (b) $\boxed{2.56}$

How to earn full marks: In part (a), clearly show your u-substitution and remember to find the constant of integration. In part (b), show your substitution and round your final answer to 3 significant figures.