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A Level Integration Worked Examples for 2026/2027 (Singapore MOE Syllabus) — Full Step-by-Step Solutions

Updated June 11, 2026A Levels
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Quick answer

When you're faced with a tough integration question at the A Level, it's normal to feel your heart sink. But don't panic — most students actually know the concept, and it's all about breaking down the problem step by step. I'll walk you through four examples, explaining each step so you can tackle these questions with confidence.

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What you need to know

Integration is like finding the area under a curve. It's a way to add up tiny pieces to get a total. You use it to solve problems where you need to find quantities like area, volume, or total change.

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Key Concepts in Integration

Recognising Integration Types

Integration questions can appear in different forms. You might see definite integrals (with limits) or indefinite integrals (without limits). The key pattern to recognise is whether you're dealing with a polynomial, trigonometric function, or something more complex.

Basic Integration Techniques

  1. Power Rule: If you see something like 𝑥𝑛𝑥^𝑛, you increase the power by 1 and divide by the new power.
  2. Substitution: Use this when you have a complicated function. You change variables to make it simpler.
  3. Integration by Parts: Use when the product of two functions is involved.

Common mistakes students make

Rushing Algebra Steps

Many students lose unnecessary marks because they rush through the algebra steps. For example, forgetting to change the limits of integration when doing substitution. Always check your steps.

Overcomplicating Simple Questions

Sometimes a question looks complex, but it's actually straightforward. Don't overthink it. If you see a simple polynomial, use the power rule immediately.

Freezing on New Question Types

When a question doesn't look like those in your tutorials, it's easy to freeze. Remember, exam questions test application, not just memorization. Break it down into smaller parts and tackle each part one at a time.

Exam tip

Always write your integration constant (+𝐶) for indefinite integrals. It's a small thing but missing it can cost you marks. Also, keep an eye on the units if the question involves real-world contexts like area or volume.

Worked examples

Question 1

Integrate (3𝑥2+2𝑥+1)dx\int (3𝑥^2 + 2𝑥 + 1) \, dx.

Solution

Step 1: Apply the power rule to each term.

  • For 3𝑥23𝑥^2: Increase the power by 1 to get 𝑥3𝑥^3, then divide by the new power: 33𝑥3=𝑥3\frac{3}{3}𝑥^3 = 𝑥^3.
  • For 2𝑥: Increase the power by 1 to get 𝑥2𝑥^2, then divide by the new power: 22𝑥2=𝑥2\frac{2}{2}𝑥^2 = 𝑥^2.
  • For 11: Integrate to get 𝑥.

Why: The power rule lets us integrate each term separately, making it simpler.

Step 2: Combine all terms: 𝑥3+𝑥2+𝑥+𝐶𝑥^3 + 𝑥^2 + 𝑥 + 𝐶.

Why: You add the results together to form the final integrated expression.

Question 2

Integrate sin(𝑥)cos(𝑥)dx\int \sin(𝑥) \cos(𝑥) \, dx.

Solution

Step 1: Use the substitution 𝑢=sin(𝑥)𝑢 = \sin(𝑥), so du=cos(𝑥)dxdu = \cos(𝑥) \, dx.

Why: Substitution simplifies the product of sine and cosine into a single variable.

Step 2: Rewrite the integral as 𝑢du\int 𝑢 \, du.

Why: Changing variables makes it easier to integrate.

Step 3: Integrate to get 12𝑢2+𝐶\frac{1}{2}𝑢^2 + 𝐶.

Why: You apply the power rule to 𝑢.

Step 4: Substitute back 𝑢=sin(𝑥)𝑢 = \sin(𝑥) to get 12sin2(𝑥)+𝐶\frac{1}{2}\sin^2(𝑥) + 𝐶.

Why: You need to express the answer in terms of the original variable.

Question 3

Find the area under the curve 𝑦=𝑥2𝑦 = 𝑥^2 from 𝑥 = 1 to 𝑥 = 3.

Solution

Step 1: Set up the definite integral: 13𝑥2dx\int_{1}^{3} 𝑥^2 \, dx.

Why: The limits tell us the range over which to find the area.

Step 2: Integrate 𝑥2𝑥^2 to get 13𝑥3\frac{1}{3}𝑥^3.

Why: Use the power rule to find the indefinite integral first.

Step 3: Evaluate from 11 to 33: 13(33)13(13)\frac{1}{3}(3^3) - \frac{1}{3}(1^3).

Why: Subtract the value at the lower limit from the value at the upper limit.

Step 4: Calculate: 27/3 - 1/3 = 26/3.

Why: This gives the total area under the curve between 𝑥 = 1 and 𝑥 = 3.

Question 4

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Integrate 𝑥𝑒2𝑥dx\int 𝑥 𝑒^{2𝑥} \, dx using integration by parts.

Solution

Step 1: Identify parts: let 𝑢 = 𝑥 and dv=𝑒2𝑥dxdv = 𝑒^{2𝑥} \, dx.

Why: Choose 𝑢 as a polynomial and dv as an exponential for easy integration.

Step 2: Differentiate and integrate: du = dx, 𝑣=12𝑒2𝑥𝑣 = \frac{1}{2}𝑒^{2𝑥}.

Why: You need these to apply the integration by parts formula.

Step 3: Use integration by parts: 𝑢dv=uv𝑣du\int 𝑢 \, dv = uv - \int 𝑣 \, du.

Why: This formula helps integrate products of functions.

Step 4: Substitute and simplify: 𝑥12𝑒2𝑥12𝑒2𝑥dx𝑥 \cdot \frac{1}{2}𝑒^{2𝑥} - \int \frac{1}{2}𝑒^{2𝑥} \, dx.

Why: Substitute the expressions for 𝑢, 𝑣, and du.

Step 5: Integrate: 12xe2𝑥14𝑒2𝑥+𝐶\frac{1}{2}xe^{2𝑥} - \frac{1}{4}𝑒^{2𝑥} + 𝐶.

Why: Finish by integrating the remaining exponential term.

Quick summary

  • Recognise integration types: polynomial, trigonometric, or complex.
  • Common mistakes: rushing steps, forgetting constants, overcomplicating.
  • Power Rule: Increase the power by 1, divide by the new power.
  • Substitution: Change variables to simplify.
  • Integration by Parts: Use for product of functions.
  • Definite integrals: Subtract lower limit from upper limit.
  • Exam tip: Always include +𝐶 for indefinite integrals.

FAQ

Q 1: What is the integration constant and why is it important?

The integration constant (+𝐶) represents any constant value that could have been differentiated to zero. It's crucial for indefinite integrals to account for all possible solutions.

Q 2: How do I know when to use substitution?

Use substitution when integrating a composite function where one part is the derivative of another. It simplifies the expression to a basic form.

Q 3: Can I skip steps if I'm confident?

It's better not to. Skipping steps often leads to careless mistakes, especially under timed exam conditions. Write each step to keep track of your logic.

Q 4: What if I get stuck during the exam?

Breathe first. Break the problem into smaller parts you can handle. Look for patterns or techniques that match the question type.

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