IGCSEFurther MathsEdexcelRevision NotesCalculusIntegrationIntroduction to Integration

Introduction to Integration (Edexcel IGCSE Further Pure Maths): Revision Note

Exam code: 4PM1

Written by: Roger B

Reviewed by: Dan Finlay

Updated on 20 October 2024

Introduction to Integration

What is integration?

Integration is the inverse operation to differentiation
- So if you differentiate a function to find its derivative
- and then integrate that derivative
- you should end up back at the original function
This can be written as $\int f^{'} (x) d x = f (x) + c$
- $\int . . . d x$ is the integral with respect to $x$ of "..."
- $c$ is the constant of integration
  - A derivative gives the rate of change of a function
    - but it doesn't give the starting point for any change
  - The constant of integration represents this 'starting point'
  - See the 'Constants of Integration' revision note for more info
This type of integral is known as an indefinite integral
- The answer to an indefinite integral is another function
  - There are also definite integrals
    - The answer to a definite integral is a number
  - See the 'Calculating Areas' revision note for more info on definite integrals
Usually you will integrate using standard formulae
- See the 'Integrating Basic Functions' revision note for these formulae

Examiner Tips and Tricks

Remember that integration and differentiation are inverse operations
- So if you differentiate your answer to an indefinite integral
  - you should end up back at the function you were integrating
- Use this to check your answers on the exam!
Don't forget the constant of integration ( $+ c$ ) when finding an indefinite integral
- Leaving it out can lose marks

Worked Example

(a) Show that the derivative of $x^{3} + \frac{1}{x}$ is $3 x^{2} - \frac{1}{x^{2}}$ .

This is a standard 'powers of $x$ 'derivative
Just remember to rewrite $\frac{1}{x}$ as a power using laws of indices

$f (x) = x^{3} + x^{- 1}$

$\begin{array}{rcl} f^{'} (x) & = & 3 x^{3 - 1} + ((- 1) x^{- 1 - 1}) \\ = & 3 x^{2} - x^{- 2} \end{array}$

Now just use laws of indices again to write the answer in the requested form

$\begin{array}{rcl} f^{'} (x) & = & 3 x^{2} - \frac{1}{x^{2}} \end{array}$

(b) Hence write down the answer to the indefinite integral $\int (3 x^{2} - \frac{1}{x^{2}}) d x$ .

We can use $\int f^{'} (x) d x = f (x) + c$ to write down the answer

This is because integration and differentiation are inverse operations

Just don't forget the constant of integration $+ c$

$\int (3 x^{2} - \frac{1}{x^{2}}) d x = x^{3} + \frac{1}{x} + c$

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