Integrating Basic Functions (Edexcel IGCSE Further Pure Maths)

Revision Note

Written by: Roger B

Reviewed by: Dan Finlay

Integrating Powers of x

How do I integrate powers of x?

Powers of $x$ are integrated according to the following formula:
- $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c$ (where $c$ is the constant of integration)
  - This is valid for any value of $n$ except $n = - 1$
  - So you cannot integrate $\int x^{- 1} d x = \int \frac{1}{x} d x$ this way

If $x^{n}$ is multiplied by a constant $a$ then
- $\int a x^{n} d x = a \int x^{n} d x = \frac{a x^{n + 1}}{n + 1} + c$
  - This also is not valid for $n = - 1$
These formulae are not on the exam formula sheet, so you need to remember them
Remember the special case:
- $\int a d x = a x + c$
  - e.g. $\int 4 d x = 4 x + c$
- This allows constant terms to be integrated
Functions involving roots will need to be rewritten as fractional powers
- e.g. $\int \sqrt[3]{x} d x$
  - rewrite $\sqrt[3]{x}$ as $x^{\frac{1}{3}}$
  - then integrate
Fractions with $x$ in the denominator will need to be rewritten as negative powers
- e.g. $\int \frac{1}{x^{2}} d x$
  - rewrite $\frac{1}{x^{2}}$ as $x^{- 2}$
  - then integrate

How do I integrate sums and differences of powers of x?

The formulae can be used to integrate sums or differences of powers of $x$
- Just integrate term by term
  - e.g. $\int (8 x^{3} - 2 x + 4) d x$
  - $= \frac{8 x^{3 + 1}}{3 + 1} - \frac{2 x^{1 + 1}}{1 + 1} + 4 x + c$
  - $= 2 x^{4} - x^{2} + 4 x + c$
Products and quotients cannot be integrated this way
- You need to expand and/or simplify first
  - e.g. $\int 8 x^{2} (2 x - 3) d x$
  - expand $8 x^{2} (2 x - 3)$ as $16 x^{3} - 24 x^{2}$
  - then integrate term by term
  - you cannot just multiply the integrals of $8 x^{2}$ and $2 x - 3$ together

What might I be asked to do once I’ve integrated?

You may be given the derivative of a function and asked to find the function
- Integration and differentiation are inverse operations so
  - $\int f^{'} (x) d x = f (x) + c$
  - $\int \frac{d y}{d x} d x = y + c$
With more information the constant of integration, $c$ , can be found
The area under a curve can also be found using integration

Examiner Tips and Tricks

Remember the basic pattern of integrating powers of x
- 'Raise the power by one and divide by the new power'
- Lots of practice will improve your speed and accuracy
It's easy to check your answer when integrating
- Just differentiate your answer
- It should turn back into the function you were integrating in the first place

Worked Example

Given that $\frac{d y}{d x} = 2 x^{2} + 3 - \frac{1}{\sqrt{x}}$ , find an expression for $y$ in terms of $x$ .

Start by rewriting $\frac{d y}{d x}$ entirely in powers of $x$

By laws of indices $\frac{1}{\sqrt{x}} = x^{- \frac{1}{2}}$

$\frac{d y}{d x} = 2 x^{2} + 3 - x^{- \frac{1}{2}}$

Remember $\int \frac{d y}{d x} d x = y + c$

We can integrate term by term using $\int a x^{n} d x = \frac{a x^{n + 1}}{n + 1} + c$

$\begin{array}{rcl} y & = & \int (2 x^{2} + 3 - x^{- \frac{1}{2}}) d x \\ = & 2 (\frac{x^{2 + 1}}{2 + 1}) + 3 x - \frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + c \\ = & 2 (\frac{x^{3}}{3}) + 3 x - \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + c \\ = & \frac{2}{3} x^{3} + 3 x - 2 x^{\frac{1}{2}} + c \end{array}$

We can't find the value of $c$ without further info, so that's the answer to the question

It's 'nice' to turn $x^{\frac{1}{2}}$ back into $\sqrt{x}$ for the final answer, but you would also get the marks without doing that

$y = \frac{2}{3} x^{3} + 3 x - 2 \sqrt{x} + c$

Integrating Trig Functions

How do I integrate sin and cos?

You can integrate $\sin x$ and $\cos x$ by using the formulae
- $\int \sin x d x = - \cos x + c$
- $\int \cos x d x = \sin x + c$
  - $c$ is the constant of integration

If $x$ is multiplied by a constant $a$ then
- $\int \sin a x d x = - \frac{1}{a} \cos a x + c$
- $\int \cos a x d x = \frac{1}{a} \sin a x + c$

None of these formulae are on the exam formula sheet, so you need to remember them
For calculus with trigonometric functions angles must be measured in radians
- Make sure you know how to change the angle mode on your calculator

Examiner Tips and Tricks

Remember to include 'c', the constant of integration, for any indefinite integrals
As soon as you see a question involving integration and trigonometry
- put your calculator into radians mode

Worked Example

Given that $f^{'} (x) = 4 \cos 3 x - \frac{1}{2} \sin 2 x$ , find an expression for $f (x)$ in terms of $x$ .

Remember that $\int f^{'} (x) d x = f (x) + c$

We can integrate term by term using the integration formulae for $\sin a x$ and $\cos a x$

$\begin{array}{rcl} f (x) & = & \int (4 \cos 3 x - \frac{1}{2} \sin 2 x) d x \\ = & 4 (\frac{1}{3} \sin 3 x) - \frac{1}{2} (- \frac{1}{2} \cos 2 x) + c \\ = & \frac{4}{3} \sin 3 x + \frac{1}{4} \cos 2 x + c \end{array}$

We can't find the value of $c$ without further info, so that's the answer to the question

$f (x) = \frac{4}{3} \sin 3 x + \frac{1}{4} \cos 2 x + c$

Integrating e^x

How do I integrate exponentials?

$e^{x}$ can be integrated using the formula
- $\int e^{x} d x = e^{x} + c$
  - $c$ is the constant of integration
If $x$ is multiplied by a constant $a$ then
- $\int e^{a x} d x = \frac{1}{a} e^{a x} + c$

These formulae are not on the exam formula sheet, so you need to remember them

Examiner Tips and Tricks

Because ' $e^{x}$ is its own integral' it is quite easy to integrate exponentials
- Just be careful dealing with the constant in $e^{a x}$

Worked Example

Given that $f^{'} (x) = \frac{e^{2 x} - e^{- 3 x}}{2}$ , find an expression for $f (x)$ in terms of $x$ .

Remember that $\int f^{'} (x) d x = f (x) + c$

We can integrate term by term using the integration formula for $e^{a x}$

You might find it easier to split the fraction into two separate fractions first

$\begin{array}{rcl} f (x) & = & \int (\frac{1}{2} e^{2 x} - \frac{1}{2} e^{- 3 x}) d x \\ = & \frac{1}{2} (\frac{1}{2} e^{2 x}) - \frac{1}{2} (\frac{1}{- 3} e^{- 3 x}) + c \\ = & \frac{1}{4} e^{2 x} + \frac{1}{6} e^{- 3 x} + c \end{array}$

We can't find the value of $c$ without further info, so that's the answer to the question

$f (x) = \frac{1}{4} e^{2 x} + \frac{1}{6} e^{- 3 x} + c$

Last updated: 20 October 2024

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Integrating Basic Functions (Edexcel IGCSE Further Pure Maths)

Revision Note

Integrating Powers of x

How do I integrate powers of x?

How do I integrate sums and differences of powers of x?

What might I be asked to do once I’ve integrated?

Examiner Tips and Tricks

Worked Example

Integrating Trig Functions

How do I integrate sin and cos?

Examiner Tips and Tricks

Worked Example

Integrating e^x

How do I integrate exponentials?

Examiner Tips and Tricks

Worked Example

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Author: Dan Finlay