Constants of Integration (Edexcel IGCSE Further Pure Maths)

Revision Note

Written by: Roger B

Reviewed by: Dan Finlay

Finding the Constant of Integration

What is the constant of integration?

The constant of integration is the $c$ used when finding an indefinite integral
- e.g. $\int (3 x^{2} - 5) d x = x^{3} - 5 x + c$
- $c$ can be any constant
Integration and differentiation are inverse operations
- That means if you differentiate your answer to an integral...
  - ...it should turn back into the original function you integrated
- The 'problem' is that the derivative of a constant is zero
  - $\frac{d}{d x} (x^{3} - 5 x) = 3 x^{2} - 5$
  - $\frac{d}{d x} (x^{3} - 5 x + 7) = 3 x^{2} - 5$
  - $\frac{d}{d x} (x^{3} - 5 x - 498) = 3 x^{2} - 5$
- So $x^{3} - 5 x$ plus or minus any constant is a valid solution to $\int (3 x^{2} - 5) d x$
But consider the graph of $y = x^{3} - 5 x + c$
- Different values of $c$ represent different vertical translations of the graph
- If we know one point on that graph we can work out the value of $c$

How do I find the constant of integration?

On an exam you may be given a derivative $\frac{d y}{d x}$ or $f^{'} (x)$
- You can integrate that to find $y$ or $f (x)$ in ' $+ c$ ' form
  - $\int \frac{d y}{d x} d x = y + c$
  - $\int f^{'} (x) d x = f (x) + c$
If you are also given a point on the graph of $y$ or of $y = f (x)$
- you can use this to find the value of $c$
- Substitute the values you know into $y + c$ or $f (x) + c$
  - Then solve for $c$
The extra information doesn't have to be a point on a graph
- As long as you know the value of $f (x)$ for one value of $x$
  - you can substitute and solve to find $c$

Examiner Tips and Tricks

An exam question probably won't tell you to 'find the constant of integration'
- Instead you'll be given the derivative of a function
  - and one value of the function or a point on its graph
- Be sure to recognise this as a 'constant of integration' question!

Worked Example

The graph of $y = f (x)$ passes through the point $(3, - 4)$ . The derivative of $f (x)$ is given by $f^{'} (x) = 3 x^{2} - 4 x - 4$ .

Find $f (x)$ .

Integrate $f^{'} (x)$ to find $f (x)$ in ' $+ c$ ' form

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

We also know the curve of $y = f (x)$ goes through $(3, - 4)$

That means the function is equal to $- 4$ when $x = 3$

${(3)}^{3} - 2 {(3)}^{2} - 4 (3) + c = - 4$

Solve for $c$

$\begin{array}{rcl} 27 - 18 - 12 + c & = & - 4 \\ c - 3 & = & - 4 \\ c & = & - 1 \end{array}$

$f (x) = x^{3} - 2 x^{2} - 4 x - 1$

Last updated: 16 October 2024

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