Integrating Powers of x

How do I integrate powers of x?

Powers of $x$ are integrated according to the following formulae:
- If $f (x) = x^{n}$ then $\int f (x) d x = \frac{x^{n + 1}}{n + 1} + c$ where $n \in ℚ, n \neq - 1$ and $c$ is the constant of integration
- For each term …
  - … increase the power (of x) by 1
  - … divide by the new power
- This does not apply when the original power is -1
  - the new power would be 0 and division by 0 is undefined

If the power of $x$ is multiplied by a constant then the integral is also multiplied by that constant
- If $f (x) = a x^{n}$ then $\int f (x) d x = \frac{a x^{n + 1}}{n + 1} + c$ where $n \in ℚ, n \neq - 1$ and $a$ is a constant and $c$ is the constant of integration
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

How do I integrate expressions containing powers of x?

The formulae for integrating powers of $x$ apply to all rational numbers so it is possible to integrate any expression that is a sum or difference of powers of $x$
- e.g. If $f (x) = 8 x^{3} - 2 x + 4$ then $\int f (x) d x = \frac{8 x^{3 + 1}}{3 + 1} - \frac{2 x^{1 + 1}}{1 + 1} + 4 x + c = 2 x^{4} - x^{2} + 4 x + c$
Functions involving roots will need to be rewritten as fractional powers of $x$ first
- eg. If $f (x) = 5 \sqrt[3]{x}$ then rewrite as $f (x) = 5 x^{\frac{1}{3}}$ and integrate
Functions involving fractions with denominators in terms of $x$ will need to be rewritten as negative powers of $x$ first
- e.g. If $f (x) = \frac{4}{x^{2}} + x^{2}$ then rewrite as $f (x) = 4 x^{- 2} + x^{2}$ and integrate
Products and quotients cannot be integrated this way so would need expanding/simplifying first
- e.g. If $f (x) = 8 x^{2} (2 x - 3)$ then $\int f (x) d x = \int (16 x^{3} - 24 x^{2}) d x = \frac{16 x^{4}}{4} - \frac{24 x^{3}}{3} + c = 4 x^{4} - 8 x^{3} + c$

Exam Tip

You can speed up the process of integration in the exam by committing the pattern of basic integration to memory
- In general you can think of it as 'raising the power by one and dividing by the new power'
- Practice this lots before your exam so that it comes quickly and naturally when doing more complicated integration questions

Worked example

Given that

$\frac{d y}{d x} = 3 x^{4} - 2 x^{2} + 3 - \frac{1}{\sqrt{x}}$

find an expression for $y$ in terms of $x$ .

Rewrite all terms as powers of x using the laws of indices for fractional and negative powers on the last term.

$\frac{d y}{d x} = 3 x^{4} - 2 x^{2} + 3 - x^{- \frac{1}{2}}$

Find y by integrating each term.

$y = \int (3 x^{4} - 2 x^{2} + 3 - x^{- \frac{1}{2}}) d x$

$y = \frac{3 x^{5}}{5} - \frac{2 x^{3}}{3} + 3 x - \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + c$

Rewrite using the same format given in the question.

$y = \frac{3}{5} x^{5} - \frac{2}{3} x^{3} + 3 x - 2 \sqrt{x} + c$

Finding the Constant of Integration

How do I find the constant of integration?

STEP 1
- Rewrite the function into a more easily integrable form
  - Each term needs to be a power of x (or a constant)
STEP 2
- Integrate each term and remember “+c”
  - Increase power by 1 and divide by new power
STEP 3
- Substitute the coordinates of a given point in to form an equation in c
  - Solve the equation to find c

Notes fig3, A Level & AS Level Pure Maths Revision Notes

Example of finding the constant of integration

Exam Tip

If a constant of integration can be found then the question will need to give you some extra information
- If this is given then make sure you use it to find the value of c

Worked example

Given $f' (x) = \frac{{(x + 3)}^{2}}{\sqrt{x}}$ and $f (1) = 25$ , find $f (x)$ .

Rewrite $f^{'} (x)$ in a form the can be integrated more easily.

$\begin{array}{rcl} f^{'} (x) & = & \frac{(x + 3) (x + 3)}{\sqrt{x}} \\ = & \frac{x^{2} + 6 x + 9}{x^{\frac{1}{2}}} \\ = & \frac{x^{2}}{x^{\frac{1}{2}}} + \frac{6 x}{x^{\frac{1}{2}}} + \frac{9}{x^{\frac{1}{2}}} \\ = & x^{\frac{3}{2}} + 6 x^{\frac{1}{2}} + 9 x^{- \frac{1}{2}} \end{array}$

Integrate each term, remember to include a constant of integration.

$f (x) = \frac{x^{\frac{5}{2}}}{\frac{5}{2}} + \frac{6 x^{\frac{3}{2}}}{\frac{3}{2}} + \frac{9 x^{\frac{1}{2}}}{\frac{1}{2}} + c$

Simplify.

$f (x) = \frac{2}{5} x^{\frac{5}{2}} + 4 x^{\frac{3}{2}} + 18 x^{\frac{1}{2}} + c$

Use f(1) =25 to find the value of c.

$f (1) = \frac{2}{5} {(1)}^{\frac{5}{2}} + 4 {(1)}^{\frac{3}{2}} + 18 {(1)}^{\frac{1}{2}} + c$

$\begin{array}{rcl} \frac{2}{5} + 4 + 18 + c & = & 25 \\ 22 \frac{2}{5} + c & = & 25 \\ c & = & \frac{13}{5} \end{array}$

$f (x) = \frac{2}{5} x^{\frac{5}{2}} + 4 x^{\frac{3}{2}} + 18 x^{\frac{1}{2}} + \frac{13}{5}$

Integrating Powers of x (CIE IGCSE Additional Maths)

Revision Note