Integrating $x^n$

In a nutshell

Integration is the inverse of differentiation. It can be used to work out a function given the gradient function, unique up to a constant of integration.

The rule for integration

Integration is the inverse of differentiation. The rule to differentiate $x^n$ is to multiply by the power, then subtract $1$ from the power.

Therefore, the rule of integration is the inverse. First, add $1$ to the power, then divide by the new power. Algebraically:

$\dfrac{dy}{dx}=ax^n \Rightarrow y=\dfrac{a}{n+1}x^{n+1}$

However, when differentiating a function, the constant term vanishes. There is no way to determine what this constant term could have been without more information.

$\begin{aligned}y=x+1 &\overset{\text{differentiate}}{\Rightarrow} \dfrac{dy}{dx}=1\overset{\text{integrate}} \Rightarrow y=x \\ \ \\ y =x-1 &\overset{\text{differentiate}}\Rightarrow \dfrac{dy}{dx}=1 \overset{\text{integrate}}\Rightarrow y=x\\ \ \\y=x+5333218 &\overset{\text{differentiate}}\Rightarrow \dfrac{dy}{dx}=1 \overset{\text{integrate}}\Rightarrow y=x\end{aligned}$

To acknowledge this loss of information, it is important to add a constant of integration after integrating, represented with a $+C$ .

$\boxed{\begin{aligned}\dfrac{dy}{dx}=ax^n &\Rightarrow y=\dfrac{a}{n+1}x^{n+1}+C &n\neq -1\\ \ \\f'(x)=ax^n &\Rightarrow f(x)=\dfrac{a}{n+1}x^{n+1}+C &n\neq -1\end{aligned}}$

Note: This formula fails for $n=-1$ . This is because if $n=-1$ , then the denominator of the function is $0$ , which leads to the function being undefined. You are not expected to know how to integrate functions of the form $x^{-1}$ .

Examples

Find $f(x)$ when:

i) $f'(x)=4x$

ii) $f'(x)=\dfrac{4}{x^3}$ 

iii) $f'(x)=16-5x$ 

Part i):

Write the function in the form $ax^n$ :

$4x=4x^1$

Add $1$ to the power and divide by this new power:

$4x^1\rightarrow 4x^2\rightarrow \dfrac{4}{2}x^2 \rightarrow2x^2$

Add the constant of integration:

$\underline{f(x)=2x^2+C}$

Part ii):

Write the function in the form $ax^n$ :

$\dfrac{4}{x^3}=4x^{-3}$

Add $1$ to the power and divide by this new power:

$4x^{-3}\rightarrow 4x^{-2}\rightarrow \dfrac{4}{-2}x^{-2}\rightarrow-2x^{-2}$

Add the constant of integration:

$\underline{f(x)=-2x^{-2}+C}$

Part iii):

Integrate each term separately:

$16=16x^0\\ \ \\16x^0\rightarrow 16x^1\rightarrow\dfrac{16}{1}x^1\rightarrow 16x^1\\ \ \\-5x=-5x^1\\ \ \\-5x^1\rightarrow -5x^2\rightarrow \dfrac{-5}{2}x^2\rightarrow -2.5x^2\\$

Add the terms together after integrating, remembering the constant of integration:

$\underline{f(x)=16x-2.5x^2+C}$ 