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Integration I

Integrating x^n

Integrating x^n

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Explainer Video

Tutor: Bilal

Summary

Integrating xnx^nxn​​

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 xnx^nxn is to multiply by the power, then subtract 111 from the power.

Therefore, the rule of integration is the inverse. First, add 111 to the power, then divide by the new power. Algebraically:


dydx=axn⇒y=an+1xn+1\dfrac{dy}{dx}=ax^n \Rightarrow y=\dfrac{a}{n+1}x^{n+1}dxdy​=axn⇒y=n+1a​xn+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.


y=x+1⇒differentiatedydx=1⇒integratey=x y=x−1⇒differentiatedydx=1⇒integratey=x y=x+5333218⇒differentiatedydx=1⇒integratey=x\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}y=x+1 y=x−1 y=x+5333218​⇒differentiatedxdy​=1⇒integratey=x⇒differentiatedxdy​=1⇒integratey=x⇒differentiatedxdy​=1⇒integratey=x​​​


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


​dydx=axn⇒y=an+1xn+1+Cn≠−1 f′(x)=axn⇒f(x)=an+1xn+1+Cn≠−1\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}}dxdy​=axn f′(x)=axn​⇒y=n+1a​xn+1+C⇒f(x)=n+1a​xn+1+C​n=−1n=−1​​​​


Note: This formula fails for n=−1n=-1n=−1. This is because if n=−1n=-1n=−1, then the denominator of the function is 000, which leads to the function being undefined. You are not expected to know how to integrate functions of the form x−1x^{-1}x−1.


Examples

Find f(x)f(x)f(x)​ when:

i) f′(x)=4xf'(x)=4xf′(x)=4x

ii) f′(x)=4x3f'(x)=\dfrac{4}{x^3}f′(x)=x34​​

iii) f′(x)=16−5xf'(x)=16-5xf′(x)=16−5x​

​

Part i):

Write the function in the form axnax^naxn:

​4x=4x14x=4x^14x=4x1​​


Add 111 to the power and divide by this new power:

​4x1→4x2→42x2→2x24x^1\rightarrow 4x^2\rightarrow \dfrac{4}{2}x^2 \rightarrow2x^24x1→4x2→24​x2→2x2​​


Add the constant of integration:

​f(x)=2x2+C‾\underline{f(x)=2x^2+C}f(x)=2x2+C​​​


Part ii):

Write the function in the form axnax^naxn:

​4x3=4x−3\dfrac{4}{x^3}=4x^{-3}x34​=4x−3​​


Add 111 to the power and divide by this new power:

​4x−3→4x−2→4−2x−2→−2x−24x^{-3}\rightarrow 4x^{-2}\rightarrow \dfrac{4}{-2}x^{-2}\rightarrow-2x^{-2}4x−3→4x−2→−24​x−2→−2x−2​​


Add the constant of integration:

f(x)=−2x−2+C‾\underline{f(x)=-2x^{-2}+C}f(x)=−2x−2+C​​​


Part iii):

Integrate each term separately:

​16=16x0 16x0→16x1→161x1→16x1 −5x=−5x1 −5x1→−5x2→−52x2→−2.5x216=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\\16=16x0 16x0→16x1→116​x1→16x1 −5x=−5x1 −5x1→−5x2→2−5​x2→−2.5x2​​


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

f(x)=16x−2.5x2+C‾\underline{f(x)=16x-2.5x^2+C}f(x)=16x−2.5x2+C​​​


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Differentiation from first principles

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FAQs - Frequently Asked Questions

Why is a constant added after integrating?

The constant of integration represents the information lost when differentiating.

How do you integrate functions of the form ax^n?

Add 1 to the power, divide by the new power and add the constant of integration?

What is integration?

Integration is the inverse of differentiation.

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