 ## IN1 Antidifferentiation Isaac Newton co-invented Calculus which comprises differentiation and antidifferentiation (integration). This portrait of Newton at age 46 was done by Godfrey Kneller in 1689.
en.wikipedia.org/wiki/Isaac_Newton

How do you antidifferentiate a function? Antidifferentiation (also called integration) is the opposite operation to differentiation.

If you have the differentiation of a function, you can then obtain the original function via integration (antidifferentiation).

Given a derivative $$f^{\prime}\left(x\right)$$ of a function we want to find the original function $$f\left(x\right)$$. The original function is called an antiderivative.

Play a short video on Antidifferentiation.

### The Basic Concept

1. If $$f\left(x\right)=\frac{x^{3}}{3}$$, then $$f'\left(x\right)=x^{2}$$ so $$\frac{x^{3}}{3}$$is an antiderivative of $$x^{2}$$.

2. If $$f\left(x\right)=\frac{x^{3}}{3}+1$$, then $$f'\left(x\right)=x^{2}$$ so $$\frac{x^{3}}{3}+1$$ is an antiderivative of $$x^{2}$$.

3. If $$f\left(x\right)=\frac{x^{3}}{3}+2$$, then $$f'\left(x\right)=x^{2}$$ so $$\frac{x^{3}}{3}+2$$ is an antiderivative of $$x^{2}$$.

Notice that adding a constant to $$\frac{x^{3}}{3}$$ does not change the fact it is an antiderivative of $$x^{2}.$$ This is because the derivative of a constant is zero.

In general an antiderivative of $$f^{\prime}\left(x\right)$$ is given by $$f\left(x\right)+c$$ where $$c$$ is a constant1 We often write $$c\in\mathbb{R}$$ which means that $$c$$ is a real number..

### Finding Antiderivatives

Antidifferentiation is more complicated than differentiation. However there are some rules to help us. One of the most important is the power rule which says2 Note that if $$n=-1$$, $$\frac{1}{n+1}$$ would be $$\frac{1}{0}$$ which has no meaning. Apart from this restriction, $$n$$ can be any number.:

If $$f'\left(x\right)=x^{n}$$, $$n\neq-1$$ the antiderivative $$f\left(x\right)=\frac{1}{n+1}x^{n+1}+c,$$ where $$c$$ is a constant.

#### Alternate Notation

If $$y=f(x)$$ then $$\frac{dy}{dx}=f'(x)$$. If $$\frac{dy}{dx}=x^{n},$$ the antiderivative $$y=$$ $$\frac{1}{n+1}x^{n+1}+c$$, where $$c$$ is a constant.

### Examples

1. Given $$\frac{dy}{dx}=x,$$find the antiderivative. 3 In this case $$n=1$$ because $$x=x^{1}.$$ \begin{alignat*}{1} y & =\frac{x^{1+1}}{1+1}+c,\;c\in\mathbb{R}\quad(\mathrm{add\,one\,to\,the\,power\,of\,\mathit{x},divide\,by\,the\,new\,power\,and\,add\,a\,constant)}\\ & =\frac{x^{2}}{2}+c. \end{alignat*}
2. Given $$\frac{dy}{dx}=1,$$find the antiderivative. 4 In this case $$n=0$$ because $$x^{0}=1.$$ \begin{alignat*}{1} y & =\frac{x^{0+1}}{0+1}+c,\;c\in\mathbb{R\quad\mathrm{(add\,one\,to\,the\,power\,of\,\mathit{x},divide\,by\,the\,new\,power\,and\,add\,a\,constant)}}\\ & =x+c. \end{alignat*}
3. Given $$\frac{dy}{dx}=x^{-3},$$find the antiderivative. 5 In this case $$n=-3$$. The new power will be $$-2$$. \begin{alignat*}{1} y & =\frac{x^{-3+1}}{-3+1}+c,\;c\in\mathbb{R}\\ & =\frac{x^{-2}}{-2}+c\\ & =-\frac{1}{2x^{2}}+c. \end{alignat*}
4. Given $$\frac{dy}{dx}=\sqrt{x},$$find the antiderivative.6 In this case, $$n=\frac{1}{2}$$ because $$\sqrt{x}=x^{\frac{1}{2}}.$$ The new power will be $$\frac{1}{2}+1=\frac{3}{2}.$$ \begin{alignat*}{1} y & =\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c,\;c\in\mathbb{R}\\ & =\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+c\\ & =\frac{2}{3}x^{\frac{3}{2}}+c. \end{alignat*}

### Exercises

Find antiderivatives for the following:

$$1.\ x^{3}\qquad2.\ s^{8}\qquad3.\ \sqrt{x}\qquad4.\ x^{-5}\qquad5.\ 6\qquad6.\ m^{-2}\qquad7.\ p^{-1/2}$$

In all cases, $$c$$ is a constant.
$$1.\ \frac{1}{4}x^{4}+c\qquad2.\ \frac{1}{9}s^{9}+c\qquad3.\ \frac{3}{4}x^{4/3}+c\qquad4.\ -\frac{1}{4}x^{-4}+c\qquad5.\ 6x+c\qquad6.\ -m^{-1}+c=-\frac{1}{m}+c\qquad7.\ 2p^{1/2}+c$$