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IN1 Antidifferentiation


Isaac Newton
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.

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

Think about the following examples:

  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.


  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 \(\frac{1}{2}+1=\frac{3}{2}.\) \[\begin{alignat*}{1} y & =\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c,\;c\in\mathbb{R}\\ & =\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+c\\ & =\frac{2}{3}x^{\frac{3}{2}}+c. \end{alignat*}\]


Find antiderivatives for the following:

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

In all cases, \(c\) is a constant.

\(1.\ \frac{1}{4}x^{4}+c\qquad2.\ \frac{1}{9}s^{9}+c\qquad3.\ \frac{3}{4}x^{4/3}+c\qquad4.\ -\frac{1}{4}x^{-4}+c\qquad5.\ 6x+c\qquad6.\ -m^{-1}+c=-\frac{1}{m}+c\qquad7.\ 2p^{1/2}+c\)

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