Learning Lab

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 constant11 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 says22 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.

Examples

1. Given \(\frac{dy}{dx}=x,\)find the antiderivative. 33 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. 44 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. 55 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.66 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:

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