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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.
Image via Wikimedia Commons, licensed in the public domain.

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.

Antiderivatives and indefinite integrals by Khan Academy, licensed under CC BY-NC-SA 3.0

We know how to take derivatives of functions. If I apply the derivative operator to x squared, I get 2x. Now, if I also apply the derivative operator to x squared plus 1, I also get 2x.

If I apply the derivative operator to x squared plus pi, I also get 2x. The derivative of x squared is 2x. Derivative, with respect to x of pi of a constant, is just 0. Derivative, with respect to x of 1, is just a constant, is just 0. So once again, this is just going to be equal to 2x.

In general, the derivative, with respect to x of x squared plus any constant, is going to be equal to 2x. The derivative of x squared, with respect to x, is 2x. Derivative of a constant, with respect to x, a constant does not change with respect to x, so it's just equal to 0. So you have– You apply the derivative operator to any of these expressions and you get 2x.

Now, let's go the other way around. Let's think about the antiderivative. And one way to think about it is we're doing the opposite of the derivative operator. The derivative operator, you get an expression and you find it's derivative.

Now, what we want to do, is given some expression, we want to find what it could be the derivative of. So if someone were to tell– or give you 2x– if someone were to say 2x– let me write this. So if someone were to ask you what is 2x the derivative of? They're essentially asking you for the antiderivative. And so you could say, well, 2x is the derivative of x squared. But you could also say 2x is the derivative of x squared plus 1. You could also say that 2x is the derivative of x squared plus pi, I think you get the general idea.

So if you wanted to write it in the most general sense, you would write that 2x is the derivative of x squared plus some constant. So this is what you would consider the antiderivative of 2x.

Now, that's all nice, but this is kind of clumsy to have to write a sentence like this, so let's come up with some notation for the antiderivative. And the convention here is to use kind of a strange looking notation, is to use a big elongated s looking thing like that, and a dx around the function that you're trying to take the antiderivative of.

So in this case, it would look something like this. This is just saying this is equal to the antiderivative of 2x, and the antiderivative of 2x, we have already seen, is x squared plus c.

Now, you might be saying, why do we use this type of crazy notation. It'll become more obvious when we study the definite integral and areas under curves and taking sums of rectangles in order to approximate the area of the curve.

Here, it really should just be viewed as a notation for antiderivative. And this notation right over here, this whole expression, is called the indefinite integral of 2x, which is another way of just saying the antiderivative of 2x.

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 constant.1 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 says:2 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:

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

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

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

Keywords