Finding antiderivatives

Key to solving integrals, as we have seen, is to find antiderivatives. Here we discuss some methods of how to find them for the basic functions. In general, however, it is not trivial at all to find for any given function an antiderivative!

Before we do that, though, note that a function $f$ has not just one, but infinitely many antiderivatives! Can you explain why?

Exercise 1

Consider the function $f(x)=x^2$ .

Find an antiderivative of $f$ .
Find another antiderivative of $f$ .
How many antiderivatives are there, and how do they differ from each other?
Does it matter, which antiderivative we take for finding the integral

\int_a^b x^2\, dx \text{ ?}

Solution

To find an antiderivative of $f$ , we have to find a function $F$ whose derivative is $f$ . That is, find $F$ with
$F'(x)=x^2$
Trying out a bit, we see that the one such possible function is
$F(x)=\frac{1}{3}x^3$
Indeed,
$F'(x)=\frac{1}{3}\cdot 3x^2 = x^2$
Clearly, the function
$G(x)=\frac{1}{3}x^3+1$
is also an antiderivative of $f$ , because the derivative of a constant, the $1$ , is $0$ :
$G'(x)=\frac{1}{3}\cdot 3x^2 + 0 =x^2$
Indeed, we can add any constant we like to $F$ , and get another antiderivative of $f$ . Thus there are infinitely many antiderivatives of $x^2$ , and they all differ by a constant from each other:
$\frac{1}{3}x^3, \frac{1}{3}x^3+1, \frac{1}{3}x^3-1.34, \frac{1}{3}x^3+100000.01, ...$
Indeed, it can be shown that all the antiderivatives of $f$ must differ by a constant. But we do not show that here.
No, it does not matter, because the constant is subtracted away. Let us denote the constant by $c$ , where $c$ can be any number, thus our antiderivative is
$F(x)=\frac{1}{3} x^3+c$
We then have
$\begin{array}{lll} \int_a^b x^2\, dx &=& F(b)-F(a)\\ &=& \left(\frac{1}{3} b^3 +c \right) - \left(\frac{1}{3} a^3 +c \right)\\ &=& \frac{1}{3} b^3 - \frac{1}{3} a^3 +c -c\\ &=& \frac{1}{3} b^3 - \frac{1}{3} a^3 \end{array}$
So you see, the constant $c$ has not effect.

Now, let's discuss some strategies to find the antiderivative of a function.

It helps to think about finding the antiderivative as going up, while finding the derivative as going down :

\begin{array}{lll} F & & F\\ \downarrow ' & & \uparrow \int\\ f & & f \\ \end{array}

Note the integral sign $\int$ to the right of the arrow. It is often used to indicate the process of finding the antiderivative.

The notion of going up for the antiderivative is enhanced by using capital letters for antiderivatives (but this is not a rule). Also, the german expressions for derivative and antiderivative are "Ableitung" and "Aufleitung", respectively.

So how do we find the antiderivative of a specific function $f$ ? The general strategy is to go through every function you know and check if its derivative is the desired function $f$ :

\begin{array}{lll} ? \\ \downarrow ' \\ f \\ \end{array}

In this way we quickly find the antiderivative of some basic functions:

\boxed{\begin{array}{cccc} e^x & \sin(x) & -\cos(x) & \ln(x)\\ '\downarrow \uparrow\int & '\downarrow\uparrow\int & '\downarrow \uparrow\int & '\downarrow \uparrow\int\\ e^x & \cos(x) & \sin(x) & x^{-1} \\ \end{array}}

And from the rules of differential calculus we see how to find the antiderivative of a function $f$ which is the weighted sum of other functions:

\boxed{\begin{array}{cccc} G & H & & a\cdot G+ b\cdot H\\ '\downarrow\uparrow\int & '\downarrow\uparrow\int & \Rightarrow & '\downarrow \uparrow\int\\ g & h & & a\cdot g+ b\cdot h \\ \end{array}}

Example 1

Find the antiderivative of the function $f(x)=3\cos(x)-5e^x$ .

Solution

\boxed{\begin{array}{cccc} \sin(x) & e^x & & 3\sin(x)-5 e^x\\ '\downarrow \uparrow\int & '\downarrow \uparrow\int & \Rightarrow & '\downarrow \uparrow\int\\ \cos(x) & e^x & & 3\cos(x)-5 e^x\\ \end{array}}

Finally, let's discuss the antiderivative of power functions $x^n$ . We know that the derivative of a power function is essentially obtained by lowering the power by one, so a candidate for the antiderivative is $x^{n+1}$ , but have to add a multiplication factor:

\boxed{\begin{array}{cccc} x^{n+1} & \frac{1}{n+1}\cdot x^{n+1}\\ '\downarrow \uparrow\int & '\downarrow \uparrow\int\\ (n+1)\cdot x^n & x^n\\ \end{array}}

So the rule for finding the antiderivative of $x^n$ is to

increase the power by 1, and
divide by this new power

(Note that adding the power by $1$ to get the antiderivative and subtracting by $1$ when taking the derivative once more fits well in our notion of going up and going down.)

There is an exception to this rule. Try to find the antiderivative of

x^{-1}

(that is, $n=-1$ ). The rule discussed above will not work, because adding the power by $1$ results in $0$ , and dividing by $0$ will not work. Indeed, we have seen above that the antiderivative of $x^{-1}$ is $\ln(x)$ :

\begin{array}{cccc} \frac{1}{0} \cdot x^{0}\,?? & \ln(x)\\ \uparrow \int & \uparrow \int\\ x^{-1} & x^{-1} &\\ \end{array}

Exercise 2

Determine the antiderivative of the following functions:

$f(x)=3$ for all $x$ (constant function)
$f(x)=2x$
$f(x)=\sqrt{x}$
$f(x)=\frac{1}{x^2}$
$f(x)=\frac{2}{x}$
$f(x)=2x^3+4x+1$
$f(x)=(x-1)^2$
$f(x)=\frac{2}{x}+3e^x$

Solution

$f(x)=3x^0$ (thus $n=0$ ). It follows $F(x)=3\cdot \frac{1}{0+1} x^{0+1}=3x$
$f(x)=2 x^1$ (thus $n=1$ ). It follows $F(x)=2\cdot \frac{1}{1+1}x^{1+1}=x^2$
$f(x)=x^{0.5}$ (thus $n=0.5$ ). It follows $F(x)=\frac{1}{1+0.5}x^{1+0.5} = \frac{2}{3}x^{3/2}$
$f(x)=x^{-2}$ (thus $n=-2$ ). It follows $F(x)=\frac{1}{-2+1}x^{-2+1} =-x^{-1}=-\frac{1}{x}$ .
$f(x)=2 x^{-1}$ (thus $n=-1$ ). It follows $F(x)=2\ln(x)$
$F(x)=2\cdot\frac{1}{3+1}x^{3+1}+4\cdot \frac{1}{1+1} x^{1+1} + 1\cdot\frac{1}{0+1}x^{0+1}=\frac{1}{2}x^4+2x^2+x$
$f(x)=x^2-2x+1$ , thus $F(x)=\frac{1}{3}x^3-2\cdot \frac{1}{2}x^2+\frac{1}{1}x=\frac{1}{3}x^3-x^2+x$
$f(x)=2\frac{1}{x}+3e^x$ , thus $F(x)=2\ln(x)+3e^x$