Differentiation of a function enables us to find another function that defines the gradient at any point on the original function. We can use the derivative to investigate the characteristics of and understand a particular function in great detail. In this lesson, we will look at the opposite process, called antidifferentiation. Antidifferentiation is a powerful tool that enables us to find the shape of a function given its derivative.
Consider the following functions and their derivatives:
$F\left(x\right)$F(x)  $F'\left(x\right)$F′(x) 

$3x^2+4x+5$3x2+4x+5  $6x+4$6x+4 
$3x^2+4x+1$3x2+4x+1  $6x+4$6x+4 
$3x^2+4x17$3x2+4x−17  $6x+4$6x+4 
$3x^2+4x3$3x2+4x−3  $6x+4$6x+4 
$3x^2+4x+a$3x2+4x+a  $6x+4$6x+4 
Notice that because all the constant terms $5,1,17,3$5,1,−17,−3 and $a$a all had a derivative of zero, we ended up getting the same gradient function for all of them $F'\left(x\right)=6x+4$F′(x)=6x+4.
If we were given the gradient function $F'\left(x\right)=6x+4$F′(x)=6x+4 and asked to find the original function, we can see above that there are infinite possibilities.
When we do the opposite to differentiation, we cannot be exactly sure which function was the original. What we find is called a primitive function or an antiderivative and we account for the different possibilities by adding a constant term, $C$C.
Let's consider our example above. We say a primitive of the function $f\left(x\right)=6x+4$f(x)=6x+4 is $F\left(x\right)=3x^2+4x+C$F(x)=3x2+4x+C, where $C$C is a constant term. In this way, we have that $F'\left(x\right)=f\left(x\right)$F′(x)=f(x).
The applet below shows a family of primitive functions. It shows $y=x^2$y=x2, $y=x^2+1$y=x2+1, $y=x^21$y=x2−1 and $y=x^23$y=x2−3. The tangent is drawn at the same value of $x$x on each function. Moving point $A$A varies this $x$x coordinate.
We can see that all the tangent lines stay parallel demonstrating that the value of the derivative is the same for the whole family of antiderivatives of $\frac{dy}{dx}=2x$dydx=2x, irrespective of the constant terms.

Remember to find the derivative of a function $F\left(x\right)=x^n$F(x)=xn, we multiplied by the index of $x$x and subtracted one from it. Antidifferentiation reverses this process. That is, we add one to the index of $x$x and divide by this new index. Importantly, we need to add a constant term, $C$C, in the primitive function for the reasons mentioned above.
Find primitive function $F(x)$F(x) for a function $y=2x$y=2x.
Think: To find the primitive function we add one to the existing index of $x$x and divide by the new index as follows.
Do:
If $\frac{dy}{dx}$dydx  $=$=  $2x$2x 
Then $F(x)$F(x)  $=$=  $\frac{2x^{1+1}}{2}+C$2x1+12+C 
And $F(x)$F(x)  $=$=  $x^2+C$x2+C 
Notice that the coefficient of $2$2 in the original function is not directly affected by the process of antidifferentiation as was the case for differentiation.
We can generalise the rule for simple functions of $x$x as follows.
A primitive function of $x^n$xn has the form $\frac{x^{n+1}}{n+1}+C$xn+1n+1+C for $n\ne1$n≠−1.
Note that, when finding the antiderivative for functions involving multiple terms, you apply the same strategy used for differentiation where each term is operated on individually, and then added or subtracted to give the final solution. The rule above also works for fractional and negative indices except for when $n=1$n=−1. This particular case will be discussed in later chapters.
Additionally, just as finding the derivative gave a function one degree less than the function being differentiated, antidifferentiation gives a primitive function one degree higher than the original function. That is, finding the antiderivative of a linear function gives a quadratic, a quadratic gives a cubic, a cubic gives a quartic and so on. Additionally, if we are given a function of $y=3$y=3 then we can consider this as $y=3x^0$y=3x0 and use the process above to find the primitive function.
Find primitive function $F\left(x\right)$F(x) of the function $f\left(x\right)=3x^2+4x$f(x)=3x2+4x.
Think: As this function includes multiple terms, we consider each term individually, raise the power and divide by the new power and then combine all the new terms together to give the antiderivative.
Do:
$F\left(x\right)$F(x)  $=$=  $\frac{3x^{2+1}}{2+1}+\frac{4x^2}{2}+C$3x2+12+1+4x22+C 
$F\left(x\right)$F(x)  $=$=  $\frac{3x^3}{3}+2x^2+C$3x33+2x2+C 
$F\left(x\right)$F(x)  $=$=  $x^3+2x^2+C$x3+2x2+C 
Find primitive function $F\left(x\right)$F(x) of the function $f\left(x\right)=\frac{1}{x^3}x^2$f(x)=1x3−x2.
Think: Just as with the process of differentiation it helps to first rewrite in terms of powers:
$f\left(x\right)=x^{3}x^2$f(x)=x−3−x2
Do: So term by term, we raise the power and divide by the new power.
$F\left(x\right)$F(x)  $=$=  $\frac{x^{2}}{2}\frac{x^3}{3}+C$x−2−2−x33+C 
$F\left(x\right)$F(x)  $=$=  $\frac{1}{2x^2}\frac{x^3}{3}+C$−12x2−x33+C 
Find an equation for $f\left(x\right)$f(x) if $f'\left(x\right)=8x$f′(x)=8x.
Use $C$C as the constant of integration.
Given that $y'=x\sqrt{x}$y′=x√x, find the primitive function $y$y.
Use $C$C as the constant of integration.
Find an equation for $y$y if $\frac{dy}{dx}=\left(5x2\right)\left(3x4\right)$dydx=(5x−2)(3x−4).
Use $C$C as the constant of integration.
Antidifferentiation involves reversing the process of differentiation. The process of integration is mathematically the same, from a process point of view, but has different notation and meaning. We will see in later lessons that integration is related to finding areas.
Let's look at how worked example $1$1 above would be represented as an integration question. The symbol $\int$∫ is used to denote the process of integration and a symbol is added, $dx$dx, to denote the variable of integration. When we integrate without resolving any limits we are determining an indefinite integral. The concept of indefinite and definite integrals will be explained in later lessons.
For example, integrating the function $y=2x$y=2x with respect to the variable $x$x would be written as follows:
$\int\ 2x\ dx$∫ 2x dx
We solve this in exactly the same way as antidifferentiation setting out our working as follows and remembering to add the constant term:
$\int\ 2x\ dx$∫ 2x dx  $=$=  $\frac{2x^{1+1}}{2}+C$2x1+12+C 
$\int\ 2x\ dx$∫ 2x dx  $=$=  $x^2+C$x2+C 
Rewriting our rule for antidifferentiation using integral notation gives us the following.
$\int\ x^n\ dx=\frac{x^{n+1}}{n+1}+C$∫ xn dx=xn+1n+1+C for $n\ne1$n≠−1.
Find $\int x^{\frac{4}{5}}dx$∫x45dx, with $C$C as the constant of integration.
Find $\int\left(3x^2+6x+3\right)dx$∫(3x2+6x+3)dx, with $C$C as the constant of integration.