Indefinite integral

We know some concepts related to the derivation: if $F(x)$ is a function, we denote $F'(x)$ to its derivative and claculate according to the rules already seen.The problem that we want to tackle now is to do the reverse, that is, from a derivative, let's call it $f(x)$, we want to find what function $F(x)$ has as a derivative $f(x)$. Or, $F'(x)=f(x)$.

In other words, we write $\displaystyle\int f(x) \ dx=F(x)$, which means that $f(x)$ is the derivative of $F(x)$ with respect to the variable $x$. Then, $F(x)$ is the indefinite integral, primitive function, or antiderivative of $f (x)$.

Let's observe that we use the symbol $\displaystyle\int$ to denote that we are integrating, and $dx$ to signify what variable we are integrating. In some cases this $dx$ might be omitted, but to avoid confusion it is better to always use it.

Let's see now some important properties of the indefinite integral:

• We know that the derivative of a constant $C$ is $\dfrac{d}{dx}C=0$. Therefore, given $f(x)$, with a primitive function $\dfrac{d}{dx}F(x)=F'(x)=f(x)$, but them also $F (x) +C$ is a valid primitive since we also know that $\dfrac{d}{dx}(F(x)+C)=f(x)$. Therefore, the primitive or antiderivative of a function is not unique. Thus, when computing an integral we will give the result as: $\displaystyle\int f(x) \ dx=F(x)+C$, where $C$ is called an integration constant. Note that we should never forget this constant.

• The integral, as well as the derivative, satisfies the properties of linearity, that is:

  • $\displaystyle \int k \cdot f(x) \ dx = k \cdot \int f(x) \ dx$
  • $\displaystyle\int (f(x)+g(x)) \ dx=\int f(x) \ dx + \int g(x) \ dx$