Antiderivatives

# Antiderivatives

 Definition: Suppose that $f$ is a continuous function on the interval $I$. We say that $F$ is an Antiderivative of $f$ if $F' = f$ for all $x \in I$, that is, the derivative of an antiderivative of $f$ equals $f$.

The idea behind antiderivatives is working backwards in a sense. Suppose that $f(x) = \cos x$. We know that $g(x) = \sin x$ has the property that $g'(x) = \cos x$, and therefore, $F(x) = g(x)$, that is, the antiderivative of $f(x) = \cos x$ must be $F(x) = \sin x$.

However, note that if $g(x) = \sin x + 1$, then $g'(x) = \cos x$… and by our definition of the antiderivative, $F(x) = \sin x + 1$ as well. In fact, $F(x) = \sin x + C$ where $C$ is any constant is an antiderivative of $f$, and thus, we will now formalize a similar definition:

 Definition: Suppose that $f$ is a continuous function on the interval $I$. We say that $F(x) + C$ where $C$ is any constant, is the Most General Antiderivative of $f$.

We will now look at some rather simple properties of antiderivatives, none of which will be proven:

• Property 1: If $f$ is a function that is continuous on $I$ and $k$ is a constant, then the most general antiderivative of $kf(x)$ is $kF(x) + C$, where $C$ is a constant.
• Property 2: If $f$ and $g$ are continuous functions on $I$, then the most general antiderivative of $f(x) + g(x)$ is $F(x) + G(x) + C$, where $C$ is a constant.

## Example 1

Find the most general antiderivative of $f(x) = 3x^2$.

We should be familiar with the derivative of $g(x) = x^3$, that is $g'(x) = 3x^2$. Therefore, the most general antiderivative $F(x) = x^3 + C$ where $C$ is any constant.

## Example 2

Find the most general antiderivative of $f(x) = \sec^2 x - \frac{1}{1 + x^2}$.

We should be familiar with the fact that if $g(x) = \tan x$, then $g'(x) = \sec ^2 x$, and if $h(x) = \tan ^{-1} x$, then $h'(x) = \frac{1}{1 + x^2}$.

Since $f(x) = g'(x) + h'(x)$, it follows that $F(x) = g(x) + h(x) + C$, or rather, our most general antiderivative of $f$ is $F(x) = \tan x + \tan ^{-1} x + C$ where $C$ is any constant.

## Example 3

If $a(t)$ represents the acceleration of a particle at time $t$, then what does the antiderivative of $a(t)$ represent?

Recall that $\frac{d}{dx} v(t) = a(t)$. Therefore, $A(t) = v(t)$, that is, the antiderivative of acceleration is velocity, so $A(t)$ represents the velocity of a particle at time $t$.

## Example 4

Find the function $f$ given that $f'(x) = 3x^2 + 4x^3$ and $f(1) = 3$.

We should first recognize that we want to find the antiderivative of $f'$. Note that if $g(x) = x^3$, then $g'(x) = 3x^2$. Furthermore, if $h(x) = x^4$, then $h'(x) = 4x^3$. Since $f'(x) = g'(x) + h'(x)$, it follows that $f(x) = g(x) + h(x) + C$, or rather:

(1)
$$f(x) = x^3 + x^4 + C$$

We now need to find a value of $C$ since we're looking for a specific antiderivative and not a general antiderivative. We're given $f(1) = 3$, so plugging that in and isolating for $C$ we get that:

(2)
\begin{align} f(1) = 1^3 + 1^4 + C \\ 3 = 2 + C \\ 1 = C \end{align}

Thus, $f(x) = x^3 + x^4 + 1$ satisfies both conditions in the problem, that is $f'(x) = 3x^2 + 4x^3$ and $f(1) = 3$.