Antiderivatives
So far, we started with a function and found its derivative. Now we go the other way: we know the derivative, and we want the function. For example, if we know the velocity of a car at every moment, can we find where it is? This reverse process leads to integrals.
What is an antiderivative?
A function $F$ is an antiderivative of $f$ on an interval if
$F'(x)=f(x)$
for every $x$ in the interval.
For example, $F(x)=x^2$ is an antiderivative of $f(x)=2x$, because the derivative of $x^2$ is $2x$.
But $x^2$ is not the only one. The functions $x^2+5$ and $x^2-\pi$ also have derivative $2x$, because the derivative of a constant is $0$.
All antiderivatives differ by a constant
In the Mean Value Theorem, we saw that two functions with the same derivative on an interval differ by a constant. So we get every antiderivative from one of them.
Theorem: If $F$ is an antiderivative of $f$ on an interval, then every antiderivative of $f$ on that interval has the form
$F(x)+C$,
where $C$ is a constant.
We call $F(x)+C$ the general antiderivative of $f$. For example, the general antiderivative of $2x$ is $x^2+C$.
On a graph, the antiderivatives form a family of curves. Each curve is the same curve shifted up or down.
Antiderivative formulas
Each derivative formula, read backward, gives an antiderivative formula. Here $F$ and $G$ are antiderivatives of $f$ and $g$:
| Function | General antiderivative |
|---|---|
| $x^n$ ($n\ne -1$) | $\dfrac{x^{n+1}}{n+1}+C$ |
| $\dfrac{1}{x}$ | $\ln|x|+C$ |
| $e^x$ | $e^x+C$ |
| $\cos x$ | $\sin x+C$ |
| $\sin x$ | $-\cos x+C$ |
| $\sec^2 x$ | $\tan x+C$ |
| $\sec x\tan x$ | $\sec x+C$ |
| $c\,f(x)$ | $c\,F(x)+C$ |
| $f(x)+g(x)$ | $F(x)+G(x)+C$ |
The first formula is the power rule backward: raise the exponent by $1$, then divide by the new exponent. It does not work for $n=-1$, because that would divide by $0$. The antiderivative of $x^{-1}=1/x$ is $\ln|x|$ instead (see derivatives of exponential and logarithmic functions).
Always check your answer by differentiating it. You should get back the original function.
Example 1: Find the general antiderivative of each function.
(a) $f(x)=x^3$
(b) $f(x)=\dfrac{1}{x^2}$
(c) $f(x)=\sqrt{x}$
Solution:
(a) Raise the exponent to $4$, and divide by $4$:
$F(x)=\dfrac{x^4}{4}+C$
(b) Write $\dfrac{1}{x^2}=x^{-2}$. Raise the exponent to $-1$, and divide by $-1$:
$F(x)=\dfrac{x^{-1}}{-1}+C=-\dfrac{1}{x}+C$
(c) Write $\sqrt{x}=x^{1/2}$. Raise the exponent to $\dfrac{3}{2}$, and divide by $\dfrac{3}{2}$:
$F(x)=\dfrac{x^{3/2}}{3/2}+C=\dfrac{2}{3}x^{3/2}+C$
Check (c):
$\begin{align*}&\dfrac{d}{dx}\left(\dfrac{2}{3}x^{3/2}\right)\\&=\dfrac{2}{3}\cdot\dfrac{3}{2}x^{1/2}\\&=\sqrt{x}\end{align*}$
Example 2: Find the general antiderivative of each function.
(a) $f(x)=3x^2-4x+5$
(b) $g(x)=2\cos x-\dfrac{3}{x}+e^x$
Solution:
Find an antiderivative of each term. One constant $C$ at the end is enough.
(a)
$\begin{align*}&F(x)\\&=3\cdot\dfrac{x^3}{3}-4\cdot\dfrac{x^2}{2}+5x+C\\&=x^3-2x^2+5x+C\end{align*}$
(b)
$G(x)=2\sin x-3\ln|x|+e^x+C$
(This works on any interval that does not contain $0$.)
Finding one particular antiderivative
Sometimes we know one value of the function, too. Then we can find the constant $C$. A problem like this is called an initial value problem.
Example 3: Find $f(x)$ if
$f'(x)=6x^2-2x$ and $f(1)=4$.
Solution:
The general antiderivative:
$f(x)=2x^3-x^2+C$
Use $f(1)=4$ to find $C$:
$\begin{align*}2(1)^3-(1)^2+C&=4\\1+C&=4\\C&=3\end{align*}$
So $f(x)=2x^3-x^2+3$.
Example 4: Find $f(x)$ if
$f''(x)=12x-4$,
$f'(0)=1$, and $f(0)=2$.
Solution:
Go back one step at a time. First, find $f'$:
$f'(x)=6x^2-4x+C_1$
Since $f'(0)=1$, we get $C_1=1$. So $f'(x)=6x^2-4x+1$.
Now find $f$:
$f(x)=2x^3-2x^2+x+C_2$
Since $f(0)=2$, we get $C_2=2$. So
$f(x)=2x^3-2x^2+x+2$.
Each step back brings a new constant. So we need two known values to find $f$ from $f''$.
Motion
Velocity is the derivative of position, and acceleration is the derivative of velocity. So going backward:
- Position $s(t)$ is an antiderivative of velocity $v(t)$.
- Velocity $v(t)$ is an antiderivative of acceleration $a(t)$.
Example 5: A ball is thrown straight up from $2$ m above the ground, with a speed of $19.6$ m/s. Gravity gives it a constant acceleration of $-9.8$ m/s$^2$ (negative because it points down). Find its height $s(t)$. When does it hit the ground?
Solution:
Start with $a(t)=-9.8$. The velocity is an antiderivative:
$v(t)=-9.8t+C_1$
At $t=0$, the velocity is $19.6$.
So $C_1=19.6$, and
$v(t)=-9.8t+19.6$.
The position is an antiderivative of the velocity:
$s(t)=-4.9t^2+19.6t+C_2$
At $t=0$, the height is $2$.
So $C_2=2$, and
$s(t)=-4.9t^2+19.6t+2$.
This is the ball from maximum and minimum values (Example 5). Now we know where its formula comes from.
It hits the ground when $s(t)=0$. Use the quadratic formula, and keep the positive answer:
$\begin{align*}&t\\&=\dfrac{-19.6-\sqrt{19.6^2+4(4.9)(2)}}{2(-4.9)}\\&=\dfrac{19.6+\sqrt{423.36}}{9.8}\\&\approx 4.10\ \text{s}\end{align*}$
Summary
- $F$ is an antiderivative of $f$ if $F'=f$.
- The general antiderivative is $F(x)+C$. All antiderivatives differ by a constant.
- Read the derivative formulas backward. For powers: raise the exponent by $1$, and divide by the new exponent ($n\ne -1$).
- Check an antiderivative by differentiating it.
- A known value of the function lets you find $C$.