3.15: Anti-derivatives
We’ll see in a future chapter that we will need to be able to be able to undo a derivative. That is, given the answer of a derivative problem, what is the original question?
Given a function \(f(x)\), \(F(x)\) is the anti-derivative of \(f(x)\) if \(F'(x) = f(x)\). We denote this by
\(\int f(x) \ dx = F(x).\)
For now, just consider \(\int\) and \(dx\) to be notation that tells you to take an anti-derivative.
Let us look at some examples with the power rule. The following is a table of derivative problems.
\(\begin{array}{cc} F(x) & F'(x) = f(x) \\ \hline 1 & 0 \\ x & 1 \\ x^2 & 2x \\ x^3 & 3x^2 \\ x^4 & 4x^3 \end{array}\)
Now let’s look at anti-derivatives. All we do is what was the original problem is now the answer, and what was the answer is now the original problem.
\(\begin{array}{cc} f(x) & \int f(x) \ dx = F(x) \\ \hline 0 & 1 \\ 1 & x \\ 2x & x^2 \\ 3x^2 & x^3 \\ 4x^3 & x^4 \end{array}\)
There is one twist we have to watch out for: anti-derivatives are not unique. Consider the following derivatives:
\(\begin{array}{cc} F(x) & F'(x) = f(x) \\ \hline 7 & 0 \\ x + 7 & 1 \\ x^2 + 7 & 2x \\ x^3 + 7 & 3x^2 \\ x^4 + 7 & 4x^3 \end{array}\)
This would lead us to the following anti-derivatives
\(\begin{array}{cc} f(x) & \int f(x) \ dx = F(x) \\ \hline 0 & 7 \\ 1 & x + 7 \\ 2x & x^2 + 7 \\ 3x^2 & x^3 + 7 \\ 4x^3 & x^4 + 7 \end{array}\)
So, suppose I ask the question “what is the anti-derivative of \(2x\)?” There are many possible solutions, including \(x^2\), \(x^2 +7\), \(x^2 + 42\), etc. However, all of these solutions are basically the same thing. To get around this problem, we say often use the notation \(x^2 + C\). Here, \(C\) is called the constant of integration . In this form, the antiderivative is called an indefinite integral .
If we are dealing with a polynomial, we know the derivative follows the power rule. With anti-derivatives, it follows the inverse power rule :
\(\boxed{\int x^m \ dx = \frac{x^{m + 1}}{m+1} + C}\)
Note that this only works if \(m \neq -1\). Can you see what goes wrong in the formula if \(m = -1\)?
We also have the rules of linearity work in reverse. So when taking anti-derivatives, you can just look at one term at a time, and the constant will stay during the anti-derivative process.
\(\boxed{\int c f \ dx = c \int f \ dx}\)
\(\boxed{\int f + g \ dx = \int f \ dx + \int g \ dx}\)
Here are some examples.
-
Find \(\int x^7 \ dx\).
Using the inverse power rule, we have \(\int x^7 \ dx = \boxed{\frac{x^8}{8} + C}\).
-
Find \(\int 6x^2 \ dx\).
The constant will “come along for the ride”, or not be affected by the anti-derivative process. Hence we see
\[\begin{align*} \int 6 x^2 \ dx & = 6 \int x^2 \ dx \\ & = 6 \left( \frac{x^3}{3} + C \right) \\ & = 6 \frac{x^3}{3} + 6C \\ & = 2x^3 + 6C. \end{align*}\]
Note that the \(6\) in the \(6C\) part of the answer is not really necessary. What is important is we are indicating that you can add any constant of integration we’d like. Therefore, we will often replace the \(6C = D\), and we have the answer
\(\int 6 x^2 \ dx = 2x^3 + D.\)
In fact, often we will just wait to add the \(C\) until the end. In that case, we would arrive at an answer of \(2x^3\), then we’d add the \(C\) at that stage, and get an answer of
\(\int 6 x^2 \ dx = \boxed{2x^3 + C}.\)
-
Find \(\int \frac{1}{2} x^3 \ dx\)
The \(\frac{1}{2}\) is a constant that does not affect the integration, so we see
\[\begin{align*} \int \frac{1}{2} x^3 \ dx & = \frac{1}{2} \int x^3 \ dx \\ & = \frac{1}{2} \left( \frac{x^4}{4} \right) \\ & = \boxed{\frac{x^4}{8} + C}. \end{align*}\]
Notice we didn’t add the constant of integration until the last step, and that’s perfectly okay.
-
Find \(\int 6x^5 + 6x \ dx\)
In a sum, we can just treat each term separately. And the constants come along for the ride.
\[\begin{align*} \int 6x^5 + 6x \ dx & = 6 \int x^5 \ dx + 6 \int x \ dx \\ & = 6 \left( \frac{x^6}{6} \right) + 6 \left(\frac{x^2}{2}\right) \\ & = \boxed{x^6 + 3x^2 + C}. \end{align*}\]
-
Find \(\int 4 e^x + \frac{3}{x} + \frac{\sin(x)}{5} \ dx\).
We treat each term separately.
-
\(\int 4 e^x \ dx\). In this term, the \(4\) is a constant multiple, and the \(e^x\) doesn’t change, so we get
\(\int 4 e^x \ dx = 4 e^x.\)
-
\(\int \frac{3}{x} \ dx\). Okay, a bit harder now. Notice the \(3\) is a constant multiple, so let’s take that out: \(3 \int \frac{1}{x} \ dx\). We recognize this as the derivative of \(\ln(x)\), so the answer is
\(\int \frac{3}{x} \ dx = 3 \ln(x).\)
-
\(\int \frac{\sin(x)}{5} \ dx\). For this one, the division by \(5\) is the same a constant multiple of \(\frac{1}{5}\). We bring this outside and get: \(\frac{1}{5} \int \sin(x) \ dx\). We know that \(\frac{d}{dx} \cos(x) = -\sin(x)\), so we know \(\frac{d}{dx} -\cos(x) = \sin(x)\). This gives the final answer of
\(\int \frac{\sin(x)}{5} \ dx = -\frac{1}{5} \cos(x).\)
Putting it altogether, we get a final answer of \(\boxed{4 e^x + 3 \ln(x) - \frac{1}{5} \cos(x) + C}\). Nice work!
-
\(\int 4 e^x \ dx\). In this term, the \(4\) is a constant multiple, and the \(e^x\) doesn’t change, so we get