Basic antiderivatives

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We introduce antiderivatives.

Computing derivatives is not too difficult. At this point, you should be able to take the derivative of almost any function you can write down. However, undoing derivatives is much harder. This process of undoing a derivative is called taking an antiderivative.

A function $F$ is called an antiderivative of $f$ on an interval if

\[ F'(x) = f(x) \]

for all $x$ in the interval.

How many antiderivatives does $f(x) = 2x$ have?

none one infinitely many

The functions $x^2$, $x^2-1.5$, $x^2+2$, $x^2+5$ and so on, are all antiderivatives of $2x$.

It is clear that any function $x^2+C$, where $C$ is a constant, is an antiderivative of $f$. Why? Because, $(x^2+C)'=2x=f(x)$.

Could there exist any other antiderivative of $f$, call it $G$, such that $G(x)$ is not a sum of $x^2$ and a constant? In that case, for all $x$ we would have

\[ (G(x)-x^2)' = G'(x)-2x=f(x)-2x= \answer [given]{0} \]

Recall: Any function whose derivative is equal to $0$ on some interval is equal to a constant on that interval. This means that the function $G(x)-x^2=C$, for some constant $C$. This implies that $G(x)=x^2+C$, for all x.

The Family of Antiderivatives

If $F$ is an antiderivative of $f$, then the function $f$ has a whole family of antiderivatives. Each antiderivative of $f$ is the sum of $F$ and some constant $C$.

When we write $F(x)+C$, we denote the entire family of antiderivatives of $f$. Alternative notation for this family of antiderivatives was introduced by G.W. Leibniz (1646-1716):

Let $f$ be a function. The family of of all antiderivatives of $f$ is denoted by

\[ \int f(x) \d x. \]

This is called the indefinite integral of $f$.

It follows that

\[\int f(x) \d x =F(x)+C, \]

where $F$ is any antiderivative of $f$ and $C$ is an arbitrary constant.

Below are a list of basic antiderivatives for our Famous Functions. Note each of these examples comes directly from our knowledge of basic derivatives.

Basic Indefinite Integrals

$\displaystyle \int k \d x= k x+C$
$\displaystyle \int \frac {1}{x} \d x= \ln |x|+C$
$\displaystyle \int x^n \d x= \frac {x^{n+1}}{n+1}+C\qquad (n\ne -1)$
$\displaystyle \int e^x \d x= e^x + C$
$\displaystyle \int a^x \d x= \frac {a^x}{\ln (a)}+C$
$\displaystyle \int \cos (x) \d x = \sin (x) + C$
$\displaystyle \int \sin (x) \d x = -\cos (x) + C$
$\displaystyle \int \sec ^2(x) \d x = \tan (x) + C$
$\displaystyle \int \csc ^2(x) \d x = -\cot (x) + C$
$\displaystyle \int \sec (x)\tan (x) \d x = \sec (x) + C$
$\displaystyle \int \csc (x)\cot (x) \d x = -\csc (x) + C$
$\displaystyle \int \frac {1}{x^2+1}\d x = \arctan {(x)} + C$
$\displaystyle \int \frac {1}{\sqrt {1-x^2}}\d x= \arcsin {(x)}+C$

It may seem that one could simply memorize these antiderivatives and antidifferentiating would be as easy as differentiating. This is not the case. The issue comes up when trying to combine these functions. When taking derivatives we have the product rule and the chain rule. The analogues of these two rules are much more difficult to deal with when taking antiderivatives. However, not all is lost.

Consider the following example.

Find an antiderivative of the function $h$ defined by expression

$h(x)=\cos (x)+\frac {1}{x}$, for all $x$ in some interval $I$.

$H(x)=\cos (x)+\frac {1}{x}$ $H(x)=\sin (x)+\frac {1}{x^2}$ $H(x)=\sin (x)+\ln (x)$

Differentiate each choice. In the last choice we get $H'(x)=\cos (x)+\frac {1}{x}=h(x)$.

It is easy to check whether an antiderivative is right: we just have to differentiate it, and check that $H'(x)=h(x)$, for all $x$ in $I$.

Notice that in the last example, the function $h$ is the sum of the two functions, $f$ and $g$, where $f(x)=\cos (x)$ and $g(x)=\frac {1}{x}$, for $x$ in $I$.

We know antiderivatives of both functions: $F(x)=\sin (x)$ and $G(x)=\ln (x)$, for $x$ in $I$, are antiderivatives of $f$ and $g$, respectively. So, in this example we see that the function $F+G$ is an antiderivative of $f+g$. In other words, "the sum of antiderivatives is an antiderivative of a sum".

Is this true in general?

Let’s check whether this rule holds for any eligible pair of functions $f$ and $g$ defined on some interval $I$.

Let $f$, $g$, $F$ and $G$ be four functions defined on some interval $I$ such that $F$ is an antiderivative of $f$ and $G$ is an antiderivative of $g$, i.e.

$F'(x)=f(x)$ and $G'(x)=g(x)$ for all $x$ in some interval $I$.

Find an antiderivative of the function $f+g$.

Differentiate the function $F+G$.

Since $\left (F(x)+G(x)\right )'=F'(x)+G'(x)=f(x)+g(x)$, it follows that $F+G$ is an antiderivative of $f+g$.

To summarize: The sum of antiderivatives is an antiderivative of a sum.

We have proved the following theorem.

The Sum Rule for Antiderivatives If $F$ is an antiderivative of $f$ and $G$ is an antiderivative of $g$, then $F+G$ is an antiderivative of $f+g$.

We can write equivalently, using indefinite integrals,

$\displaystyle \int \left (f(x)+g(x)\right ) \d x= \int f(x)\d x +\int g(x) \d x$.

Next, we will try to prove an analogue of the constant multiple rule for derivatives. Let’s consider the following example.

Find an antiderivative of the function $h$, where $h(x)=5\sec ^{2}(x)$, for all $x$ in some interval $I$.

$H(x)=5\tan (x)$ $H(x)=5\sec ^{2}(x)$ $H(x)=\sec ^{2}(5x)$

Differentiate each choice for $H(x)$. In the first choice we get $H'(x)=5\sec ^{2}(x)=h(x)$.

It is easy to recognize an antiderivative: we just have to differentiate it, and check whether $H'(x)=h(x)$, for all $x$ in $I$. Notice, in this example the function $h$ is a constant multiple of $f$, where $f(x)=\sec ^{2}(x)$. On the other hand, we know that the function $F$, defined by $F(x)=\tan (x)$ is an antiderivative of $f$.

If we differentiate the function $5F$, we get that $\left (5F(x)\right )'=\left (5\tan (x)\right )'=5\left (\tan (x)\right )'=5\sec ^2(x)=5f(x)$, for $x$ in $I$.

In other words, “a constant multiple of an antiderivative is an antiderivative of a constant multiple of a function.” Is this always true?

Let’s check whether this rule holds for any constant $k$ and any eligible function $f$ defined on some interval $I$.

Let $k$ be a constant, let $f$ be a function defined on some interval $I$ and let $F$ be an antiderivative of $f$, i.e.

$F'(x)=f(x)$, for all $x$ in some interval $I$.

Find an antiderivative of the function $k f$.

Differentiate the function $kF$.

Since $\left (kF(x)\right )'=kF'(x)=kf(x)$, it follows that $kF$ is an antiderivative of $kf$. To summarize: The constant multiple of an antiderivative is an antiderivative of a constant multiple of a function.

We have proved the following theorem.

The Constant Multiple Rule for Antiderivatives If $F$ is an antiderivative of $f$, and $k$ is a constant, then $kF$ is an antiderivative of $kf$.

We can write equivalently, using indefinite integrals, $\displaystyle \int kf(x) \d x= k\int f(x)\d x$.

Let’s put these rules and our knowledge of basic derivatives to work.

Find the antiderivative of $3 x^7$.

By the theorems above , we see that

\begin{align*} \int 3 x^7 \d x &= 3 \int x^7 \d x\\ &= 3 \cdot \answer [given]{\frac {x^8}{8}}+C. \end{align*}

The sum rule for antiderivatives allows us to integrate term-by-term. Let’s see an example of this.

Compute:

\[ \int \left (x^4 + 5x^2 - \cos (x)\right ) \d x \]

Let’s start by simplifying the problem using the sum rule and constant multiple rule for antiderivatives,

\begin{align*} \int &\left (x^4 + 5x^2 - \cos (x)\right ) \d x\\ &= \int x^4 \d x + 5\int x^2 \d x - \int \cos (x) \d x. \end{align*}

Now we may integrate term-by-term to find

\[ = \answer [given]{\frac {x^5}{5} + \frac {5x^3}{3} - \sin (x)}+C. \]

While the sum rule for antiderivatives allows us to integrate term-by-term, we cannot integrate factor-by-factor, meaning that in general

\[ \int f(x)g(x) \d x \ne \int f(x) \d x\cdot \int g(x) \d x. \]

A student claims that $\int 2x \cos (x) \d x = x^2 \sin (x) +C$. Determine whether the student is correct or incorrect.

If the student were correct, then the derivative of $x^2 \sin (x) +C$ with respect to $x$ would have to be $2x \cos (x)$. However:

\[\ddx x^2 \sin (x) = \answer [given]{2x \sin (x) +x^2 \cos (x)} \]

Hence, the student is incorrect!

1 Computing antiderivatives

Unfortunately, we cannot tell you how to compute every antiderivative; we view them as a sort of puzzle. Later we will learn a hand-full of techniques for computing antiderivatives. However, a robust and simple way to compute antiderivatives is guess-and-check.

Tips for guessing antiderivatives

(a): If possible, express the function that you are integrating in a form that is convenient for integration.
(b): Make a guess for the antiderivative.
(c): Take the derivative of your guess.
(d): Note how the above derivative is different from the function whose antiderivative you want to find.
(e): Change your original guess by multiplying by constants or by adding in new functions.

Compute:

\[ \int \frac {\sqrt {x}+1+x}{x} \d x \]

Before guessing the solution, let’s express the function in the form convenient for integration. Due to Sum Rule, it is more convenient to have a sum of functions, instead of a single, complicated term.

\begin{align*} \int \frac {\sqrt {x}+1+x}{x} \d x &=\int \left (\frac {\sqrt {x}}{x}+\frac {1}{x}+\frac {x}{x}\right ) \d x\\ &=\int \left (\frac {1}{\sqrt {x}}+\frac {1}{x}+1\right ) \d x\\ &=\int \left (x^{-\frac {1}{2}}+\frac {1}{x}+1\right ) \d x \end{align*}

Now, we can apply the Sum Rule.

\[ \int \left (x^{-\frac {1}{2}}+\frac {1}{x}+1\right ) \d x=\int x^{-\frac {1}{2}}\d x+\int \frac {1}{x}\d x+\int 1 \d x \]

Now we can finish the problem.

\[ \int \frac {\sqrt {x}+1+x}{x} \d x= \answer [given]{2}\sqrt {x}+\ln (x)+ \answer [given]{x} + C. \]

Compute:

\[ \int \cos {(4x)} \d x \]

Start with a guess of

\[ \int \cos {(4x)} \d x \approx \sin {(4x)}. \]

Here, we used the $\approx $ symbol to indicate that this is a guess, based on the integrand. Take the derivative of your guess to see if it is correct:

\[ \ddx \sin {(4x)} = 4 \cos {(4x)}. \]

We’re off by a factor of $\frac {1}{4}$, so multiply your guess by this constant.

\[ \ddx \dfrac {1}{4} \sin {(4x)} = \cos {(4x)}. \]

That gives our answer.

\[ \int \cos {(4x)} \d x = \answer [given]{\frac {1}{4}\sin {(4x)}}+C. \]

Compute:

\[ \int 6 e^{5x+1} \d x \]

Since the integrand here is an exponential, we’ll start with a guess of an exponential.

\[ \int 6e^{5x+1} \d x \approx e^{5x+1}. \]

Differentiate our guess to check it:

\[ \ddx e^{5x+1} = 5 e^{5x+1}. \]

It appears that we are off by a constant factor. We will divide by $5$ to get rid of the coefficient on the right-side.

\[ \ddx \dfrac {1}{5}e^{5x+1} = e^{5x+1}. \]

We need a coefficient of $6$ for the exponential, so we multiply by $6$.

\[ \ddx \dfrac {6}{5}e^{5x+1} = 6e^{5x+1}. \]

That gives our result.

\[ \int 6e^{5x+1} \d x = \dfrac {6}{5}\answer [given]{ e^{5x+1}} + C. \]

Compute:

\[ \int \dfrac {1}{7}e^{2x}\cos {(3e^{2x}-5)} \d x \]

Start with a guess of

\[ \int \dfrac {1}{7}e^{2x}\cos {(3e^{2x}-5)} \d x \approx \sin (3e^{2x}-5). \]

Take the derivative of your guess to see if it is correct:

\[ \ddx \sin (3e^{2x}-5) = 6e^{2x}\cos (3e^{2x}-5). \]

To get rid of the coefficient $6$, we divide both sides.

\[ \ddx \dfrac {1}{6}\sin (3e^{2x}-5) = e^{2x}\cos (3e^{2x}-5). \]

Multiply both sides by $\frac {1}{7}$ to get the correct coefficient.

\[ \ddx \dfrac {1}{42}\sin (3e^{2x}-5) = \dfrac {1}{7}e^{2x}\cos (3e^{2x}-5). \]

Therefore,

\[ \int \dfrac {1}{7}e^{2x}\cos {(3e^{2x}-5)} \d x = \dfrac {1}{42}\answer [given]{\sin (3e^{2x}-5)} + C. \]

Compute:

\[ \int x^2\sin {(x^3 -6)} \d x \]

Start with a guess of

\[ \int x^2\sin {(x^3 -6)} \d x \approx -\cos {(x^3 -6)}. \]

Take the derivative of your guess to see if it is correct:

\[ \ddx \left (-\cos {(x^3 -6)}\right ) = 3x^2\sin {(x^3 -6)}. \]

We’re off by a factor of $1/3$, so multiply our guess by this constant to get the solution,

\[ \int x^2\sin {(x^3 -6)} \d x =- \answer [given]{\frac {1}{3}}\cos {(x^3 -6)}+C. \]

Compute:

\[ \int \frac {x^3}{\sqrt {x^4 -6}} \d x \]

Start by rewriting the indefinite integral as

\[ \int x^3\left (x^4 -6\right )^{-1/2} \d x. \]

Now start with a guess of

\[ \int x^3\left (x^4 -6\right )^{-1/2} \d x \approx \left (x^4 -6\right )^{1/2}. \]

Take the derivative of your guess to see if it is correct:

\[ \ddx \left (x^4 -6\right )^{1/2} = (4/2)x^3\left (x^4 -6\right )^{-1/2}. \]

We’re off by a factor of $2/4$, so multiply our guess by this constant to get the solution,

\[ \int \frac {x^3}{\sqrt {x^4 -6}} \d x = \answer [given]{\frac {1}{2}}(x^4 -6)^{1/2}+C. \]

Compute:

\[ \int xe^{x^2} \d x \]

We try to guess the antiderivative. Start with a guess of

\[ \int xe^{x^2} \d x \approx e^{x^2}. \]

Take the derivative of your guess to see if it is correct:

\[ \ddx e^{x^2} = 2xe^{x^2} . \]

Ah! So we need only multiply our original guess by $\frac {1}{2}$. We now find

\[ \int xe^{x^2} \d x =\answer [given]{\frac {1}{2}}e^{x^2} + C. \]

Compute:

\[ \int \frac {21x^2}{7x^3 + 3} \d x \]

We notice that the numerator is the derivative of the denominator, i.e., the function that we are integrating has the form $\frac {f'(x)}{f(x)}$.

\[ \int \frac {21x^2}{7x^3 + 3} \d x = \ln (7x^3+3). \]

Take the derivative of your guess to see if it is correct:

\[ \ddx \ln (7x^3+3) = \frac {21x^2}{7x^3 + 3}. \]

Therefore,

\[ \int \frac {21x^2}{7x^3 + 3} \d x = \ln {(\answer [given]{7x^3+3})}+C. \]

Compute:

\[ \int \frac {2x^2}{7x^3 + 3} \d x \]

We’ll start with a guess of

\[ \int \frac {2x^2}{7x^3 + 3} \d x \approx \ln (7x^3+3). \]

Take the derivative of your guess to see if it is correct:

\[ \ddx \ln (7x^3+3) = \frac {21x^2}{7x^3 + 3}. \]

We are only off by a factor of $2/21$, so we need to multiply our original guess by this constant to get the solution,

\[ \int \frac {2x^2}{7x^3 + 3} \d x = \answer [given]{(2/21)\ln (7x^3+3)}+C. \]

2 Final thoughts

Computing antiderivatives is a place where insight and rote computation meet. We cannot teach you a method that will always work. Moreover, merely understanding the examples above will probably not be enough for you to become proficient in computing antiderivatives. You must practice, practice, practice!

3 Differential equations

Differential equations show you relationships between rates of functions.

A differential equation is simply an equation with a derivative in it. Here is an example:

\[ a\cdot f'(x)+ b\cdot f(x) = g(x). \]

Which one is a differential equation?

$x^2+3x-1=0$ $f'(x)=x^2+3x-1$ $f(x)=x^2+3x-1$

When a mathematician solves a differential equation, they are finding functions satisfying the equation.

Which of the following functions solve the differential equation

\[ f'(x) = f(x)? \]

$f(x) = \sin (x)$ $f(x) = x^2$ $f(x) = e^x$ $f(x) =6 e^x$ $f(x) = e^{-x}$ $f(x) = \tan (x)$

We can directly check that any function $f(x)=Ce^x$ is a solution to our differential equation $f'(x)=f(x)$. Could there be any others? It turns out that these are the only solutions. But showing that we didn’t miss any is a bit tricky.

explanation

Well, suppose we have some mysterious function $f$ and all we know is that $f'(x)=f(x)$. Let’s define a new function $g(x)=f(x)/e^x$. Since our denominator is never 0, the quotient rule tells us that

\[ g'(x)= \frac {e^x f'(x)- f(x) e^x}{(e^x)^2} = \frac {e^x\left (f'(x)- f(x)\right )}{(e^x)^2} = \frac {e^x\cdot 0}{(e^x)^2}=0. \]

But we know that a function with a 0 derivative is constant, so $g(x)=C$. Plugging this back into our formula for $g(x)$ tells us that $C=f(x)/e^x$. Now we rearrange to get $f(x)=Ce^x$.

This shows us that any solution to our differential equation must be one of the functions we’ve already found. Our argument relies on some features that are special to this problem; proving that we’ve found all the possible solutions to an arbitrary differential equation is a very difficult task!

A function $Ce^x$ is called a general solution of the differential equation. Since there are infinitely many solutions to a differential equation, we can impose an additional condition (say $f(0)=1$), called an initial condition. When we are asked to find a function $f$ that satisfies both the differential equation (DE) and the initial condition (IC), this is called an initial value problem (IVP). Let’s try one out.

Solve the initial value problem (IVP):

\begin{align*} f'(x) & = f(x) && \text {(DE)} \\ f(0) & = 1 && \text {(IC)} \end{align*}

The figure below shows several solutions of the differential equation.

The figure suggests that the only solution to the initial value problem is the function $f(x)=e^x$. We can verify this result.

Since all solutions of the differential equation have the form $f(x)=Ce^x$ and $f(0)=1$, it follows that

\[ Ce^0 = 1. \]

Therefore,

\[ C = \answer [given]{1}. \]

So, the unique solution to the given initial value problem is the function $f(x)=e^x$.

Solve the initial value problem (IVP):

\begin{align*} f'(x) & = \sin (x) \\ f(0) & = -1 \end{align*}

First, we have to solve the differential equation. The solution is clearly an antiderivative of $\sin (x)$.

\[ f(x)=-\answer [given]{\cos (x)}+C, \]

where $C$ is an arbitrary constant. We have found the general solution of the DE and this family of solutions is illustrated by the figure below.

Now we must find the solution that also satisfies the initial condition $f(0)=-1$.

Since

\[ f(0)=\answer [given]{-1}+C, \]

it follows that

\[ -1=\answer [given]{-1}+C. \]

Therefore,

\[ C=\answer [given]{0}. \]

The function

\[ f(x)=\answer [given]{-\cos (x)}. \]

is the solution of the initial value problem and it is shown in the figure above.